A114828 Numbers k such that the k-th octagonal number has 9 prime factors counted with multiplicity.
64, 96, 128, 144, 162, 182, 198, 216, 224, 234, 246, 270, 278, 288, 304, 310, 320, 324, 352, 390, 414, 416, 432, 438, 480, 504, 528, 544, 550, 558, 584, 594, 600, 646, 648, 654, 662, 684, 694, 702, 710, 729, 750, 752, 756, 798, 810, 834, 850, 870, 888, 900
Offset: 1
Examples
a(1) = 64 because OctagonalNumber(64) = Oct(64) = 64*(3*64-2) = 12160 = 2^7 * 5 * 19 has exactly 9 prime factors (seven are all equally 2; factors need not be distinct). a(2) = 96 because Oct(96) = 96*(3*96-2) = 27456 = 2^6 * 3 * 11 * 13 is 9-almost prime [also 27456 = Oct(96) = Oct(Oct(6)) is an iterated octagonal number]. a(3) = 128 because Oct(128) = 128*(3*128-2) = 48896 = 2^8 * 191.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Eric Weisstein's World of Mathematics, Almost Prime.
- Eric Weisstein's World of Mathematics, Octagonal Number.
Crossrefs
Programs
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Magma
A000567:=func< n | n*(3*n-2) >; Is9almostprime:=func< n | &+[k[2]: k in Factorization(n)] eq 9 >; [ n: n in [2..1000] | Is9almostprime(A000567(n)) ]; // Klaus Brockhaus, Dec 22 2010
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Mathematica
Select[Range[900],PrimeOmega[PolygonalNumber[8,#]]==9&] (* James C. McMahon, Jul 30 2024 *)
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PARI
isok(k) = bigomega(k*(3*k-2)) == 9; \\ Michel Marcus, Aug 02 2024
Formula
Integers k such that k*(3*k-2) has exactly nine prime factors (with multiplicity).
Integers k such that (3*k-2)*(3*k-1)*(3*k)/((3*k-2)+(3*k-1)+(3*k)) is in A046310.
Extensions
Missing terms inserted by R. J. Mathar, Dec 22 2010
a(40)-a(52) from James C. McMahon, Jul 30 2024
Name edited by David A. Corneth, Jul 31 2024
Comments