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A264448 a(n) = n*(n + 11)*(n + 22)*(n + 33)/24.

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%I A264448 #10 Sep 08 2022 08:46:14
%S A264448 0,391,910,1575,2405,3420,4641,6090,7790,9765,12040,14641,17595,20930,
%T A264448 24675,28860,33516,38675,44370,50635,57505,65016,73205,82110,91770,
%U A264448 102225,113516,125685,138775,152830,167895,184016,201240,219615,239190,260015,282141,305620,330505
%N A264448 a(n) = n*(n + 11)*(n + 22)*(n + 33)/24.
%C A264448 It is well-known, and easy to prove, that the product of 4 consecutive integers n*(n + 1)*(n + 2)*(n + 3) is divisible by 4!. It can be shown that the product of 4 integers in arithmetic progression n*(n + r)*(n + 2*r)*(n + 3*r) is divisible by 4! if and only if r is not divisible by 2 or 3 (see A007310 for these numbers). This is the case r = 11.
%H A264448 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A264448 O.g.f.: x*(391 - 1045*x + 935*x^2 - 280*x^3)/(1 - x)^5.
%F A264448 a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5) for n>4. - _Vincenzo Librandi_, Nov 16 2015
%p A264448 seq( n*(n + 11)*(n + 22)*(n + 33)/24, n = 0..38 );
%t A264448 Table[n (n + 11) (n + 22) (n + 33)/24, {n, 0, 40}] (* _Vincenzo Librandi_, Nov 16 2015 *)
%o A264448 (PARI) vector(100, n, n--; n*(n+11)*(n+22)*(n+33)/24) \\ _Altug Alkan_, Nov 15 2015
%o A264448 (Magma) [n*(n+11)*(n+22)*(n+33)/24: n in [0..40]]; // _Vincenzo Librandi_, Nov 16 2015
%Y A264448 Cf. A007310, A264443, A264444, A264445, A264446, A264447, A264449, A264450.
%K A264448 nonn,easy
%O A264448 0,2
%A A264448 _Peter Bala_, Nov 13 2015