A357567 a(n) = 5*A005259(n) - 14*A005258(n).

-9, -17, 99, 5167, 147491, 3937483, 105834699, 2907476527, 81702447651, 2342097382483, 68273597307599, 2018243113678027, 60365426282638091, 1823553517258576723, 55557712038989195099, 1705170989220937925167, 52672595030914982754851, 1636296525812843554700323

Offset

0,1

Comments

Conjectures:

1) a(p) == a(1) (mod p^5) for all primes p >= 5.

2) For r >= 2, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for all primes p >= 5.

These are stronger supercongruences than those satisfied separately by the two types of Apéry numbers A005258 and A005259.

From Peter Bala, Oct 25 2022: (Start)

Additional conjectures:

3) the sequence {u(n): n>= 1} defined by u(n) = (3^42)*A005259(n)^25 - (5^25)* A005258(n)^42 also satisfies the congruences in 1) and 2) above.

4) u(n) == 0 (mod n^5) for integer n of the form 3^i*5^j (see A003593). (End)

Links

Table of n, a(n) for n=0..17.

Formula

a(n) = 5*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2 - 14*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k).

For positive integers n and r, a(n*p^r) == a(n*p^(r-1)) ( mod p^(3*r) ) for all primes p >= 5.

Example

a(11) - a(1) = 2018243113678027 + 17 = (2^2)*(3^2)*(11^5)*17*20476637 == 0 (mod 11^5).

Maple

seq(add(5*binomial(n, k)^2*binomial(n+k, k)^2 - 14*binomial(n, k)^2*binomial(n+k, k), k = 0..n), n = 0..20);

Crossrefs

Cf. A005258, A005259, A212334, A352655, A357506, A357507, A357508, A357509, A357568, A357569, A357956, A357957, A357958, A357959, A357960.

Sequence in context: A253212 A217965 A121442 * A049440 A177200 A177166

Adjacent sequences: A357564 A357565 A357566 * A357568 A357569 A357570

Keyword

sign,easy

Author

Peter Bala, Oct 19 2022

Status

approved