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A114634
Numbers k such that the k-th octagonal number is 6-almost prime.
2
6, 14, 16, 18, 34, 36, 40, 42, 44, 46, 50, 52, 56, 60, 62, 74, 88, 98, 100, 122, 124, 130, 132, 135, 138, 142, 148, 152, 156, 158, 170, 178, 186, 189, 194, 196, 209, 226, 232, 242, 243, 244, 258, 260, 266, 274, 282, 292, 296, 297, 302, 308, 314, 315, 316, 322
OFFSET
1,1
COMMENTS
It is necessary but not sufficient that k must be prime (A000040), semiprime (A001358), 3-almost prime (A014612), 4-almost prime (A014613), or 5-almost prime (A014614).
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)
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 six prime factors (with multiplicity).
Numbers k such that A000567(k) is a term of A046306.
Numbers k such that A001222(A000567(k)) = 6.
Numbers k such that A001222(k) + A001222(3*k-2) = 6.
Numbers k such that [(3*k-2)*(3*k-1)*(3*k)]/[(3*k-2)+(3*k-1)+(3*k)] is a term of A046306.
EXAMPLE
a(1) = 6 because OctagonalNumber(6) = Oct(6) = 6*(3*6-2) = 96 = 2^5 * 3 has exactly 6 prime factors (five are all equally 2; factors need not be distinct).
a(2) = 14 because Oct(14) = 14*(3*14-2) = 560 = 2^4 * 5 * 7 is 6-almost prime.
a(3) = 16 because Oct(16) = 16*(3*16-2) = 736 = 2^5 * 23.
a(7) = 40 because Oct(40) = 40*(3*40-2) = 4720 = 2^4 * 5 * 59 [also, 4720 = Oct(40) = Oct(Oct(4)), an iterated octagonal number].
a(19) = 100 because Oct(100) = 100*(3*100-2) = 29800 = 2^3 * 5^2 * 149.
MATHEMATICA
Flatten[Position[Table[n(3n-2), {n, 400}], _?(PrimeOmega[#]==6&)]] (* Harvey P. Dale, Jun 17 2013 *)
Select[Range[400], PrimeOmega[PolygonalNumber[8, #]]==6&] (* Harvey P. Dale, Feb 23 2022 *)
PROG
(PARI) is(n)=my(t=bigomega(3*n-2)); t<6 && (t<5 || !isprime(n)) && t+bigomega(n)==6 \\ Charles R Greathouse IV, Feb 01 2017
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Feb 17 2006
STATUS
approved