A222716 Numbers which are both the sum of n+1 consecutive triangular numbers and the sum of the n-1 immediately following triangular numbers.

0, 10, 100, 460, 1460, 3710, 8120, 15960, 28920, 49170, 79420, 122980, 183820, 266630, 376880, 520880, 705840, 939930, 1232340, 1593340, 2034340, 2567950, 3208040, 3969800, 4869800, 5926050, 7158060, 8586900, 10235260, 12127510, 14289760, 16749920, 19537760, 22684970, 26225220, 30194220, 34629780, 39571870, 45062680, 51146680

Offset: 1

Author

Jonathan Sondow, Mar 02 2013

Keywords

• nonn
• easy

Comments

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).

Similar sums of consecutive integers are A059270.

Similar sums of consecutive squares are A059255.

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]

Examples

			T(1) + T(2) + T(3) = 1 + 3 + 6 = 10 = T(4) and 4 = 2^2, so a(2) = 10.
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.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000
• Roger B. Nelsen and Hasan Unal, Proof Without Words: Runs of Triangular Numbers, Math. Mag., 85 (2012), 373.
• Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Crossrefs

Cf. A000217, A000292, A024166, A059255, A059270.

Programs

Mathematica

Table[ n/6 (2 - 5 n^2 + 3 n^4), {n, 1, 40}]
LinearRecurrence[{6,-15,20,-15,6,-1},{0,10,100,460,1460,3710},40] (* Harvey P. Dale, Apr 19 2016 *)

Formula

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.

a(n) = (3*n^5 - 5*n^3 + 2*n)/6 = (n-1)*n*(n+1)*(3*n^2 - 2)/6.

G.f.: 10*x^2*(1+4*x+x^2)/(1-x)^6. [Bruno Berselli, Mar 04 2013]

a(n) = -a(-n) = 10*A024166(n-1) = A000292(n-1)*A100536(n). [Bruno Berselli, Mar 04 2013]

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]