A363984 a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n,k)*binomial(n+k,k)*A363983(k).

1, 3, 73, 2163, 75001, 2835003, 113329945, 4711519347, 201638246905, 8824346685003, 393088036809073, 17764622316152715, 812477640612743977, 37535247213943518315, 1749047441756088054073, 82108960863923963522163, 3879675478363506548275705

Offset

0,2

Comments

The Legendre transform of a sequence {b(n)} is the sequence {c(n)} defined by c(n) = Sum_{k = 0..n} binomial(n,k)*binomial(n+k,k)*b(k).

It is known that the sequence of Apéry numbers A005259 is the Legendre transform of the sequence of Franel numbers A000172. Note that A000172(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(2*k,n) (the first Strehl identity). Here we consider the Legendre transform of (a signed version of) A363983, which by analogy with Strehl's identity for A000172 is given by A363981(n) = (-1)^n * Sum_{k = 0..n} binomial(-n,k)* binomial(n,k)*binomial(2*k,n).

The Apéry numbers A005259 and also A005258 satisfy the pair of supercongruences (see, for example, Straub, Introduction)

(1) u(n*p^r - 1) == u(n*p^(r-1) - 1) (mod p^(3*r)) and

(2) u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)),

with both congruences valid for all primes p >= 5 and all positive integers n and r.

Straub, Example 3.4, observes that, among the known Apéry-like numbers, the Apéry numbers A005258 and A005259 are the only ones to satisfy shifted supercongruences of the form (1) in addition to the supercongruences of the form (2).

We conjecture that the present sequence satisfies the same pair of congruences.

Links

Table of n, a(n) for n=0..16.

Armin Straub, Multivariate Apéry numbers and supercongruences of rational functions, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; arXiv preprint, arXiv:1401.0854 [math.NT], 2014.

Eric W. Weisstein's World of Mathematics, Legendre Transform

Eric W. Weisstein's World of Mathematics, Strehl identities

Example

Examples of supercongruences:
p = 11:
a(11) - a(1) = 17764622316152715 - 3 = (2^3)*(3^2)*7*(11^3)*13*2037061001 == 0 (mod 11^3).
a(11 - 1) - a(0) = 393088036809073 - 1 = (2^4)*3*(11^3)*29*67*1381*2293 == 0 (mod 11^3).
p = 5:
a(5^2) - a(5) = 5545311482504558271924122566108960335003 - 2835003 = (2^4)*3*(5^7)*11*31*91546780597609*23684663949545369 == 0 (mod 5^7).
a(5^2 - 1) - a(5 - 1) = 113353062539459038723143413569578825001 - 75001 = (2^4)*3*(5^7)*(11^2)*29*53*162533449533306503812325773 == 0 (mod 5^7).

Maple

A363983 := proc(n) option remember: add((-1)^(n+k)*binomial(n, k)*binomial(n+k-1, k)*binomial(2*k, n), k = 0..n) end:
seq(add((-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A363983(k), k = 0..n), n = 0..20);

Crossrefs

Cf. A000172, A005259, A363983.

Sequence in context: A173807 A093165 A189306 * A012810 A364300 A020517

Adjacent sequences: A363981 A363982 A363983 * A363985 A363986 A363987

Keyword

nonn,easy

Author

Peter Bala, Jul 01 2023

Status

approved