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%I #50 Apr 24 2021 03:49:23
%S 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,
%T 54,21,23,495,42,14,26,8,49,27,15,20,98,30,32,80,9,19,35,62,45,17,20,
%U 14,99,39,10,18,54,24,44,78,81,45,25,85,153,11,50,125,20,29,53,94,97
%N Smallest k > 0 such that T(n) + T(k) = T(m), for some m, T(i) being the triangular numbers, n > 1.
%C 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
%C 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
%C 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
%C Comments from Richard C. Schroeppel, Feb 19 2020: (Start)
%C 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.
%C 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).
%C 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.
%C 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).
%C Example: n=5, T(n)=15, 2T(n)=30 = 3*10, d=3, q=10, k=3, m=6, 15+6 = 21. (End)
%H Chai Wah Wu, <a href="/A082183/b082183.txt">Table of n, a(n) for n = 2..10000</a>
%H J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, <a href="http://arxiv.org/abs/2004.14000">Three Cousins of Recaman's Sequence</a>, arXiv:2004:14000 [math.NT], April 2020.
%p f:= proc(n) local e,t,te;
%p t:= n*(n+1);
%p e:= padic:-ordp(t,2);
%p te:= 2^e;
%p min(map(d -> (abs(te*d-t/(te*d))-1)/2, numtheory:-divisors(t/te)) minus {0}):
%p map(f, [$2..100]); # _Robert Israel_, Sep 15 2017
%t 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_ *)
%o (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)))
%o (Python)
%o from __future__ import division
%o from sympy import divisors
%o def A082183(n):
%o t = n*(n+1)
%o ds = divisors(t)
%o for i in range(len(ds)//2-2,-1,-1):
%o x = ds[i]
%o y = t//x
%o a, b = divmod(y-x,2)
%o if b:
%o return a
%o return -1 # _Chai Wah Wu_, Sep 12 2017
%Y Cf. A000217, A072522, values of m are in A082184, A332547.
%Y A332554 is an upper bound on a(n).
%Y See A055527 for a very similar sequence involving Pythagorean triples. - _Bradley Klee_, Feb 20 2020
%Y See also A309332 (number of ways to write a triangular number as a sum of two triangular numbers), A309507 (... as a difference ...).
%K nonn
%O 2,1
%A _Ralf Stephan_, Apr 06 2003
%E Entry updated by _N. J. A. Sloane_, Feb 22 2020
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