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%I #8 Aug 09 2015 00:28:25
%S 3,7,8,9,17,18,20,21,22,23,25,26,28,30,31,37,44,49,50,61,62,65,66,69,
%T 71,74,76,78,79,85,89,93,97,98,113,116,121,122,129,130,133,137,141,
%U 146,148,151,154,157,158,161,164,166,170,173,174,178,185,186,188,190,193,194
%N Indices of 3-almost prime pentagonal numbers.
%C P(2) = 5 is the only prime pentagonal number, all other factor as P(k) = (k/2)*(3*k-1) or k*((3*k-1)/2) and thus have at least 2 prime factors. P(k) is semiprime iff [k prime and (3*k-1)/2 prime] or [k/2 prime and 3*k-1 prime].
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PentagonalNumber.html">Pentagonal Number.</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AlmostPrime.html">Almost Prime.</a>
%F {a(n)} = {k such that A001222(A000326(k)) = 3}. {a(n)} = {k such that k*(3*k-1)/2 has exactly 3 prime factors}. {a(n)} = {k such that A000326(k) is an element of A014612}.
%e a(1) = 3 because P(3) = PentagonalNumber(3) = 3*(3*3 -1)/2 = 12 = 2^2 * 3 is a 3-almost prime.
%e a(2) = 7 because P(7) = 7*(3*7 -1)/2 = 70 = 2 * 5 * 7 is a 3-almost prime.
%p A000326 := proc(n) n*(3*n-1)/2 ; end: isA014612 := proc(n) option remember ; RETURN( numtheory[bigomega](n) = 3) ; end: for n from 1 to 400 do if isA014612(A000326(n)) then printf("%d,",n) ; fi; od: # _R. J. Mathar_, Jan 27 2009
%Y Cf. A000326, A001222, A014612, A115709.
%K easy,nonn
%O 1,1
%A _Jonathan Vos Post_, Feb 14 2006
%E 125 removed, 145 replaced with 146 by _R. J. Mathar_, Jan 27 2009
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