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A264444
a(n) = n*(n + 7)*(n + 14)/6.
7
0, 20, 48, 85, 132, 190, 260, 343, 440, 552, 680, 825, 988, 1170, 1372, 1595, 1840, 2108, 2400, 2717, 3060, 3430, 3828, 4255, 4712, 5200, 5720, 6273, 6860, 7482, 8140, 8835, 9568, 10340, 11152, 12005, 12900, 13838, 14820, 15847, 16920
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 = 7.
FORMULA
O.g.f.: x*(13*x^2 - 32*x + 20)/(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 + 7)*(n + 14)/6, n = 0..40 );
MATHEMATICA
Table[n (n + 7) (n + 14)/6, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
PROG
(PARI) vector(100, n, n--; n*(n+7)*(n+14)/6) \\ Altug Alkan, Nov 15 2015
(Magma) [n*(n+7)*(n+14)/6: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 13 2015
STATUS
approved