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A222716 Numbers which are both the sum of n+1 consecutive triangular numbers and the sum of the n-1 immediately following triangular numbers. 8
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 (list; graph; refs; listen; history; text; internal format)
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 A208074 A092707

Adjacent sequences:  A222713 A222714 A222715 * A222717 A222718 A222719

KEYWORD

nonn,easy

AUTHOR

Jonathan Sondow, Mar 02 2013

STATUS

approved

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Last modified September 24 07:30 EDT 2018. Contains 315308 sequences. (Running on oeis4.)