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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.
3
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
1,10
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.
LINKS
G. H. Hardy and J. E. Littlewood, Some problems of 'Partitio numerorum'; III: on the expression of a number as a sum of primes, Acta Mathematica, Vol. 44, pp. 1-70, 1923.
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Sep 19 2016
STATUS
approved