Quoting Neil Sloane <njasloane@gmail.com>:
Let T(i) = i(i+1)/2. Given n, let k be smallest number such that T(n) + T(k) = T(m) for some m. The k and m values are in A082183 and A082184. It must be classical that k and m always exist. - can someone supply a reference or a proof?
The graph of the k values is quite irregular. Is there an upper bound? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
2T(n) = 2T(m) - 2T(k) = m^2 + m - k^2 - k = (m-k) (m + k + 1) (m-k) and (m+k+1) are of opposite parity because their sum is 2m+1, which is odd. So, factor 2T(n) into the product of an odd number times an even number. One of these is m-k, and the other is m+k+1. 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, there will be a non-trivial odd divisor of n(n+1) other than 1, n, or n+1. Select the odd divisor d logarithmicly closest to n+.5 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. Example: n=5, T(n)=15, 2T(n)=30 = 3*10, d=3, q=10, k=3, m=6, 15+6 = 21. Rich