User johann cigler - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T12:36:36Z http://mathoverflow.net/feeds/user/5585 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102229/are-the-following-q-genocchi-numbers-known Are the following q-Genocchi numbers known? Johann Cigler 2012-07-14T14:33:00Z 2013-03-30T05:22:00Z <p>The sequence of Genocchi numbers<br> ${({G_{2n}})_{n \ge 0}}=$ $(0,1,1,3,17,155,2073,...)$ </p> <p>can be defined by the generating function $z\frac{{1 - {e^z}}}{{1 + {e^z}}} = \sum {{{( - 1)}^n}{G_{2n}}\frac{{{z^{2n}}}}{{(2n)!}}} .$</p> <p>Many different q-analogs of these numbers have been studied. Does anyone know if the following q-analog ${({G_{2n}(q)})_{n \ge 0}}$ is known? It is intimately related with q-Chebyshev polynomials.</p> <p>Let $(a;q)_n=(1-a)(1-qa) \cdots (1-q^{n-1}a)$, $[n]=1+q+\cdots+q^{n-1}$ and $[n]!=[1][2] \cdots[n].$</p> <p>The q-analog can defined by the generating function</p> <p>$\sum\limits_{n \ge 1} {\frac{{{{( - 1)}^{n - 1}}{G_{2n}}(q){{( - q;q)}_{2n - 1}}}}{{[2n]!}}} {z^{2n}} =$</p> <p>$\sum\limits_{n \ge 1} {\frac{{{{( - q;q)}_{2n - 1}}}}{{[2n]!}}} {z^{2n}}$ divided by</p> <p>$\sum\limits_{n \ge 0} {\frac{{{{( - q;q)}_{2n}}}}{{[2n + 1]!}}} {z^{2n}}.$</p> http://mathoverflow.net/questions/123542/polynomials-satisfying-a-three-term-recurrence Polynomials satisfying a three-term recurrence Johann Cigler 2013-03-04T16:18:22Z 2013-03-04T16:18:22Z <p>Let ${p_n}(x) = x{p_{n - 1}}(x) - {a_{n - 2}}{p_{n - 2}}(x)$ for some numbers ${a_n}$ with initial values ${p_{ - 1}}(x) = 0$ and ${p_0}(x) = 1.$ By Favard’s theorem about orthogonal polynomials there exists a linear functional $F$ on the vector space of polynomials such that<br> $$F\left( {{p_n}(x){x^k}} \right) = 0$$ for $0 \le k &lt; \deg {p_n}$ </p> <p>and </p> <p>$$\det \left( {F\left( {{x^{i + j}}} \right)} \right)_{i,j = 0}^{\deg {p_n} - 1} \ne 0.$$</p> <p>It seems that analogous results also hold if ${p_n}(x) = {x^{{r_{n - 1}}}}{p_{n - 1}}(x) - {a_{n - 2}}{p_{n - 2}}(x),$ where ${r_0} = 1$ and ${r_n}$ are arbitrary positive integers for $n > 0.$<br> I do not know where to look for such results. Therefore I would be very grateful for references.</p> http://mathoverflow.net/questions/120323/an-identity-for-hankel-determinants An identity for Hankel determinants Johann Cigler 2013-01-30T14:00:56Z 2013-01-31T22:17:23Z <p>Is the following result about Hankel determinants known or a simple consequence of some known results? Let</p> <p>$f(x) = \frac{\displaystyle 1}{{\displaystyle 1 - \frac{{a x^{m + 2}}}{\displaystyle {1 - \frac{{b{x^{m + 2}}}}{\displaystyle {1 - {x^2}g(x)}}}}}} = \sum\limits_{n \ge 0} {{f_n}{x^n}}$</p> <p>and </p> <p>$g(x) = \sum\limits_{n \ge 0} {{g_n}{x^n}}$ </p> <p>be formal power series. Then the Hankel determinants </p> <p>${d_f}(n) = \det \left( {{f_{i + j}}} \right)_{i,j = 0}^n$ and</p> <p>${d_g}(n) = \det \left( {{g_{i + j}}} \right)_{i,j = 0}^n$</p> <p>are closely related. To simplify notation let </p> <p>${d_g}(n) = 1$ for $n &lt; 0.$</p> <p>Then for $n \ge 0$</p> <p>$${d_f}(n + m + 1) = (-1)^{m+1\choose 2} {a^{m+n+1}}{b^n}{d_g}(n - 2)$$</p> <p>Remark: For $m = 0$ this result is known and follows from the fact that the moment generating function of monic orthogonal polynomials can be represented as a continued fraction.</p> http://mathoverflow.net/questions/103003/reference-request-for-an-identity-for-tangent-numbers Reference request for an identity for tangent numbers Johann Cigler 2012-07-24T13:40:18Z 2012-07-25T16:00:29Z <p>The tangent numbers $(T_{2n+1})=(1,2,16,272,7936,...)$ (cf. OEIS: A000182) satisfy many recurrences. I would be interested to find references for the following which I think must be very old: $T_3 -2T_1=0$, $T_5 -8T_3 =0,$ $T_7 -18T_5 +8T_3 =0,...$ or more generally</p> <p>$${T_{2n + 1}} = \sum\limits_{j \ge 1} {}{(-1)}^{j-1}{2^{2j}} {\binom{n+1}{2j}} {\frac{n+1-j}{n+1}}T_{2n - 2j + 1}.$$</p> http://mathoverflow.net/questions/96546/lucky-chance-or-combinatorial-cause Lucky chance or combinatorial cause? Johann Cigler 2012-05-10T07:08:07Z 2012-06-01T09:07:02Z <p>Consider an $n \times 1 -$rectangle where the $n$ squares are numbered $1$ to $n$. Cover this rectangle with white squares, black squares, and dominoes. To each covering of the rectangle associate the following weight: Each white square has weight 1, each black square at position $i$ and each domino at position ${(i,i + 1)}$ have weight $q^i.$ As usual the weight of a covering is the product of its components and the weight of a set of coverings is the sum of their weights. Then it is easy to verify that the weight $u(n,k)$ of all coverings with precisely $k$ dominoes is the product $u(n,k)=a(n,k)b(n,k)$ with $a(n,k)= {q^{k^2}}{n-k\brack k}$ and $b(n,k) = (1 + {q^{k + 1}})(1 + {q^{k + 2}}) \cdots (1 + {q^{n - k}}).$</p> <p>Here ${n\brack k}=[1][2]\dots[n]/(([1]\dots[k])([1]\dots[n-k]))$ with $[n]=(1-q^n)/(1-q)$ denotes a $q-$binomial coefficient.</p> <p>It is often claimed that there are no accidents in mathematics. Therefore my question is: Is there a simple combinatorial reason for the fact that $u(n,k)$ is the <strong>product</strong> of two terms with simple combinatorial interpretations or is it an accident after all?