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 = 7.
LINKS
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
O.g.f.: x*(110 - 274*x + 230*x^2 - 65*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 + 7)*(n + 14)*(n + 21)/24, n = 0..40 );
MATHEMATICA
Table[n (n + 7) (n + 14) (n + 21)/24, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
Table[Times@@(n+7*Range[0, 3])/24, {n, 0, 40}] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 110, 276, 510, 825}, 50] (* Harvey P. Dale, May 01 2017 *)
PROG
(PARI) vector(100, n, n--; n*(n+7)*(n+14)*(n+21)/24) \\ Altug Alkan, Nov 15 2015
(Magma) [n*(n+7)*(n+14)*(n+21)/24: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 13 2015
STATUS
approved