All Questions
259 questions
69
votes
2
answers
4k
views
Function that produces primes
For any $n\geq 2$ consider the recursion
\begin{align*}
a(0,n)&=n;\\
a(m,n)&=a(m-1,n)+\operatorname{gcd}(a(m-1,n),n-m),\qquad m\geq 1.
\end{align*}
I conjecture that $a(n-1,n)$ is always ...
- 4,938
57
votes
0
answers
3k
views
On the first sequence without triple in arithmetic progression
In this Numberphile video (from 3:36 to 7:41), Neil Sloane explains an amazing sequence:
It is the lexicographically first among the sequences of positive integers without triple in arithmetic ...
46
votes
5
answers
4k
views
Fibonacci series captures Euler $e=2.718\dots$
The Fibonacci recurrence $F_n=F_{n-1}+F_{n-2}$ allows values for all indices $n\in\mathbb{Z}$. There is an almost endless list of properties of these numbers in all sorts of ways. The below question ...
- 43.2k
42
votes
2
answers
2k
views
Numbers that are generic w.r.t. exponentiation
This is a follow-up to my old question Number of distinct values taken by $x\hat{\phantom{\hat{}}}x\hat{\phantom{\hat{}}}\dots\hat{\phantom{\hat{}}}x$ with parentheses inserted in all possible ways.
...
- 6,819
37
votes
3
answers
2k
views
How to prove the identity $L(2,(\frac{\cdot}3))=\frac2{15}\sum\limits_{k=1}^\infty\frac{48^k}{k(2k-1)\binom{4k}{2k}\binom{2k}k}$?
For the Dirichlet character $\chi(a)=(\frac a3)$ (which is the Legendre symbol), we have
$$L(2,\chi)=\sum_{n=1}^\infty\frac{(\frac n3)}{n^2}=0.781302412896486296867187429624\ldots.$$
Note that this ...
- 15.6k
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
$$a(n)=\frac{1}{n+1}\binom{2n}{...
- 4,938
23
votes
4
answers
2k
views
Identity for an infinite product
Here is an experimental "result" exhibiting the difference of two (formal) infinite products that "almost factorizes".
QUESTION. Is this true?
$$\prod_{n\geq1}(1+x^{2n-1})^{24} - \...
- 43.2k
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 ...
- 3,443
16
votes
0
answers
784
views
How to explain the picturesque patterns in François Brunault's matrix?
How to explain the patterns in the matrix defined in François Brunault's
answer to the question Freeness of a Z[x] module depicted below? --
Choosing colors according to the highest power of 2 which ...
- 19.6k
15
votes
2
answers
601
views
Integer but not Laurent sequences
Are there any sequence given by a recurrence relation:
$x_{n+t}=P(x_t,\cdots,x_{t+n-1})$, where $P$ is a positive Laurent Polynomial, satisfy:
if $x_0=\cdots=x_{n-1}=1$, then the sequence is only ...
15
votes
3
answers
1k
views
Does anyone remember what happened to the experimental search for polynomial identities for $\pi$?
So a while back I was on the internet and had encountered a website containing an experimental search for identities for $\pi$. My memory was that the page belonged to either Jonathan Sondow or ...
- 2,689
14
votes
1
answer
755
views
Generating function of the Thue-Morse sequence
Let $T$ be the generating function of the Thue-Morse sequence; thus,
$T(x)=x+x^2+x^4+x^7+\dotsb$. It is known that $T$ satisfies the nice
congruence
$$ (1+x)^3 T^2(x) + (1+x)^2 T(x) + x \equiv 0 \...
- 23k
13
votes
1
answer
468
views
Four new series for $\pi$ and related identities involving harmonic numbers
Recently, I discovered the following four new (conjectural) series for $\pi$:
\begin{align}\sum_{k=1}^\infty\frac{(5k^2-4k+1)8^k\binom{3k}k}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}}&=\frac{3\pi}2,\...
- 15.6k
12
votes
1
answer
406
views
Looking for a "clever" argument for a $q$-series identity
Consider the below $q$-series identity. One of the things I like about this expansion is how nicely the difference on the left hand side factors to the right hand side of the equation.
$$\prod_{k\geq1}...
- 43.2k
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 ...
- 479
11
votes
3
answers
1k
views
Integrality of a binomial sum
The following sequence appears to be always an integer, experimentally.
QUESTION. Let $n\in\mathbb{Z}^{+}$. Are these indeed integers?
$$\sum_{k=1}^n\frac{(4k - 1)4^{2k - 1}\binom{2n}n^2}{k^2\binom{...
- 43.2k
10
votes
1
answer
625
views
Generating function for A261041
Let $a(n)$ be A261041 (i.e., number of partitions of subsets of $\{1,2,\dotsc,n\}$, where consecutive integers are required to be in different parts).
Let $b(n)$ be an integer sequence with generating ...
- 4,938
10
votes
1
answer
694
views
Prime numbers from permutation
Let $P(n)$ of a sequence $s(1),s(2),s(3),...$ be obtained by leaving $s(1),...,s(n)$ fixed and reverse-cyclically permuting every $n$ consecutive terms thereafter; apply $P(2)$ to $1,2,3,...$ to get $...
