A371010 Powerful numbers that are the sum of 2 squares.
1, 4, 8, 9, 16, 25, 32, 36, 49, 64, 72, 81, 100, 121, 125, 128, 144, 169, 196, 200, 225, 256, 288, 289, 324, 361, 392, 400, 441, 484, 500, 512, 529, 576, 625, 648, 676, 729, 784, 800, 841, 900, 961, 968, 1000, 1024, 1089, 1125, 1152, 1156, 1225, 1296, 1352, 1369
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Rafael Jakimczuk, Generalizations of Mertens's Formula and k-Free and s-Full Numbers with Prime Divisors in Arithmetic Progression, ResearchGate, 2024.
Crossrefs
Programs
-
Mathematica
Select[Range[1500], SquaresR[2, #] > 0 && (# == 1 || Min[FactorInteger[#][[;; , 2]]] > 1) &]
-
PARI
is(n) = {my(f=factor(n)); for(i=1, #f~, if(f[i, 2] == 1 || (f[i, 2]%2 && f[i, 1]%4 == 3), return(0))); 1;}
Formula
The number of terms that do not exceed x is ~ c * sqrt(x), where c = (6/Pi^2) * (1 + 1/(3*(sqrt(2)-1))) * Product_{primes p == 1 (mod 4)} (1 + 1/((sqrt(p)-1)*(p+1))) * Product_{primes p == 3 (mod 4)} (1 + 1/(p^2-1)) = 1.58769... (Jakimczuk, 2024, Theorem 4.7, p. 50).
Sum_{n>=1} 1/a(n) = (3/2) * Product_{primes p == 1 (mod 4)} (1 + 1/(p*(p-1))) * Product_{primes p == 3 (mod 4)} (1 + 1/(p^2-1)) = (3*Pi^2/16) * A334424 = 1.86676402705119927669... .
Comments