|
%I #23 Sep 08 2022 08:45:50
%S 8,120,3680,144760,6427008,306745824,15364514880,796663553400,
%T 42395640372800,2302336805317120,127078484504270208,
%U 7108177964254013920,402042028998035350400,22954860061516225396800
%N The arithmetic mean of (21*k + 8)*binomial(2*k,k)^3 (k=0..n-1).
%C 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 + 1*6*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.
%H G. C. Greubel, <a href="/A173774/b173774.txt">Table of n, a(n) for n = 1..500</a>
%H Kasper Andersen, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind1002&L=nmbrthry&T=0&P=1395">Re: A somewhat surprising conjecture</a>
%H Zhi-Wei Sun, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;a39bf149.1002">A somewhat surprising conjecture</a>
%H Zhi-Wei Sun, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;4d3d48b9.1002">Re: A somewhat surprising conjecture</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/0911.5665">Open conjectures on congruences</a>, preprint, arXiv:0911.5665 [math.NT], 2009-2011.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1001.4453">Super congruences and Euler numbers</a>, preprint, arXiv:1001.4453 [math.NT], 2010-2011.
%F a(n) = (1/n)*Sum_{k=0..n-1} (21*k + 8)*binomial(2*k,k)^3.
%F (n+1)*a(n+1) = n*a(n) + 8*(21*n + 8)*binomial(2*n-1, n)^3, n > 0, with a(1) = 8.
%F a(n) ~ 2^(6*n) / (3 * (Pi*n)^(3/2)). - _Vaclav Kotesovec_, Jan 24 2019
%F a(n) = (1/n)*Sum_{j=0..n-1} (21*j + 8)*(j+1)^3*Catalan(j)^3. - _G. C. Greubel_, Jul 06 2021
%e For n=2 we have a(2)=120 since (8*binomial(0,0)^3 + (21+8)*binomial(2,1)^3)/2 = 120.
%t a[n_]:= Sum[(21*k+8)*Binomial[2*k,k]^3, {k,0,n-1}]/n; Table[a[n], {n, 25}]
%o (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
%o (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
%Y Cf. A000984, A112029, A122045.
%K nonn
%O 1,1
%A _Zhi-Wei Sun_, Feb 24 2010
|
