A046232 Cubes which are palindromes in base 4.
0, 1, 125, 4913, 274625, 16974593, 1076890625, 68769820673, 4398851866625, 281487861809153, 18014604668698625, 1152924803144876033, 73787029071408922625, 4722367327294625677313, 302231468414456377114625, 19342813330006849714388993, 1237940042744144791940890625, 79228162569604569827557507073
Offset: 1
Links
- Patrick De Geest, World!Of Numbers, Palindromic cubes in bases 2 to 17.
Formula
It seems that a(n+2) = 64^n + 3*16^n + 3*4^n + 1 for n > 0.
Conjectured g.f.: x^2*(1 + 40*x - 4284*x^2 + 30080*x^3 - 28672*x^4)/((1 - x)*(1 - 4*x)*(1 - 16*x)*(1 - 64*x)). - Stefano Spezia, Jul 31 2022
Extensions
Offset changed to 1 by Stefano Spezia, Jul 31 2022
More terms using A046231 added by Michel Marcus, Aug 02 2022