OFFSET
1,4
FORMULA
It appears that a(n) is congruent to n!*h(n) (mod {n+1}) where h(n) = (1/2)*H(n/2) for even n and h(n) = H(n) - (1/2)*H(floor(n/2)) for odd n.
MATHEMATICA
Table[ Mod[ Sum[ HarmonicNumber[k]k!(n - k)!, {k, 1, n}], n + 1], {n, 1, 95}] (* or *) (* Robert G. Wilson v, Jan 14 2004 *)
h[n_] := If[ EvenQ[n], (1/2)HarmonicNumber[n/2], HarmonicNumber[n] - (1/2)HarmonicNumber[ Floor[n/2]]]; Table[ Mod[ n!h[n], n + 1], {n, 1, 95}]
(* or *) h[n_] := Sum[1/(2k - If[ EvenQ[n], 0, 1]), {k, 1, Floor[(n + 1)/2]}]; Table[ Mod[ n!h[n], n + 1], {n, 1, 95}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Leroy Quet, Jan 08 2004
EXTENSIONS
Edited and extended by Robert G. Wilson v, Jan 14 2004
STATUS
approved
