OFFSET
0,2
COMMENTS
It is well-known, and easy to prove, that the product of 3 consecutive integers n*(n + 1)*(n + 2) is divisible by 6. It can be shown that the product of 3 integers in arithmetic progression n*(n + r)*(n + 2*r) is divisible by 6 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 (4,-6,4,-1).
FORMULA
O.g.f.: x*(35*x^2 - 80*x + 46)/(1 - x)^4.
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3. - Vincenzo Librandi, Nov 16 2015
MAPLE
seq( n*(n + 11)*(n + 22)/6, n = 0..40 );
MATHEMATICA
Table[n (n + 11) (n + 22)/6, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
LinearRecurrence[{4, -6, 4, -1}, {0, 46, 104, 175}, 50] (* Harvey P. Dale, Dec 11 2018 *)
PROG
(PARI) vector(100, n, n--; n*(n+11)*(n+22)/6) \\ Altug Alkan, Nov 15 2015
(Magma) [n*(n+11)*(n+22)/6: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 13 2015
STATUS
approved