Recent content by OlderOwl

I don't see how you get pim ≡ (±3)im ≡ (±3)i(8q)i since p and m both ≡ (±3) mod 8

I found that the series represented by An = n∗(n+1)/2 - k never includes a perfect square if and only if the prime factorization of 8∗k+1 includes a prime factor, p, to the ith power where p = +/−3 mod 8 and i is odd and that this can be proved mod p(i + 1) For instance, 8* 4 + 1 = 3^1 * 11^1...

My big typo,. I should have typed n*(n-1)/2 - k where I typed n*(n+1)/2+k for instance 0 is a triangular number while 0 + 4 is a square 0 - 4 is not a square. Neither is n*(n+n)/2 - 4 a square for any integer n. Proof since 3^1 is part of the prime factorization of 8*4 + 1 and equals +/- 3...

I retired from the US Patent and Trademark Office after serving 35+ years there. I had a passion for Math in elementary and high school but never had much more than the minimum required calculus and differential equations etc. required for engineering, having taken nighttime instruction at...

I think the recusive algorithm is based on the fact that it takes Sk-2 steps to move k-2 discs to one pole then you have 2 free poles to move the remaining 2 discs. But you also have to prove that this is the minumum possible number of moves i.e. that Sk >= 2*(Sk-2) + 3 !

I say the latter route, Solve moving 2 disks, 4 disks, 6 disks, 8 disks and note the algorithm that solves the problem in the fewest moves each time to see the relation of that series. Then solve for moving 1 disk, 3 disks, 5 disks, 7 disks etc and note that relation. In the end the relations...