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

%I #16 Apr 26 2020 07:40:51

%S 1,1,2,4,4,12,12,35,63,63,173,368,734,1448,2884,5607,11340,16947,

%T 39627,79301,118928,271750,543500,1092066,2184858,4358317,8727848,

%U 17455759,34911652,61095259,130918366,244381036,506138640,1012353685,2024551664

%N a(1)=1; a(n+1) = Sum_{k=0..a(n) mod n} a(n-k).

%H Vaclav Kotesovec, <a href="/A130618/b130618.txt">Table of n, a(n) for n = 1..3300</a>

%H Vaclav Kotesovec, <a href="/A130618/a130618.jpg">Plot of a(n) / (2^n/n) for n = 1..50000</a>

%e a(10) mod 10 = 63 mod 10 = 3. So a(11) = Sum_{k=0..3} a(10-k) = a(10) + a(9) + a(8) + a(7) = 63 + 63 + 35 + 12 = 173.

%p a[1] := 1; for n to 35 do a[n+1] := add(a[n-k], k = 0 .. `mod`(a[n], n)) end do; seq(a[n], n = 1 .. 35); # _Emeric Deutsch_, Jun 21 2007

%t a[1] = 1; a[n_] := a[n] = Sum[a[n-1-k], {k, 0, Mod[a[n-1], n-1]}]; Table[a[n], {n, 1, 50}] (* _Vaclav Kotesovec_, Apr 26 2020 *)

%Y Cf. A057176.

%K nonn

%O 1,3

%A _Leroy Quet_, Jun 18 2007

%E More terms from _Jon E. Schoenfield_, Jun 21 2007

%E More terms from _Emeric Deutsch_, Jun 21 2007