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A088468 a(0) = 1, a(n) = a(floor(n/2)) + a(floor(n/3)) for n > 0.

%I A088468 #20 Sep 27 2023 10:06:51
%S A088468 1,2,3,4,5,5,7,7,8,9,9,9,12,12,12,12,13,13,16,16,16,16,16,16,20,20,20,
%T A088468 21,21,21,21,21,22,22,22,22,28,28,28,28,28,28,28,28,28,28,28,28,33,33,
%U A088468 33,33,33,33,37,37,37,37,37,37,37,37,37,37,38,38,38,38,38,38,38,38,48
%N A088468 a(0) = 1, a(n) = a(floor(n/2)) + a(floor(n/3)) for n > 0.
%C A088468 Record values greater than 1 occur at 3-smooth numbers: A160519(n)=a(A003586(n)) and A160519(m)<a(m) for m < A003586(n). - _Reinhard Zumkeller_, May 16 2009
%H A088468 Reinhard Zumkeller, <a href="/A088468/b088468.txt">Table of n, a(n) for n = 0..10000</a>
%H A088468 A. R. Lebeck, <a href="http://www.cs.duke.edu/courses/spring03/cps104/homework/hw3.html">CPS 104: Homework #3</a>.
%H A088468 Michael Penn, <a href="https://www.youtube.com/watch?v=u7ATQEA4Hbw">Erdős but simpler</a>, Youtube video.
%F A088468 Limit_{n->oo} a(n)/n = 0, as proved in Michael Penn's Youtube video (see Links). Michael Penn states in the video that this is a simplification of a problem of Paul Erdős, where the original problem is to show that limit_{n->oo} b(n)/n = 12/log(432) for b(0) = 1, b(n) = b(floor(n/2)) + b(floor(n/3)) + b(floor(n/6)) for n > 0 ({b(n)} is the sequence A007731). - _Jianing Song_, Sep 27 2023
%t A088468 a[0]=1;a[n_]:=a[n]=a[Floor[n/2]]+a[Floor[n/3]];Array[a,75,0] (* _Harvey P. Dale_, Aug 23 2020 *)
%o A088468 (PARI) a(n)=if(n<1,n==0,a(n\2)+a(n\3))
%Y A088468 Equals A061984(n) + 1.
%Y A088468 Cf. A007731, A083662, A165704, A165706. - _Reinhard Zumkeller_, Sep 26 2009
%K A088468 nonn
%O A088468 0,2
%A A088468 _Michael Somos_, Oct 02 2003