- 4,938
9
votes
2
answers
440
views
How to prove this sum involving powers of cosec is an integer?
It is claimed that the following function produces only integer values for all integer $m \geq 1$, $N \geq 2$.
$F(m,N)=\frac{N^m}{2^m}\displaystyle \sum_{j=1}^{N-1} \operatorname{cosec} ^{2m}\left(\...
- 201
9
votes
0
answers
258
views
On a continued fraction and vector $\nu$ of length $n$
Please note that this question has been completely reworked in order not to overload it with unnecessary and useless information.
Let $f(n)$ be an arbitrary function with integer values.
Let $a(n)$ ...
- 4,938
9
votes
0
answers
225
views
On the first sequence without collinear triple
Let $u_n$ be the sequence lexicographically first among the sequences of nonnegative integers with graphs without collinear three points (as for $a_n=n^2$ or $b_n=2^n$). It is a variation of that one.
...
8
votes
1
answer
671
views
Infinite series and sum of two squares
Consider the following infinite sequence $a(n)$ generated by
$$\sum_{n\geq0} a(n)q^n
=\frac{\sum_{k\geq0}F(2k+1)q^{\binom{k+1}2}}{\sum_{k\geq0} q^{\binom{k+1}2}}$$
where the $F(2k+1)$ are the odd ...
- 43.2k
8
votes
2
answers
565
views
integral transform of Fibonacci polynomials is integral
The Fibonacci polynomials are defined recursively by $F_0(x)=0, F_1(x)=1$ and $F_n(x)=xF_{n-1}(x)+F_{n-2}(x)$, for $n\geq2$.
While computing certain integrals, I observe the following (numerically) ...
- 43.2k
8
votes
2
answers
2k
views
5n+1 sequence starting at 7
Consider the following variant of the Collatz function: $f:\mathbb N\rightarrow\mathbb N$ is defined by
\begin{equation}
f(n):=\begin{cases}
n/2 & \text{if $n$ is even}\\
5n+1 & \...
- 654
8
votes
0
answers
237
views
Sequences for which $\prod (1-z^n)^{a(n)}$ is a polynomial
This is mostly a reference request.
I'm working with complex coefficients, although all I have in mind have integer coefficients.
Let $a=(a(n))_{n\ge 1}$ be a sequence, say of integers (I have non-...
- 63.9k
7
votes
3
answers
933
views
In search of an alternative proof of a series expansion for $\log 2$
We all know the series expansion
$$\log 2=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n. \tag1$$
I also am able to use the method of Wilf-Zeilberger to the effect that
$$\log 2=3\sum_{n=1}^{\infty}\frac{(-1)^{...
- 43.2k
7
votes
1
answer
701
views
One conjecture by sequencedb.net
Let $a(n)$ be A214973, number of terms in greedy representation of $n$ using Fibonacci and Lucas numbers.
Let $b(n)$ be A329320, sequence which arises from attempts to simplify computing of A329319. ...
- 4,938
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 ...
- 3,443
7
votes
2
answers
2k
views
series expansion of the q-Pochhammer symbol
The following identity arose while I was working on a recent MO question:
$-\sum_{n=1}^{\infty}\frac{1}{n}\frac{(-x)^n}{1-x^n}=\sum_{n=1}^{\infty}\frac{1}{n}\frac{x^n}{1-x^{2n}}.$
I have no doubt ...
- 188k
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
$$
1, 1, 2, 2, 4, 5, 8, 12, 19, 28, 45, 70, 110, ...
- 4,938
7
votes
1
answer
792
views
Remarkable recursions for the A204262
Let $a(n)$ be A204262 i.e. permanent of the matrix $n\times n$ with elements $\min(i,j)$.
Let
$$
f_{n,\ell}(x)=g_{n,\ell}(x)+f_{n,\ell-1}(\ell)-g_{n,\ell}(\ell), \\
g_{n,\ell}(x)=\int (n-\ell)^2 f_{n-...
- 4,938
7
votes
3
answers
735
views
Expanding in Fibonacci powers
Let $F_n$ denote the all-familiar Fibonacci numbers, with $F_0=0, F_1=1, F_2=1$, etc.
There is a plethora of properties for these numbers involving their sums, products, convolutions and so on. Here, ...
- 43.2k
7
votes
1
answer
145
views
How to prove the relationship between Stern's diatomic series and Lucas sequence $U_n(x,1)$ over the field GF(2)?
I found the bit count of Lucas sequence $U_n(x,1)$ over the field GF(2) is Stern's diatomic series, I want to know the reason?
https://oeis.org/A002487 : Stern's diatomic series
https://oeis.org/...
- 317
7
votes
1
answer
439
views
Two conjectural series for $\pi$ involving the central trinomial coefficients
For each $n=0,1,2,\ldots$, the central trinomial coefficient $T_n$ is defined as the coefficient of $x^n$ in the expansion of $(x^2+x+1)^n$. It is easy to see that $T_n=\sum_{k=0}^{\lfloor n/2\rfloor}\...
