OFFSET
1,2
COMMENTS
Conjecture:
(i) n+1 | a(n) if and only if n is not congruent to 1 modulo 4.
(ii) For any Fermat prime p, we have a(p-1) == ((p-1)/2)!*p (mod p^2). If n > 1 is neither congruent to 2 modulo 4 nor a Fermat prime, then a(n-1) == 0 (mod n^2).
The author has proved that p | a(p-1) for any odd prime p. This implies that n | a(n) for all n > 2.
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 1..33 (terms 1..15 from Zhi-Wei Sun, terms 16..30 from Jinyuan Wang)
Zhi-Wei Sun, On the permanent per[i^(j-1)]_{i,j=1..n} modulo p^2, Question 316839 on Mathoverflow, Dec. 3, 2018.
Zhi-Wei Sun, On restricted permutations of {1,...,n}, arXiv:1811.10503 [math.CO], 2018.
EXAMPLE
a(2) = 3 since permanent[i^(j-1)]_{i,j=1,2} = 1*2 + 1*1 = 3.
MATHEMATICA
Permanent[m_List]:=With[{v = Array[x, Length[m]]}, Coefficient[Times @@ (m.v), Times @@ v]];
a[n_]:=a[n]=Permanent[Table[i^(j-1), {i, 1, n}, {j, 1, n}]];
Do[Print[n, " ", a[n]], {n, 1, 15}]
PROG
(PARI) a(n) = matpermanent(matrix(n, n, i, j, i^(j-1))); \\ Michel Marcus, Dec 05 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 04 2018
STATUS
approved