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A214697 Least k > 1 such that tri(n)+ ... + tri(n+k-1) is a triangular number. 2
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 (list; graph; refs; listen; history; text; internal format)
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: A219339 A048112 A001042 * A035089 A045313 A321910

Adjacent sequences:  A214694 A214695 A214696 * A214698 A214699 A214700

KEYWORD

nonn,look

AUTHOR

Alex Ratushnyak, Jul 26 2012

STATUS

approved

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Last modified March 26 04:32 EDT 2019. Contains 321481 sequences. (Running on oeis4.)