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A352653 a(n) = Sum_{k = 0..n-1} binomial(n,k)*binomial(n-1,k)*binomial(n+k,k)*binomial(n-1+k,k).

%I A352653 #56 Dec 26 2024 04:24:26
%S A352653 0,1,13,253,5741,142001,3713305,100961365,2825369965,80843126905,
%T A352653 2354354542013,69555586698533,2079506849058809,62797822901091409,
%U A352653 1912714261589863633,58691012753161717253,1812592783819686728045,56298976785759622597385,1757493098495181029912485
%N A352653 a(n) = Sum_{k = 0..n-1} binomial(n,k)*binomial(n-1,k)*binomial(n+k,k)*binomial(n-1+k,k).
%C A352653 Compare with A005259(n) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2.
%C A352653 Main superdiagonal (or main subdiagonal) of the symmetric square array A143007.
%C A352653 This is the sequence (A(n,n-1,n,n-1)) in the notation of Straub 2014, where it is proved that the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.
%C A352653 More generally, for nonnegative integers A, B and C with A >= 2 let S(n;A,B,C) = Sum_{k = 1..n} binomial(n,k)^A * binomial(n+k-1,k-1)^B * binomial(2*k,n)^C. Then we conjecture the sequence (S(n;A,B,C))n>=1 satisfies the same supercongruences. This is the case A = B = 2, C = 0. Compare with Osburn et al., Theorem 1.2.
%H A352653 Robert Osburn, Brundaban Sahu and Armin Straub, <a href="https://arxiv.org/abs/1312.2195">Supercongruences for sporadic sequences</a>, arXiv:1312.2195 [math.NT], 2013-2014.
%H A352653 Armin Straub, <a href="http://dx.doi.org/10.2140/ant.2014.8.1985">Multivariate Apéry numbers and supercongruences of rational functions</a>, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; <a href="https://arxiv.org/abs/1401.0854">arXiv preprint</a>, arXiv:1401.0854 [math.NT], 2014.
%H A352653 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/StrehlIdentities.html">Strehl identities</a>
%F A352653 The sequence can be extended to negative values of n:
%F A352653 a(-n) = Sum_{k} binomial(-n,k)*binomial(-n+k,k)*binomial(-n-1,k)*binomial(-n-1+k,k) = a(n), since binomial(-m,k) = (-1)^k*binomial(m+k-1,k) for nonnegative k.
%F A352653 a(n) = A177316(n)/2 for n >= 1.
%F A352653 a(n) = Sum_{k = 1..n} binomial(n,k)^2*binomial(n+k-1,k-1)^2.
%F A352653 a(n) = (1/2)*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k-1,k)^2 for n >= 1.
%F A352653 a(n) = hypergeom([n, n+1, -n, -n+1], [1, 1, 1], 1) for n >= 1.
%F A352653 a(n) = (1/2)*hypergeom([n, n, -n, -n], [1, 1, 1], 1) for n >= 1.
%F A352653 a(n) = (1/4)*binomial(2*n,n)^2*hypergeom([-n+1, -n+1, -n, -n], [1, -2*n+1, -2*n+1], 1) for n >= 1
%F A352653 a(n) = [x^n*y^(n-1)*z^n*t^(n-1)] 1/((1 - x - y)(1 - z - t) - x*y*z*t).
%F A352653 a(n) = [x^n] 1/(1 - x) * P(n-1,(1 + x)/(1 - x))^2 = [x^(n-1)] 1/(1 - x) * P(n,(1 + x)/(1 - x))^2 for n >= 1, where P(n,x) denotes the n-th Legendre polynomial. Cf. A005259(n) = [x^n] 1/(1 - x) * P(n,(1 + x)/(1 - x))^2.
%F A352653 (n + 1)^3*(2*n - 1)(3*n^2 - 3*n + 1)*a(n+1) = 2*(102*n^6 - 68*n^4 + 21*n^2 - 3)*a(n) - (n - 1)^3*(2*n + 1)*(3*n^2 + 3*n + 1)*a(n-1) with a(0) = 0 and a(1) = 1.
%F A352653 The Gauss congruences a(n*p^r) == a(n*p^(r-1)) (mod p^r) hold for all primes p and all positive integers n and r. It follows that the expansion of exp( Sum_{n >= 1} a(n)/n*x^n ) = 1 + x + 7*x^2 + 91*x^3 + 1544*x^4 + 30448*x^5 + 661506*x^6 + 15377010*x^7 + ... has integer coefficients.
%F A352653 a(n) = n^2*hypergeom([1 - n, 1 - n, 1 + n, 1 + n], [1, 2, 2], 1). - _Peter Luschny_, Apr 17 2022
%F A352653 From _Peter Bala_, Mar 18 2023: (Start)
%F A352653 a(n) = Sum_{k = 0..n-1} (n-k)/(n+k) * binomial(n,k)^2 * binomial(n+k,k)^2.
%F A352653 a(n) = 1/6*(A005259(n) + A005259(n-1)). (End)
%F A352653 a(n) ~ (1 + sqrt(2))^(4*n) / (2^(9/4) * Pi^(3/2) * n^(3/2)). - _Vaclav Kotesovec_, Mar 29 2023
%F A352653 a(n) = (1/3)*Sum_{k = 0..n} binomial(n,k)*binomial(n+k-1,k)*f(k) for n >= 1, where f(n) = A000172(n) denotes the n-th Franel number. Cf. A005259(n) = Sum_{k = 0..n} binomial(n,k)*binomial(n+k,k)*f(k) (the second Strehl identity). - _Peter Bala_, Jun 26 2023
%F A352653 a(n) = Sum_{0 <= j <= k <= n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)* binomial(n+j-1, j)^2*binomial(n-1, k-j) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n-1, k). Cf. Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, k) = A005259(n). - _Peter Bala_, Sep 22 2024
%e A352653 Examples of supercongruences:
%e A352653 a(11) - a(1) = 69555586698533 - 1 = (2^2)*(11^3)*1321*2521*3923 == 0 (mod 11^3).
%e A352653 a(2*7) - a(2) = 1912714261589863633 - 13 = (2^2)*(3^2)*5*(7^3)*776221* 39911503 == 0 (mod 7^3).
%p A352653 seq(add(binomial(n,k)^2*binomial(n+k-1,k-1)^2, k = 1..n), n = 0..20);
%t A352653 a[n_] := n^2 HypergeometricPFQ[{1 - n, 1 - n, 1 + n, 1 + n}, {1, 2, 2}, 1];
%t A352653 Table[a[n], {n, 0, 18}] (* _Peter Luschny_, Apr 17 2022 *)
%o A352653 (PARI) a(n) = sum(k=1, n, binomial(n,k)^2*binomial(n+k-1,k-1)^2); \\ _Michel Marcus_, Apr 19 2022
%o A352653 (Python)
%o A352653 def A352653(n):
%o A352653     if n == 0: return 0
%o A352653     m, g = 1, 0
%o A352653     for k in range(n+1):
%o A352653         g += m*n**2//(n+k)**2
%o A352653         m *= ((n+k+1)*(n-k))**2
%o A352653         m //= (k+1)**4
%o A352653     return g>>1 # _Chai Wah Wu_, Oct 03 2022
%Y A352653 Cf. A002002, A005259, A108628, A143007, A177316.
%K A352653 nonn,easy
%O A352653 0,3
%A A352653 _Peter Bala_, Apr 06 2022