cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A362722 a(n) = [x^n] ( E(x)/E(-x) )^n where E(x) = exp( Sum_{k >= 1} A005258(k)*x^k/k ).

Original entry on oeis.org

1, 6, 72, 1266, 23232, 445506, 8740728, 174366114, 3519799296, 71696570010, 1470795168072, 30344633110710, 628994746308288, 13089254107521234, 273292588355096760, 5722454505166750266, 120119862431845048320, 2526922404360157374738, 53260275108329790626952
Offset: 0

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Author

Peter Bala, May 01 2023

Keywords

Comments

It is known that the sequence of Apéry numbers A005258 satisfies the Gauss congruences A005258(n*p^r) == A005258(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
One consequence is that the power series expansion of E(x) = exp( Sum_{k
>= 1} A005258(k)*x^k/k ) = 1 + 3*x + 14*x^2 + 82*x^3 + 551*x^4 + ... has integer coefficients (see, for example, Beukers, Proposition, p. 143). Therefore, the power series expansion of E(x)/E(-x) also has integer coefficients and so a(n) = [x^n] ( E(x)/E(-x) )^n is an integer.
In fact, the Apéry numbers satisfy stronger congruences than the Gauss congruences known as supercongruences: A005258(n*p^r) == A005258(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r (see Straub, Section 1).
We conjecture below that {a(n)} satisfies supercongruences similar to (but weaker than) the above supercongruences satisfied by the Apéry numbers.

Crossrefs

Programs

  • Maple
    A005258 := proc(n) add(binomial(n, k)^2*binomial(n+k,k), k = 0..n) end proc:
    E(n,x) := series(exp(n*add(2*A005258(2*k+1)*x^(2*k+1)/(2*k+1), k = 0..10)), x, 21):
    seq(coeftayl(E(n,x), x = 0, n), n = 0..20);

Formula

a(n) = [x^n] exp( Sum_{k >= 1} n*( 2*A005258(2*k+1)*x^(2*k+1) )/(2*k+1) ).
Conjectures:
1) the supercongruence a(p^r) == a(p^(r-1)) (mod p^(2*r+1)) holds for all primes p >= 5.
2) for n >= 2, a(n*p) == a(n) (mod p^2) holds for all primes p >= 3.
3) for r >= 2, the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for all primes p >= 3 and n >= 1.

A362725 a(n) = [x^n] E(x)^n, where E(x) = exp( Sum_{k >= 1} A005259(k)*x^k/k ).

Original entry on oeis.org

1, 5, 123, 3650, 118059, 4015380, 141175410, 5082313276, 186243853995, 6920379988871, 260030830600748, 9860709859708350, 376821110248674594, 14494688046084958080, 560708803489098556632, 21797478402692370515400, 851057798310071946207915, 33356751162583463626417872
Offset: 0

Views

Author

Peter Bala, May 02 2023

Keywords

Comments

Compare with A362723.
It is known that the sequence of Apéry numbers A005259 satisfies the Gauss congruences A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
One consequence is that the power series expansion of E(x) = exp( Sum_{k
>= 1} A005259(k)*x^k/k ) = 1 + 5*x + 49*x^2 + 685*x^3 + 11807*x^4 + ... has integer coefficients (see, for example, Beukers, Proposition, p. 143), and therefore a(n) = [x^n] E(x)^n is an integer.
In fact, the Apéry numbers satisfy stronger congruences than the Gauss congruences known as supercongruences: A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r (see Straub, Section 1).
We conjecture below that the present sequence satisfies supercongruences similar to (but weaker than) the above supercongruences satisfied by the Apéry numbers.
More generally, we inductively define a family of sequences {a(i,n) : n >= 0}, i >= 0, by setting a(0,n) = A005259(n) and, for i >= 1, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. In this notation the present sequence is {a(1,n)}.
We conjecture that the sequences {a(i,n) : n >= 0}, i >= 1, satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(2*r)) for all primes p >= 3, and positive integers n and r.

Crossrefs

Programs

  • Maple
    A005259 := proc(n) add(binomial(n,k)^2*binomial(n+k,k)^2, k = 0..n) end proc:
    E(n,x) := series(exp(n*add((A005259(k)*x^k)/k, k = 1..20)), x, 21):
    seq(coeftayl(E(n,x), x = 0, n), n = 0..20);

Formula

Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for all primes p >= 3 and positive integers n and r.
Showing 1-2 of 2 results.