|
%I #5 Mar 30 2012 18:58:12
%S 1,0,2,1,2,1,1,1,1,0,3,1,2,1,2,0,2,0,2,1,2,1,1,0,0,1,1,0,4,1,2,2,2,1,
%T 1,1,1,1,3,1,3,1,1,2,1,0,3,1,1,1,1,0,2,1,2,2,1,1,4,0,1,1,0,0,2,0,2,2,
%U 3,0,4,1,2,2,1,1,3,1,2,1,2,1,3,1,3,2,3,1,3,0,1,0,2,1,2,0,2,0,2
%N Number of solutions (n,k) of p(k+1)=p(n+1) (mod n), where 1<=k<n.
%C Related to A206588, which includes differences p-2.
%e For k=1 to 5, the numbers p(7)-p(k+1) are 14,12,10,6,4, so that a(6)=2.
%t f[n_,k_]:=If[Mod[Prime[n+1]-Prime[k+1],n]==0,1,0];
%t t[n_] := Flatten[Table[f[n, k], {k, 1, n - 1}]]
%t a[n_] := Count[Flatten[t[n]], 1]
%t Table[a[n], {n, 2, 120}] (* A206589 *)
%Y Cf. A206588.
%K nonn
%O 2,3
%A _Clark Kimberling_, Feb 09 2012
|