</p> <p>Since this seems to be rather elementary I have posted this question in <a href="http://math.stackexchange.com/questions/142158/lucky-chance-or-combinatorial-cause" rel="nofollow">http://math.stackexchange.com/questions/142158/lucky-chance-or-combinatorial-cause</a> but did not get an answer.</p> <p><strong>Edit</strong></p> <p>Some days ago Ilse Fischer has shown me a simple bijection. Associate with a tiling of an $n -$board with $k$ dominoes , $\ell$ black squares and $n - 2k - \ell$ white squares the word ${c_1}{c_2} \cdots {c_{n - k}}$ in the letters $w,b,d,$ where $d$ occurs $k$ times, $b$ occurs $\ell$ times and $w$ occurs $n - 2k - \ell$ times. Let $W({c_1}{c_2} \cdots {c_{n - k}})$ be the weight of the tiling. </p> <p>First reverse in ${c_1}{c_2} \cdots {c_{n - k}}$ the order of the letters $b,d$ and obtain a word ${C_1}{C_2} \cdots {C_{n - k}}.$ </p> <p>Let e.g. $(n,k,\ell ) = (12,3,2)$ and ${c_1}{c_2} \cdots {c_9} = wbdwwdbwd.$ Then ${C_1}{C_2} \cdots {C_9} = wdbwwddwb.$ </p> <p>Then replace in ${C_1}{C_2} \cdots {C_{n - k}}$ all $b$ by $w.$ This gives a word $A$ with $k$ letters $d$ and $n - 2k$ letters $w.$ In our example we get $A = wdwwwddww.$ </p> <p>Then delete in ${C_1}{C_2} \cdots {C_{n - k}}$ all letters $d$ and get a word $B$ with $n - 2k$ letters $w,b.$ In our example $B = wbwwwb.$</p> <p>Then $W({c_1}{c_2} \cdots {c_{n - k}}) = {q^{k\ell }}W(A)W(B).$</p> <p>In our example we have $W({c_1}{c_2} \cdots {c_9}) = W(wbdwwdbwd) = {q^{2 + 3 + 7 + 9 + 11}} = {q^{32}},$ $W(A) = W(wdwwwddww) = {q^{2 + 7 + 9}} = {q^{18}},$ $W(B) = W(wbwwwb) = {q^{2 + 6}} = {q^8}.$</p> <p>If $u(n,k,\ell )$ denotes the weighted enumeration of all tilings this implies $u(n,k,\ell ) = {q^{k\ell }}u(n,k,0)u(n - 2k,0,\ell ).$</p> http://mathoverflow.net/questions/91648/simple-and-general-relation-between-continuant-polynomials/91695#91695 Answer by Johann Cigler for Simple and general relation between continuant polynomials Johann Cigler 2012-03-20T08:51:45Z 2012-03-20T08:51:45Z <p>As Gerry remarked it follows from the recursion of the continuants. By the recurrence of the continuants you have ${K_{k + 3}}({a_0}, \cdots ,{a_k},1,1) = {K_{k + 2}}({a_0}, \cdots ,{a_k},1) + {K_{k + 1}}({a_0}, \cdots ,{a_k}).$ Therefore ${K_{k + 4}}({a_0}, \cdots ,{a_k},1,1,{a_{k + 1}}) = {a_{k + 1}}{K_{k + 3}}({a_0}, \cdots ,{a_k},1,1) + {K_{k + 2}}({a_0}, \cdots ,{a_k},1)$, ${K_{k + 3}}({a_0}, \cdots ,{a_k},1,{a_{k + 1}}) = {a_{k + 1}}{K_{k + 2}}({a_0}, \cdots ,{a_k},1) + {K_{k + 1}}({a_0}, \cdots ,{a_k}),$ ${K_{k + 2}}({a_0}, \cdots ,{a_k},{a_{k + 1}}) = {a_{k + 1}}{K_{k + 1}}({a_0}, \cdots ,{a_k}) + {K_k}({a_0}, \cdots ,{a_{k - 1}}).$ This implies ${K_{k + 4}}({a_0}, \cdots ,{a_k},1,1,{a_{k + 1}}) = {K_{k + 3}}({a_0}, \cdots ,{a_k},1,{a_{k + 1}}) + {K_{k + 2}}({a_0}, \cdots ,{a_k},{a_{k + 1}}).$ By induction your result follows.</p> http://mathoverflow.net/questions/91161/eigenvectors-and-eigenvalues-of-tridiagonal-matrix/91358#91358 Answer by Johann Cigler for Eigenvectors and eigenvalues of Tridiagonal matrix Johann Cigler 2012-03-16T08:58:17Z 2012-03-18T16:55:55Z <p>@ Emilio: Could you please show how you get $\det(\mathcal{T}_n(p,q)-\lambda)=2(pq)^n T_n(-\lambda/2)$?</p> <p>Sorry I wanted to comment another answer. I do not know how to delete this post.</p> <p><strong>Edit</strong></p> <p>I think the first problem can be reduced to Chebyshev polynomials of the second kind ${U_n}(x)$ because </p> <p>$$\det \left( {{T_n}(p,q) - \lambda } \right) = - \lambda \det \left( {{T_{n - 1}}(p,q) - \lambda } \right) - pq\det \left( {{T_{n - 2}}(p,q) - \lambda } \right).$$</p> <p>implies</p> <p>$$\det \left( {{T_n}(p,q) - \lambda } \right) ={(\sqrt {pq} )^n}{U_n}\left( {-\frac{\lambda }{{2\sqrt {pq} }}} \right).$$</p> <p>From the fact that the zeros of ${U_n}(x)$ are $\cos \frac{{k\pi }}{{n + 1}}$ the eigenvalues are<br> $2\sqrt {pq} \cos \frac{{k\pi }}{{n + 1}}.$</p> http://mathoverflow.net/questions/89996/why-are-some-q-analogues-more-canonical-than-others/90060#90060 Answer by Johann Cigler for Why are some q-analogues more canonical than others? Johann Cigler 2012-03-02T17:37:27Z 2012-03-02T17:37:27Z <p>As I said above I have some difficulty to denote specific $q-$analogues as canonical. Consider as example the Catalan numbers $\frac{1}{{n + 1}}{2n\choose n}$ . They have a simple generating function $f(z)$ which satisfies $f(z) = 1 + zf(z)^2$ or equivalently $f(z) = \frac{{1 - \sqrt {1 - 4z} }}{{2z}}$ and they are characterized by the fact that all Hankel determinants are 1.</p> <p>Which of the following simple $q -$analogues should be called "canonical"? </p> <p>a) The polynomials $C_n (q)$ introduced by Carlitz with generating function $F(z) = 1 + zF(z)F(qz)$. They also have very simple Hankel determinants, but there is no known formula for the polynomials themselves.</p> <p>b) The polynomials $\frac{1}{{[n + 1]}}{2n\brack n}$ . They have a simple formula but no simple formula for their generating function and no simple Hankel determinants.