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a(0)=1. a(n) = number of earlier terms a(k), 0<=k<=n-1, such that (k+n) is divisible by a(k).
3

%I #14 Jul 23 2020 19:29:23

%S 1,1,2,2,3,4,3,4,4,5,3,7,5,5,5,5,6,8,5,4,8,8,5,9,6,6,8,8,7,9,7,9,10,9,

%T 4,10,9,8,12,7,7,11,9,9,9,12,9,13,11,7,9,12,10,12,9,9,12,12,13,13,8,

%U 10,13,14,6,12,12,12,17,10,11,14,12,12,9,14,11,16,13,12,14,13,9,16,16,12,13

%N a(0)=1. a(n) = number of earlier terms a(k), 0<=k<=n-1, such that (k+n) is divisible by a(k).

%H Rémy Sigrist, <a href="/A127432/b127432.txt">Table of n, a(n) for n = 0..10000</a>

%H Rémy Sigrist, <a href="/A127432/a127432.txt">C program for A127432</a>

%e (0+9) is a multiple of a(0)=1; (1+9) is a multiple of a(1)=1; (3+9) is a multiple of a(3)=2; (6+9) is a multiple of a(6)=3 and (7+9) is a multiple of a(7)=4. These five cases are the only cases where (k+n) is divisible by a(k), for 0<=k<=8. So a(9) = 5.

%t f[l_List] := Block[{n = Length[l]},Append[l, Count[Table[Mod[k + n - 1, l[[k]]], {k, n}], 0]]];Nest[f, {1}, 87] (* _Ray Chandler_, Jan 22 2007 *)

%o (C) See Links section.

%Y Cf. A127431, A127434.

%K nonn,look

%O 0,3

%A _Leroy Quet_, Jan 14 2007

%E Extended by _Ray Chandler_, Jan 22 2007