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Search: id:A112801
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| A112801 |
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Number of ways of representing 2n-1 as sum of three integers with 2 distinct prime factors. |
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+0 5
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| 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 2, 2, 2, 4, 4, 4, 8, 7, 8, 11, 11, 13, 15, 16, 18, 23, 23, 26, 30, 31, 33, 40, 40, 45, 51, 53, 56, 62, 66, 66, 76, 79, 82, 88, 94, 96, 105, 111, 111, 124, 127, 132, 141, 145, 148, 164, 166, 170, 180, 187, 187, 206, 204, 208
(list; graph; listen)
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OFFSET
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1,17
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COMMENT
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Meng proves a remarkable generalization of the Goldbach-Vinogradov classical result that every sufficiently large odd integer N can be partitioned as the sum of three primes N = p1 + p2 + p3. The new proof is that every sufficiently large odd integer N can be partitioned as the sum of three integers N = a + b + c where each of a, b, c has k distinct prime factors for the same k.
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REFERENCES
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Xianmeng Meng, On sums of three integers with a fixed number of prime factors, Journal of Number Theory, Vol. 114 (2005), pp. 37-65.
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FORMULA
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Number of ways of representing 2n-1 as sum of three semiprimes (A001358) or products of two powers of primes (A000961 except 1). Number of ways of representing 2n-1 as a + b + c where omega(a) = omega(b) = omega(c) = 2. Number of ways of representing 2n-1 as a + b + c where A001221(a) = A001221(b) A001221(c) = 2.
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EXAMPLE
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a(14) = 1 because the only partition into three integers each with 2 distinct prime factors of (2*14)-1 = 27 is 27 = 6 + 6 + 15 = (2*3) + (2*3) + (3*5).
a(16) = 1 because the only partition into three integers each with 2 distinct prime factors of (2*16)-1 = 31 is 31 = 6 + 10 + 15 = (2*3) + (2*5) + (3*5).
a(17) = 2 because the two partitions into three integers each with 2 distinct prime factors of (2*17)-1 = 33 are 33 = 6 + 6 + 21 = 6 + 12 + 15.
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CROSSREFS
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Cf. A000961, A001358, A112799, A112800, A112802.
Sequence in context: A007294 A053282 A001584 this_sequence A089873 A096323 A035682
Adjacent sequences: A112798 A112799 A112800 this_sequence A112802 A112803 A112804
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KEYWORD
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nonn
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AUTHOR
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Jonathan Vos Post (jvospost2(AT)yahoo.com) and Ray Chandler (rayjchandler(AT)sbcglobal.net), Sep 19 2005
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