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A082183
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Smallest k > 0 such that T(n) + T(k) = T(m), for some m, T(i) being the triangular numbers, n > 1.
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17
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2, 5, 9, 3, 5, 27, 10, 4, 8, 14, 17, 9, 5, 21, 135, 12, 14, 35, 6, 9, 17, 30, 12, 18, 10, 7, 54, 21, 23, 495, 42, 14, 26, 8, 49, 27, 15, 20, 98, 30, 32, 80, 9, 19, 35, 62, 45, 17, 20, 14, 99, 39, 10, 18, 54, 24, 44, 78, 81, 45, 25, 85, 153, 11, 50, 125, 20, 29, 53, 94, 97
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OFFSET
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2,1
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COMMENTS
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For 16 years this entry stood with no upper bound, and indeed with no proof that a(n) always existed. In February 2020 the following three bounds and formulas arrived. They are listed in chronological order. Here k = k(n) denotes the smallest number such that T(n)+T(k) is a triangular number T(m) for some m = m(n). - N. J. A. Sloane, Feb 22 2020
k = T(n) - 1 is an upper bound on k(n) = a(n). For T(k) makes a huge triangle; all the elements of the T(n) triangle can be thinly plated onto the side of the big one as a single additional row, producing T(k+1) with m = k+1. - Allan C. Wechsler, Feb 19 2020
Let Q be the largest odd number < n dividing T(n). Then T(n) is the sum of Q consecutive integers, the last Q rows of the triangle T(m) with m = T(n)/Q + (Q-1)/2, giving the upper bound k <= T(n)/Q - (Q+1)/2. [This bound is now A332554, the values of Q are in A332547.] This bound is not tight: for n=9 it gives a(9) <= 6 when in fact a(9) = 4. - Michael J. Collins, Feb 19 2020
Comments from Richard C. Schroeppel, Feb 19 2020: (Start)
2T(n) = 2T(m) - 2T(k) = m^2 + m - k^2 - k = (m-k) (m + k + 1). Now (m-k) and (m+k+1) are of opposite parity. Factor 2T(n) into the product of an odd number times an even number. We can take one of these to be m-k, and the other to be m+k+1.
The factorization 2T(n) = n^2 + n gives two obvious solutions, n * (n+1) and 1 * (n^2+n). Equating these to (m-k) * (m+k+1) gives the two "trivial" solutions k=0, m=n and k=T(n)-1, m=T(n).
Unless n is a Mersenne prime, or n+1 is a Fermat prime [these are the n such that Q=1, see A068194] there will be a nontrivial odd divisor of n(n+1) other than 1, n, or n+1. Select the odd divisor d logarithmicly closest to n + 1/2 that isn't n or n+1.
Let q be the quotient n(n+1)/q. Then m-k = min(d,q) and m+k+1 = max(d,q). Solve for k, which is the required minimum k(n) = a(n).
Example: n=5, T(n)=15, 2T(n)=30 = 3*10, d=3, q=10, k=3, m=6, 15+6 = 21. (End)
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LINKS
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MAPLE
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f:= proc(n) local e, t, te;
t:= n*(n+1);
e:= padic:-ordp(t, 2);
te:= 2^e;
min(map(d -> (abs(te*d-t/(te*d))-1)/2, numtheory:-divisors(t/te)) minus {0}):
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MATHEMATICA
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Table[SelectFirst[Range[10^3], Function[m, PolygonalNumber@ Floor@ Sqrt[2 m] == m][PolygonalNumber[n] + PolygonalNumber[#]] &], {n, 2, 72}] (* Michael De Vlieger, Sep 19 2017, after Maple by Robert Israel *)
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PROG
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(PARI) for(n=2, 100, t=n*(n+1)/2; for(k=1, 10^9, u=t+k*(k+1)/2; v=floor(sqrt(2*u)); if(v*(v+1)/2==u, print1(k", "); break)))
(Python)
from __future__ import division
from sympy import divisors
t = n*(n+1)
ds = divisors(t)
for i in range(len(ds)//2-2, -1, -1):
x = ds[i]
y = t//x
a, b = divmod(y-x, 2)
if b:
return a
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CROSSREFS
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See A055527 for a very similar sequence involving Pythagorean triples. - Bradley Klee, Feb 20 2020
See also A309332 (number of ways to write a triangular number as a sum of two triangular numbers), A309507 (... as a difference ...).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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