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A277060 a(n) = (1/2) * Sum_{k=0..n} (binomial(n,k) * binomial(n+k,k+1))^2 for n >= 0.

%I A277060 #40 Mar 23 2023 03:34:50
%S A277060 0,1,28,729,19376,529575,14835780,424231465,12338211520,363931754949,
%T A277060 10862528888300,327501958094003,9959845931792784,305175084350065267,
%U A277060 9412306255856822388,291982561878565118025,9104382992541189221120
%N A277060 a(n) = (1/2) * Sum_{k=0..n} (binomial(n,k) * binomial(n+k,k+1))^2 for n >= 0.
%C A277060 Conjecture: the supercongruences a(p-1) == 1 (mod p^4) holds for all primes p >= 5 and a(p^2-1) == 1 (mod p^5) holds for all primes p >= 3. - _Peter Bala_, Mar 22 2023
%H A277060 Seiichi Manyama, <a href="/A277060/b277060.txt">Table of n, a(n) for n = 0..656</a>
%F A277060 a(n) = n^2 * A074635(n)/2.
%F A277060 From _Peter Bala_, Mar 22 2023: (Start)
%F A277060 a(n) = Sum_{k = 0..n-1} binomial(n+1,k)*binomial(n-1,k)*binomial(n+k,k)^2.
%F A277060 P-recursive: (n-1)^2*(3*n^2-6*n+2)*(n+1)^3*a(n) = (2*n-1)*(51*n^4-102*n^3+19*n^2+ 32*n-14)*n^2*a(n-1) - n^2*(n-2)*(3*n^2-1)*(n-1)^2*a(n-2) with a(0) = 0 and a(1) = 1.
%F A277060 a(n) ~ sqrt(12 + 17*sqrt(2)/2)*(17 + 12*sqrt(2))^n/(4*n^(3/2)*Pi^(3/2)). (End)
%p A277060 a := proc(n) option remember; if n = 0 then 0 elif n = 1 then 1 else ( (2*n-1)*(51*n^4-102*n^3+19*n^2+ 32*n-14)*n^2*a(n-1) - n^2*(n-2)*(3*n^2-1)*(n-1)^2*a(n-2) )/( (n-1)^2*(3*n^2-6*n+2)*(n+1)^3 ) end if; end:
%p A277060 seq(a(n), n = 0..20); # _Peter Bala_, Mar 22 2023
%o A277060 (PARI) a(n)=my(t=n); if(n<2, return(n)); sum(k=1,n, t*=(n-k+1)*(n+k)/k/(k+1); t^2, n^2)/2 \\ _Charles R Greathouse IV_, Nov 07 2016
%Y A277060 Cf. 1/2 * Sum_{k=0..n} (binomial(n,k) * binomial(n+k,k+1))^m: A050151 (m=1), this sequence (m=2).
%Y A277060 Cf. A005259, A074635.
%K A277060 nonn,easy
%O A277060 0,3
%A A277060 _Seiichi Manyama_, Nov 07 2016