A173998 For n>=1, a(n) = n + 2 + sum(i=1..n-1, a(i)*a(n-i) ).

3, 13, 83, 673, 6203, 61613, 642683, 6940673, 76930803, 870136013, 10002590883, 116521027873, 1372486213803, 16318813519213, 195599588228683, 2360929398934273, 28671940652447203, 350089944825571213, 4295280755452388083, 52926654021145267873

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

Cf. A030431.

Sequence in context: A219906 A000904 A201304 * A135743 A123114 A104032

Adjacent sequences: A173995 A173996 A173997 * A173999 A174000 A174001

Keyword

nonn

Author

Vladimir Shevelev, Mar 05 2010

Status

approved