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A364507
a(n) = (5*n)!*(4*n)! / ((3*n)!^2 * (2*n)! * n!).
4
1, 40, 5880, 1101100, 229265400, 50678855040, 11641642112100, 2746924727976000, 661097260785195000, 161538994454795003200, 39949572934939198410880, 9976687616280042928424700, 2511716999955421326631644900, 636662322699394050738883008000
OFFSET
0,2
COMMENTS
Row 2 of A364506.
FORMULA
a(n) = Sum_{k = -n..n} (-1)^k * binomial(2*n, n + k) * binomial(4*n, 2*n + k)^2 (showing a(n) to be integral). Compare with Dixon's identity, Sum_{k = -n..n} (-1)^k * binomial(2*n, n + k)^3 = (3*n)!/n!^3 = A006480(n).
P-recursive: a(n) = (20/9)*(4*n-1)*(4*n-3)*(5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)/((3*n-1)^2*(3*n-2)^2*n^2) * a(n-1) with a(0) = 1.
a(n) ~ c^n * sqrt(10)/(6*Pi*n), where c = (2^6)*(5^5)/(3^6).
a(n) = [x^n] G(x)^(20*n), where the power series G(x) = 1 + 2*x + 69*x^2 + 5647*x^3 + 618860*x^4 + 79241349*x^5 + 11177111981*x^6 + 1684171189810*x^7 + 266238907746252*x^8 + ... appears to have integer coefficients.
exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^20, where the power series F(x) = 1 + 2*x + 149*x^2 + 18647*x^3 + 2913620*x^4 + 515276389*x^5 + 98628630997*x^6 + 19944410220744*x^7 + 4199273746072180*x^8 + ... appears to have integer coefficients.
Conjecture: 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 [added Aug 04 2023: the conjecture follows from Meštrović. equation 39].
a(n) = binomial(4*n,n)*binomial(5*n,2*n). - Christian Krause, Aug 03 2023
EXAMPLE
Examples of supercongruences:
a(7) - a(1) = 2746924727976000 - 40 = (2^3)*5*(7^4)*28601881799 == 0 (mod 7^4).
a(11) - a(1) = 9976687616280042928424700 - 40 = (2^2)*5*(11^3)*18397*3568463* 5708869513 == 0 (mod 11^3).
MAPLE
seq( (5*n)!*(4*n)!*(2*n)! / ((3*n)!^2 * (2*n)!^2 * n!), n = 0..15);
MATHEMATICA
A364507[n_]:=(5n)!(4n)!/((3n)!^2(2n)!n!); Array[A364507, 15, 0] (* Paolo Xausa, Oct 06 2023 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Jul 27 2023
STATUS
approved