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34 votes
1 answer
3k views

A remarkable almost-identity

OEIS sequence A210247 gives the signs of $\text{li}(-n,-1/3) = \sum_{k=1}^\infty (-1)^k k^n/3^k$, also the signs of the Maclaurin coefficients of $4/(3 + \exp(4x))$. Mikhail Kurkov noticed that it ...
Robert Israel's user avatar
32 votes
0 answers
2k views

A question related to the Hofstadter–Conway \$10000 sequence

The Hofstadter–Conway \$10000 sequence is defined by the nested recurrence relation $$c(n) = c(c(n-1)) + c(n-c(n-1))$$ with $c(1) = c(2) = 1$. This sequence is A004001 and it is well-known that this ...
Alkan's user avatar
  • 701
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}{...
Notamathematician's user avatar
24 votes
1 answer
2k views

Parity of the multiplicative order of 2 modulo p

Let $\operatorname{ord}_p(2)$ be the order of 2 in the multiplicative group modulo $p$. Let $A$ be the subset of primes $p$ where $\operatorname{ord}_p(2)$ is odd, and let $B$ be the subset of primes $...
Shahab's user avatar
  • 429
24 votes
0 answers
1k views

Is A276175 integer-only?

The terms of the sequence A276123, defined by $a_0=a_1=a_2=1$ and $$a_n=\dfrac{(a_{n-1}+1)(a_{n-2}+1)}{a_{n-3}}\;,$$ are all integers (it's easy to prove that for all $n\geq2$, $a_n=\frac{9-3(-1)^n}{2}...
uvdose's user avatar
  • 655
20 votes
2 answers
1k views

A possibly surprising appearance of $\sqrt{2}.$

Define $A=(a_n)$ and $B=(b_n)$ as follows: $a_0=1$, $a_1=2$, $b_0=3$, $b_1=4$, and $$a_n=a_1b_{n-1}-a_0b_{n-2} + 2n$$ for $n \geq 2$, where $A$ and $B$ are increasing and every positive integer occurs ...
Clark Kimberling's user avatar
12 votes
1 answer
427 views

Subwords of the infinite Fibonacci word

Let $W = 01001010010010 \ldots$ be the infinite Fibonacci word, A003849 in the OEIS. Let $B(m)$ be the set of $m+1$ subwords of $W$ that have length $m$, and for each such subword $u$, let $p(u)$ be ...
Clark Kimberling's user avatar
8 votes
0 answers
237 views

Sequences for which $\prod (1-z^n)^{a(n)}$ is a polynomial

This is mostly a reference request. I'm working with complex coefficients, although all I have in mind have integer coefficients. Let $a=(a(n))_{n\ge 1}$ be a sequence, say of integers (I have non-...
YCor's user avatar
  • 63.9k
8 votes
0 answers
145 views

Minimum length of sequence such that every integer from 1 to n can be achieved as the sum of some contiguous subsequence

This question literally came to me in a fever dream last night, and it's frustrating me to no end. I'll try to explain it as best I can, but there may not be a satisfying answer; the best outcome ...
Joachim Worthington's user avatar
7 votes
2 answers
428 views

Limit associated with complementary sequences

Define $A=(a_n)$ and $B=(b_n)$ as follows: $a_0=1$, $a_1=2$, $b_0=3$, $b_1=4$, and $$a_n=a_0b_{n-1}+a_1b_{n-2}$$ for $n \geq 2$, where $A$ and $B$ are increasing and every positive integer occurs ...
Clark Kimberling's user avatar
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, ...
Notamathematician's user avatar
7 votes
1 answer
283 views

On one class of Somos-like sequences

This question is motivated by integrability of the sequence mistakenly arisen in the question Does this sequence always give an integer? Let $m_1,\ldots, m_{k-1}$ be positive integers and sequence $\{...
Alexey Ustinov's user avatar
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: $0 \to 01$, $1 \to 12$...
Notamathematician's user avatar
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 $P$...
De Costa's user avatar
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 $A = n$, $B = 1$, $...
Notamathematician's user avatar
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)...
joro's user avatar
  • 25.4k
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,...
Notamathematician's user avatar
6 votes
1 answer
240 views

