A245329 a(n) = sum_{k=0..n}C(n,k)^3*(-8)^k with C(n,k) = n!/(k!(n-k)!).

1, -7, 1, 1001, -15359, 30233, 3126529, -61392247, 259448833, 11970181433, -287815672319, 1854020654633, 48800262650881, -1443188813338279, 12410505050039041, 198977188596692681, -7472188661349285887, 80331498114096555641

Offset: 0

Author

Zhi-Wei Sun, Jul 18 2014

Keywords

• sign

Comments

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.

(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).

Examples

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

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..150
• Zhi-Wei Sun, Congruences involving g_n(x) = sum_{k=0}^n C(n,k)^2*C(2k,k)*x^k, preprint, arXiv:1407.0967 [math.NT], 2014-2016.
• Zhi-Wei Sun, Congruences for Franel numbers, preprint, arXiv:1112.1034 [math.NT], 2011-2013.
• Z.-W. Sun, Congruences for Franel numbers, Adv. in Appl. Math. 51(2013), 524-535.

Crossrefs

Cf. A000172, A245089.

Programs

Mathematica

a[n_]:=Sum[Binomial[n,k]^3*(-8)^k,{k,0,n}]
Table[a[n],{n,0,17}]

Formula

Recurrence (obtained via the Zeilberger algorithm):

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.