OFFSET
1,1
COMMENTS
Using induction, it is easy to prove that a(n)==3 (mod 10).
The largest prime factors of these terms are large (they start 3, 13, 83, 673, 6203, 61613, 642683, 161411, 9221, 870136013, 751453, 4016443, 6267060337, 16318813519213,..)
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..200
FORMULA
Recurrence: n*a(n) = 3*(5*n-7)*a(n-1) - (23*n-48)*a(n-2) + 9*(n-3)*a(n-3). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ sqrt(13*sqrt(10)-40)*(7+2*sqrt(10))^n/(4*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 20 2012
MATHEMATICA
aa=ConstantArray[0, 20]; aa[[1]]=3; Do[aa[[n]]=n+2+Sum[aa[[i]]*aa[[n-i]], {i, 1, n-1}], {n, 2, 20}]; aa (* Vaclav Kotesovec, Oct 20 2012 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Mar 05 2010
STATUS
approved