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Generating functions related to generating function of Catalan numbers

Let $C_n$ be A000108 (i.e., Catalan numbers). Here generating function is $C(x)$ such that $$ C(x) = \frac{1-\sqrt{1-4x}}{2x}. $$ Let $a(n)$ be an integer sequence with generating function $A(x)$ such ...
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1 vote
0 answers
84 views

Coarse well-distributedness/equidistribution of Pell sequence prefixes

I am interested in the distributedness or "mixing" behavior of certain linear recurrences modulo powers of $2$. In particular, consider the Pell sequence (https://oeis.org/A000129), modulo $...
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2 votes
0 answers
62 views

Algorithm for main diagonal of integer coefficients associated with Schroeder numbers

Let $T_q(n, k)$ be an integer table such that $$T_q(n, k) = \begin{cases} 1 & \textrm{if } n = 0 \vee k = 0 \\ qT_q(n-1, n-1) + T_q(n, n-1) & \textrm{if } n = k > 0 \\ T_q(n, k-1) + T_q(n-1,...
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6 votes
1 answer
282 views

Integer sequences with a periodic pattern

Let $A$ and $B$ be two different integers. Let $S$ be a finite integer sequence with exactly $n_A$ $A$s and $n_B$ $B$s. By repeating $S$ infinitely many times we obtain an infinite integer sequence $P$...
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2 votes
0 answers
28 views

On doubling or addition formulas for the sequence $a(n)=(b_1 n +b_2)a(n-1)+(b_3 n + b_4)a(n-2)$

We are interested which integer sequences are efficiently computable possibly over finite rings. Define the integer sequence $a(n)=(b_1 n +b_2)a(n-1)+(b_3 n + b_4)a(n-2)$ with initial terms $a(0),a(1)$...
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6 votes
2 answers
389 views

Conjectured Somos-like closed form of recurrences with polynomial coefficients

From Our short paper For polynomial $F$ with integer coefficients, define the recurrence $f(n)=F(n,f(n-1),f(n-2),...,f(n-d))$. We conjecture that $f(n)$ satisfy Somos like sequence $f(n)=\frac{G(f(n-1)...
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  • 25.4k
1 vote
0 answers
168 views

Integer coefficients and integrals

Let $a(n,p,q)$ be the family of integer sequences such that exponential generating functions for it satisfy $$ A_1(x)=\exp\left(x + p\int\int (A_1(x))^q \, dx \, dx\right). $$ Let $b(n,p,q)$ be the ...
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0 votes
0 answers
28 views

Short periods modulo primes of linear recurrences with polynomial coefficients

Let $f_i(x)$ be polynomials with integer coefficients. Define the integer linear recurrence with polynomial coefficients: $$ a(n)=f_1(n) a(n-1)+f_2(n)a(n-2)+\cdots +f_d(n) a(n-d) $$ and the initial ...
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0 votes
0 answers
55 views

Sequences that sum up to sums of integer coefficients

Let $$ T(n,k,p,q,r,s) = (q(k-1)+1)T(n-1,k,p,q,r,s) + s(n+r(k-1)+p-2)T(n-1,k-1,p,q,r,s), \\ T(n,1,p,q,r,s) = 1, \\ T(n,0,p,q,r,s) = T(0,k,p,q,r,s) = 0 $$ Let $$ \ell(n) = \left\lfloor\log_2 n\right\...
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2 votes
1 answer
216 views

Simplification of the closed form for the A329369

Let $s(n,k)$ be a (signed) Stirling number of the first kind. Let ${n \brace k}$ be a Stirling number of the second kind. Let $$ f(n,m,i) = (-1)^{m-i+1}\sum\limits_{j=i}^{m+1}j^n s(j,i) {m+1 \brace ...
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1 vote
0 answers
133 views

Sequence that sums up to A000153

Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \leqslant k ...
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1 vote
0 answers
106 views

Simpler recursion for the A358612

Let $T(n,k)$ be an integer coefficients (A358612) such that $$ T(2n+1, k) = kT(n, k) + T(n, k-1), \\ T(2n, k) = kT(n, k) + T(n, k-1) - \frac{T(2n, k-1) + T(n, k-1)}{k-1}, \\ T(n, 1) = T(0, 2) = 1 $$ ...
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1 vote
0 answers
115 views

Representing A329369 using A358612

Let $a(n)$ be A329369 (i.e., number of permutations of $\{1,2,\dotsc,m\}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \...
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4 votes
1 answer
148 views