</p> <p>c) The $q -$Catalan numbers $c_n (q)$ introduced by George Andrews. Their generating function $A(z)$ is a $q -$analogue of $\frac{{1 - \sqrt {1 - 4z} }}{{2z}}.$ Let $h(z)$ be the $q -$analogue of $\sqrt {1 + z}$ defined by $h(z)h(qz)=1+z$. Then $A(z)= \frac{1+q}{{4qz}}(1-h(-4qz))$. They have both simple formulas and a simple formula for the generating function and also simple Hankel determinants. But they are not polynomials in $q.$ </p> http://mathoverflow.net/questions/89934/sum-of-gaussian-binomial-coefficients/89944#89944 Answer by Johann Cigler for Sum of Gaussian binomial coefficients. Johann Cigler 2012-03-01T09:44:35Z 2012-03-01T09:55:33Z <p>There are many possibilities, e.g. $\sum_{i=0}^{n}q^i{n \choose i}_{q^2}=(1+q)(1+q^2)...(1+q^n)$</p> <p>or</p> <p>$\sum_{i=0}^{n}q^{i(i+1)/2}{n \choose i}_{q }=(1+q)(1+q^2)...(1+q^n).$</p> http://mathoverflow.net/questions/87040/existence-of-rodrigues-type-formulae Existence of Rodrigues-type formulae? Johann Cigler 2012-01-30T16:59:32Z 2012-01-30T16:59:32Z <p>For many orthogonal polynomials ${p_n}(x)$ so-called Rodrigue type formulae hold, i.e representations of the form ${p_n}(x) = {a_n}w(x){D^n}({h_n}(x))$ where $D$ denotes the differentiation or $q -$ differentiation operator. </p> <p>Are there general results known about the existence of such representations?</p> <p>More concretely: Does such a formula exist for the $q -$ Fibonacci polynomials defined by the recursion $f(n,x,s,q) = xf(n - 1,x,s,q) + {q^{n - 2}}sf(n - 2,x,s,q)$ with initial values $f(0,x,s,q) = 0$ and $f(1,x,s,q) = 1?$</p> <p>For $q = 1$ it is well known that $f(n,x,s,1) = n!/(2n - 1)!/\sqrt {{x^2} + 4s} {\left( {\frac{d}{{dx}}} \right)^{n - 1}}\left( {{{({x^2} + 4s)}^{n - \frac{1}{2}}}} \right)$.</p> http://mathoverflow.net/questions/75751/reference-request-vanishing-coefficients-in-power-series Reference request: Vanishing coefficients in power series. Johann Cigler 2011-09-18T14:13:07Z 2011-09-18T23:11:25Z <p>Let $$\alpha = \frac{{1 - az + \sqrt {(1 - az)^2 - 4bz^2 } }}{2}.$$</p> <p>In the power series expansion of $\alpha ^n$ the coefficients of $z^k$ vanish for $n + 1 \le k \le 2n - 1.$<br> Similar results hold for $$\alpha (m) = \frac{{1 - az + \sqrt {(1 - az)^2 - 4bz^m } }}{2}.$$ Since the proofs are very simple such results must be well known. I would be very grateful for references.</p> http://mathoverflow.net/questions/75306/usefulness-of-symbolic-devices Usefulness of symbolic devices Johann Cigler 2011-09-13T12:37:39Z 2011-09-13T16:06:32Z <p>Each mathematician knows that good notation or symbolism – which seems to be irrelevant from a purely logical point of view – makes theorems more plausible and motivates results which would otherwise be overlooked. Examples abound. Let me only mention such diverse things as decimal notation, $B^A$ for the set of mappings from $A$ to $B$, $\frac{{df}}{{dx}}$ for differentiation, commutative diagrams, umbral calculus, etc. I would be interested in a list of examples of the most useful notations or symbolic devices together with hints for the reason of their usefulness.</p> http://mathoverflow.net/questions/75059/recognizing-a-measure-whose-moments-are-the-motzkin-numbers/75080#75080 Answer by Johann Cigler for Recognizing a measure whose moments are the motzkin numbers Johann Cigler 2011-09-10T09:58:39Z 2011-09-11T12:41:56Z <p>The formula for the measure of the Motzkin numbers as stated by Gjergij follows from the formula for the Catalan numbers if we write the formula for the Catalan numbers in the form $\frac{1}{{2\pi }}\int_{ - 2}^2 {x^{2n} \sqrt {4 - x^2 } } dx = C_n$ and observe that the corresponding orthogonal polynomials are the Fibonacci polynomials $F_{n + 1} (x, - 1)$ defined by $F_n (x, - 1) = xF_{n - 1} (x, - 1) - F_{n - 2} (x, - 1)$ with initial values $F_0 (x, - 1) = 0$ and $F_1 (x, - 1) = 1.$ For the Motzkin numbers the corresponding orthogonal polynomials are $F_{n + 1} (x - 1, - 1).$ Therefore a simple transformation of the integral gives the result.</p> <p><strong>Edit:</strong> More generally (as answer to the comments by Gjergji and Brendan): Let $f(z) = \sum {r(n,c,d)z^n }$ satisfy $f(z) = 1 + czf(z) + d^2 z^2 f(z)^2.$ Then $r(n,c,d) = \frac{1}{{2\pi d^2 }}\int_{c – 2d}^{c + 2d} {x^n \sqrt {4d^2 - (x - c)^2 } } dx.$</p> http://mathoverflow.net/questions/74841/an-example-of-a-beautiful-proof-that-would-be-accessible-at-the-high-school-level/74858#74858 Answer by Johann Cigler for An example of a beautiful proof that would be accessible at the high school level? Johann Cigler 2011-09-08T10:59:23Z 2011-09-08T10:59:23Z <p>1) Many elementary binomial identities or identities with Fibonacci numbers have beautiful proofs. Let me only mention the matrix representation of Fibonacci numbers whose determinant gives Cassini's identity.</p> <p>2) Another elementary problem is the following: Is it possible to cover a checkerboard from which one white and one black square have been removed with dominoes? To show that this is possible run through the board in a cyclical way. Observe that on this path between a white and a black square are an even number of squares. Since I don't know how to make figures I indicate such a path for a 4x4-board: ((1,1),(1,2),(1,3),(1,4),(2,4),(2,3),(2,2),(3,2),(3,3),(3,4),(4,4),(4,3),(4,2),(4,1),(3,1),(2,1)).