A364118 a(n) = 3*A364114(n) - 11*A364114(n-1).

10, 412, 15076, 643900, 30440010, 1541377330, 81983235064, 4524150828092, 256902133600630, 14924997512212912, 883403610976880740, 53105747607145638706, 3234568078911042493578, 199234128948556264779390, 12391648147019445115584576, 777286417688953098495554620

Offset

1,1

Comments

It is conjectured that the sequence {A364114(n)} and the shifted sequence {A364114(n-1)} both satisfy the supercongruences A364114(p^r) == A364114(p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers r. Stronger supercongruences may hold for the present sequence, a linear combination of A364114(n) and A364114(n-1).

Conjectures: 1) the supercongruences a(p) == a(1) (mod p^5) hold for all primes p >= 7 (checked up to p = 101).

2) for r >= 2, the supercongruences a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) hold for all primes p >= 7. Cf. A212334.

There is also a multiplicative version of this sequence. Define a sequence of rational numbers {b(n) : n >= 1} by b(n) = A364114(n)^21 / A364114(n-1)^11. Then we conjecture that the above pair of supercongruences also hold for the sequence {b(n)}.

Links

Table of n, a(n) for n=1..16.

Peter Bala, Some conjectural supercongruences for the Apéry numbers.

Maple

A364114 := n -> coeff(series( 1/(1-x)* LegendreP(n, (1+x)/(1-x))^3, x, 21), x, n):
seq(3*A364114(n) - 11*A364114(n-1), n = 1..20);

Crossrefs

Cf. A212334, A357506, A357507, A357568, A364114, A364119.

Sequence in context: A041767 A130557 A231052 * A085000 A248553 A374229

Adjacent sequences: A364115 A364116 A364117 * A364119 A364120 A364121

Keyword

nonn,easy

Author

Peter Bala, Jul 12 2023

Status

approved