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A099741 a(1) = a(2) = 1; a(n) = a([n/2])+a([n/3]) (n >= 3).

%I A099741 #25 Oct 10 2021 03:41:05
%S A099741 1,1,2,2,2,3,3,3,4,4,4,5,5,5,5,5,5,7,7,7,7,7,7,8,8,8,9,9,9,9,9,9,9,9,
%T A099741 9,12,12,12,12,12,12,12,12,12,12,12,12,13,13,13,13,13,13,16,16,16,16,
%U A099741 16,16,16,16,16,16,16,16,16,16,16,16,16,16,20,20,20,20,20,20,20,20,20,21
%N A099741 a(1) = a(2) = 1; a(n) = a([n/2])+a([n/3]) (n >= 3).
%C A099741 Let f = f[x,y] be a Fibonacci variant with recurrence f(1) = f(2) = 1; f(n) = f(ceiling((n-1)/x))+f(ceiling((n-2)/y)). This sequence is f[2,3].
%C A099741 Nondecreasing. Increases only when n is of the form 2^x*3^y.
%C A099741 By the Akra-Bazzi theorem, we have a(n) = Theta(n^e), where e ~ 0.78788491102586978 is the root of the equation (1/2)^e + (1/3)^e = 1. - _Jeffrey Shallit_, Mar 15 2018
%H A099741 Robert Israel, <a href="/A099741/b099741.txt">Table of n, a(n) for n = 1..10000</a>
%H A099741 Wikipedia, <a href="https://en.wikipedia.org/wiki/Akra%E2%80%93Bazzi_method">Akra-Bazzi method</a>
%F A099741 G.f. g(x) satisfies g(x) = x + (1+x)*g(x^2) + (1+x+x^2)*g(x^3). - _Robert Israel_, Mar 15 2018
%e A099741 a(19) = a([19/2])+a([19/3]) = a(9)+a(6) = 4+3 = 7.
%p A099741 f:= proc(n) option remember; procname(floor(n/2))+procname(floor(n/3)) end proc:
%p A099741 f(1):= 1: f(2):= 1:
%p A099741 map(f, [$1..100]); # _Robert Israel_, Mar 15 2018
%t A099741 a[1] = a[2] = 1;
%t A099741 a[n_] := a[n] = a[Floor[n/2]] + a[Floor[n/3]];
%t A099741 Array[a, 100] (* _Jean-François Alcover_, Aug 28 2020 *)
%Y A099741 Cf. A088468.
%K A099741 easy,nonn
%O A099741 1,3
%A A099741 _Darrell Minor_, Nov 09 2004
%E A099741 Name corrected by _Robert Israel_, Mar 15 2018