A228289 Determinant of the p_n X p_n matrix with (i,j)-entry equal to D(i+j) for all i,j = 0,...,p_n-1, where D(k) = A002895(k) is the k-th Domb number and p_n is the n-th prime.

12, 2448, 428587718400, 4994319435309277891448832, 191901511752240055024005979549622856313555581586068578283027431424, 637213222716753775758429677219909335140503764595701930312765250413280716374852064945052319744

Offset: 1

Author

Zhi-Wei Sun, Aug 19 2013

Keywords

• nonn

Comments

Conjecture: If p_n == 1 (mod 3) and p_n = x^2 + 3*y^2 with x and y integers, then we have a(n) == (-1)^{(p_n-1)/2}*(4*x^2-2*p_n) (mod p_n^2). In the case p_n == 2 (mod 3), we have a(n) == 0 (mod p_n^2).

Zhi-Wei Sun also made the following similar conjecture:

If p is an odd prime and b(p) is the p X p determinant with (i,j)-entry equal to A053175(i+j) for all i,j = 0,...,p-1, then we have the congruence b(p) == (-1)^{(p-1)/2} (mod p^2).

References

• Zhi-Wei Sun, Conjectures and results on x^2 mod p^2 with 4*p = 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.

Crossrefs

Cf. A002895, A053175, A225776, A228143.

Programs

Mathematica

d[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k]Binomial[2(n-k),n-k],{k,0,n}]
a[n_]:=Det[Table[d[i+j],{i,0,Prime[n]-1},{j,0,Prime[n]-1}]]
Table[a[n],{n,1,8}]