A352275 a(0) = 1 and a(n) = Sum_{k = 0..2*n} n/(n + 2*k)*binomial(n + 2*k,k) for n >= 1.

1, 4, 64, 1429, 35072, 898129, 23571781, 628750217, 16965558016, 461752375705, 12652302369439, 348552604899778, 9644571491252069, 267852878928912034, 7462156684641697991, 208446714456132946429, 5836259481820028112640, 163741162073796817779389, 4602160147618819467316159

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, A005809(n) = binomial(3*n,n) = Sum_{k = 0..2*n} n/(n + k)*binomial(n + k,k).

The binomial coefficients A005809(n) are known to satisfy the supercongruences A005809(n*p^r) == A005809(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 = 2. See A333093 (case m = 1) and A352276 case (m = 3).

Examples

			Examples of supercongruences:
a(3*5) - a(3) = 208446714456132946429 - 1429 = (2^3)*3*(5^4)*13*41* 26072134391011 == 0 (mod 5^4)
a(17) - a(1) = 163741162073796817779389 - 4 = 5*(17^3)*1506943* 4423278397003 == 0 (mod 17^3)
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..674
• 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. A005809, A333093, A352276.

Programs

Maple

seq(add(n/(n + 2*k)*binomial(n + 2*k,k), k = 0..2*n), n = 1..25);

Mathematica

nterms=25;Join[{1},Table[Sum[n/(n+2k)Binomial[n+2k,k],{k,0,2n}],{n,nterms-1}]] (* Paolo Xausa, Apr 11 2022 *)

PARI

a(n) = if (n==0, 1, sum(k=0, 2*n, binomial(n + 2*k,k)*n/(n+2*k))); \\ Michel Marcus, Mar 17 2022

Formula

a(n) ~ 5^(5*n + 3/2) / (19 * sqrt(Pi*n) * 2^(2*n + 1) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Mar 15 2022