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A180082 Semiprime centered cube numbers: m^3 + (m+1)^3.

%I A180082 #38 Jun 13 2021 22:55:02
%S A180082 9,35,91,341,559,1241,6119,7471,17261,19909,75241,143009,257651,
%T A180082 323839,671509,860851,967591,1433969,1482571,1970299,2348641,2772559,
%U A180082 3413159,4548059,5313691,5666509,7233841,7520291,9568441
%N A180082 Semiprime centered cube numbers: m^3 + (m+1)^3.
%C A180082 There are no prime centered cube numbers because m^3 + (m+1)^3 = (2m+1)*(m^2+m+1). - _Zak Seidov_, Feb 08 2011
%C A180082 Products of two primes p and q = (p^2+3)/4 with p's in A118939. - _Zak Seidov_, Feb 08 2011
%C A180082 From _Lamine Ngom_, Apr 17 2021: (Start)
%C A180082 Also numbers which are products of two primes whose sum and difference are both promic (A002378).
%C A180082 Subsequence of A217843 (sums of consecutive nonnegative cubes) limited to the terms that have only two prime factors (multiplicity counted).
%C A180082 As stated in A217843, any number that is the sum of consecutive nonnegative cubes can also be expressed as the product of two integers whose sum and difference are both promic. (Therefore, it cannot be prime.) It can also be expressed as the difference of the squares of two triangular numbers (A000217); thus, the two primes that are the factors of any term of this sequence are respectively the sum and difference of two triangular numbers. (End)
%H A180082 Harvey P. Dale, <a href="/A180082/b180082.txt">Table of n, a(n) for n = 1..1000</a>
%F A180082 A001358 INTERSECTION A005898.
%e A180082 a(1) = 1^3 + (1+1)^3 = 9 = 3^2 is semiprime.
%e A180082 a(2) = 2^3 + (2+1)^3 = 35 = 5 * 7.
%e A180082 a(3) = 3^3 + (3+1)^3 = 91 = 7 * 13.
%t A180082 Select[Total/@Partition[Range[200]^3,2,1],PrimeOmega[#]==2&] (* _Harvey P. Dale_, Feb 02 2019 *)
%Y A180082 Cf. A001222, A001358, A005898.
%Y A180082 Cf. A217843, A000217, A002378.
%K A180082 nonn,easy
%O A180082 1,1
%A A180082 _Jonathan Vos Post_, Feb 06 2011
%E A180082 More terms from _Vincenzo Librandi_, Feb 06 2011