OFFSET
0,2
COMMENTS
It is well-known, and easy to prove, that the product of 5 consecutive integers n*(n + 1)*(n + 2)*(n + 3)*(n + 4) is divisible by 5!. It can be shown that the product of 5 integers in arithmetic progression n*(n + r)*(n + 2*r)*(n + 3*r)*(n + 4*r) is divisible by 5! if and only if r is not divisible by 2, 3 or 5 (see A007775 for these numbers). This is the case r = 11.
LINKS
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).
FORMULA
O.g.f.: x*(2408*x^4 - 10542*x^3 + 17358*x^2 - 12742*x + 3519)/(1 - x)^6.
a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6) for n>5. - Vincenzo Librandi, Nov 16 2015
MAPLE
seq( n*(n + 11)*(n + 22)*(n + 33)*(n + 44)/120, n = 0..33 );
MATHEMATICA
Table[n (n + 11) (n + 22) (n + 33) (n + 44)/120, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 3519, 8372, 14805, 23088, 33516}, 40] (* Harvey P. Dale, Nov 27 2015 *)
PROG
(PARI) vector(100, n, n--; n*(n+11)*(n+22)*(n+33)*(n+44)/120) \\ Altug Alkan, Nov 15 2015
(Magma) [n*(n+11)*(n+22)*(n+33)*(n+44)/120: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 13 2015
STATUS
approved