

A216414


a(n) = (1)^(n3)*binomial(n,3)  1.


1



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

3,2


COMMENTS

a(n1) 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.


LINKS

Andrew Howroyd, Table of n, a(n) for n = 3..1000
Index entries for linear recurrences with constant coefficients, signature (3,2,2,3,1).


FORMULA

a(n) = (1)^(n+1)*binomial(n,3) + 1.
a(n+1)a(n)  = A000330(n).
G.f. x*(5+6*x+4*x^2+x^3) / ( (x1)*(1+x)^4 ).  R. J. Mathar, Sep 07 2012
a(n) = sum(k=1..n3, (1)^k*C(n+1,k)).  Mircea Merca, Feb 07 2013


EXAMPLE

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).


PROG

(PARI) for(n=3, n=50, print((1)^(n3)*binomial(n, 3)1))


CROSSREFS

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).
Sequence in context: A068481 A146827 A297360 * A147018 A081883 A049744
Adjacent sequences: A216411 A216412 A216413 * A216415 A216416 A216417


KEYWORD

sign,easy


AUTHOR

V. Raman, Sep 07 2012


STATUS

approved