Closed form for the A110501 (unsigned Genocchi numbers (of first kind) of even index)

Let $a(n)$ be A110501 (i.e., unsigned Genocchi numbers (of first kind) of even index). Here $$ a(n) = \sum\limits_{i=1}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{n}{2i}a(n-i)(-1)^{i-1}, \\ a(1) = 1 ...
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3 votes
0 answers
128 views

Fast and simple algorithm for the A329369

Let $a(n)$ be A329369 (i.e., number of permutations of $\{1,2,\cdots,m\}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \...
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6 votes
0 answers
245 views

Searching for a proof of the pattern and identification of integer coefficients for the A329369

Please see the update given below. Everything you need to know from the old version of the question are the functions $a(n), \ell(n), s(n), t(n), r(n)$. Let $a(n)$ be A329369 (i.e, number of ...
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1 vote
1 answer
221 views

Correctness of the algorithm for the A329369, A347205 and related sequences

Let $a(n)$ be A347205. It is enough for us to know that $$ a(2^m(2k+1)) = \sum\limits_{j=0}^{m}a(2^jk), \\ a(0) = 1 $$ Let $b(n)$ be A329369. It is enough for us to know that $$ b(2^m(2k+1)) = \sum\...
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6 votes
1 answer
393 views

Test for pair of odd primes $(p, 2p^2-1)$

Let $a(n)$ be A106483 (i.e., primes $p$ such that $2p^2-1$ is also prime). Let $b(n)$ be an integer sequence such that $b(n) = B$ after the whole transformation where we start with $A = n$, $B = 1$, $...
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6 votes
1 answer
367 views

On A057985 and A287066

Let $a(n)$ be A057985 (i.e., start with $0$ and repeatedly substitute: $0 \to 01$, $1 \to 12$, $2 \to 0$). Let $b(n)$ be A287066 (i.e., start with $1$ and repeatedly substitute: $0 \to 01$, $1 \to 12$...
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1 vote
1 answer
77 views

Sequence derived from transform of a given vector (with Fibonacci as partial sums)

Let F_n be A000045 (i.e., Fibonacci numbers). Here $$ F_n = F_{n-1} + F_{n-2}, \\ F_0 = 0, F_1 = 1 $$ Let $\operatorname{wt}(n)$ be A000120 (i.e., number of ones in the binary expansion of $n$). ...
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2 votes
1 answer
131 views

Sequence that sums up to A224071

Let $a(n)$ be A224071 (i.e., number of Schroeder paths of semilength $n$ in which there are no $(2,0)$-steps at level $1$). Here $$ a(n) = \frac{1}{2(n+1)}\sum\limits_{k=0}^{n}(k+1)((-1)^{\left\...
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2 votes
2 answers
242 views

Negated Fibonacci and the floor function

Let $F_n$ be A000045 (i.e., Fibonacci numbers). Here $$ F_n = F_{n-1} + F_{n-2}, \\ F_0 = 0, F_1 = 1, \\ F_{-n} = (-1)^{n-1}F_n $$ I conjecture that $$ F_{-n} = \left\lfloor\frac{n+1}{2}\right\rfloor ...
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0 votes
0 answers
190 views

On a A057985 without recursion

Let $a(n)$ be A057985 (i.e., start with $0$ and repeatedly substitute: $0\to01, 1\to12, 2\to0$). Let $\operatorname{wt}(n)$ be A000120 (i.e., number of ones in the binary expansion of $n$). Here $$ \...
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7 votes
1 answer
527 views

Suitable closed form for the A079501

Let $a(n)$ be A079501 (i.e., number of compositions of the integer $n$ with strictly smallest part in the first position). The sequence begins with $$ 1, 1, 2, 2, 4, 5, 8, 12, 19, 28, 45, 70, 110, ...
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2 votes
1 answer
120 views

Recursion for the Chebyshev transform of $m^n$

Let $$ R(n, q, m) = R(n-1, q+1, m) + \sum\limits_{j=0}^{q} (-1)^{q-j}R(n-1, j, m), \\ R(0, q, m) = (m-1)^q $$ I conjecture that $R(n, 0, m)$ is a Chebyshev transform of $m^n$. Examples of Chebyshev ...
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2 votes
0 answers
163 views

Interesting conjecture by Sequence Machine

Let $a(n)$ be A344960 (i.e., position of binary complement of $n$-th word in A341258). By definition, in order to calculate $a(n)$, we need to know A341258. Below we will correspond this sequence with ...
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0 votes
0 answers
63 views

