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%I #21 Apr 18 2024 06:11:09
%S 4,9,13,25,34,38,49,74,83,87,121,169,170,195,204,208,289,290,339,361,
%T 364,373,377,458,529,579,628,650,653,662,666,819,841,890,940,961,989,
%U 1014,1023,1027,1179,1348,1369,1370,1469,1518,1543,1552,1556,1681,1731,1802,1849
%N Numbers which are the sum of some number of consecutive prime squares.
%H Cathal O'Sullivan, Jonathan P. Sorenson, and Aryn Stahl, <a href="https://arxiv.org/abs/2204.10930">An Algorithm to Find Sums of Consecutive Powers of Primes</a>, arXiv:2204.10930 [math.NT], 2022-2023. See S2 p. 10.
%H Janyarak Tongsomporn, Saeree Wananiyakul, and Jörn Steuding, <a href="https://arxiv.org/abs/2101.07558">Sums of Consecutive Prime Squares</a>, arXiv:2101.07558 [math.NT], 2021.
%H Janyarak Tongsomporn, Saeree Wananiyakul, and Jörn Steuding, <a href="http://math.colgate.edu/~integers/w9/w9.pdf">Sums of Consecutive Prime Squares</a>, #A9 INTEGERS 22 (2022).
%e The initial terms are 2^2, 3^2, 2^2+3^2, 5^2, 3^2+5^2, ...
%p N:= 10000: # for terms <= N
%p PS:= [0,seq(ithprime(i)^2,i=1..numtheory:-pi(floor(sqrt(N))))]:
%p SPS:= ListTools:-PartialSums(PS):
%p sort(convert(select(`<=`,{seq(seq(SPS[t]-SPS[s],s=1..t-1),t=2..nops(SPS))},N),list)); # _Robert Israel_, Jan 20 2021
%o (PARI) lista(nn) = {my(list = List(), ip = primepi(nn), vp = primes(ip)); for(i=1, ip, my(s=vp[i]^2); listput(list, s); for (j=i+1, ip, s += vp[j]^2; if (s >vp[ip]^2, break); listput(list, s););); Vec(vecsort(list,,8));} \\ _Michel Marcus_, Jan 20 2021
%Y Cf. A001248, A034707.
%K nonn,changed
%O 1,1
%A _Michel Marcus_, Jan 20 2021
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