A352276 a(0) = 1 and a(n) = Sum_{k = 0..3*n} n/(n + 2*k)*binomial(n + 2*k,k) for n >= 1.

1, 9, 625, 58785, 5986993, 633580634, 68611922731, 7545931449401, 839183314181297, 94112350842056469, 10623982584664109750, 1205644823097085684641, 137414820511792364274091, 15718880489100999321142976, 1803621273322664188151352631, 207499462144488863314062180035

Offset: 0

Author

Peter Bala, Mar 10 2022

Keywords

• nonn
• easy

Comments

The following identity can be easily verified using Maple's SumTools:-Summation procedure: for n >= 1, A005810(n) = binomial(4*n,n) = Sum_{k = 0..3*n} n/(n + k)*binomial(n + k,k).

The binomial coefficients A005810(n) are known to satisfy the supercongruences A005810(n*p^r) == A005810(n*p^(r-1)) (mod p^(3*r)) for primes p >= 5 and positive integers n and r (see Meštrović, equation 39).

Calculation suggests that the present sequence satisfies the same congruences.

Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r.

More generally, for m a positive integer, define a sequence u_m by setting u_m(n) = Sum_{k = 0..m*n} n/(n + 2*k)*binomial(n + 2*k,k) for n >= 1.

Then we conjecture that each sequence u_m satisfies the above supercongruences. This is the case m = 3. See A333093 (case m = 1) and A352275 (m = 2).

Examples

			Examples of supercongruences:
a(11) - a(1) = 1205644823097085684641 - 9 = (2^3)*3*(11^3)*43*2887*5059* 60096637 == 0 (mod 11^3)
a(3*5) - a(3) = 207499462144488863314062180035 - 58785 = 2*(5^4)*1801* 4959701*18583938263214197 == 0 (mod 5^4)
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..475
• R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.

Crossrefs

Cf. A005810, A333093, A352275.

Programs

Mathematica

nterms=25;Join[{1},Table[Sum[n/(n+2k)Binomial[n+2k,k],{k,0,3n}],{n,nterms-1}]] (* Paolo Xausa, Apr 10 2022 *)

Formula

a(n) ~ 7^(7*n + 3/2) / (37 * sqrt(Pi*n) * 2^(8*n + 3/2) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Mar 15 2022