OFFSET
1,3
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
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
Wikipedia, Akra-Bazzi method
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
EXAMPLE
a(19) = a([19/2])+a([19/3]) = a(9)+a(6) = 4+3 = 7.
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 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Darrell Minor, Nov 09 2004
EXTENSIONS
Name corrected by Robert Israel, Mar 15 2018
STATUS
approved