A272610 a(1)=5, a(2)=9, a(3)=4, a(4)=6; thereafter a(n) = a(n-a(n-1)) + a(n-a(n-2)).

5, 9, 4, 6, 5, 5, 18, 4, 5, 10, 10, 18, 4, 10, 15, 10, 27, 4, 15, 15, 10, 36, 4, 15, 20, 15, 36, 4, 20, 25, 15, 45, 4, 25, 25, 20, 45, 4, 25, 35, 15, 54, 4, 35, 25, 20, 72, 4, 25, 40, 25, 54, 4, 40, 40, 20, 72, 4, 40, 35, 25, 81, 4, 35, 50, 25, 81, 4, 50, 40, 25, 117

Offset: 1

Author

Nathan Fox, May 03 2016

Keywords

• nonn

Comments

In calculating terms of this sequence, use the convention that a(n)=0 for n<=0.

Similar to Hofstadter's Q-sequence A005185 but with different starting values.

This sequence exists as long as A272611 and A272612 exist.

No other term of this sequence changes if a(4) is replaced by a number greater than 6.

If a(2) is replaced by a number N greater than 9, then every other term of the form a(5n+2) is replaced by a(5n+2)*N/9.

Links

• Nathan Fox, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A005185, A272611, A272612, A272613.

Programs

Maple

A272610:=proc(n) option remember:
    if n <= 0 then
        return 0:
    elif n = 1 then
        return 5:
    elif n = 2 then
        return 9:
    elif n = 3 then
        return 4:
    elif n = 4 then
        return 6:
    else
        return A272610(n-A272610(n-1))+A272610(n-A272610(n-2)):
    fi:
end:

Mathematica

a[n_] := a[n] = Switch[n, ?NonPositive, 0, 1, 5, 2, 9, 3, 4, 4, 6, ,
   a[n - a[n - 1]] + a[n - a[n - 2]]];
Table[a[n], {n, 1, 80}] (* Jean-François Alcover, Jul 24 2022 *)

Python

from functools import cache
@cache
def a(n):
    if n < 0: return 0
    if n < 5: return [0, 5, 9, 4, 6][n]
    return a(n - a(n-1)) + a(n - a(n-2))
print([a(n) for n in range(1, 73)]) # Michael S. Branicky, Sep 20 2021

Formula

a(1)=5, a(2)=9, a(3)=4, a(4)=6; thereafter a(5n)=5*A272611(n), a(5n+1)=5*A272612(n), a(5n+2)=9*A272613(n), a(5n+3)=4, a(5n+4)=5*A272611(n).