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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

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