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A219562 a(n) = Sum_{k=0..n} binomial(n+k,k)^4.

%I A219562 #47 Dec 03 2024 12:33:22
%S A219562 1,17,1378,170257,25561876,4294835666,776487013506,147812510671121,
%T A219562 29234435383857304,5955068493838815892,1241820686691538181636,
%U A219562 263946916625793118532050,56996643356459050103185444,12473214064899644269110156626,2760963661677614009262282769378
%N A219562 a(n) = Sum_{k=0..n} binomial(n+k,k)^4.
%H A219562 Vincenzo Librandi, <a href="/A219562/b219562.txt">Table of n, a(n) for n = 0..200</a>
%H A219562 F. Baldassarri, S. Bosch, B. Dwork, (eds), <a href="https://doi.org/10.1007/BFb0091139">p-adic Analysis. Lecture Notes in Mathematics</a>, vol. 1454, pp. 194 - 204,  Springer, Berlin, Heidelberg.
%H A219562 Matthijs Coster, <a href="https://core.ac.uk/download/pdf/301642554.pdf">Supercongruences</a>.
%H A219562 Vaclav Kotesovec, <a href="https://oeis.org/wiki/User:Vaclav_Kotesovec">Asymptotic of generalized Apery sequences with powers of binomial coefficients</a>, Nov 04 2012
%F A219562 a(n) ~ 2^(8*n+4)/(15*Pi^2*n^2).
%F A219562 Recurrence: 4*(n-1)*(4*n-1)*(4*n+1)*(279825*n^6 - 2240985*n^5 + 7416081*n^4 - 12962383*n^3 + 12597634*n^2 - 6438500*n + 1347304)*n^4*a(n) = 2*(n-1)*(2290647450*n^12 - 22926837585*n^11 + 100717526436*n^10 - 254986993727*n^9 + 410380920831*n^8 - 435959897978*n^7 + 305660392723*n^6 - 134977315842*n^5 + 31413259700*n^4 + 2833672*n^3 - 2076143616*n^2 + 500898816*n - 39813120)*a(n-1) + (859902225*n^13 - 10755967005*n^12 + 60090860763*n^11 - 197381561581*n^10 + 422055067481*n^9 - 613861172995*n^8 + 615013106513*n^7 - 418396400175*n^6 + 182810864162*n^5 - 42759392772*n^4 + 146171272*n^3 + 2813432832*n^2 - 691172352*n + 55738368)*a(n-2) - 16*(n-2)^3*(2*n-3)^4*(279825*n^6 - 562035*n^5 + 408531*n^4 - 111409*n^3 - 5504*n^2 + 7968*n - 1024)*a(n-3).
%F A219562 G.f. as an expression in terms of 2F1 and 4F3 functions is given in the Maple program below. - _Mark van Hoeij_, Apr 02 2013
%F A219562 From _Peter Bala_, Nov 29 2024: (Start)
%F A219562 Conjecture: a(p-1) == 1 (mod p^5) for prime p >= 7 (checked up to p = 499). Coster, Theorem 4, proves that a(p-1) == 1 (mod p^3) for primes p >= 5.
%F A219562 Conjecture: for r >= 2, the supercongruence a(p^r - 1) == a(p^(r-1) - 1) (mod p^(3*r+3)) may hold for all primes p >= 5. Coster, Theorem 4, proves that a(p^r -1) == a(p^(r-1) - 1) (mod p^(3*r)) for r >= 2 and all primes p >= 5. (End)
%p A219562 q := x-4+I*((x+4)*(16-x))^(1/2);
%p A219562 f := x*(q/8)^4;
%p A219562 s := ((q-2)/(8*I-6))^(1/4);
%p A219562 y1 := hypergeom([1/8, 1/8], [3/4], f) * s / x^(1/8);
%p A219562 r := 2/(x*((x+4)*(16-x))^(1/2)*y1^2);
%p A219562 h := hypergeom([1/2, 1/2, 1/2, 1/2],[1, 1, 1],256*x);
%p A219562 u := (15*(223*x+72)*x^2*diff(h,x,x,x)+(14579*x+3226)*x*diff(h,x,x)
%p A219562 +(9969*x+1002)*diff(h,x)+320*h)/(16*(16-x)*(x+4)*x^2);
%p A219562 ogf := y1^2*Int(r*(1+Int(r*Int(u/(r*y1)^2,x),x)),x) ;
%p A219562 # Check o.g.f. by computing a series expansion:
%p A219562 SER := proc(a,x) series(a,x,20) end:
%p A219562 INT := proc(a,x) int(SER(a,x),x) end:
%p A219562 SER(eval(ogf, Int = INT),x); # _Mark van Hoeij_, Apr 02 2013
%t A219562 Table[Sum[Binomial[n+k,k]^4, {k,0,n}], {n,0,20}]
%o A219562 (PARI) a(n) = sum(k=0, n, binomial(n+k,k)^4); \\ _Michel Marcus_, Jul 15 2022
%Y A219562 Cf. A001700, A112029, A112028, A219563, A219564.
%K A219562 nonn,easy
%O A219562 0,2
%A A219562 _Vaclav Kotesovec_, Nov 23 2012