%I #28 Jun 29 2023 09:30:12
%S 1,98,9606,941780,92352070,9058034748,888610349724,87192397723368,
%T 8557276143987270,840005101192014380,82474083957903064756,
%U 8099197733721011526168,795527368821049695145756,78154959591300863484042200,7679729103551077344613236600,754784236214755050742369782480,74197094919316919158188333048390
%N Coefficient of x^n in (x^2 + 98*x + 1)^n.
%C On Jan 29 2011, _Zhi-Wei Sun_ conjectured that Sum_{k>=0} (40k + 3)*a(k)*binomial(4k, 2k)*binomial(2k, k)/112^(2k) = 70*sqrt(21)/(9*Pi).
%C He also conjectured that 2n(2n + 1)*binomial(2n, n) divides Sum_{k=0..n-1} (40k + 3)*a(k)*binomial(4k, 2k)*binomial(2k, k)112^{2(n - 1 - k)} for each n = 2, 3, 4, ... and that for any prime p different from 2 and 7 we have the congruence Sum_{k=0..p-1} (40k + 3)*a(k)*binomial(4k, 2k)*binomial(2k, k)/112^(2k) == p(-21/p)(5 - 2(-2/p)) (mod p^2).
%C Another conjecture of his states that for any prime p == 1, 3 (mod 8) with p = x^2 + 2y^2 (x, y integers) we have Sum_{k=0..p-1} a(k)*binomial(4k, 2k)*binomial(2k, k)/112^(2k) == (7/p)(4x^2 - 2p) (mod p^2). By a result of Sun, Sum_{k=0..p-1} (40k + 3)*a(k)*binomial(4k, 2k)*binomial(2k, k)/112^(2k) == 0 (mod p^2) for any prime p = 5,7 (mod 8).
%H G. C. Greubel, <a href="/A181553/b181553.txt">Table of n, a(n) for n = 0..495</a>
%H Zhi-Wei Sun, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;895b1a21.1101">One more conjectural series for 1/pi</a>, a message to Number Theory List, Jan. 29, 2011.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1101.0600">On sums related to central binomial and trinomial coefficients</a>, arXiv:1101.0600 [math.NT], 2014.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/0911.5665">Open Conjectures on Congruences</a>, arXiv:0911.5665 [math.NT], 2009.
%F a(n) = Sum_{k = 0..floor(n/2)} binomial(n, 2*k)*binomial(2*k, k) *(98)^{n - 2*k}.
%F G.f.: (9600*x^2 - 196*x + 1)^(-1/2).
%e For n = 2 we have a(2) = coefficient of x^2 in (x^2 + 98x + 1)^2 = 9606.
%t A[n_] := If[n > 0, Coefficient[(x^2 + 98x + 1)^n, x^n], 1]; Table[A[n], {n, 0, 20}]
%o (PARI) x='x+O('x^20); Vec((9600*x^2 - 196*x + 1)^(-1/2)) \\ _G. C. Greubel_, Mar 06 2017
%o (Magma) m:=20; R<x>:=PowerSeriesRing(Rationals(), m); Coefficients(R!( 1/Sqrt(9600*x^2 - 196*x + 1) )); // _G. C. Greubel_, Nov 10 2018
%Y Cf. A000984, A002426.
%K nonn
%O 0,2
%A _Zhi-Wei Sun_, Jan 29 2011