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%I #14 Sep 08 2022 08:46:14
%S 0,46,104,175,260,360,476,609,760,930,1120,1331,1564,1820,2100,2405,
%T 2736,3094,3480,3895,4340,4816,5324,5865,6440,7050,7696,8379,9100,
%U 9860,10660,11501,12384,13310,14280,15295,16356,17464,18620,19825,21080
%N a(n) = n*(n + 11)*(n + 22)/6.
%C 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.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F O.g.f.: x*(35*x^2 - 80*x + 46)/(1 - x)^4.
%F 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
%p seq( n*(n + 11)*(n + 22)/6, n = 0..40 );
%t Table[n (n + 11) (n + 22)/6, {n, 0, 40}] (* _Vincenzo Librandi_, Nov 16 2015 *)
%t LinearRecurrence[{4,-6,4,-1},{0,46,104,175},50] (* _Harvey P. Dale_, Dec 11 2018 *)
%o (PARI) vector(100, n, n--; n*(n+11)*(n+22)/6) \\ _Altug Alkan_, Nov 15 2015
%o (Magma) [n*(n+11)*(n+22)/6: n in [0..40]]; // _Vincenzo Librandi_, Nov 16 2015
%Y Cf. A007310, A264443, A264444, A264446, A264447, A264448, A264449, A264450.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Nov 13 2015
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