Discussion
Tue Apr 19
16:33
Bernard Schott: 2^n-1is never a square; otherwise, you should have 2^n-1= m^2 for some m; then 2^n - m^2 = 1, and by Mihăilescu - Catalan theorem, only consecutive perfect powers are 8 and 9.
20:30
Chai Wah Wu: 2^n-1 is a square for n = 1. The Mihăilescu - Catalan theorem assumes the exponent > 1.
Wed Apr 20
00:27
Michel Marcus: could we write 2^n-1 rather than -1+2^n ??
09:47
Chai Wah Wu: I prefer 2^n-1 over -1+2^n.
15:08
Chai Wah Wu: @Bernard, perhaps you can add a comment that the Mihăilescu - Catalan theorem implies that a(n) > 0 for n > 1.
