login
a(1) = 1; a(n) = 1 + Product_{k=1..n-1} a(k) (mod n-1).
0

%I #15 Sep 08 2020 03:29:23

%S 1,1,2,3,3,4,1,3,1,1,7,6,1,12,1,10,1,12,1,3,1,1,21,12,1,6,21,1,1,15,1,

%T 20,1,31,15,1,1,32,13,1,1,18,1,7,25,1,17,38,1,1,1,1,1,26,1,6,1,1,29,

%U 47,1,42,1,1,1,1,61,26,1,25,1,21,1,64,21,1,1,29,1,18

%N a(1) = 1; a(n) = 1 + Product_{k=1..n-1} a(k) (mod n-1).

%C Note that the running product for each a(n) is incrementally computed mod n-1.

%t a[1] = 1; a[n_] := a[n] = 1 + Mod[Product[a[k], {k, 1, n - 1}], n - 1]; Array[a, 100] (* _Amiram Eldar_, Aug 22 2020 *)

%o (Python)

%o def f(n):

%o if n == 1: return 1

%o a = 1

%o for k in range(1, n):

%o a = a * f(k) % (n - 1)

%o return a + 1

%o (PARI) lista(nn) = {my(va = vector(nn)); va[1] = 1; for (n=2, nn, va[n] = 1 + prod(k=1, n-1, va[k]) % (n-1);); va;} \\ _Michel Marcus_, Aug 23 2020

%Y Inspired by A066910.

%Y Cf. A129871 (without the mod operation).

%K nonn

%O 1,3

%A _Matt Donahoe_, Aug 22 2020