%I #20 Aug 05 2019 05:32:54

%S 1,-7,1,1001,-15359,30233,3126529,-61392247,259448833,11970181433,

%T -287815672319,1854020654633,48800262650881,-1443188813338279,

%U 12410505050039041,198977188596692681,-7472188661349285887,80331498114096555641

%N a(n) = sum_{k=0..n}C(n,k)^3*(-8)^k with C(n,k) = n!/(k!(n-k)!).

%C Conjecture: (i) For any prime p > 3, we have sum_{k=0}^{p-1}(-1)^k*a(k) == (p/3) - p^2/12*B_{p-2}(1/3) (mod p^3) and sum_{k=0}^{p-1}(-1)^k*a(k)*H(k,2) == B_{p-2}(1/3) (mod p), where B_m(x) denotes the Bernoulli polynomial of degree m and H(k,2) stands for sum_{j=1..k} 1/j^2. Also, sum_{k=1}^{p-1}(-1)^k*a(k)/k == - 3*(2^p-2)/p (mod p) for any prime p.

%C (ii) For any positive integer n, the arithmetic mean (sum_{k=0}^{n-1}(6*k+5)(-1)^k*a(k))/n is an odd integer. Moreover, if p is a prime then sum_{k=0}^{p-1}(6*k+5)(-1)^k*a(k) == 3*p (mod p^2).

%H Zhi-Wei Sun, <a href="/A245329/b245329.txt">Table of n, a(n) for n = 0..150</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1407.0967">Congruences involving g_n(x) = sum_{k=0}^n C(n,k)^2*C(2k,k)*x^k</a>, preprint, arXiv:1407.0967 [math.NT], 2014-2016.

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1112.1034">Congruences for Franel numbers</a>, preprint, arXiv:1112.1034 [math.NT], 2011-2013.

%H Z.-W. Sun, <a href="http://dx.doi.org/10.1016/j.aam.2013.06.004">Congruences for Franel numbers</a>, Adv. in Appl. Math. 51(2013), 524-535.

%F Recurrence (obtained via the Zeilberger algorithm):

%F 343*(3n+7)*(n+1)^2*a(n) + (3n+5)*(363n^2+1331n+1113)*a(n+1) + 7*(9n^3+57n^2+116n+74)*a(n+2) + (3n+4)*(n+3)^2*a(n+3) = 0.

%e a(2) = 1 since sum_{k=0}^2 C(2,k)^3*(-8)^k = 1 + 2^3*(-8) + (-8)^2 = 1.

%t a[n_]:=Sum[Binomial[n,k]^3*(-8)^k,{k,0,n}]

%t Table[a[n],{n,0,17}]

%Y Cf. A000172, A245089.

%K sign

%O 0,2

%A _Zhi-Wei Sun_, Jul 18 2014