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

0,1

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

Links

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

Example

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.

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 *)

Prog

(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<sum:
            tn_gte_sum = ti*(ti+1)
            ti+=1
    print i-n,
(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

Crossrefs

Cf. A000217, A000292, A214648, A214696.

Sequence in context: A360695 A048112 A001042 * A035089 A045313 A321910

Adjacent sequences: A214694 A214695 A214696 * A214698 A214699 A214700

Keyword

nonn,look

Author

Alex Ratushnyak, Jul 26 2012

Status

approved