Pairs of permutations such that $p(n)<2^k$ iff $n<2^k$

Let $p(n)$ be an arbitrary permutation of natural numbers such that $p(n)<2^k$ iff $n<2^k$. Let $q(n)$ be an inverse permutation of $p(n)$. Let $$ \ell(n)=\left\lfloor\log_2 n\right\rfloor $$ ...
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5 votes
0 answers
133 views

Formula and smallest solution for the A260711

Let $a(n)$ be A260711 without initial $0$ (i.e., numbers of the form $x^2 - y^2$ with $x > y$ where $x$ and $y$ are odd, $x + y$ is a power of $2$). The sequence begins with $$ 8, 16, 32, 48, 64, ...
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1 vote
0 answers
68 views

On a numbers $k$ with specific $2$-adic valuation

Let $a(n)$ be A002326 (i.e., multiplicative order of $2 \operatorname{mod} 2n+1$). Let $b(n)$ be A179382 (i.e., the smallest period of pseudo-arithmetic progression with initial term $1$ and ...
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0 votes
0 answers
107 views

Formula for individual term of the Proth numbers

Let $a(n)$ be A080075 i.e. Proth numbers: of the form $k2^m + 1$ for $k$ odd, $m \geqslant 1$ and $2^m > k$. The sequence begins with $$ 3, 5, 9, 13, 17, 25, 33, 41, 49, 57, 65, 81, 97, 113, 129 $$...
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2 votes
0 answers
72 views

Possible subsequence of the A110978

Let $a(n)$ be A110978 i.e. odd integers that are nonprime, such that there exist two factors of each number that when multiplied together in binary base, do not ever require the use of a "carry&...
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1 vote
0 answers
125 views

On a Fibonacci and binary

Let F(n) be A000045 i.e. Fibonacci numbers. Here $$ F(n) = F(n-1) + F(n-2), \\ F(0) = 0, F(1) = 1 $$ Let $$ \ell(n) = \left\lfloor\log_2 n\right\rfloor $$ Let $$ T(n, k) = \left\lfloor\frac{n}{2^k}\...
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3 votes
0 answers
120 views

Sequence which is related to the binary expansion of $n$ and partition numbers

Let $p(n)$ be A000041 i.e. the number of partitions of $n$ (the partition numbers). Let $$ \ell(n)=\left\lfloor\log_2 n\right\rfloor $$ Let $\operatorname{wt}(n)$ be A000120 i.e. number of $1$'s in ...
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0 votes
1 answer
122 views

Permutation of the natural numbers from operation related to binary expansion of $n$

Let $$ \ell(n) = \left\lfloor\log_2 n\right\rfloor $$ Let $T(n,k)$ be a $(k+1)$-th bit from the right side in the binary expansion of $n$. Here $$ T(n, k) = \left\lfloor\frac{n}{2^k}\right\rfloor \...
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2 votes
0 answers
199 views

Not a twin prime pair test using $\gcd$ only

Let $m$ be an odd positive integer such that $m=2k+1$, $k\in\mathbb{N}$. Let $v$ be a vector of $n$ positive integers. Let $v(i)$ be the $i$-th element of the vector. Then we start with $v(i)=m(i+1)-2$...
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3 votes
0 answers
69 views

Sequence that sum up to A343685

Let $a(n)$ be A343685 i.e. $$ a(n)=2na(n-1)+\sum\limits_{j=0}^{n-1}\binom{n}{j}(n-j-1)!a(j), \\ a(0)=1 $$ Here the exponential generating function $A(x)$ satisfy $$ A(x)=\frac{1}{1-2x+\log(1-x)} $$ ...
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1 vote
0 answers
111 views

Recursion for the Bessel polynomial $y_n(x)$

Let $a(n)$ be A001515 i.e. the Bessel polynomial $y_n(x)$ evaluated at $x=1$. Here $$ a(n) = (2n-1)a(n-1) + a(n-2), \\ a(0) = 1, a(1) = 2 $$ The closed form is $$ a(n)=\sum\limits_{k=0}^{n}\binom{n+k}{...
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1 vote
0 answers
94 views

Combinatorial interpretation for the more general case of $R(n,0)$

Let $f(n), g(n,m), h(n)$ be an arbitrary functions which equal to the non-negative integers. Let $$ R(n,q) = \sum\limits_{j=0}^{f(q)}g(q,j)R(n-1,j),\\ R(0,q) = h(q) $$ In the comment to the one of ...
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0 votes
1 answer
140 views

