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a(n) = n*(n + 11)*(n + 22)*(n + 33)*(n + 44)/120.
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%I #13 Sep 08 2022 08:46:14

%S 0,3519,8372,14805,23088,33516,46410,62118,81016,103509,130032,161051,

%T 197064,238602,286230,340548,402192,471835,550188,638001,736064,

%U 845208,966306,1100274,1248072,1410705,1589224,1784727,1998360,2231318,2484846,2760240,3058848,3382071

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

%C 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.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

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

%F 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

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

%t Table[n (n + 11) (n + 22) (n + 33) (n + 44)/120, {n, 0, 40}] (* _Vincenzo Librandi_, Nov 16 2015 *)

%t LinearRecurrence[{6,-15,20,-15,6,-1},{0,3519,8372,14805,23088,33516},40] (* _Harvey P. Dale_, Nov 27 2015 *)

%o (PARI) vector(100, n, n--; n*(n+11)*(n+22)*(n+33)*(n+44)/120) \\ _Altug Alkan_, Nov 15 2015

%o (Magma) [n*(n+11)*(n+22)*(n+33)*(n+44)/120: n in [0..40]]; // _Vincenzo Librandi_, Nov 16 2015

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

%K nonn,easy

%O 0,2

%A _Peter Bala_, Nov 13 2015