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

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, 9, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 20, 20, 20, 20, 20, 20, 20, 20, 20, 21

Offset: 1

Author

Darrell Minor, Nov 09 2004

Keywords

• easy
• nonn

Comments

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].

Nondecreasing. Increases only when n is of the form 2^x*3^y.

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

Examples

			a(19) = a([19/2])+a([19/3]) = a(9)+a(6) = 4+3 = 7.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Wikipedia, Akra-Bazzi method

Crossrefs

Cf. A088468.

Programs

Maple

f:= proc(n) option remember; procname(floor(n/2))+procname(floor(n/3)) end proc:
f(1):= 1: f(2):= 1:
map(f, [$1..100]); # Robert Israel, Mar 15 2018

Mathematica

a[1] = a[2] = 1;
a[n_] := a[n] = a[Floor[n/2]] + a[Floor[n/3]];
Array[a, 100] (* Jean-François Alcover, Aug 28 2020 *)

Formula

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

Extensions

Name corrected by Robert Israel, Mar 15 2018