A114634 Numbers k such that the k-th octagonal number is 6-almost prime.
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
Examples
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.
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.
Programs
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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 *) -
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
Formula
Numbers k such that k*(3*k-2) has exactly six prime factors (with multiplicity).
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.
Comments