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A114621
Numbers k such that the k-th octagonal number is 5-almost prime.
2
8, 10, 12, 20, 26, 28, 45, 58, 63, 68, 76, 81, 82, 92, 99, 106, 115, 116, 129, 146, 159, 165, 171, 172, 188, 195, 202, 212, 213, 218, 225, 236, 255, 259, 261, 268, 273, 279, 298, 309, 325, 339, 343, 351, 362, 375, 387, 395, 399
OFFSET
1,1
COMMENTS
It is necessary but not sufficient that k must be prime (A000040), semiprime (A001358), 3-almost prime (A014612), or 4-almost prime (A014613).
LINKS
Eric Weisstein's World of Mathematics, Almost Prime.
Eric Weisstein's World of Mathematics, Octagonal Number.
FORMULA
Numbers k such that k*(3*k-2) has exactly five prime factors (with multiplicity).
Numbers k such that A000567(k) is a term of A014614.
Numbers k such that A001222(A000567(k)) = 5.
Numbers k such that A001222(k) + A001222(3*k-2) = 5.
Numbers k such that [(3*k-2)*(3*k-1)*(3*k)]/[(3*k-2)+(3*k-1)+(3*k)] is a term of A014614.
EXAMPLE
a(1) = 8 because OctagonalNumber(8) = Oct(8) = 8*(3*8-2) = 176 = 2^4 * 11 has exactly 5 prime factors (four are all equally 2; factors need not be distinct). Also, 176 = Oct(8) = Oct(Oct(2)), an iterated octagonal number. Also, 176 is a pentagonal number, hence a term of A046189 octagonal pentagonal numbers.
a(2) = 10 because Oct(10) = 10*(3*10-2) = 280 = 2^3 * 5 * 7 is 5-almost prime.
a(4) = 20 because Oct(20) = 20*(3*20-2) = 1160 = 2^3 * 5 * 29.
a(5) = 26 because Oct(26) = 26*(3*26-2) = 1976 = 2^3 * 13 * 19.
a(19) = 129 because Oct(129) = 129*(3*129-2) = 49665 = 3 * 5 * 7 * 11 * 43 is 5-almost prime (in this case, the 5 prime factors are distinct).
MATHEMATICA
Select[Range[500], PrimeOmega[PolygonalNumber[8, #]] == 5 &] (* Amiram Eldar, Oct 07 2024 *)
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Feb 17 2006
EXTENSIONS
12, 63, 99 inserted and 117 removed by R. J. Mathar, Dec 22 2010
STATUS
approved