A173774 The arithmetic mean of (21*k + 8)*binomial(2*k,k)^3 (k=0..n-1).

8, 120, 3680, 144760, 6427008, 306745824, 15364514880, 796663553400, 42395640372800, 2302336805317120, 127078484504270208, 7108177964254013920, 402042028998035350400, 22954860061516225396800

Offset: 1

Author

Zhi-Wei Sun, Feb 24 2010

Keywords

• nonn

Comments

On Feb 10 2010, Zhi-Wei Sun introduced the sequence and conjectured that each term a(n) is an integer divisible by 4*binomial(2*n,n). On Feb 11 2011, Kasper Andersen confirmed this conjecture by noting that the sequence b(n) = a(n)/(4*binomial(2*n,n)), for n > 0, coincides with A112029. It was proved that for every prime p and positive integer a we have a(p^a) == 8 + 16*p^3*B_(p-3) (mod p^4), where B_0, B_1, B_2, ... are Bernoulli numbers. Given a prime p, it has been conjectured that Sum_{k=0..(p-1)/2} (21*k + 8)*binomial(2*k,k)^3 == 8*p + (-1)^((p-1)/2)*32*p^3*E_(p-3) (mod p^4) if p > 3 (where E_0, E_1, E_2, ... are Euler numbers), and that Sum_{k=0..floor(2p^a/3)} (21*k + 8)*binomial(2*k,k)^3 == 8*p^a (mod p^(a + 5 + (-1)^p)) if a is a positive integer with p^a == 1 (mod 3). He also observed that b(n) = a(n)/(4*binomial(2*n,n)) is odd if and only if n is a power of two.

Examples

			For n=2 we have a(2)=120 since (8*binomial(0,0)^3 + (21+8)*binomial(2,1)^3)/2 = 120.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..500
• Kasper Andersen, Re: A somewhat surprising conjecture
• Zhi-Wei Sun, A somewhat surprising conjecture
• Zhi-Wei Sun, Re: A somewhat surprising conjecture
• Zhi-Wei Sun, Open conjectures on congruences, preprint, arXiv:0911.5665 [math.NT], 2009-2011.
• Zhi-Wei Sun, Super congruences and Euler numbers, preprint, arXiv:1001.4453 [math.NT], 2010-2011.

Crossrefs

Cf. A000984, A112029, A122045.

Programs

Magma

[(&+[(21*j+8)*(j+1)^3*Catalan(j)^2: j in [0..n-1]])/n: n in [1..30]]; // G. C. Greubel, Jul 06 2021

Mathematica

a[n_]:= Sum[(21*k+8)*Binomial[2*k,k]^3, {k,0,n-1}]/n; Table[a[n], {n, 25}]

Sage

[(1/n)*sum((21*j+8)*binomial(2*j,j)^3 for j in (0..n-1)) for n in (1..30)] # G. C. Greubel, Jul 06 2021

Formula

a(n) = (1/n)*Sum_{k=0..n-1} (21*k + 8)*binomial(2*k,k)^3.

(n+1)*a(n+1) = n*a(n) + 8*(21*n + 8)*binomial(2*n-1, n)^3, n > 0, with a(1) = 8.

a(n) ~ 2^(6*n) / (3 * (Pi*n)^(3/2)). - Vaclav Kotesovec, Jan 24 2019

a(n) = (1/n)*Sum_{j=0..n-1} (21*j + 8)*(j+1)^3*Catalan(j)^3. - G. C. Greubel, Jul 06 2021