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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A276825 Number of ways to write n as x^3 + P_2, where x and P_2 are positive integers with P_2 a product of at most two primes.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 2, 3, 2, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 3, 1, 2, 1, 2, 2, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 1, 3, 4, 2, 3, 2, 4, 1, 1, 3, 3, 3, 1, 2, 3, 2, 2
Offset: 1

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Author

Zhi-Wei Sun, Sep 19 2016

Keywords

Comments

Conjecture: Any integer n > 1 can be written as x^3 + P_2, where x and P_2 are positive integers with P_2 a product of at most two primes.
We have verified this for n up to 10^8, and we guess that a(n) > 1 for all n > 3275.
It seems that any integer n > 1 also can be written as x^2 + P_2, where x and P_2 are positive integers with P_2 a product of at most two primes. Goldbach's conjecture implies that for each integer n > 1 we can write 2*n as p + q with p <= n and q >= n both prime, and hence n^2 - (n-p)^2 = p*(2n-p) = p*q is a product of two primes. In 1923 Hardy and Littlewood conjectured that if an integer n is large enough and not a square then it can be written as the sum of a prime and a square.

Examples

			a(7) = 1 since 7 = 1^3 + 2*3 with 2 and 3 both prime.
a(17) = 1 since 17 = 2^3 + 3^2 with 3 prime.
a(28) = 1 since 28 = 3^3 + 1.
a(76) = 1 since 76 = 3^3 + 7^2 with 7 prime.
a(995) = 1 since 995 = 6^3 + 19*41 with 19 and 41 both prime.
a(1072) = 1 since 1072 = 5^3 + 947 with 947 prime.
a(1252) = 1 since 1252 = 9^3 + 523 with 523 prime.
a(1574) = 1 since 1574 = 7^3 + 1231 with 1231 prime.
a(1637) = 1 since 1637 = 7^3 + 2*647 with 2 and 647 both prime.
a(2458) = 1 since 2458 = 5^3 + 2333 with 2333 prime.
a(2647) = 1 since 2647 = 12^3 + 919 with 919 prime.
a(2752) = 1 since 2752 = 5^3 + 37*71 with 37 and 71 both prime.
a(2764) = 1 since 2764 = 11^3 + 1433 with 1433 prime.
a(3275) = 1 since 3275 = 1^3 + 2*1637 with 2 and 1637 both prime.
		

Crossrefs

Programs

  • Mathematica
    P2[n_]:=P2[n]=PrimeQ[Sqrt[n]]||(SquareFreeQ[n]&&Length[FactorInteger[n]]<=2)
    Do[r=0;Do[If[P2[n-k^3],r=r+1],{k,1,(n-1)^(1/3)}];Print[n," ",r];Continue,{n,1,80}]