OFFSET
0,2
COMMENTS
Conjecture: (i) a(n) is always positive and odd, and not congruent to 7 modulo 8.
(ii) For any odd prime p, if p == 1 (mod 3) and p = x^2 + 3*y^2 with x == 1 (mod 3), then a(p-1) == (-1)^{(p-1)/2}*(2*x-p/(2*x)) (mod p^2); if p == 2 (mod 3) then a(p-1) == (-1)^{(p+1)/2}*3p/binomial((p+1)/2,(p+1)/6) (mod p^2).
REFERENCES
Zhi-Wei Sun, Conjectures and results on x^2 mod p^2 with 4p = x^2+d*y^2, in: Number Theory and Related Area (eds., Y. Ouyang, C. Xing, F. Xu and P. Zhang), Higher Education Press & International Press, Beijing and Boston, 2013, pp. 147-195.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..25
Zhi-Wei Sun, Conjectures and results on x^2 mod p^2 with 4p = x^2+d*y^2, arXiv:1103.4325 [math.NT], 2011-2014.
Z.-W. Sun, On sums of Apéry polynomials and related congruences, J. Number Theory 132(2012), 2673-2699.
Zhi-Wei Sun, Connections between p = x^2 + 3*y^2 and Franel numbers, J. Number Theory 133(2013), 2914-2928.
Zhi-Wei Sun, On some determinants with Legendre symbol entries, preprint, arXiv:1308.2900 [math.NT], 2013-2018.
EXAMPLE
a(0) = 1 since A086618(0) = 1.
MATHEMATICA
f[n_]:=Sum[Binomial[n, k]^2*Binomial[2k, k]/(k+1), {k, 0, n}]
a[n_]:=Det[Table[f[i+j], {i, 0, n}, {j, 0, n}]]
Table[a[n], {n, 0, 10}]
PROG
(PARI) f(n)=sum(k=0, n, binomial(n, k)^2*binomial(2*k, k)/(k+1))
a(n)=my(v=vector(2*n+1, k, f(k-1))); matdet(matrix(n+1, n+1, i, j, v[i+j-1])) \\ Charles R Greathouse IV, Jul 30 2016
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
Zhi-Wei Sun, Aug 22 2013
STATUS
approved