A357568 - OEIS

%I #34 Jul 17 2024 09:00:28

%S 1,12,204,2928,40140,547512,7535472,105077376,1484848332,21237645000,

%T 306972655704,4477160465856,65802123629424,973487343836448,

%U 14483651478207360,216550246159148928,3251660678391659724,49011343741651501800,741221951008966181160,11243583961952559386400

%N a(n) = 9*binomial(2*n,n)^2 - 8*binomial(3*n,n).

%C Conjectures:

%C 1) a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for r >= 2 and all primes p >= 3.

%C These are stronger supercongruences than those satisfied separately by the sequences {binomial(2*n,n)} = A000984 and {binomial(3*n,n)} = A005809.

%C 2) More generally, for k >= 1, the sequence {9*binomial(2*n,n)^k - k*(2^k)*binomial(3*n,n): n >= 0} may satisfy the same supercongruences. This is the case k = 2. See A357509 for the case k = 1.

%H Paolo Xausa, <a href="/A357568/b357568.txt">Table of n, a(n) for n = 0..800</a>

%H C. Helou and G. Terjanian, <a href="https://doi.org/10.1016/j.jnt.2007.06.008">On Wolstenholme’s theorem and its converse</a>, J. Number Theory 128 (2008), 475-499.

%H Romeo Meštrović, <a href="http://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011)</a>, arXiv preprint arXiv:1111.3057 [math.NT], 2011.

%F a(n) = 9*A002894(n) - 8*A005809(n) = 9*A000984(n)^2 - 8*A005809(n). .

%F a(k*p^r) == a(k*p^(r-1)) ( mod p^(3*r) ) for positive integers k and r and for all primes p >= 5 (see Meštrović, Section 6, equation 39).

%F a(p) == a(1) (mod p^5) for all primes p >= 7 (apply Helou and Terjanian, Section 3, Proposition 2).

%F From _Stefano Spezia_, Jul 17 2024: (Start)

%F G.f.: 16*cos(arccos(1-27*x/2)/6)/sqrt(4-27*x) + 18*EllipticK(16*x)/Pi.

%F E.g.f.: 9*hypergeom([1/2, 1/2], [1, 1], 16*x) - 8*hypergeom([1/3, 2/3], [1/2, 1], 27*x/4).

%F a(n) ~ 9*2^(4*n)/(n*Pi). (End)

%e Examples of supercongruences:

%e a(11) - a(1) = 4477160465856 - 12 = (2^2)*3*(11^5)*101*22937 == 0 (mod 11^5).

%e a(5^2) - a(5) = 143816772358933669354266172512 - 547512 = (2^3)*3*(5^9)*167191* 194659*94271599039 == 0 (mod 5^9).

%p seq(9*binomial(2*n,n)^2 - 8*binomial(3*n,n), n = 0..20);

%t A357568[n_] := 9*Binomial[2*n, n]^2 - 8*Binomial[3*n, n];

%t Array[A357568, 25, 0] (* _Paolo Xausa_, Jul 17 2024 *)

%Y Cf. A000984, A002894, A005809, A357509, A357567, A357569, A357955.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Oct 21 2022