

A270424


Numbers m such that m^2 is the sum of the squares of two or more consecutive primes.


1



586, 6088, 8174, 11585, 11707, 270106, 288818, 375661, 724909, 732910, 937423, 1141509, 1326970, 1619934, 1776809, 1930140, 2239367, 2489647, 3063687, 3649371, 3790381, 3941615, 4193988, 4821615, 4887146, 5572173, 6047246, 6192322, 8088524, 9158347
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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 jk 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., jk+1) vary widely, with the smallest being 28 and largest being 10360, for those m values listed above.
Also note jk+1 mod 8 = {0,1,4}, as expected, since prime(i)^2 mod 24 = 1, for i > 2.


LINKS



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



STATUS

approved



