

A082183


Smallest k > 0 such that T(n) + T(k) = T(m), for some m, T(i) being the triangular numbers, n > 1.


17



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

2,1


COMMENTS

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 + (Q1)/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 = (mk) (m + k + 1). Now (mk) 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 mk, 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 (mk) * (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 mk = 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)


LINKS

Chai Wah Wu, Table of n, a(n) for n = 2..10000
J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, Three Cousins of Recaman's Sequence, arXiv:2004:14000, April 2020


MAPLE

f:= proc(n) local e, t, te;
t:= n*(n+1);
e:= padic:ordp(t, 2);
te:= 2^e;
min(map(d > (abs(te*dt/(te*d))1)/2, numtheory:divisors(t/te)) minus {0}):
map(f, [$2..100]); # Robert Israel, Sep 15 2017


MATHEMATICA

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 *)


PROG

(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
def A082183(n):
t = n*(n+1)
ds = divisors(t)
for i in range(len(ds)//22, 1, 1):
x = ds[i]
y = t//x
a, b = divmod(yx, 2)
if b:
return a
return 1 # Chai Wah Wu, Sep 12 2017


CROSSREFS

Cf. A000217, A072522, values of m are in A082184, A332547.
A332554 is an upper bound on a(n).
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 ...).
Sequence in context: A248934 A011432 A324835 * A332554 A333530 A111474
Adjacent sequences: A082180 A082181 A082182 * A082184 A082185 A082186


KEYWORD

nonn,changed


AUTHOR

Ralf Stephan, Apr 06 2003


EXTENSIONS

Entry updated by N. J. A. Sloane, Feb 22 2020


STATUS

approved



