OFFSET
0,2
COMMENTS
It is well-known, and easy to prove, that the product of 4 consecutive integers n*(n + 1)*(n + 2)*(n + 3) is divisible by 4!. It can be shown that the product of 4 integers in arithmetic progression n*(n + r)*(n + 2*r)*(n + 3*r) is divisible by 4! if and only if r is not divisible by 2 or 3 (see A007310 for these numbers). This is the case r = 11.
LINKS
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
O.g.f.: x*(391 - 1045*x + 935*x^2 - 280*x^3)/(1 - x)^5.
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5) for n>4. - Vincenzo Librandi, Nov 16 2015
MAPLE
seq( n*(n + 11)*(n + 22)*(n + 33)/24, n = 0..38 );
MATHEMATICA
Table[n (n + 11) (n + 22) (n + 33)/24, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
PROG
(PARI) vector(100, n, n--; n*(n+11)*(n+22)*(n+33)/24) \\ Altug Alkan, Nov 15 2015
(Magma) [n*(n+11)*(n+22)*(n+33)/24: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 13 2015
STATUS
approved