A264445 a(n) = n*(n + 11)*(n + 22)/6.

0, 46, 104, 175, 260, 360, 476, 609, 760, 930, 1120, 1331, 1564, 1820, 2100, 2405, 2736, 3094, 3480, 3895, 4340, 4816, 5324, 5865, 6440, 7050, 7696, 8379, 9100, 9860, 10660, 11501, 12384, 13310, 14280, 15295, 16356, 17464, 18620, 19825, 21080

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

Table of n, a(n) for n=0..40.

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

Cf. A007310, A264443, A264444, A264446, A264447, A264448, A264449, A264450.

Sequence in context: A119417 A287767 A050961 * A248529 A204769 A333040

Adjacent sequences: A264442 A264443 A264444 * A264446 A264447 A264448

Keyword

nonn,easy

Author

Peter Bala, Nov 13 2015

Status

approved