cp's OEIS Frontend

Since (by definition) a(n) = n + A082183(n) - A082184(n) = - (n^2 + A082183(n)^2 - A082184(n)^2), this can be described as the distance of (n, A082183(n), A082184(n)) from a Pythagorean triple. Also a(n) > 0 for all n. See the Myers et al. link. - Bradley Klee, Feb 19 2020

Comments from N. J. A. Sloane, Feb 23 2020: (Start)

To study the lowest values taken by a(n), consider the record high values of n/a(n). The data suggests two conjectures.

Conjecture 1: The record high values of n/a(n) are j/2 + 1 for j = 2,3,4,5,... and occur at n = j*(j+1)/2 - 1.

This would imply:

Conjecture 2: Let j = 2,3,4,5,... For 1 <= n < T_j - 1, a(n) > 2*n/(j+2). (End)

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)

Consists of 1, Mersenne primes (A000668) and Fermat primes (A019434) minus 1. Proof: The sum of r consecutive integers starting with j is r*(r + 2*j - 1)/2, so k*(k+1)/2 has an extra representation of the desired form iff k*(k+1) = r*s where 1 < r, r+1 < s, and r and s have opposite parity. If k is even, let k = 2^e*m with m odd and let p be a prime divisor of k+1. Then we may take r = 2^e and s = m*(k+1) unless m=1 and we may take r = (k+1)/p and s = k*p unless k+1 is prime. Thus an even number k is in the sequence iff k+1 is a Fermat prime. Similarly an odd number k is in the sequence iff k=1 or k is a Mersenne prime.

Indices of partial maxima of A082184. - Ralf Stephan, Sep 01 2004

Consists of 1 and numbers m such that A001227(m) + A001227(m+1) = 3. - Juri-Stepan Gerasimov, Oct 06 2023

Used by Michael J. Collins in the analysis of A082183 and A082184.

Consists of 1, even perfect numbers and numbers of the form n(n+1)/2 where n+1 is a Fermat prime. (See A068194 for proof.)