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 A101437 a(1) = 1; a(n+1) = sum{k=1 to n} a(n+1-k)(mod k). 0

%I #8 Mar 14 2015 17:08:04

%S 1,0,1,1,2,3,3,6,6,8,11,13,17,15,25,30,34,34,46,47,42,61,72,82,76,86,

%T 93,95,102,130,145,163,153,148,172,230,205,226,214,182,230,231,272,

%U 322,329,315,340,373,354,356,378,431,510,457,498,565,558,468,528,591,613,578

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

%e a(7) = a(6)(mod 1) + a(5)(mod 2) + a(4)(mod 3) + a(3)(mod 4) + a(2)(mod 5) + a(1)(mod 6) = 0 + 0 + 1 + 1 + 0 + 1 = 3.

%t a[1] = 1; a[n_] := a[n] = Sum[Mod[a[n - k], k], {k, 1, n - 1}]; Table[ a[n], {n, 62}] (* _Robert G. Wilson v_, Jan 25 2005 *)

%K nonn

%O 1,5

%A _Leroy Quet_, Jan 18 2005

%E More terms from _Robert G. Wilson v_, Jan 25 2005

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Last modified February 21 01:11 EST 2024. Contains 370219 sequences. (Running on oeis4.)