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A264448
a(n) = n*(n + 11)*(n + 22)*(n + 33)/24.
7
0, 391, 910, 1575, 2405, 3420, 4641, 6090, 7790, 9765, 12040, 14641, 17595, 20930, 24675, 28860, 33516, 38675, 44370, 50635, 57505, 65016, 73205, 82110, 91770, 102225, 113516, 125685, 138775, 152830, 167895, 184016, 201240, 219615, 239190, 260015, 282141, 305620, 330505
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.
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
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 13 2015
STATUS
approved