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A143201
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Product of distances between prime factors in factorization of n.
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16
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1, 1, 1, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 6, 3, 1, 1, 2, 1, 4, 5, 10, 1, 2, 1, 12, 1, 6, 1, 6, 1, 1, 9, 16, 3, 2, 1, 18, 11, 4, 1, 10, 1, 10, 3, 22, 1, 2, 1, 4, 15, 12, 1, 2, 7, 6, 17, 28, 1, 6, 1, 30, 5, 1, 9, 18, 1, 16, 21, 12, 1, 2, 1, 36, 3, 18, 5, 22, 1, 4, 1, 40, 1, 10, 13, 42, 27, 10, 1, 6, 7, 22
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OFFSET
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1,6
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COMMENTS
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a(n) is the product of the sum of 1 and first differences of prime factors of n with multiplicity, with a(n) = 1 for n = 1 or prime n. - Michael De Vlieger, Nov 12 2023.
a(n) = 1 iff n is a prime power: a(A000961(n))=1;
a(n) = 2 iff n has exactly 2 and 3 as prime factors:
a(n) = 4 iff n has exactly 2 and 5 as prime factors:
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LINKS
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FORMULA
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a(n) = f(n,1,1) where f(n,q,y) = if n=1 then y else if q=1 then f(n/p,p,1)) else f(n/p,p,y*(p-q+1)) with p = A020639(n) = smallest prime factor of n.
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EXAMPLE
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a(86) = a(43*2) = 43-2+1 = 42;
a(138) = a(23*3*2) = (23-3+1)*(3-2+1) = 42;
a(172) = a(43*2*2) = (43-2+1)*(2-2+1) = 42;
a(182) = a(13*7*2) = (13-7+1)*(7-2+1) = 42;
a(276) = a(23*3*2*2) = (23-3+1)*(3-2+1)*(2-2+1) = 42;
a(330) = a(11*5*3*2) = (11-5+1)*(5-3+1)*(3-2+1) = 42.
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MATHEMATICA
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Table[Times@@(Differences[Flatten[Table[First[#], {Last[#]}]&/@ FactorInteger[ n]]]+1), {n, 100}] (* Harvey P. Dale, Dec 07 2011 *)
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PROG
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(Haskell)
a143201 1 = 1
a143201 n = product $ map (+ 1) $ zipWith (-) (tail pfs) pfs
where pfs = a027748_row n
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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