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a(n) = n-th partial sum of A094339 divided by A094339(n+1).
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%I #8 Aug 08 2015 20:14:09

%S 2,1,1,3,2,2,4,9,5,4,3,5,5,9,5,4,8,15,8,12,27,52,7,4,2,5,3,6,106,14,5,

%T 9,5,4,6,107,180,21,362,121,183,176,69,59,150,28,151,232,19,10,2,11,9,

%U 233,360,247,304,155,244,195,98,231,174,196,50,591,296,198,51,199,160,115

%N a(n) = n-th partial sum of A094339 divided by A094339(n+1).

%C Conjecture: Every natural number occurs in this sequence.

%p A094339 := proc(nmax) local a,n,sprev,i; a := [2] ; while nops(a) < nmax do sprev := add(i,i=a) ; n := 1 ; while sprev mod n <> 0 or n in a do n := n+1 ; od ; a := [op(a),n] ; od ; RETURN(a) ; end: A094340 := proc(a094339,n) add( op(i,a094339),i=1..n)/op(n+1,a094339) ; end: a094339 := A094339(100) ; for n from 1 to nops(a094339)-1 do printf("%d, ", A094340(a094339,n)) ; od ; # _R. J. Mathar_, Apr 30 2007

%Y Cf. A094339, A094341.

%K nonn

%O 1,1

%A _Amarnath Murthy_, May 17 2004

%E Corrected and extended by _R. J. Mathar_, Apr 30 2007