A097417 a(1)=1; a(n+1) = Sum_{k=1..n} a(k) a(floor(n/k)).

1, 1, 2, 5, 13, 34, 90, 236, 621, 1629, 4274, 11193, 29337, 76818, 201173, 526730, 1379178, 3610804, 9453695, 24750281, 64798235, 169644626, 444138288, 1162770238, 3044180080, 7969770106, 20865148382, 54625676431, 143011928942

Offset

1,3

Comments

4 is the only composite number n such that a(n+1) = 3a(n) - a(n-1) and if n is a composite number greater than 4 then a(n+1) > 3a(n) - a(n-1). - Farideh Firoozbakht, Feb 05 2005

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Formula

Ratio a(n+1)/a(n) seems to tend to 1 + Golden Ratio = 2.61803398... = 1 + A001622. - Mark Hudson (mrmarkhudson(AT)hotmail.com), Aug 23 2004

Satisfies the "partial linear recursion": a(prime(n)+1) = 3*a(prime(n)) - a(prime(n)-1). This explains why we get a(n+1)/a(n) -> 1 + phi. Also, lim_{n->oo} a(n)/(1 + phi)^n exists but should not have a simple closed form. - Benoit Cloitre, Aug 29 2004

Limit_{n->oo} a(n)/(1 + phi)^n = 0.108165624886204570982244311730754895284041534583990405146651275318889227986... - Vaclav Kotesovec, May 28 2021

Maple

a[1]:=1: for n from 1 to 50 do: a[n+1]:=sum(a[k]*a[floor(n/k)], k=1..n): od: seq(a[i], i=1..51) # Mark Hudson, Aug 21 2004

Mathematica

a[1] = 1; a[n_] := a[n] = Sum[ a[k]*a[Floor[(n - 1)/k]], {k, n - 1}]; Table[ a[n], {n, 29}] (* Robert G. Wilson v, Aug 21 2004 *)

Prog

(PARI) {m=29; a=vector(m); print1(a[1]=1, ", "); for(n=1, m-1, print1(a[n+1]=sum(k=1, n, a[k]*a[floor(n/k)]), ", "))} \\ Klaus Brockhaus, Aug 21 2004

Crossrefs

Cf. A038044, A078346, A097919.

Sequence in context: A367658 A114299 A112842 * A367657 A006801 A329674

Adjacent sequences: A097414 A097415 A097416 * A097418 A097419 A097420

Keyword

easy,nonn

Author

Leroy Quet, Aug 19 2004

Extensions

More terms from Klaus Brockhaus, Robert G. Wilson v and Mark Hudson (mrmarkhudson(AT)hotmail.com), Aug 21 2004

Status

approved