</p> http://mathoverflow.net/questions/66075/the-half-life-of-a-theorem-or-arnolds-principle-at-work/66203#66203 Answer by Johann Cigler for The half-life of a theorem, or Arnold's principle at work Johann Cigler 2011-05-27T17:15:39Z 2011-05-27T17:15:39Z <p>I gave an example in mathoverflow.net/questions/63386, where I asked a similar question.</p> http://mathoverflow.net/questions/65281/are-these-two-definitions-of-uniformly-distributed-equivalent/65298#65298 Answer by Johann Cigler for Are these Two Definitions of Uniformly Distributed" Equivalent? Johann Cigler 2011-05-18T07:16:33Z 2011-05-19T11:18:25Z <p>A sequence is $\mu-$ uniformly distributed -B iff the limit relation A holds for each Borel set $M$ whose boundary has $\mu-$ measure $0.$</p> <p><strong>Edit</strong></p> <p>In the meantime I found the following books which have proofs of this theorem:</p> <p>"Uniform distribution of sequences" by L. Kuipers and H. Niederreiter, Wiley 1974 </p> <p>and P. Billingsley, "Convergence of probability measures", Wiley 1999.</p> http://mathoverflow.net/questions/64817/identifying-the-generating-function-ga-z-sum-n0-infty-an-zn1n/64893#64893 Answer by Johann Cigler for Identifying the generating function $G(a,z) = \sum_{n=0}^{\infty} a^n z^{(n+1)(n+2)/2}.$ Johann Cigler 2011-05-13T12:23:22Z 2011-05-13T12:23:22Z <p>Your generating function is related to a simple continued fraction expansion due to Touchard: </p> <p>$\sum\limits_{k \ge 0} ( - 1)^k q^{k+1\choose2} v^k$ =$\frac{1}{{1 + v - \frac{{(1 - q)v}}{{1 + v - \frac{{(1 - q^2 )v}}{ \cdots }}}}}.$</p> <p>A simple proof can be found in a paper by H. Prodinger <a href="http://de.arxiv.org/abs/1102.5186" rel="nofollow">http://de.arxiv.org/abs/1102.5186</a></p> http://mathoverflow.net/questions/63386/two-different-theorems-but-only-one-fact Two different theorems but only one fact? Johann Cigler 2011-04-29T08:05:11Z 2011-04-29T12:53:20Z <p>Let me first state an example: Let $X$ be the multiplication operator on the polynomials in $x$ defined by $Xf(x)=xf(x)$ and let $D$ be the differentiation operator defined by $Df(x) = f'(x).$ Recently I noticed that apparently there are two different "traditions" to express $(X + D)^n$. In one tradition, concerned with special polynomials, a theorem of Burchnall relates it to a variant of the Hermite polynomials in the form $$(X + D)^n = \sum_{j = 0}^{n }{n\choose j} H_{n-j}(X)D^j.$$ Another tradition is concerned with "normal ordering" of operators. Here a theorem of Mikhailov states that </p> <p>$$(X + D)^n = \sum_{m = 0}^{n }\sum_{j = 0}^{Min(m,n-m) }{n\brack m}_{j} X^{m-j}D^{n-m-j},$$</p> <p>where ${n\brack m}_{j}$ are called "Weyl binomial coefficients". It turns out that both theorems say precisely the same thing. But seemingly the different traditions are unaware of this fact. </p> <p>I would be interested in other examples of this sort.</p> <p><strong>Edit</strong></p> <p>I think the real reason for this situation is the use of different "languages" describing the same things. Originally the theorem about normal ordering has been derived for "abstract" operators satisfying $DX=XD+1$ instead of using the equivalent "concrete" differentiation and multiplication operators for which the situation seems rather evident.</p> <p>So my question could also be formulated to look for examples of different "languages" describing the same facts.</p> http://mathoverflow.net/questions/32775/curious-q-analogues Curious $q$-analogues Johann Cigler 2010-07-21T09:56:46Z 2011-04-11T11:50:43Z <p>Consider the Fibonacci polynomials<br> $$F_n (x) = \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor }\binom{n-j}{j} x^{n - 2j}$$ and the Lucas polynomials<br> $$L_n (x) = \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor }\frac{n}{n-j }\binom{n-j}{j} x^{n - 2j} .$$</p> <p>Let $X$ be the multiplication operator $Xp(x)=xp(x)$ on the polynomials, $D_q$ the $q$-differentiation operator defined by $$D_q p(x)=\frac{p(qx)-p(x)}{qx-x}$$ and $A=X+(q-1)D_q$.</p> <p>Applying the operator $F_n(A)$ to the constant polynomial $1$ has the curious effect that all binomial coefficients $\binom{n}{j}$ are changed to $q$-binomial coefficients ${n\brack j}$ together with some $q$-power. More specifically, we get $$F_n(A)1=F_n(x|q)= \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor } {q^{j+1\choose 2}}{n-j\brack j} x^{n - 2j}$$ and $$L_n(A)1=L_n(x|q)= \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor } {q^{j\choose 2}}{\frac{[n]}{[n-j]}}{n-j\brack j} x^{n - 2j} .$$ </p> <p>Is this an isolated phenomenon or do there exist similar formulae for polynomials? In other words, are there polynomials $p_n(x)$ in one variable $x$ and operators $B(q)$ satisfying $B(1)=X$ such that $p_n(B(q))1$ is a simple or beautiful $q$-analogue of $p_n(x)$? </p> <p>A trivial example would be $p_n(x)=x^n$ and $B(q)=\epsilon(q)X$ with $\epsilon(q)p(x)=p(qx)$.</p> <p><strong>Edit</strong></p> <p>Recently I noticed that the Rogers-Szegö polynomials $$R_n(x,s)= \sum_{j = 0}^{n }{n\brack j} x^{j}s^{n-j}$$ can also be represented in this way: $$R_n(x,s)=(x+s+(q-1)xsD_q)^n1.