Someone asked me to give more examples, so here are the first 10 cases of the identity: 1: 1 = 1 2: 2^2 = 1+3 =4 3: 3^2-1^2 = 1+3+4 = 8 4: 4^2-1^2 = " + 7 = 15 5: 5^2-2^2 = " + 6 = 21 6: 6^2-2^2+1^2 = " + 12 = 33 7: 7^2-3^2+1^2 = " + 8 = 41 8: 8^2-3^2+1^2 = " + 15 = 56 9: 9^2-4^2+2^2 = " + 13 = 69 10: 10^2-4^2+2^2-1^2 = " + 18 = 87 For the LHS (before squaring) see A235791, and the RHS is A024916 (partial sums of sigma(n)) On Thu, Nov 19, 2020 at 10:54 PM Neil Sloane <njasloane@gmail.com> wrote:
Dear Math Fun, The OEIS is full of assertions whose status is unclear - are they theorems or conjectures? This one is stated unconditionally but without a proof. If we had a proof it would help with quite a lot of other sequences. It is surely true, and maybe not difficult to prove.
Can someone help? T(k) = k*(k+1)/2, k >= 0, is a triangular number.
Claim: For n >= 1, we have
Sum_{ k >= 1, stop when T(k-1) >= n } (-1)^(k+1) * ( floor ( (n - T(k-1))/k ) )^2
= Sum_{k=1..n} sigma(k).
The RHS is the sum of the divisors of all numbers from 1 to n, A024916(n), which can also be written as Sum_{d=1..n} d*floor(n/d) = n^2 - Sum_{d=1..n} n%d.
The summands on the LHS look like [ (n-T(k-1))/k ]^2, and the sum stops when the quantity n - T(k-1) becomes zero or goes negative.
For example, when n=8, the assertion is that 8^2 - [ (8-1)/2 ]^2 + [ (8-3)/3 ]^2 = 8^2 - 3^2 + 1^2 = 56 = 1+3+4+7+6+12+8+15 .
Neil