A276825 - OEIS

%I #16 Jul 04 2018 05:50:54

%S 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,

%T 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,

%U 2,4,1,1,3,3,3,1,2,3,2,2

%N 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.

%C 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.

%C We have verified this for n up to 10^8, and we guess that a(n) > 1 for all n > 3275.

%C 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.

%H Zhi-Wei Sun, <a href="/A276825/b276825.txt">Table of n, a(n) for n = 1..10000</a>

%H G. H. Hardy and J. E. Littlewood, <a href="https://dx.doi.org/10.1007/BF02403921">Some problems of 'Partitio numerorum'; III: on the expression of a number as a sum of primes</a>, Acta Mathematica, Vol. 44, pp. 1-70, 1923.

%e a(7) = 1 since 7 = 1^3 + 2*3 with 2 and 3 both prime.

%e a(17) = 1 since 17 = 2^3 + 3^2 with 3 prime.

%e a(28) = 1 since 28 = 3^3 + 1.

%e a(76) = 1 since 76 = 3^3 + 7^2 with 7 prime.

%e a(995) = 1 since 995 = 6^3 + 19*41 with 19 and 41 both prime.

%e a(1072) = 1 since 1072 = 5^3 + 947 with 947 prime.

%e a(1252) = 1 since 1252 = 9^3 + 523 with 523 prime.

%e a(1574) = 1 since 1574 = 7^3 + 1231 with 1231 prime.

%e a(1637) = 1 since 1637 = 7^3 + 2*647 with 2 and 647 both prime.

%e a(2458) = 1 since 2458 = 5^3 + 2333 with 2333 prime.

%e a(2647) = 1 since 2647 = 12^3 + 919 with 919 prime.

%e a(2752) = 1 since 2752 = 5^3 + 37*71 with 37 and 71 both prime.

%e a(2764) = 1 since 2764 = 11^3 + 1433 with 1433 prime.

%e a(3275) = 1 since 3275 = 1^3 + 2*1637 with 2 and 1637 both prime.

%t P2[n_]:=P2[n]=PrimeQ[Sqrt[n]]||(SquareFreeQ[n]&&Length[FactorInteger[n]]<=2)

%t 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}]

%Y Cf. A000040, A000290, A000578, A037143, A045917, A235645.

%K nonn

%O 1,10

%A _Zhi-Wei Sun_, Sep 19 2016