$$ Since this is the most natural $q-$ analogue of the binomial theorem I would be astonished if nobody has as yet seen this fact. The proof is the same as that for the well-known recursion $R_{n+1}(x,s)=xR_{n}(x,s)+(q^n-1)xs R_{n-1}(x,s).$</p> <p>It follows immediately from the recurrence of the $q-$ binomial coefficients:</p> <p>$(x+s+(q-1)xsD_q) \sum_{j = 0}^{n }{n\brack j} x^{j}s^{n-j}$ $=\sum_{j = 0}^{n }{n\brack j} x^{j+1}s^{n-j}$ $+\sum_{j = 0}^{n }{n\brack {j}} x^{j}s^{n-j+1}$ $+\sum_{j = 0}^{n }(q^{j}-1){n\brack j} x^{j}s^{n-j+1}$ $= \sum_{j = 0}^{n+1 }({n\brack j}+{n\brack {j-1}}+(q^{j}-1) {n\brack {j}}) x^{j} s^{n-j+1}$ $=\sum_{j = 0}^{n +1}{{n+1}\brack j} x^{j}s^{n-j+1}.$</p> http://mathoverflow.net/questions/59077/bivariate-polynomials-with-special-properties/59150#59150 Answer by Johann Cigler for Bivariate polynomials with special properties Johann Cigler 2011-03-22T08:10:30Z 2011-03-22T13:39:39Z <p>Your sequence $p_n (u,v)$ can be defined by $p_n (u,v) = vp_{n - 1} (u,v) - up_{n - 2} (u,v) + p_{n - 3} (u,v)$ with initial values $p_0 (u,v) = 1,p_1 (u,v) = v,p_2 (u,v) = v^2 - u.$ In my above remark I have overlooked a term in the fifth polynomial. This is now the same as the formula given by ARupinski.</p> <p><strong>Added later:</strong> Extend the sequence $p_n (u,v)$ to negative indices by $p_{ - 1} (u,v) = p_{ - 2} (u,v) = 0$ and $p_{ - n} (u,v) = p_{ n - 3} (v,u)$ for $n >2 .$ Define a new sequence of polynomials $r_n (u,v)$ by the same recurrence and initial values $r_{ - 1} (u,v) = 1,r_0 (u,v) = 0,r_1 (u,v) = - u.$ Extend it to negative values by $r_{ - n} (u,v) = r_{n - 2} (v,u).$</p> <p>Let $A$ be the matrix with rows $(0,1,0),(0,0,1),(1, - u,v).$</p> <p>Then $A^n$ is the matrix with the following rows: $\left( {p_{n - 3 + j} (u,v),r_{n - 2 + j} (u,v),p_{n - 2 + j} (u,v} \right)$ for $0 \le j \le 2.$</p> <p>It seems that the sequence $r_n (u,v)$ or the sequence $r_{2n} (u,v)/(1 - uv)$ has analogous properties with respect to the zeroes. Is it also related to the group representation?</p> http://mathoverflow.net/questions/51767/stirling-and-genocchi-numbers Stirling and Genocchi numbers Johann Cigler 2011-01-11T15:08:31Z 2011-01-11T16:56:23Z <p>Define a variant of Stirling numbers (of the second kind) by $S(n,k)=S(n-1,k-1)+k^2 S(n-1,k)$ with $S(n,0)=[n=0]$ and $S(0,k)=[k=0]$ or equivalently by the generating function $\sum {S(n,k)x^n } = \frac{{x^k }}{{\prod\limits_{j = 1}^k {(1 - j^2 x)} }}.$ Have these numbers been studied in the literature?</p> <p>There are various connections with the Genocchi numbers $$(G_{2n} )_{n \ge 1} = (1,1,3,17,155,2073, \cdots ),$$</p> <p>defined by the generating function $z\frac{{1 - e^z }}{{1 + e^z }} = \sum {( - 1)^n G_{2n} } \frac{{z^{2n} }}{{(2n)!}}.$ E.g. the identity $\sum\limits_{k = 0}^n {( - 1)^k } k!(k + 1)!S(n,k+1) = ( - 1)^{n - 1} G_{2n} .$ Are such formulas known?</p> <p><strong>Edit:</strong> I withdraw this question, since I have observed that it has already been answered in Problem 5.8 of Stanley, Enumerative combinatorics 2.</p> http://mathoverflow.net/questions/49878/inverse-of-a-matrix-with-binomial-coefficients Inverse of a matrix with binomial coefficients Johann Cigler 2010-12-19T13:39:31Z 2010-12-20T12:49:16Z <p>Let $a(n,k)=(-1)^k {{2n-k}\choose k}$ for $0 \le k \le n$ and $a(n,k)=0$ else. Then it is known (cf. OEIS A005439 and A098435) that the first column of the inverse matrix of $(a(i,j))_{i,j\ge0}$ is the sequence $1,1,2,8,56,608,9440,…$ of median Genocchi numbers. </p> <p>Now let $b(n,k)=(-1)^k {{2n+1-k}\choose k}$ for $0 \le k \le n$ and $b(n,k)=0$ else. </p> <p>My question is: Is a formula known for the entries of the first column of the inverse matrix of $(b(i,j))_{i,j\ge0}?$ The first terms of this sequence are $1, 1/2, 1/3, 1/3, 8/15, 4/3, 512/105, 368/15…$. Multiplying the $n-$th term with $n!$ gives the sequence $1,1,2,8,64,960,24576,989184,…$.</p> <p><strong>Edit.</strong> Motivated by the comments below and analogous results by D. Dumont and J. Zeng about Genocchi numbers I found the following connections with Bernoulli numbers <code>$B_{2n}.$</code> </p> <p>1) Let <code>$\left( {c(n)} \right)_{n \ge 0}=\left({1, 1/2, 1/3, 1/3, 8/15, 4/3, 512/105, 368/15…}\right).$</code> Then there is an expansion into a formal power series <code>$\sum\limits_{n \ge 0} {( - 1)^n c(n)\frac{{z^{2n} }}{{(1 - z)^{n + 1} }}} = z + \sum\limits_{n \ge 0} {(2n + 1)B_{2n} z^{2n} } .$</code></p> <p>2) <code>$c(2n + 1) = \sum\limits_{j = 0}^n {{2n+1}\choose{2j+1} } (2n + 2j + 3)B_{2n + 2j + 2}$</code> and <code>$c(2n ) = -\sum\limits_{j = 0}^{n-1} {{2n}\choose{2j+1} } (2n + 2j + 3)B_{2n + 2j + 2}.$</code></p> <p>There remains the question whether the sequence <code>$(c(n))$</code> also occurs in a natural way in other contexts.</p> http://mathoverflow.net/questions/48444/recurrent-sequences-and-bernoulli-like-numbers Recurrent sequences and Bernoulli-like numbers Johann Cigler 2010-12-06T14:19:50Z 2010-12-13T13:03:52Z <p>Consider the Fibonacci polynomials defined by $$F_n(s)=F_{n-1}(s)+sF_{n-2}(s)$$ with initial values $F_0(s)=0$ and $F_1(s)=1$ and define a linear functional $L$ on the polynomials in $s$ by $$L(F_{2n})=\delta_{n,1}.