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%I #24 Dec 07 2024 08:02:23
%S 0,1,125,4913,274625,16974593,1076890625,68769820673,4398851866625,
%T 281487861809153,18014604668698625,1152924803144876033,
%U 73787029071408922625,4722367327294625677313,302231468414456377114625,19342813330006849714388993,1237940042744144791940890625,79228162569604569827557507073
%N Cubes which are palindromes in base 4.
%H Patrick De Geest, <a href="https://www.worldofnumbers.com/nobase10pg4.htm">World!Of Numbers</a>, Palindromic cubes in bases 2 to 17.
%F It seems that a(n+2) = 64^n + 3*16^n + 3*4^n + 1 for n > 0.
%F 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
%Y Cf. A020531, A046231.
%K nonn,base
%O 1,3
%A _Patrick De Geest_, May 15 1998
%E Offset changed to 1 by _Stefano Spezia_, Jul 31 2022
%E More terms using A046231 added by _Michel Marcus_, Aug 02 2022