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A264446
a(n) = n*(n + 5)*(n + 10)*(n + 15)/24.
7
0, 44, 119, 234, 399, 625, 924, 1309, 1794, 2394, 3125, 4004, 5049, 6279, 7714, 9375, 11284, 13464, 15939, 18734, 21875, 25389, 29304, 33649, 38454, 43750, 49569, 55944, 62909, 70499, 78750, 87699, 97384, 107844, 119119, 131250, 144279, 158249, 173204, 189189, 206250
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 = 5.
FORMULA
O.g.f.: x*(4 - 3*x)*(7*x^2 - 17*x + 11)/(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 + 5)*(n + 10)*(n + 15)/24, n = 0..40 );
MATHEMATICA
Table[n (n + 5) (n + 10) (n + 15)/24, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
PROG
(PARI) vector(100, n, n--; n*(n+5)*(n+10)*(n+15)/24) \\ Altug Alkan, Nov 15 2015
(Magma) [n*(n+5)*(n+10)*(n+15)/24: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 13 2015
STATUS
approved