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) = 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
KEYWORD
nonn,easy
AUTHOR
Jonathan Sondow, Mar 02 2013
STATUS
approved
