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A114618
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Numbers n such that n-th octagonal number is 4-almost prime.
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1
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4, 9, 27, 39, 49, 57, 59, 69, 75, 85, 87, 105, 109, 117, 119, 121, 125, 143, 147, 153, 161, 169, 175, 177, 185, 187, 199, 207, 217, 219, 231, 235, 239, 245, 249, 265, 267, 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
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OFFSET
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1,1
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COMMENTS
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It is necessary but not sufficient that n must be prime (A000040), semiprime (A001358), or 3-almost prime (A014612).
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LINKS
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FORMULA
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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.
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EXAMPLE
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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).
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)].
a(3) = 27 because Oct(27) = 27*(3*27-2) = 2133 = 3^3 * 79.
a(4) = 39 because Oct(39) = 39*(3*39-2) = 4485 = 3 * 5 * 13 * 23 has exactly 4 prime factors, in this case distinct.
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].
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MATHEMATICA
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Select[Range[400], PrimeOmega[#(3#-2)]==4&] (* Harvey P. Dale, Sep 07 2011 *)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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265 inserted by R. J. Mathar, Dec 22 2010
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STATUS
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approved
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