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A364115 a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^4 for n >= 0.

%I A364115 #12 Jul 12 2023 11:03:39
%S A364115 1,9,289,14409,908001,65898009,5246665201,445752724041,39731504675041,
%T A364115 3674479246416009,349918540195094289,34125049533650776281,
%U A364115 3394306634561379583281,343284252364774351717641,35215197976859176290014289,3657148830889736882170190409
%N A364115 a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^4 for n >= 0.
%C A364115 Row 4 of A364113.
%C A364115 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 A364115 Both types of Apéry numbers satisfy the supercongruences
%C A364115 1) u (n*p^r) == u(n*p^(r-1)) (mod p^(3*r))
%C A364115 and the shifted supercongruences
%C A364115 2) u (n*p^r - 1) == u(n*p^(r-1) - 1) (mod p^(3*r))
%C A364115 for all primes p >= 5 and positive integers n and r.
%C A364115 We conjecture that the present sequence also satisfies the supercongruences 1) and 2).
%F A364115 a(n) ~ phi^(10*n + 5) / (2^(3/2) * 5^(1/4) * Pi^(5/2) * n^(5/2)), where phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Jul 09 2023
%e A364115 Examples of supercongruences:
%e A364115 a(11) - a(1) = 34125049533650776281 - 9  = (2^4)*(3^2)*(11^3)*13*97*11423* 12360541 == 0 (mod 11^3).
%e A364115 a(11 - 1) - a(0) = 349918540195094289 - 1 = (2^4)*(11^3)*103*159526079101 == 0 (mod 11^3).
%e A364115 a(5^2) - a(5) = 823068999686576893970482230168234294266351898009 - 65898009 = (2^7)*(3^2)*(5^6)*11*17*31*311*35978539*2371705409*297232149579326831 == 0 (mod 5^6).
%e A364115 a(5^2 - 1) - a(5 - 1) = 7402345246022867712987394168675984358488158001- 908001 = (2^4)*(5^6)*13*29*911*1459*26046751*925152076787*2452153330349 == 0 (mod 5^6).
%p A364115 a(n) := coeff(series(1/((1-x))* LegendreP(k,(1+x)/(1-x))^4,x, 21):
%p A364115 seq(a(n), n = 0..20);
%t A364115 Table[SeriesCoefficient[1/(1 - x) * LegendreP[n, (1 + x)/(1 - x)]^4, {x,0,n}], {n,0,20}] (* _Vaclav Kotesovec_, Jul 09 2023 *)
%o A364115 (PARI) a(n) = my(x='x+O('x^(n+1))); polcoef((1/(1-x))*pollegendre(n, (1+x)/(1-x))^4, n); \\ _Michel Marcus_, Jul 12 2023
%Y A364115 Cf. A005258, A005259, A364113, A364114, A364116.
%K A364115 nonn,easy
%O A364115 0,2
%A A364115 _Peter Bala_, Jul 08 2023