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

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

Author

Alex Ratushnyak, Jul 26 2012

Keywords

• nonn
• look

Comments

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

Examples

			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.
		

Links

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

Crossrefs

Cf. A000217, A000292, A214648, A214696.

Programs

Maple

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

Mathematica

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 *)

Python

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
				

Python

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