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A364114 a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^3 for n >= 0.

%I A364114 #12 Jul 12 2023 11:04:00
%S A364114 1,7,163,5623,235251,11009257,554159719,29359663991,1615702377331,
%T A364114 91558286583757,5310712888211413,313940484249068761,
%U A364114 18853030977961798359,1147317139889540758509,70618205829113737707663,4389482803713232076789623,275190242843266217113413491
%N A364114 a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^3 for n >= 0.
%C A364114 Row 3 of A364113.
%C A364114 Compare with the two types of Apéry numbers A005258 and A005259, which are related to the Legendre polynomials by A005258(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x)) and A005259(n) = [x^k] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^2.
%C A364114 Both types of Apéry numbers satisfy the supercongruences
%C A364114 1) u (n*p^r) == u(n*p^(r-1)) (mod p^(3*r))
%C A364114 and the shifted supercongruences
%C A364114 2) u (n*p^r - 1) == u(n*p^(r-1) - 1) (mod p^(3*r))
%C A364114 for all primes p >= 5 and positive integers n and r.
%C A364114 We conjecture that the present sequence also satisfies the supercongruences 1) and 2).
%F A364114 a(n) ~ (1921 + 533*sqrt(13))^(n + 1/2) / (13^(1/4) * Pi^2 * n^2 * 2^(n + 7/2) * 3^(3*n + 1/2)). - _Vaclav Kotesovec_, Jul 09 2023
%F A364114 Conjectures:
%F A364114 1) 3*a(p) - 11*a(p-1) == 10 (mod p^5) for all primes p >= 7 (checked up to p = 101).
%F A364114 2) a(p)^21 == (7^21)*a(p-1)^11 (mod p^5) for all primes p >= 7 (checked up to p = 101).
%e A364114 Examples of supercongruences:
%e A364114 a(7) - a(1) = 29359663991 - 7 = (2^4)*(7^3)*37*144589 == 0 (mod 7^3).
%e A364114 a(7 - 1) - a(0) = 554159719 - 1 = 2*(3^4)*(7^3)*9973 == 0 (mod 7^3).
%e A364114 a(5^2) - a(5) = 5343160378366596176372561346633696195759257 - 11009257 = (2^4)*(5^6)*21372641513466384705490245386534784739 == 0 (mod 5^6).
%e A364114 a(5^2 - 1) - a(5 - 1) = 81394273032250674032560324508765757297751 - 235251 = (2^2)*(5^6)*7*13*29*6317*78120239161449483411026081851 == 0 (mod 5^6).
%p A364114 a(n) := coeff(series( 1/(1-x)* LegendreP(n,(1+x)/(1-x))^3, x, 21), x, n):
%p A364114 seq(a(n), n = 0..20);
%t A364114 Table[SeriesCoefficient[1/(1 - x) * LegendreP[n, (1 + x)/(1 - x)]^3, {x,0,n}], {n,0,20}] (* _Vaclav Kotesovec_, Jul 09 2023 *)
%Y A364114 Cf. A005258, A005259, A364113, A364115, A364116.
%K A364114 nonn,easy
%O A364114 0,2
%A A364114 _Peter Bala_, Jul 07 2023