On the growth and bounds for a certain sequence of integers known as Bogotá numbers

A Bogotá number is a non-negative integer equal to some smaller number, or itself, times its digital product, i.e. the product of its digits. For example, 138 is a Bogotá number because 138 = 23 x (2 ...
Bernardo Recamán Santos's user avatar
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 ...
Notamathematician's user avatar
5 votes
5 answers
677 views

Can an integer or rational sequence satisfy some bounded order recurrence $\mod \ $ almost all primes but doesn't satisfy such in $\mathbb{Q}$?

Can an integer or rational sequence satisfy some bounded order recurrence $\mod \ $ almost all primes but doesn't satisfy such in $\mathbb{Q}$? The recurrences $\mod p$ can be different, possibly ...
joro's user avatar
  • 25.4k
5 votes
2 answers
1k views

Proof that $3^ns + \sum_{k=0}^{n-1} 3^{n-k-1}2^{a_k}=2^m.$

How would I go about proving the following: For any odd positive integer $s$, there exists a sequence of nonnegative integers $( a_0, a_1, \cdots, a_{n-1})$ and a nonnegative integer $m$ such that, $...
ReverseFlowControl's user avatar
5 votes
1 answer
303 views

Simply generated sequences with mysterious differences

Suppose that $a_0 < a_1,$ $b_0 < b_1,$ and $$a_n=a_1b_{n-1}+a_0b_{n-2}+qn+r$$ for $n \geq 2$, where $a_0,a_1,b_0,b_1,q,r$ are integers such that $(a_n)$ and $(b_n)$ are increasing and ${(|a_n|)}$...
Clark Kimberling's user avatar
5 votes
1 answer
310 views

In the Oldenburger-Kolakoski sequence, is #1s = #2s infinitely many times?

The Oldenburger-Kolakoski sequence, $OK$, is the unique sequence of $1$s and $2$s that starts with $1$ and is its own runlength sequence: $$OK = (1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2,1,1,\ldots).$...
Clark Kimberling's user avatar
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, ...
Notamathematician's user avatar
4 votes
1 answer
168 views

An inequality involving $k$-generalized Fibonacci numbers

I have worked on a Diophantine equation by using transcendental and reduction methods given by Baker and Davenport. However, to solve completely the equation I have one complicated case and I proved ...
Davi's user avatar
  • 41
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 ...
Notamathematician's user avatar
4 votes
1 answer
219 views

Numbers $n$ whose representation as the product of two divisors require more digits than that of $n$

Note: Posting in MO since it was unanswered in MSE Let $f(x)$ be the number of digits in the decimal representation of $x$ e.g. $, f(0) = 1, f(1729) = 4$. If $n = ab$ then we can show that $f(ab) > ...
Nilotpal Kanti Sinha's user avatar
4 votes
1 answer
217 views

Why do convoluted convolved Fibonacci numbers pop up from this triangle?

Start with this triangle (OEIS A118981). This triangle is simple to generate with the following recurrence relation (though $T(0,0)$ ends up different from the OEIS version): $$ T(0,0) = 2;T(1,0) = 1;...
Mitch's user avatar
  • 194
4 votes
0 answers
97 views

When is $\lfloor C^n \rfloor \mod b$ efficiently computable?

