All Questions
93 questions
1
vote
0
answers
82
views
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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6
votes
0
answers
171
views
An inequality involving integer partitions
For integers $n\ge k\ge0$, let $p(n,k)$ denote the number of ways to write $n$ as a sum of $k$ positive integers (repetition allowed). For example, $p(6,3)=3$ since
$$6=1+1+4=1+2+3=2+2+2.$$
QUESTION. ...
- 15.6k
2
votes
0
answers
61
views
Algorithm for main diagonal of integer coefficients associated with Schroeder numbers
Let $T_q(n, k)$ be an integer table such that
$$T_q(n, k) = \begin{cases}
1 & \textrm{if } n = 0 \vee k = 0 \\
qT_q(n-1, n-1) + T_q(n, n-1) & \textrm{if } n = k > 0 \\
T_q(n, k-1) + T_q(n-1,...
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6
votes
1
answer
282
views
Integer sequences with a periodic pattern
Let $A$ and $B$ be two different integers. Let $S$ be a finite integer sequence with exactly $n_A$ $A$s and $n_B$ $B$s. By repeating $S$ infinitely many times we obtain an infinite integer sequence $P$...
- 93
1
vote
0
answers
168
views
Integer coefficients and integrals
Let $a(n,p,q)$ be the family of integer sequences such that exponential generating functions for it satisfy
$$
A_1(x)=\exp\left(x + p\int\int (A_1(x))^q \, dx \, dx\right).
$$
Let $b(n,p,q)$ be the ...
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0
votes
0
answers
55
views
Sequences that sum up to sums of integer coefficients
Let
$$
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
214
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
$$
f(n,m,i) = (-1)^{m-i+1}\sum\limits_{j=i}^{m+1}j^n s(j,i) {m+1 \brace ...
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1
vote
0
answers
132
views
Sequence that sums up to A000153
Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \leqslant k ...
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1
vote
0
answers
105
views
Simpler recursion for the A358612
Let $T(n,k)$ be an integer coefficients (A358612) such that
$$
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
$$
...
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1
vote
0
answers
113
views
Representing A329369 using A358612
Let $a(n)$ be A329369 (i.e., number of permutations of $\{1,2,\dotsc,m\}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \...
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4
votes
1
answer
148
views
Closed form for the A110501 (unsigned Genocchi numbers (of first kind) of even index)
Let $a(n)$ be A110501 (i.e., unsigned Genocchi numbers (of first kind) of even index). Here
$$
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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5
votes
0
answers
307
views
On $s$-additive sequences
For a non-negative integer $s$, a strictly increasing sequence of positive integers $\{a_n\}$ is called $s$-additive if for $n>2s$, $a_n$ is the least integer exceeding $a_{n-1}$ which has ...
- 791
1
vote
1
answer
217
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
$$R(n,q)=2(q+2)R(n-1,q+1)+\sum\...
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3
votes
1
answer
138
views
Sequences that sum up to Dowling numbers
Let $a(n,k)$ be the sequence of $k$-Dowling numbers (for more information see A007405 and its CROSSREFS section) with e.g.f.
$$\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}$$
Here square brackets denote Iverson brackets.
There are many sequences in the OEIS that are ...
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6
votes
1
answer
267
views
Sequence that sums up to the number of permutations avoiding the pattern $1-23-4$
Let $a(n)$ be A113227, i.e., the number of permutations on $[n]\equiv \{1, \ldots, n\}$ avoiding the pattern $1-23-4$.
The sequence begins with
$$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
$$a(n) = a(n-1) + (n-1)a(n-2), a(...
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1
vote
0
answers
56
views
Recurrence for the number of permutations with a given excedance set
Let
$$\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
$$1, 2, ...
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1
vote
0
answers
81
views
Infiniteness of the pairs of sequences with a given conditions
Let
$$\varphi=\frac{1+\sqrt{5}}{2}$$
Let
$$a_1(n)=\left\lfloor n\varphi \right\rfloor, a_2(n)=n+a_1(n)$$
Let $\operatorname{tr}(n)$ be A007814, i.e., the number of trailing zeros in the binary ...
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0
votes
0
answers
61
views
Stolarsky array and Stolarsky representation
Let $T(n,k)$ be A035506, i.e., Stolarsky array read by antidiagonals. Here we consider that $T(n,k)=0$ for $n<1, k<1$.
Let $a(n)$ be A200714, i.e., Stolarsky representation interpreted as binary ...
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