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A279895 a(n) = n*(5*n + 11)/2.

%I A279895 #31 Nov 14 2022 17:04:05
%S A279895 0,8,21,39,62,90,123,161,204,252,305,363,426,494,567,645,728,816,909,
%T A279895 1007,1110,1218,1331,1449,1572,1700,1833,1971,2114,2262,2415,2573,
%U A279895 2736,2904,3077,3255,3438,3626,3819,4017,4220,4428,4641,4859,5082,5310,5543,5781,6024,6272,6525
%N A279895 a(n) = n*(5*n + 11)/2.
%H A279895 Bruno Berselli, <a href="/A279895/b279895.txt">Table of n, a(n) for n = 0..1000</a>
%H A279895 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A279895 O.g.f.: x*(8 - 3*x)/(1 - x)^3.
%F A279895 E.g.f.: x*(16 + 5*x)*exp(x)/2.
%F A279895 a(n+h) - a(n-h) = h*A017281(n+1), with h>=0. A particular case:
%F A279895 a(n) - a(-n) = 11*n = A008593(n).
%F A279895 a(n+h) + a(n-h) = 2*a(n) + A033429(h), with h>=0. A particular case:
%F A279895 a(n) + a(-n) = A033429(n).
%F A279895 a(n) - a(n-2) = A017281(n) for n>1. Also:
%F A279895 40*a(n) + 121 = A017281(n+1)^2.
%F A279895 a(n) = A000566(n) + 7*n, also a(n) = A000566(n) + A008589(n). - _Michel Marcus_, Dec 22 2016
%t A279895 Table[n (5 n + 11)/2, {n, 0, 60}]
%t A279895 LinearRecurrence[{3,-3,1},{0,8,21},60] (* _Harvey P. Dale_, Nov 14 2022 *)
%o A279895 (PARI) vector(60, n, n--; n*(5*n+11)/2)
%o A279895 (Python) [n*(5*n+11)/2 for n in range(60)]
%o A279895 (Sage) [n*(5*n+11)/2 for n in range(60)]
%o A279895 (Magma) [n*(5*n+11)/2: n in [0..60]];
%Y A279895 Cf. A000566, A008589, A017281.
%Y A279895 Second bisection of A165720.
%Y A279895 The first differences are in A016885.
%Y A279895 Cf. similar sequences provided by P(s,m)+s*m, where P(s,m) = ((s-2)*m^2-(s-4)*m)/2 is the m-th s-gonal number: A008585 (s=2), A055999 (s=3), A028347 (s=4), A140091 (s=5), A033537 (s=6), this sequence (s=7), A067725 (s=8).
%K A279895 nonn,easy
%O A279895 0,2
%A A279895 _Bruno Berselli_, Dec 22 2016