%I #11 Oct 23 2012 21:59:03
%S 1,5,18,48,117,304,823,2224,6113,16974,48614,139349
%N Number of primes up to 10^n representable as sums of consecutive squares.
%C There are no common representations of two, three or six squares for n < 13, so
%C a(n) = A218208(n) + A218210(n) + A218212(n); n < 13.
%F a(n) = sum(A218213(k),k=1..n)
%e a(1) = 1 because only one prime less than 10 can be represented as a sum of consecutive squares, namely 5 = 1^2 + 2^2.
%e a(2) = 5 because there are five primes less than 100 representable as a sum of consecutive squares: the aforementioned 5, as well as 13 = 2^2 + 3^2, 29 = 2^2 + 3^2 + 4^2, 41 = 4^2 + 5^2 and 61 = 5^2 + 6^2.
%t nn = 8; nMax = 10^nn; t = Table[0, {nn}]; Do[k = n; s = 0; While[s = s + k^2; s <= nMax, If[PrimeQ[s], t[[Ceiling[Log[10, s]]]]++]; k++], {n, Sqrt[nMax]}]; Accumulate[t] (* _T. D. Noe_, Oct 23 2012 *)
%Y Cf. A027861, A027862, A027863, A027864, A027866, A027867, A163251, A174069, A218208, A218210, A218212, A218213.
%K nonn,base
%O 1,2
%A _Martin Renner_, Oct 23 2012