A214648 -id:A214648 - cp's OEIS frontend

A214696 Sum of next a(n) > 1 positive triangular numbers is a triangular number.

3, 7, 13, 17, 4919, 73789919

Offset: 1

Alex Ratushnyak, Jul 26 2012

Two or more triangular numbers in the sum.

Because an oblong number is twice a triangular number, this sequence also gives: the sum of next a(n) positive oblong numbers is an oblong number.

			1+3+6 = 10 is a triangular number, so a(1)=3, then 10+15+21+28+36+45+55 = 210 is a triangular number, seven summands, so a(2)=7.

Cf. A000217, A073684, A214648.

a(n) is the smallest integer k > 1 such that (4*k^3 + 12*s*k^2 + 4*(3*s^2-1)*k)/3 + 1 is a square, where s = 1 + a(1) + a(2) + ... + a(n-1). - Max Alekseyev, Jan 30 2014

A214697 Least k > 1 such that tri(n)+ ... + tri(n+k-1) is a triangular number.

Original entry on oeis.org

2, 3, 5, 17, 7, 2, 89, 125, 3, 215, 269, 13, 10, 8, 11, 27, 719, 815, 21, 57, 316, 11, 26, 1517, 17, 1799, 30, 26, 7, 5, 2609, 11, 2975, 10, 2, 76, 3779, 1251, 208, 4445, 115, 4919, 1045, 5417, 11, 17, 1205, 6485, 38, 2860, 7349, 18, 25, 8267, 8585, 8909

Offset: 0

Alex Ratushnyak, Jul 26 2012

tri(n) = n*(n+1)/2 is the n-th triangular number, A000217(n).
a(n) is how many consecutive triangular numbers starting from tri(n) are needed to sum up to tri(x) for some x. The requirement a(n) > 1 is needed, because otherwise all a(n) = 1.
Because an oblong number (A002378) is twice a triangular number, this sequence is also the least k > 1 such that oblong(n) + ... + oblong(n+k-1) is an oblong number.
a(n) is least k > 1 such that 12*k^3 + 36*n*k^2 + 36*k*n^2 - 12*k + 9 is a perfect square. - Chai Wah Wu, Mar 01 2016
a(n) <= 3*n^2 - 3*n - 1 for n > 1, since 12*k^3 + 36*n*k^2 + 36*k*n^2 - 12*k + 9 is a square when k = 3*n^2 - 3*n - 1. - Robert Israel, Mar 03 2016

			0+1 = 1 is a triangular number, two summands, so a(0)=2.
1+3+6 = 10 is a triangular number, three summands, so a(1)=3.
3+6+10+15+21 = 55 is a triangular number, five summands, so a(2)=5.
Starting from Triangular(5)=15:  15+21=36 is a triangular number, two summands, so a(5)=2.

Chai Wah Wu, Table of n, a(n) for n = 0..5000

Cf. A000217, A000292, A214648, A214696.

f:= proc(n) local k;
    for k from 2 do if issqr(12*k^3+36*k^2*n+36*k*n^2-12*k+9) then return k fi od
end proc:
map(f, [$0..100]); # Robert Israel, Mar 03 2016

triQ[n_] := IntegerQ[Sqrt[1+8*n]]; Table[k = n+1; s = k^2; While[! triQ[s], k++; s = s + k*(k+1)/2]; k - n + 1, {n, 0, 55}] (* T. D. Noe, Jul 26 2012 *)

for n in range(77):
    i = ti = n
    sum = 0
    tn_gte_sum = 0  # least oblong  number >= sum
    while i-n<=1 or tn_gte_sum!=sum:
        sum += i*(i+1)
        i+=1
        while tn_gte_sum

from math import sqrt
def A214697(n):
    k, a1, a2, m = 2, 36*n, 36*n**2 - 12, n*(72*n + 144) + 81
    while int(round(sqrt(m)))**2 != m:
        k += 1
        m = k*(k*(12*k + a1) + a2) + 9
    return k # Chai Wah Wu, Mar 01 2016

cp's OEIS Frontend

A214696 Sum of next a(n) > 1 positive triangular numbers is a triangular number.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A214697 Least k > 1 such that tri(n)+ ... + tri(n+k-1) is a triangular number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Python

Python