OFFSET
1,3
COMMENTS
1/abs(a(n)) + 1/abs(a(n+1)) = 1/(n-1)!, n = 3,5,7,... hence Sum_{n>1} 1/abs(a(n)) = cosh(1). - Peter McNair, Mar 04 2022
FORMULA
For n>1, a(n) = (-1)^floor(n/2) * A001710(n) / floor(n/2). - Vaclav Kotesovec, Jan 26 2016
MATHEMATICA
a[1]=1; a[n_] := a[n] = Sum[(-1)^i*i*a[i], {i, 1, n - 1}]; Array[a, 33]
CROSSREFS
KEYWORD
sign
AUTHOR
José María Grau Ribas, Dec 07 2015
STATUS
approved
