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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.

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

Original entry on oeis.org

1, 3, 73, 2163, 75001, 2835003, 113329945, 4711519347, 201638246905, 8824346685003, 393088036809073, 17764622316152715, 812477640612743977, 37535247213943518315, 1749047441756088054073, 82108960863923963522163, 3879675478363506548275705
Offset: 0

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Author

Peter Bala, Jul 01 2023

Keywords

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.

Examples

			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).

Links

Crossrefs

Cf. A000172, A005259, A363983.

Programs

  • 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);
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