|
%I #27 Apr 30 2023 14:15:20
%S 1,1,2,5,13,34,90,236,621,1629,4274,11193,29337,76818,201173,526730,
%T 1379178,3610804,9453695,24750281,64798235,169644626,444138288,
%U 1162770238,3044180080,7969770106,20865148382,54625676431,143011928942
%N a(1)=1; a(n+1) = Sum_{k=1..n} a(k) a(floor(n/k)).
%C 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
%H Vincenzo Librandi, <a href="/A097417/b097417.txt">Table of n, a(n) for n = 1..1000</a>
%F 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
%F 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
%F Limit_{n->oo} a(n)/(1 + phi)^n = 0.108165624886204570982244311730754895284041534583990405146651275318889227986... - _Vaclav Kotesovec_, May 28 2021
%p 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
%t 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 *)
%o (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
%Y Cf. A038044, A078346, A097919.
%K easy,nonn
%O 1,3
%A _Leroy Quet_, Aug 19 2004
%E More terms from _Klaus Brockhaus_, _Robert G. Wilson v_ and Mark Hudson (mrmarkhudson(AT)hotmail.com), Aug 21 2004
|
