A371011 Cubefull numbers that are the sum of 2 squares.

1, 8, 16, 32, 64, 81, 125, 128, 256, 512, 625, 648, 729, 1000, 1024, 1296, 2000, 2048, 2197, 2401, 2592, 3125, 4000, 4096, 4913, 5000, 5184, 5832, 6561, 8000, 8192, 10000, 10125, 10368, 11664, 14641, 15625, 16000, 16384, 17576, 19208, 20000, 20736, 23328, 24389, 25000

Offset: 1

Author

Amiram Eldar, Mar 08 2024

Keywords

• nonn
• easy

Comments

For each term k, let p^e be the highest power of a prime p that divides k. Then e >= 4 is even for p == 3 (mod 4), or e >= 3 otherwise.

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

Intersection of A001481 and A036966.

Subsequence of A371010.

Cf. A020893, A064533.

Programs

Mathematica

Select[Range[25000], SquaresR[2, #] > 0 && (# == 1 || Min[FactorInteger[#][[;; , 2]]] > 2) &]

PARI

is(n) = {my(f=factor(n)); for(i=1, #f~, if(f[i, 2] < 3 || (f[i, 2]%2 && f[i, 1]%4 == 3), return(0))); 1;}

Formula

The number of terms that do not exceed x is ~ c * x^(1/3)/sqrt(log(x)), where c = (6/Pi^2) * sqrt(3) * (1 + 1/(3*(2^(1/3)-1))) * B * Product_{primes p == 1 (mod 4)} (1 + 1/((p^(1/3)-1)*(p+1))) * Product_{primes p == 3 (mod 4)} (1 + p^(1/3)/(p*(p^(2/3)-1))), where B is the Landau-Ramanujan constant (A064533) (Jakimczuk, 2024, Theorem 4.5, p. 47).

Sum_{n>=1} 1/a(n) = (5/4) * Product_{primes p == 1 (mod 4)} (1 + 1/(p^2*(p-1))) * Product_{primes p == 3 (mod 4)} (1 + 1/(p^2*(p^2-1))) = 1.281719491797642498... .