$$ Then $$L(F_{2n+1})=(2n+1)B_n,$$ where $B_n$ are the Bernoulli numbers defined by $B_n={\sum{n\choose k} B_k}$ for $n\ge2$ and $B_0=1.$<br> Choosing the linear functional $M$ defined by $$M(F_{2n+1})=\delta_{n,0},$$ gives $$M(F_{2n})=(-1)^n G_{2n},$$ where $G_{2n}$ are the Genocchi numbers $G_{2n}=(-1)^n 2 (1-4^n) B_{2n}.$</p> <p>Finally let $H_n$ be a variant of the Hermite polynomials defined by $$H_n(s)=H_{n-1}(s)-(n-1)s H_{n-2}(s)$$ and the linear functional $N$ defined by $$N(H_{2n})=\delta_{n,0},$$ then we get $$N(H_{2n-1})=(-1)^{n-1} T_{2n-1},$$ where $T_{2n-1}=(-1)^{n-1} \frac{4^n (4^n-1)}{2n} B_{2n}$ are the tangent numbers.</p> <p>My question is: Are these isolated results or special cases of a more general theorem? Does anyone know other such examples?</p> <p><strong>Edit.</strong> To make my question somewhat more precise: Define the Fibonacci polynomials by $F_n(x,s)=xF_{n-1}(x,s)+sF_{n-2}(x,s)$ and the Hermite polynomials by $H_n(x,s)=xH_{n-1}(x,s)-(n-1)s H_{n-2}(x,s).$ The above results follow from the identities $$(e^{xz} + 1)\sum {\frac{{F_{2n} (x,s)}}{{(2n)!}}z^{2n} =(e^{x z}-1) \sum {\frac{{F_{2n + 1} (x,s)}}{{(2n + 1)!}}z^{2n + 1} } }$$ and $$(e^{2xz} + 1)\sum {\frac{{H_{2n + 1} (x,s)}}{{(2n + 1)!}}z^{2n + 1} = (e^{2xz} - 1)\sum {\frac{{H_{2n} (x,s)}}{{(2n)!}}z^{2n} } }.$$</p> <p>Thus a more precise question would be: Are there polynomial sequences which satisfy similar identities?</p> <p><strong>Further edit.</strong> A more precise question: Are there "naturally occurring sequences" $A_n(x,s)$ satisfying $A_n(x,s)=xA_{n-1}(x,s)+c(n,s)A_{n-2}(x,s)$ such that $$\sum {\frac{{A_{2n} (x,s)}}{{(2n)!}}z^{2n} =b(z,x) \sum {\frac{{A_{2n + 1} (x,s)}}{{(2n + 1)!}}z^{2n + 1} } },$$ where $b(z,x)$ does not depend on $s?$ The only examples I know besides the Fibonacci and Hermite polynomials are the Lucas polynomials $L_n(x,s)$ defined by $L_n(x,s)=xL_{n-1}(x,s)+sL_{n-2}(x,s)$ with initial values $L_0(x,s)=2$ and $L_1(x,s)=x.$</p> http://mathoverflow.net/questions/47520/spread-polynomials/47566#47566 Answer by Johann Cigler for Spread polynomials Johann Cigler 2010-11-28T08:19:10Z 2010-11-29T09:06:09Z <p>Factorization of the spread polynomials can be reduced to the factorization of the Chebyshev polynomials by observing that $$1-T_{2n+2}(x)=2 U_n(x)^2 (1-x^2)$$ and $$1-T_{2n+3}(x)=(U_{n+1}(x)+U_n(x))^2 (1-x).$$ Edit: The following formulae show the connection between the factorization of spread polynomials and Chebyshev polynomials most clearly: $$S_{2n}(x^2)= (1-x^2)U_{2n-1}(x)^2$$ and $$S_{2n+1}(x^2)= T_{2n+1}(x)^2.$$</p> http://mathoverflow.net/questions/47561/deriving-inverse-of-hilbert-matrix/47592#47592 Answer by Johann Cigler for Deriving Inverse of Hilbert Matrix Johann Cigler 2010-11-28T16:11:54Z 2010-11-28T16:51:09Z <p>A simple proof is in the paper <a href="http://uk.arxiv.org/abs/math/0609283" rel="nofollow">"Fibonacci numbers and orthogonal polynomials"</a> by Christian Berg.</p> http://mathoverflow.net/questions/11885/nontrivial-question-about-fibonacci-numbers/46601#46601 Answer by Johann Cigler for Nontrivial question about fibonacci numbers? Johann Cigler 2010-11-19T08:03:48Z 2010-11-19T08:03:48Z <p>My favorite identities are the formulae $$F_{n+1}=\sum_{2k\le{n}}{n-k\choose k}\ =\sum_{i\in\mathbb{Z}}(-1)^i {n\choose{\lfloor {(n+5i)}/2\rfloor}}$$ and $$F_{n}=\sum_{2k\le{n-1}}{n-1-k\choose k}\ =\sum_{i\in\mathbb{Z}}(-1)^i {n\choose{\lfloor {(n+5i-1)}/2\rfloor}}$$. They have been found by I. Schur in 1917. In fact he has proved a q-analogue which immediately implies the Rogers-Ramanujan identities.</p> http://mathoverflow.net/questions/12085/experimental-mathematics/37515#37515 Answer by Johann Cigler for Experimental Mathematics Johann Cigler 2010-09-02T17:00:40Z 2010-09-02T17:00:40Z <p>In some simple cases it is even possible to both guess some results and prove them too experimentally. I have done this with some Hankel determinants. Suppose you want to compute the Hankel determinant $\det \left( {a_{i + j} } \right)_{i,j = 0}^{n - 1}$ of a sequence $(a_n )$ . If no Hankel determinant vanishes then you can compute polynomials $p_n (x)$ which are orthogonal with respect to the linear functional $F$ on the polynomials defined by $F(x^n ) = a_n .$ By Favard's theorem there exist numbers $s_n ,t_n$ such that $p_{n + 2} (x) = (x - s_{n + 1} )p_{n + 1} (x) - t_n p_n (x).$ In some cases it is easy to guess a formula for these numbers after computing some of them. Thus also the Hankel determinant can be computed. In order to show that this guess gives the correct answer we can do the following: Define $a(n,j) = a(n - 1,j - 1) + s_j a(n - 1,j) + t_j a(n - 1,j + 1)$ with $a(0,j) = \delta _{j,0} .$ By the underlying theory it suffices to show that $a(n,0) = a_n .$ In order to do this we again compute $a(n,j)$ for small values and try to guess a closed formula for them. In many cases we succeed. Then it suffices to verify that $a(n,j) = a(n - 1,j - 1) + s_j a(n - 1,j) + t_j a(n - 1,j + 1)$ holds for the conjectured formula.