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Semiprimes of the form (2*n^3+n)/3.
1

%I #13 Jul 17 2014 22:31:10

%S 6,85,146,489,1469,3281,4579,6181,8119,19871,23969,99269,238631,

%T 439031,470009,536269,961969,1240619,1365631,2579981,2887219,3105031,

%U 3696881,3953221,5096981,5413801,7002379,8006069,8874781,22050881,23310631,27854731,34596869,40465769

%N Semiprimes of the form (2*n^3+n)/3.

%C Semiprimes in A005900.

%C The n-th octahedral number = (2*n^3+n)/3.

%H K. D. Bajpai, <a href="/A245232/b245232.txt">Table of n, a(n) for n = 1..8050</a>

%e n=5: (2*n^3 + n)/3 = 85 = 5 * 17 which is semiprime. Hence 85 appears in the sequence.

%e n=9: (2*n^3 + n)/3 = 489 = 3 * 163 which is semiprime. Hence 489 appears in the sequence.

%p select(t -> numtheory:-bigomega(t)=2, [seq((2*n^3+n)/3, n=1..1000)]); # _Robert Israel_, Jul 15 2014

%t Select[Table[(2*n^3 + n)/3, {n, 500}], PrimeOmega[#] == 2 &]

%o (PARI) s=[]; for(n=1, 500, m=(2*n^3+n); if(m%3==0 && bigomega(m\3)==2, s=concat(s, m\3))); s \\ _Colin Barker_, Jul 15 2014

%Y Cf. A001358, A005900.

%K nonn

%O 1,1

%A _K. D. Bajpai_, Jul 14 2014