%I #10 Sep 08 2022 08:46:14
%S 0,638,1656,3162,5280,8151,11934,16807,22968,30636,40052,51480,65208,
%T 81549,100842,123453,149776,180234,215280,255398,301104,352947,411510,
%U 477411,551304,633880,725868,828036,941192,1066185,1203906,1355289,1521312,1702998,1901416,2117682
%N a(n) = n*(n + 7)*(n + 14)*(n + 21)*(n + 28)/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 = 7.
%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*(351*x^4 - 1612*x^3 + 2796*x^2 - 2172*x + 638)/(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 + 7)*(n + 14)*(n + 21)*(n + 28)/120, n = 0..35 );
%t Table[n (n + 7) (n + 14) (n + 21) (n + 28)/120, {n, 0, 40}] (* _Vincenzo Librandi_, Nov 16 2015 *)
%o (PARI) vector(100, n, n--; n*(n+7)*(n+14)*(n+21)*(n+28)/120) \\ _Altug Alkan_, Nov 15 2015
%o (Magma) [n*(n+7)*(n+14)*(n+21)*(n+28)/120: n in [0..40]]; // _Vincenzo Librandi_, Nov 16 2015
%Y Cf. A007775, A264443, A264444, A264445, A264446, A264447, A264448, A264450.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Nov 13 2015