A240169 Numbers n such that (6n)^3 is the sum of a twin prime pair.

1, 29, 65, 81, 99, 136, 165, 174, 176, 191, 200, 266, 295, 301, 319, 346, 351, 370, 400, 411, 431, 434, 436, 456, 491, 494, 526, 541, 599, 651, 676, 714, 746, 790, 924, 956, 991, 1011, 1131, 1161, 1194, 1259, 1274, 1280, 1304, 1374, 1550, 1641, 1644, 1649, 1714, 1715, 1739, 1804, 1811, 1814, 1830, 1879, 1941, 2000

Offset: 1

Author

Zak Seidov, Aug 02 2014

Keywords

• nonn

Comments

No terms end with 2, 3, 7, 8. Minimal differences are 1; e,g., a(52) - a(51) = 1715 - 1714. There are no three consecutive terms with common difference 1.

Distribution of last digits for first 61000 terms: W(0..9) = (10190, 10162, 0, 0, 10178, 10222, 10027, 0, 0, 10221).

For "existing" digits distribution is rather uniform.

Examples

			m = 1: (6m)^3 = 216 = 107 + 109, m = 29: (6m)^3 = 5268024 = 2634011 + 2634013.
		

Links

• Jens Kruse Andersen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A245591.

Programs

Maple

select(n -> isprime(108 * n^3 - 1) and isprime(108 * n^3 + 1), [$1..1000]); # Robert Israel, Aug 03 2014

Mathematica

Select[Range[1000], PrimeQ[216#^3/2 - 1] && PrimeQ[216#^3/2 + 1] &] (* Alonso del Arte, Aug 02 2014 *)

PARI

N=2*10^3; for(k=1,N,p=216*k^3; if(isprime(p/2-1)&&isprime(1+p/2), print1(k, ", ")))

Formula

a(n) = (1/6)*A245591(n+1)^(1/3).