OFFSET
1,1
COMMENTS
m^2 = Sum_{i=k..j} prime(i)^2 is a square, for some k,j, j > k.
The 30 numbers given above are the only m values for all possible summations where the resulting m^2 < 10^14 (m <10^7). This requires searching from k values up to ~482,000, but with decreasing j-k ranges for efficiency.
Values of k that yield results begin: 13, 37, 101, 183, 235, 588, 805, 891, 1066, ... but do not correspond fully to the order of the m values shown.
Number of sequential summands (i.e., j-k+1) vary widely, with the smallest being 28 and largest being 10360, for those m values listed above.
Also note j-k+1 mod 8 = {0,1,4}, as expected, since prime(i)^2 mod 24 = 1, for i > 2.
LINKS
Giovanni Resta, Table of n, a(n) for n = 1..164 (terms < 4*10^9)
Giovanni Resta, Details of the sums, for a(n) < 4*10^9
EXAMPLE
586 is in the sequence because 586^2 = 343396 = Sum_{i=13..40} prime(i)^2.
MATHEMATICA
lim = 20000^2; L={}; P=Prime[Range[2 + PrimePi@ Sqrt[lim/2]]]^2; i = 1; While[ P[[i]] + P[[i+1]] <= lim, s = P[[i]]; j = i+1; While[(s += P[[j++]]) <= lim, If[IntegerQ@ Sqrt@ s, AppendTo[L, Sqrt@ s]]]; i++]; Union@L (* Giovanni Resta, Apr 13 2016 *)
result = {}; k = 3; While[k <= 481167, resultk = {}; sump = 0;
count = 0; i = k; While[sump < 10^14, sump += Prime[i]^2;
If[Mod[i - k + 1, 8] == 1 || Mod[i - k + 1, 8] == 0 ||
Mod[i - k + 1, 8] == 4, If[i != k && IntegerQ[Sqrt[sump]], count++;
AppendTo[resultk, {k, i - k + 1, sump}]]]; i++];
If[count > 0, AppendTo[result, resultk]]; k++]; result (* Only for k>2, so as to use index values to reduce repeated checking Sqrt - Richard R. Forberg, Apr 14 2016 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Richard R. Forberg, Mar 30 2016
STATUS
approved