A340771 Numbers which are the sum of some number of consecutive prime squares.
4, 9, 13, 25, 34, 38, 49, 74, 83, 87, 121, 169, 170, 195, 204, 208, 289, 290, 339, 361, 364, 373, 377, 458, 529, 579, 628, 650, 653, 662, 666, 819, 841, 890, 940, 961, 989, 1014, 1023, 1027, 1179, 1348, 1369, 1370, 1469, 1518, 1543, 1552, 1556, 1681, 1731, 1802, 1849
Offset: 1
Keywords
Examples
The initial terms are 2^2, 3^2, 2^2+3^2, 5^2, 3^2+5^2, ...
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
- Cathal O'Sullivan, Jonathan P. Sorenson, and Aryn Stahl, An Algorithm to Find Sums of Consecutive Powers of Primes, arXiv:2204.10930 [math.NT], 2022-2023. See S2 p. 10.
- Janyarak Tongsomporn, Saeree Wananiyakul, and Jörn Steuding, Sums of Consecutive Prime Squares, arXiv:2101.07558 [math.NT], 2021.
- Janyarak Tongsomporn, Saeree Wananiyakul, and Jörn Steuding, Sums of Consecutive Prime Squares, #A9 INTEGERS 22 (2022).
Programs
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Maple
N:= 10000: # for terms <= N PS:= [0,seq(ithprime(i)^2,i=1..numtheory:-pi(floor(sqrt(N))))]: SPS:= ListTools:-PartialSums(PS): sort(convert(select(`<=`,{seq(seq(SPS[t]-SPS[s],s=1..t-1),t=2..nops(SPS))},N),list)); # Robert Israel, Jan 20 2021
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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