Series reversion using something like continued fraction

Let $f(n)$ be an arbitrary function such that $f(n)\in\mathbb{Z}$. Let $$ F(x)=\sum\limits_{m\geqslant 0}f(m)x^m $$ Define the operator $\operatorname{SR}$, which is associated with the series ...
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26 votes
1 answer
7k views

Elegant recursion for A301897

Let $a(n)$ be A301897, i.e., number of permutations $b$ of length $n$ that satisfy the Diaconis-Graham inequality $I_n(b) + EX_n(b) \leqslant D_n(b)$ with equality. Here $$a(n)=\frac{1}{n+1}\binom{2n}{...
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2 votes
0 answers
126 views

Recurrence for A004208

Let $a(n)$ be A004208. Here $$a(n)=n\prod\limits_{j=1}^{n}(2j-1)-\sum\limits_{i=1}^{n-1}a(i)\prod\limits_{j=1}^{n-i}(2j-1)$$ I conjecture that $$a(n)=R(n-1,0)$$ where $$R(n,q)=2(q+2)R(n-1,q+1)+\sum\...
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3 votes
1 answer
140 views

Sequences that sum up to Dowling numbers

Let $a(n,k)$ be the sequence of $k$-Dowling numbers (for more information see A007405 and its CROSSREFS section) with e.g.f. $$\operatorname{exp}\left(x + \frac{\operatorname{exp}(kx) - 1}{k}\right)$$ ...
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2 votes
0 answers
105 views

Sequences that sum up to the many sequences in the OEIS

Let $$a(n,m,k)=\frac{1}{n}\sum\limits_{j=0}^{n}[n+kj\geqslant 0]\binom{n}{j}\binom{n+kj}{j-1}(m-1)^{j-1}$$ Here square brackets denote Iverson brackets. There are many sequences in the OEIS that are ...
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6 votes
1 answer
268 views

Sequence that sums up to the number of permutations avoiding the pattern $1-23-4$

Let $a(n)$ be A113227, i.e., the number of permutations on $[n]\equiv \{1, \ldots, n\}$ avoiding the pattern $1-23-4$. The sequence begins with $$1, 1, 2, 6, 23, 105, 549, 3207, 20577, 143239, 1071704,...
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2 votes
0 answers
70 views

Integer coefficients such that $T(n,k)=R(n,k)-R(n,k-1)$

Let $a(n)$ be A000085, i.e., the number of self-inverse permutations on $n$ letters, also known as involutions; number of standard Young tableaux with $n$ cells. Here $$a(n) = a(n-1) + (n-1)a(n-2), a(...
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1 vote
0 answers
57 views

Recurrence for the number of permutations with a given excedance set

Let $$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$ $$f(n)=n-2^{\ell(n)}$$ $$T(n,k)=\left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2$$ $$\operatorname{wt}(2n+1)=\operatorname{wt}(n)+1, \...
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1 vote
0 answers
134 views

Recurrence for the A284005

Let $$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$ $$f(n)=n-2^{\ell(n)}$$ $$T(n,k)=\left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2$$ $$\operatorname{wt}(2n+1)=\operatorname{wt}(n)+1, \...
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2 votes
0 answers
76 views

Uniqueness of the permutation

Let $f(n)$ be A000045(n), i.e., Fibonacci numbers: $f(n)=f(n-1)+f(n-2)$ for $n>1$ with $f(0)=0$ and $f(1)=1$. Let $g(n)$ be A072649, i.e., $n$ occurs $f(n)$ times. The sequence begins with $$1, 2, ...
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1 vote
0 answers
109 views

Existence of binary permutations with a given property

Let $$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$ Let $$f(n)=n-2^{\ell(n)}$$ Let $a(n)$ be a permutation of the nonnegative integers such that $a(0)=0$, $a(n)=n$ if $n$ is a power of $2$ and ...
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2 votes
1 answer
172 views

Permutation and its binary analog

Let $f(n)$ be A000045(n), i.e., Fibonacci numbers: $f(n)=f(n-1)+f(n-2)$ for $n>1$ with $f(0)=0$ and $f(1)=1$. Let $g(n)$ be A072649, i.e., $n$ occurs $f(n)$ times. The sequence begins with $$1, 2, ...
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