|
%I #29 Sep 20 2017 08:34:42
%S 0,10,100,460,1460,3710,8120,15960,28920,49170,79420,122980,183820,
%T 266630,376880,520880,705840,939930,1232340,1593340,2034340,2567950,
%U 3208040,3969800,4869800,5926050,7158060,8586900,10235260,12127510,14289760,16749920,19537760,22684970,26225220,30194220,34629780,39571870,45062680,51146680
%N Numbers which are both the sum of n+1 consecutive triangular numbers and the sum of the n-1 immediately following triangular numbers.
%C The n+1 consecutive triangular numbers start with the A028387(n-2)-th triangular number A000217(n^2-n-1), while the n-1 consecutive triangular numbers start with the A000290(n)-th triangular number A000217(n^2).
%C Similar sums of consecutive integers are A059270.
%C Similar sums of consecutive squares are A059255.
%C Berselli points out that a(n) = 10*A024166(n-1) = A000292(n-1)*(3*n^2 - 2). Since a(n) is a sum of triangular numbers, 10=1+2+3+4 is the 4th triangular number, A024166 is a sum of cubes, and A000292 is a tetrahedral number, is there a geometric proof of Berselli's formula? (Compare Nelsen and Unal's "Proof Without Words: Runs of Triangular Numbers.") [_Jonathan Sondow_, Mar 04 2013]
%H Seiichi Manyama, <a href="/A222716/b222716.txt">Table of n, a(n) for n = 1..10000</a>
%H Roger B. Nelsen and Hasan Unal, <a href="http://maa.publisher.ingentaconnect.com/content/maa/mm/2012/00000085/00000005/art00010">Proof Without Words: Runs of Triangular Numbers</a>, Math. Mag., 85 (2012), 373.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%F a(n) = T(n^2-n-1)+T(n^2-n)+...+T(n^2-1) = T(n^2)+T(n^2+1)+...+T(n^2+n-2), where T = A000217.
%F a(n) = (3*n^5 - 5*n^3 + 2*n)/6 = (n-1)*n*(n+1)*(3*n^2 - 2)/6.
%F G.f.: 10*x^2*(1+4*x+x^2)/(1-x)^6. [_Bruno Berselli_, Mar 04 2013]
%F a(n) = -a(-n) = 10*A024166(n-1) = A000292(n-1)*A100536(n). [_Bruno Berselli_, Mar 04 2013]
%F a(n) = TP(n^2-1)-TP(n^2-n-2) = TP(n^2+n-2)-TP(n^2-1) = TP(n-1)*(3*n^2-2), where TP = A000292. [_Jonathan Sondow_, Mar 04 2013]
%e T(1) + T(2) + T(3) = 1 + 3 + 6 = 10 = T(4) and 4 = 2^2, so a(2) = 10.
%e T(5) + T(6) + T(7) + T(8) = 15 + 21 + 28 + 36 = 100 = 45 + 55 = T(9) + T(10) and 9 = 3^2, so a(3) = 100.
%t Table[ n/6 (2 - 5 n^2 + 3 n^4), {n, 1, 40}]
%t LinearRecurrence[{6,-15,20,-15,6,-1},{0,10,100,460,1460,3710},40] (* _Harvey P. Dale_, Apr 19 2016 *)
%Y Cf. A000217, A000292, A024166, A059255, A059270.
%K nonn,easy
%O 1,2
%A _Jonathan Sondow_, Mar 02 2013
|