</p> http://mathoverflow.net/questions/23437/are-the-q-catalan-numbers-q-holonomic Are the q-Catalan numbers q-holonomic? Johann Cigler 2010-05-04T14:48:59Z 2010-05-20T02:37:37Z <p>The generating function $f(z)$ of the Catalan numbers which is characterized by $f(z)=1+zf(z)^2$ is D-finite, or holonomic, i.e. it satisfies a linear differential equation with polynomial coefficients. The generating function $F(z)$ of the $q$-Catalan numbers is analogously characterized by the functional equation $F(z)=1+z F(z) F(qz)$. I suspect that $F(z)$ is not $q$-holonomic, i.e. does not satisfy a linear $q$-differential equation with polynomial coefficients. But I have no proof. Is there a proof in the literature or references which may lead to a proof?</p> <p>Since there were some misunderstandings I want to clarify the situation. A power series $F(z)$ is called $q -$holonomic if there exist polynomials $p_i (z)$ such that $\sum\limits_{i = 0}^r {p_i (z)D_q^i } F(z) = 0$ where $D_q$ denotes the $q -$differentiation operator defined by $D_q F(z) = \frac{{F(z) - F(qz)}}{{z - qz}}.$ Equivalently if there exist (other) polynomials such that $\sum\limits_{i = 0}^r {p_i (z)F(q^i } z) = 0.$</p> <p>Let $f(z)$ be the generating function of the Catalan numbers $\frac{1}{{n + 1}}{2n\choose n}$ . Then $f(z) = 1 + zf(z)^2$ or equivalently $f(z) = \frac{{1 - \sqrt {1 - 4z} }}{{2z}}.$ There are 3 simple $q -$analogues of the Catalan numbers: a) The polynomials $C_n (q)$ introduced by Carlitz with generating function $F(z) = 1 + zF(z)F(qz)$. My question is about these polynomials. Their generating function satisfies a simple equation, but there is no known formula for the polynomials themselves. b) The polynomials $\frac{1}{{[n + 1]}}{2n\brack n}$ . They have a simple formula but no simple formula for their generating function. c) The $q -$Catalan numbers $c_n (q)$ introduced by George Andrews. Their generating function $A(z)$ is a $q -$analogue of $\frac{{1 - \sqrt {1 - 4z} }}{{2z}}.$ Let $h(z)$ be the $q -$analogue of $sqrt {(1 + z)}$ defined by $h(z)h(qz)=1+z$. Then $A(z)= \frac{1+q}{{4qz}}(1-h(-4qz))$. They have both simple formulas and a simple formula for the generating function. But they are not polynomials in $q.$ Both b) and c) are $q$-holonomic. My question is a proof that a) is not $q$-holonomic.</p> http://mathoverflow.net/questions/22927/why-worry-about-the-axiom-of-choice/22974#22974 Answer by Johann Cigler for Why worry about the axiom of choice? Johann Cigler 2010-04-29T12:40:13Z 2010-04-29T13:50:07Z <p>Many years ago I have been a professor for abstract analysis and had no reservations whatever against using the axiom of choice or "equivalent" statements as e.g. Zorn's lemma since many results in functional analysis depend heavily on it. I have also been very impressed by nonstandard analysis especially in the version of Nelson. But in the mean time I have become more skeptical. I am still very impressed by the results mathematics has obtained by treating the infinite world analogous to the finite world, but I have the feeling that there are some sorts of levels in mathematics which should not be confused. This is of course a very vague assertion. But consider the general statement that every vector space has a basis. This belongs to the infinite world which is "far away". Here Zorn's lemma seems to be the appropriate means. Then consider the statement that each continuous solution of the functional equation f(x+y)=f(x)+f(y) over the reals has the form f(x)=ax. This belongs to the finite world. But the statement that there exist discontinuous solutions mixes both worlds. It uses the existence of a Hamelbasis for the reals over the rationals. Here I am not sure what this really means. It seems that this gives an explanation of what discontinuous solutions look like. But in fact it gives only an illusion of what such solutions look like.</p> http://mathoverflow.net/questions/9218/probabilistic-proofs-of-analytic-facts/22819#22819 Answer by Johann Cigler for Probabilistic Proofs of Analytic Facts Johann Cigler 2010-04-28T07:10:25Z 2010-04-28T07:10:25Z <p>Some examples can be found in the book "Statistical independence in probability, analysis and number theory" by Mark Kac.</p> http://mathoverflow.net/questions/123542/polynomials-satisfying-a-three-term-recurrence Comment by Johann Cigler Johann Cigler 2013-03-04T17:56:58Z 2013-03-04T17:56:58Z @ Peter: For the Chebyshev polynomials of the first kind $F$ is given as an integral on the interval [-1,1] with weight $\frac{1}{{\sqrt {1 - {x^2}} }}$, for the Chebyshev polynomials of the second kind with weight $\sqrt {1 - {x^2}} .