|
%I #10 Nov 21 2013 12:48:47
%S 4,9,27,39,49,57,59,69,75,85,87,105,109,117,119,121,125,143,147,153,
%T 161,169,175,177,185,187,199,207,217,219,231,235,239,245,249,265,267,
%U 269,275,283,285,289,291,299,301,305,311,319,321,327,329,333,335,345,349,357,359,361,363,371,381,385
%N Numbers n such that n-th octagonal number is 4-almost prime.
%C It is necessary but not sufficient that n must be prime (A000040), semiprime (A001358), or 3-almost prime (A014612).
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/OctagonalNumber.html">Octagonal Number.</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AlmostPrime.html">Almost Prime.</a>
%F n such that n*(3*n-2) has exactly four prime factors (with multiplicity). n such that A000567(n) is an element of A014613. n such that A001222(A000567(n)) = 4. n such that A001222(n) + A001222(3*n-2) = 4. n such that [(3*n-2)*(3*n-1)*(3*n)]/[(3*n-2)+(3*n-1)+(3*n)] is an element of A014613.
%e a(1) = 4 because OctagonalNumber(4) = Oct(4) = 4*(3*4-2) = 40 = 2^3 * 5 has exactly 4 prime factors (3 are all equally 2; factors need not be distinct).
%e a(2) = 9 because Oct(9) = 9*(3*9-2) = 225 = 3^2 * 5^2, a 4-almost prime [225 is also a square, hence a square octagonal number A036428, as is Oct(121)].
%e a(3) = 27 because Oct(27) = 27*(3*27-2) = 2133 = 3^3 * 79.
%e a(4) = 39 because Oct(39) = 39*(3*39-2) = 4485 = 3 * 5 * 13 * 23 has exactly 4 prime factors, in this case distinct.
%e a(26) = 187 because Oct(187) = 187*(3*187-2) = 104533 = 11 * 13 * 17 * 43 [a 4-brilliant number, that is with 4 prime factors that are each the same number of digits in length].
%t Select[Range[400],PrimeOmega[#(3#-2)]==4&] (* _Harvey P. Dale_, Sep 07 2011 *)
%Y Cf. A000040, A000567, A001222, A001358, A014612, A014613, A036428.
%K easy,nonn
%O 1,1
%A _Jonathan Vos Post_, Feb 17 2006
%E 265 inserted by R. J. Mathar, Dec 22 2010
|
