A263613 Palindromic numbers in base 4 that are cubes.

0, 1, 1331, 1030301, 1003003001, 1000300030001, 1000030000300001, 1000003000003000001, 1000000300000030000001, 1000000030000000300000001, 1000000003000000003000000001, 1000000000300000000030000000001, 1000000000030000000000300000000001, 1000000000003000000000003000000000001

Offset: 1

Author

N. J. A. Sloane, Oct 22 2015

Keywords

• nonn
• base

Comments

There is an obvious pattern here, but it is not known how long it will continue.

The cube roots (written in base 4) are 0, 1, 11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001 (coinciding with A056810 to this point).

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..18
• G. J. Simmons, On palindromic squares of non-palindromic numbers, J. Rec. Math., 5 (No. 1, 1972), 11-19. [Annotated scanned copy]

Crossrefs

Cf. A056810, A168575.

Formula

Conjectures from Chai Wah Wu, Dec 25 2023: (Start)

a(n) = 1111*a(n-1) - 112110*a(n-2) + 1111000*a(n-3) - 1000000*a(n-4) for n > 6.

G.f.: x^2*(-7000000*x^4 + 6446000*x^3 - 336330*x^2 + 220*x + 1)/((x - 1)*(10*x - 1)*(100*x - 1)*(1000*x - 1)). (End)

Extensions

a(11) from Chai Wah Wu, Oct 23 2015

a(12)-a(14) from Chai Wah Wu, Sep 26 2017