A128863 a(0)=1. For n >= 1, a(n) = number of positive divisors of (n+a(n-1)).

1, 2, 3, 4, 4, 3, 3, 4, 6, 4, 4, 4, 5, 6, 6, 4, 6, 2, 6, 3, 2, 2, 8, 2, 4, 2, 6, 4, 6, 4, 4, 4, 9, 8, 8, 2, 4, 2, 8, 2, 8, 3, 6, 3, 2, 2, 10, 4, 6, 4, 8, 2, 8, 2, 8, 6, 4, 2, 12, 2, 4, 4, 8, 2, 8, 2, 6, 2, 8, 4, 4, 6, 8, 5, 2, 4, 10, 4, 4, 2, 4, 4, 4, 4, 8, 4, 12, 6, 4, 4, 4, 4, 12, 8, 8, 2, 6, 2, 9, 12, 10, 4

Offset: 0

Author

Leroy Quet, Apr 17 2007

Keywords

• nonn

Comments

Starting the sequence at a(1)=1 instead and using the recursion to get all terms after this, gets the sequence beginning: 1,2,2,4,3,3,4,6,4,4,4,5,... The sequences are the same from term a(4) on, whether starting with a(0)=1 or with a(1)=1.

Examples

			a(11)+12 = 16. So a(12) is the number of positive divisors of 16, which is 5.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000

Programs

Maple

with(numtheory): a:=proc(n) if n=0 then 1 else tau(n+a(n-1)) fi end: seq(a(n),n=0..130); # Emeric Deutsch, Apr 26 2007

Mathematica

nxt[{n_,a_}]:={n+1,DivisorSigma[0,a+n+1]}; NestList[nxt,{0,1},110][[All,2]] (* Harvey P. Dale, Mar 12 2019 *)

Extensions

More terms from Emeric Deutsch, Apr 26 2007