A264450 a(n) = n*(n + 11)*(n + 22)*(n + 33)*(n + 44)/120.

0, 3519, 8372, 14805, 23088, 33516, 46410, 62118, 81016, 103509, 130032, 161051, 197064, 238602, 286230, 340548, 402192, 471835, 550188, 638001, 736064, 845208, 966306, 1100274, 1248072, 1410705, 1589224, 1784727, 1998360, 2231318, 2484846, 2760240, 3058848, 3382071

Offset

0,2

Comments

It is well-known, and easy to prove, that the product of 5 consecutive integers n*(n + 1)*(n + 2)*(n + 3)*(n + 4) is divisible by 5!. It can be shown that the product of 5 integers in arithmetic progression n*(n + r)*(n + 2*r)*(n + 3*r)*(n + 4*r) is divisible by 5! if and only if r is not divisible by 2, 3 or 5 (see A007775 for these numbers). This is the case r = 11.

Links

Table of n, a(n) for n=0..33.

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Formula

O.g.f.: x*(2408*x^4 - 10542*x^3 + 17358*x^2 - 12742*x + 3519)/(1 - x)^6.

a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6) for n>5. - Vincenzo Librandi, Nov 16 2015

Maple

seq( n*(n + 11)*(n + 22)*(n + 33)*(n + 44)/120, n = 0..33 );

Mathematica

Table[n (n + 11) (n + 22) (n + 33) (n + 44)/120, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 3519, 8372, 14805, 23088, 33516}, 40] (* Harvey P. Dale, Nov 27 2015 *)

Prog

(PARI) vector(100, n, n--; n*(n+11)*(n+22)*(n+33)*(n+44)/120) \\ Altug Alkan, Nov 15 2015
(Magma) [n*(n+11)*(n+22)*(n+33)*(n+44)/120: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015

Crossrefs

Cf. A007775, A264443, A264444, A264445, A264446, A264447, A264448, A264449.

Sequence in context: A263302 A205278 A185920 * A187726 A267125 A241240

Adjacent sequences: A264447 A264448 A264449 * A264451 A264452 A264453

Keyword

nonn,easy

Author

Peter Bala, Nov 13 2015

Status

approved