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A216414
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a(n) = (-1)^(n-3)*binomial(n,3) - 1.
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1
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0, -5, 9, -21, 34, -57, 83, -121, 164, -221, 285, -365, 454, -561, 679, -817, 968, -1141, 1329, -1541, 1770, -2025, 2299, -2601, 2924, -3277, 3653, -4061, 4494, -4961, 5455, -5985, 6544, -7141, 7769, -8437, 9138, -9881, 10659, -11481, 12340, -13245, 14189
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OFFSET
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3,2
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COMMENTS
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a(n-1) gives the number of times the sum of all numbers divisible by all combinations of n fixed, distinct primes below a certain limit have to be added in counting the numbers divisible by at least 4 distinct primes below a certain limit.
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LINKS
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FORMULA
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a(n) = (-1)^(n+1)*binomial(n,3) + 1.
G.f. x*(5+6*x+4*x^2+x^3) / ( (x-1)*(1+x)^4 ). - R. J. Mathar, Sep 07 2012
a(n) = sum(k=1..n-3, (-1)^k*C(n+1,k)). - Mircea Merca, Feb 07 2013
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EXAMPLE
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Let k be the number of values below 10^16 that are divisible by at least four distinct primes below 100. Let b(n) be the sum of all values below 10^16 that are divisible by all combinations of n fixed, distinct primes below 100. Then k = b(4) - 5*b(5) + 9*b(6) - 21*b(7) + 34*b(8) - 57*b(9) + 83*b(10) - 121*b(11) + 164*b(12) - 221*b(13).
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PROG
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(PARI) for(n=3, n=50, print((-1)^(n-3)*binomial(n, 3)-1))
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CROSSREFS
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Cf. A002663 (number of times an element with n distinct prime factors will be counted, while taking the count for all the combinations of elements with some 4 fixed, distinct prime factors).
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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