For real irrational $C > 1 $ and natural $n,b$, define $a(C,n,b)=\lfloor C^n \rfloor \mod b$ Q1 For which $C,b$ is $a(C,n,b)$ computable in time polynomial in $\log{n}$? Searching in OEIS ...
joro's user avatar
  • 25.4k
3 votes
1 answer
240 views

The sequence $a(n)=(2^n \bmod p)^{p-1} \bmod p^2$

Related to this question. Let $p$ be prime and $n$ positive integer. Define $a(n)=(2^n \bmod p)^{p-1} \bmod p^2$ Let $D(n)$ be the base $2$ discrete logarithm of $a(n)$, i.e. given $p,a(n)$ we have $2^...
joro's user avatar
  • 25.4k
3 votes
1 answer
138 views

Properties of a certain sequence

During research I came to the following sequence: Let $\lambda>1$ and define $n_{k+1}=\text{IntergerPart}[\lambda\cdot n_k]$ where we assume that $n_0$ is sufficently large integer, so that the ...
Luka Thaler's user avatar
3 votes
1 answer
92 views

Partition of $(2^{n+1}+1)2^{2^{n-1}+n-1}-1$ into parts with binary weight equals $2^{n-1}+n$

Let $\operatorname{wt}(n)$ be A000120, i.e., number of $1$'s in binary expansion of $n$ (or the binary weight of $n$). Let $a(n,m)$ be the sequence of numbers $k$ such that $\operatorname{wt}(k)=m$. ...
Notamathematician's user avatar
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. $$\operatorname{exp}\left(x + \frac{\operatorname{exp}(kx) - 1}{k}\right)$$ ...
Notamathematician's user avatar
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 \...
Notamathematician's user avatar
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 ...
Notamathematician's user avatar
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 $$ A(x)=\frac{1}{1-2x+\log(1-x)} $$ ...
Notamathematician's user avatar
3 votes
0 answers
165 views

Closed form for $a(2^m(2^n-2^p-1))$

Let $q(n)$ be A007814, i.e., the number of trailing zeros in the binary representation of $n$. Here $$q(2n+1)=0, q(2n)=q(n)+1$$ Let $a(n)$ be A329369. Here $$a(2n+1)=a(n), a(2n)=a(n)+a(n-2^{q(n)})+a(...
Notamathematician's user avatar
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 ...
Notamathematician's user avatar
2 votes
1 answer
740 views

Power tower made of $2$s and $3$s: too high, too soon?

A power tower of a number $x$ is typified by $$ x^{x^{x^{x^{x^{x^{x^{x^{x^x}}}}}}}}.$$ Here, however, we take the liberty of referring to the set $T$ of "$\{2,3\}$-power towers"; i.e., numbers $$...
Clark Kimberling's user avatar
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 ...
Notamathematician's user avatar
2 votes
1 answer
153 views

Bounds for the sequence $a(n)=a(n-1)+a(\lfloor n-n^A \rfloor)$

Related to the question about a(n)=a(n-1)+a(floor(n/2)) Let $A$ be real constant $ 0 < A < 1$. Define the sequence $a(n)$ by $a(1)=1, a(n)=a(n-1)+a(\lfloor n-n^A \rfloor)$ (if you prefer take $a'...
joro's user avatar
  • 25.4k
2 votes
1 answer
913 views

Numbers that are the sum of 2 distinct nonzero squares in exactly 1 way [closed]

I need to emulate this sequence for a program: http://oeis.org/A025302 Stuff that I've taken into account: After finding the prime divisors of a number. I take any divisor as p and apply the ...
Greggz's user avatar
  • 123
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\...
Notamathematician's user avatar
2 votes
1 answer
532 views

The Euler's totient function and the product of distinct primes dividing $n$ versus the Heronian means

For integers $n\geq 1$ with $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p$$ we denote the squarefree kernel or radical of an integer $n$ (see if you want this Wikipedia). And $\...
user avatar
2 votes
1 answer
154 views

GCD for two Cullen numbers

The $n$'th Cullen number is $C_n = n\cdot2^n+1$. If $m$ and $n$ are natural numbers, what can one say about $\gcd(C_n,C_m)$, where $m$ and $n$ are different positive integers?
dalibor's user avatar
  • 21
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 ...
Notamathematician's user avatar
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, ...
Notamathematician's user avatar
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,...
Notamathematician's user avatar
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 $a(0),a(1)$...
joro's user avatar
  • 25.4k
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 ...
Notamathematician's user avatar