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

1,2

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]

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).

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]

Example

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.

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 *)

Crossrefs

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

Sequence in context: A208365 A208144 A207713 * A111434 A364342 A208074

Adjacent sequences: A222713 A222714 A222715 * A222717 A222718 A222719

Keyword

nonn,easy

Author

Jonathan Sondow, Mar 02 2013

Status

approved