- 15.6k
7
votes
1
answer
240
views
$q$-Eulerian type B enjoy symmetry
Let $(q;q)_n=(1-q)(1-q^2)\cdots(1-q^n)$ with $(q;q)_0:=1$. Define a $q$-exponential by
$$e(z;q)=\sum_{n\geq0}\frac{z^n}{(q;q)_n}.$$
There is a notion of $q$-Eulerian polynomials, see the reference. I ...
- 43.2k
7
votes
0
answers
429
views
Dynamics of a curious bijection of $\mathbb N$
The two sequences A48680 and A48679 of the OEIS define two mutually inverse bijections on the set of all strictly positive natural numbers given (for the comfort of the reader) as follows:
Given an ...
- 17.9k
7
votes
0
answers
124
views
in search of intepretations and connections for $k$-central binomials
Fix a positive integer $k$. Then, the sequences
$$c(n,k)=\frac{k^n}{n!}\prod_{m=1}^{n-1}(1+km)=[x^n]\left(\frac1{1-k^2x}\right)^{1/k}$$
are referred to as "$k$-central binomial coefficients",...
- 43.2k
6
votes
4
answers
1k
views
Approximating $e$ with 2s and 3s
How can I generate a series of 2s and 3s such that the average of the generated values (so far) is as close to $e$ as possible?
For example:
...
- 161
6
votes
2
answers
547
views
2-adic valuation of a certain binomial sum
Consider the sequence (of rational numbers) given by
$$a_n=\sum_{k=1}^n\binom{n}k\frac{k}{n+k}.$$
Let $s(n)$ be the sum of binary digits of $n$, i.e. the total number of $1$'s.
QUESTION. Is it true ...
- 43.2k
6
votes
2
answers
371
views
Sequence of $k^2$ and $2k^2$ ordered in ascending order
Let $\eta(n)$ be A006337, an "eta-sequence" defined as follows:
$$\eta(n)=\left\lfloor(n+1)\sqrt{2}\right\rfloor-\left\lfloor n\sqrt{2}\right\rfloor$$
Sequence begins
$$1, 2, 1, 2, 1, 1, 2, ...
- 4,938
6
votes
1
answer
367
views
On A057985 and A287066
Let $a(n)$ be A057985 (i.e., start with $0$ and repeatedly substitute: $0 \to 01$, $1 \to 12$, $2 \to 0$).
Let $b(n)$ be A287066 (i.e., start with $1$ and repeatedly substitute: $0 \to 01$, $1 \to 12$...
- 4,938
6
votes
1
answer
282
views
Integer sequences with a periodic pattern
Let $A$ and $B$ be two different integers. Let $S$ be a finite integer sequence with exactly $n_A$ $A$s and $n_B$ $B$s. By repeating $S$ infinitely many times we obtain an infinite integer sequence $P$...
- 93
6
votes
1
answer
484
views
Three conjectural series for $\pi^2$ and related identities
Recently, I found the following three (conjectural) identities for $\pi^2$:
$$\sum_{k=1}^\infty\frac{145k^2-104k+18}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{\pi^2}3,\tag{1}$$
$$\sum_{k=1}^\infty\frac{...
- 15.6k
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 $A = n$, $B = 1$, $...
- 4,938
6
votes
1
answer
302
views
A 3rd formula for the central Delannoy numbers?
There are several in the literature proving the two alternative formulas for the (diagonal) Delannoy numbers; namely that
$$d_n=\sum_{k=0}^n\binom{n}k\binom{n+k}k=\sum_{k=0}^n\binom{n}k^22^k.$$
Each ...
- 43.2k
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
$$1, 1, 2, 6, 23, 105, 549, 3207, 20577, 143239, 1071704,...
- 4,938
6
votes
1
answer
184
views
Does asymptotic behavior of $\left|\sum_{d|n}f(d)\right|$ imply asymptotic properties of $f(d)$?
The classic example of a function that has a drastic cancelation when summed over divisors is $\mu(n)$, with complete cancellation for every number other than $1$. Another such function is the ...
- 2,902
6
votes
0
answers
752
views
For all $n\in \mathbb{N}$, How to find $\min\{ m+k\}$ such that $ \binom{m}{k}=n$?
I asked this question on MSE here.
Most numbers in pascal triangle appear only once (excluding the duplicates in the same row of the Pascal's triangle) but certain numbers appear multiple times. ...
- 541
6
votes
0
answers
245
views
Searching for a proof of the pattern and identification of integer coefficients for the A329369
Please see the update given below. Everything you need to know from the old version of the question are the functions $a(n), \ell(n), s(n), t(n), r(n)$.
Let $a(n)$ be A329369 (i.e, number of ...
- 4,938
6
votes
0
answers
235
views
A curious series for $L(2,(\frac{-3}{\cdot}))$
Let
$$K:=L\left(2,\left(\frac{-3}{\cdot}\right)\right)=\sum_{k=1}^\infty\frac{(\frac k3)}{k^2}=\sum_{j=0}^\infty\left(\frac1{(3j+1)^2}-\frac1{(3j+2)^2}\right),$$ where $(\frac k3)$ is the Legendre ...
- 15.6k