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a(n)=sum(k=1,n,C(n,n reduced (mod k))).
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%I #7 Jul 16 2017 14:04:29

%S 1,2,5,7,22,25,79,124,276,444,1267,1657,4435,7753,17327,28309,72728,

%T 110949,279663,476090,1071672,1834419,4512625,7194685,17326148,

%U 30229021,68385742,117709413,280395840,466385404,1099376097,1923155966

%N a(n)=sum(k=1,n,C(n,n reduced (mod k))).

%F Observation: lim n -> infinity a(n+2)/a(n) = 4 and it seems that a(n+2)/a(n) < 4 or >4 for infinitely many values of n.

%t Table[Sum[Binomial[n,Mod[n,k]],{k,n}],{n,40}] (* _Harvey P. Dale_, Jul 16 2017 *)

%o (PARI) a(n)=sum(k=1,n,binomial(n,n%k))

%K nonn

%O 1,2

%A _Benoit Cloitre_, Aug 20 2002