|
%I #9 Jul 07 2015 09:39:17
%S 1,1,1,2,1,1,2,1,3,1,3,2,2,1,1,2,2,2,2,5,2,3,2,4,2,1,3,2,1,1,2,2,1,5,
%T 1,4,2,2,1,3,1,2,1,1,4,1,2,1,1,1,1,3,2,2,2,1,1,5,3,1,1,1,2,2,2,1,1,3,
%U 3,1,4,1,2,2,1,3,3,1,3,1,2,1,1,3,1,1,1,2,1,3,1,1,1,2,3,1,1,2,4,4,2,4,1,3,2
%N a(n) = number of sets of consecutive primes whose arithmetic mean is A060863(n).
%C Essentially A122821 with the 0's removed.
%F a(n) = A122821(A060863(n)).
%e For n=4; A060863(4) = 5. the two sets are 5/1 = 5, (3+5+7)/3 = 5. so a(4)=2.
%t f[n_]:=Block[{i=1,j,c=0,m},While[Prime[i]<=n, j=1; While[m=Sum[Prime[k],{k,i,i+j-1}]/j; If[m==n,c++ ]; m<n, j++ ]; i++ ]; c]; Select[Table[f[n],{n,160}],#>0&] (* _Ray Chandler_, Oct 03 2006 *)
%Y Cf. A060863, A122821.
%K easy,nonn
%O 1,4
%A _Naohiro Nomoto_, May 08 2003
%E Extended by _Ray Chandler_, Oct 03 2006
|