$ http://mathoverflow.net/questions/103003/reference-request-for-an-identity-for-tangent-numbers/103109#103109 Comment by Johann Cigler Johann Cigler 2012-07-25T20:17:01Z 2012-07-25T20:17:01Z @Vladimir: Thank you for the idea to reformulate the identity with other known numbers. I now tried it with the Genocchi numbers $G_{2n}$. They satisfy $nT_{2n-1}=2^{2n-2}G_{2n}$. Then the identity reduces to $\sum{(-1)^j}{\binom{n}{2j}}G_{2n-2j}$. This is Seidel’s identity for the Genocchi numbers. So I accept your comments as answer to my question. http://mathoverflow.net/questions/103003/reference-request-for-an-identity-for-tangent-numbers/103011#103011 Comment by Johann Cigler Johann Cigler 2012-07-24T17:42:49Z 2012-07-24T17:42:49Z No, these formulae are different. Equation (4) of the cited paper uses all terms up to $2n+1$ whereas the above formula needs only half of them. http://mathoverflow.net/questions/102229/are-the-following-q-genocchi-numbers-known Comment by Johann Cigler Johann Cigler 2012-07-20T08:57:28Z 2012-07-20T08:57:28Z In the mean time I have seen that these q-Genocchi numbers are related to the usual $q-$tangent numbers ${T_{2n - 1}}(q)$ by ${(- q;q)_ {2n - 1}} {G_{2n}}(q) = [2n] {T_{2n - 1}}(q).$ http://mathoverflow.net/questions/102229/are-the-following-q-genocchi-numbers-known Comment by Johann Cigler Johann Cigler 2012-07-15T06:45:30Z 2012-07-15T06:45:30Z The corresponding Seidel identity is $$\sum{(-1)^k}q^{\binom{2k}{2}} {n\brack{2k}}{{(-q^{n-2k+1};q)_{2k}}}/ {{(-q^{2n-2k};q)_{2k}}}{G_{2n-2k}}(q) =[n=1].$$ http://mathoverflow.net/questions/96546/lucky-chance-or-combinatorial-cause/98465#98465 Comment by Johann Cigler Johann Cigler 2012-06-01T11:16:44Z 2012-06-01T11:16:44Z Now I understand what you mean. But this is somewhat confusing. Sometimes you simply ignore 0 parts, so that (2,1,0,0,0) is the same partition as (2,1). At other times you say 0 is a distinct part of a partition. http://mathoverflow.net/questions/96546/lucky-chance-or-combinatorial-cause/98465#98465 Comment by Johann Cigler Johann Cigler 2012-06-01T10:11:04Z 2012-06-01T10:11:04Z Perhaps I have misunderstood some things. Consider e.g. the tiling bdwwdd . Here P=(2,2,0,0), Q=(6,4,2,1) and Q&#180;=(2,1,0,0). What is meant by the transpose of Q&#180;? What is R? http://mathoverflow.net/questions/96546/lucky-chance-or-combinatorial-cause/98465#98465 Comment by Johann Cigler Johann Cigler 2012-06-01T08:34:49Z 2012-06-01T08:34:49Z Excuse me, but I still have some difficulties. I am not familiar with partitions where some parts may be 0 and how in this case the transpose is defined. Both P and Q&#180; can have 0-parts but B(x,y) refers to partitions with positive parts. (BTW there seems to be a typo in the definition of Q). http://mathoverflow.net/questions/96546/lucky-chance-or-combinatorial-cause/98465#98465 Comment by Johann Cigler Johann Cigler 2012-05-31T10:24:23Z 2012-05-31T10:24:23Z Sorry, I did not understand Proposition B. Can you sketch this procedure with a typical example? http://mathoverflow.net/questions/97920/a-certain-sum-with-q-by-the-power-of-binomial-n-2 Comment by Johann Cigler Johann Cigler 2012-05-25T15:22:22Z 2012-05-25T15:22:22Z I don't think that there is a closed form, but there is a simple continued fraction, see formula (1.1) in <a href="http://math.sun.ac.za/~hproding/pdffiles/touchard-2011.pdf" rel="nofollow">math.sun.ac.za/~hproding/pdffiles/&hellip;</a>. http://mathoverflow.net/questions/96546/lucky-chance-or-combinatorial-cause Comment by Johann Cigler Johann Cigler 2012-05-25T14:54:54Z 2012-05-25T14:54:54Z @martin: yes, I don't even know a bijection if there is only one black square. http://mathoverflow.net/questions/91161/eigenvectors-and-eigenvalues-of-tridiagonal-matrix/91229#91229 Comment by Johann Cigler Johann Cigler 2012-03-17T07:35:29Z 2012-03-17T07:35:29Z @Emilio: Perhaps I should add a proof. Let $a(n, x)=\det(\mathcal{T}_n(p,q)-x)$. As you have shown this sequence satisfies the recurrence $a(n, x)=-x a(n-1, x)-pqa(n-2,x)$ with initial values $a(1, x)=-x$ and $a(2, x)= x^2-pq.$ Therefore $b(n,x)= (1/{\sqrt {pq}) ^n}a(n, - 2 \sqrt {pq} x)$ satisfies $b(n,x)=2xb(n-1,x)-b(n-2,x)$ with initial values $b(1,x)=2x$ and $b(2,x)=4x^2-1$ and therefore $b(n,x) = {U_n}(x).$ http://mathoverflow.net/questions/91161/eigenvectors-and-eigenvalues-of-tridiagonal-matrix/91229#91229 Comment by Johann Cigler Johann Cigler 2012-03-16T20:37:05Z 2012-03-16T20:37:05Z @Emilio: I think that the correct formula is as in my answer $\det(\mathcal{T}_n(p,q)-\lambda)=(\sqrt{pq})^n U_n\left(\frac{-\lambda}{2\sqrt{pq}}\right)$ with the Chebyshev polynomials of the second kind. http://mathoverflow.net/questions/91161/eigenvectors-and-eigenvalues-of-tridiagonal-matrix/91229#91229 Comment by Johann Cigler Johann Cigler 2012-03-16T09:02:10Z 2012-03-16T09:02:10Z @ Emilio: Could you please show how you get $\det(\mathcal{T}_n(p,q)-\lambda)=2(pq)^n T_n(-\lambda/2)$? http://mathoverflow.net/questions/89996/why-are-some-q-analogues-more-canonical-than-others Comment by Johann Cigler Johann Cigler 2012-03-02T07:53:44Z 2012-03-02T07:53:44Z I am sceptical that there are “canonical” $q-$analogues. So I think the powers of $q$ do not make things canonical. For example the formula $$\prod_{i=0}^{n-1} (1+xq^i) = \sum_{k=0}^n q^{{k\choose 2}}{n\choose k}_qx^k$$ seems to me “more canonical” than the recursion ${r_n}(x) = (1 + x){r_{n - 1}}(x) + ({q^{n - 1}} - 1)x{r_{n - 2}}(x)$ for $r(n)= \sum_{k=0}^n {n\choose k}_qx^k.$ Very often there are two different recurrences as for the $q-$binomial coefficients. Some $q-$analogues have nice formulae, others nice recurrences, but only seldom they have both properties.