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a(n) = n*(n + 1)*(3*n^2 + 3*n - 2)/8.
11

%I #41 Sep 08 2022 08:46:06

%S 0,1,12,51,145,330,651,1162,1926,3015,4510,6501,9087,12376,16485,

%T 21540,27676,35037,43776,54055,66045,79926,95887,114126,134850,158275,

%U 184626,214137,247051,283620,324105,368776,417912,471801,530740,595035,665001,740962

%N a(n) = n*(n + 1)*(3*n^2 + 3*n - 2)/8.

%C After 0, first trisection of A011779 and right border of A177708.

%H Bruno Berselli, <a href="/A236770/b236770.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F G.f.: x*(1 + 7*x + x^2)/(1 - x)^5.

%F a(n) = a(-n-1) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).

%F a(n) = A000326(A000217(n)).

%F a(n) = A000217(n) + 9*A000332(n+2).

%F Sum_{n>=1} 1/a(n) = 2 + 4*sqrt(3/11)*Pi*tan(sqrt(11/3)*Pi/2) = 1.11700627139319... . - _Vaclav Kotesovec_, Apr 27 2016

%t Table[n (n + 1) (3 n^2 + 3 n - 2)/8, {n, 0, 40}]

%t LinearRecurrence[{5,-10,10,-5,1},{0,1,12,51,145},40] (* _Harvey P. Dale_, Aug 22 2016 *)

%o (PARI) for(n=0, 40, print1(n*(n+1)*(3*n^2+3*n-2)/8", "));

%o (Magma) [n*(n+1)*(3*n^2+3*n-2)/8: n in [0..40]];

%Y Partial sums of A004188.

%Y Cf. A000217, A000332, A011779, A177708.

%Y Cf. similar sequences on the polygonal numbers: A002817(n) = A000217(A000217(n)); A000537(n) = A000290(A000217(n)); A037270(n) = A000217(A000290(n)); A062392(n) = A000384(A000217(n)).

%Y Cf. sequences of the form A000217(m)+k*A000332(m+2): A062392 (k=12); A264854 (k=11); A264853 (k=10); this sequence (k=9); A006324 (k=8); A006323 (k=7); A000537 (k=6); A006322 (k=5); A006325 (k=4), A002817 (k=3), A006007 (k=2), A006522 (k=1).

%Y Cf. A232713, A260810.

%K nonn,easy

%O 0,3

%A _Bruno Berselli_, Jan 31 2014