cp's OEIS Frontend

A216414 a(n) = (-1)^(n-3)*binomial(n,3) - 1.

%I A216414 #29 Feb 25 2018 22:49:32
%S A216414 0,-5,9,-21,34,-57,83,-121,164,-221,285,-365,454,-561,679,-817,968,
%T A216414 -1141,1329,-1541,1770,-2025,2299,-2601,2924,-3277,3653,-4061,4494,
%U A216414 -4961,5455,-5985,6544,-7141,7769,-8437,9138,-9881,10659,-11481,12340,-13245,14189
%N A216414 a(n) = (-1)^(n-3)*binomial(n,3) - 1.
%C A216414 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.
%H A216414 Andrew Howroyd, <a href="/A216414/b216414.txt">Table of n, a(n) for n = 3..1000</a>
%H A216414 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (-3,-2,2,3,1).
%F A216414 a(n) = (-1)^(n+1)*binomial(n,3) + 1.
%F A216414 |a(n+1)-a(n) | = A000330(n).
%F A216414 G.f. x*(5+6*x+4*x^2+x^3) / ( (x-1)*(1+x)^4 ). - _R. J. Mathar_, Sep 07 2012
%F A216414 a(n) = sum(k=1..n-3, (-1)^k*C(n+1,k)). - _Mircea Merca_, Feb 07 2013
%e A216414 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).
%o A216414 (PARI) for(n=3,n=50,print((-1)^(n-3)*binomial(n,3)-1))
%Y A216414 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).
%K A216414 sign,easy
%O A216414 3,2
%A A216414 _V. Raman_, Sep 07 2012