A186080 Fourth powers that are palindromic in base 10.

0, 1, 14641, 104060401, 1004006004001, 10004000600040001, 100004000060000400001, 1000004000006000004000001, 10000004000000600000040000001, 100000004000000060000000400000001, 1000000004000000006000000004000000001, 10000000004000000000600000000040000000001, 100000000004000000000060000000000400000000001

Offset

1,3

Comments

See A056810 (the main entry for this problem) for further information, including the search limit. - N. J. A. Sloane, Mar 07 2011

Conjecture: If k^4 is a palindrome > 0, then k begins and ends with digit 1, all other digits of k being 0.

The number of zeros in 1x1, where the x are zeros, is the same as (the number of zeros)/4 in (1x1)^4 = 1x4x6x4x1.

Links

Table of n, a(n) for n=1..13.

P. De Geest, Palindromic cubes (The Simmons test is mentioned here) [broken link]

G. J. Simmons, Palindromic powers, J. Rec. Math., 3 (No. 2, 1970), 93-98. [Annotated scanned copy]

Formula

a(n) = A056810(n)^4.

Mathematica

Do[If[Module[{idn = IntegerDigits[n^4, 10]}, idn == Reverse[idn]], Print[n^4]], {n, 100000001}]

Prog

(Magma) [ p: n in [0..10000000] | s eq Reverse(s) where s is Intseq(p) where p is n^4 ];

Crossrefs

Cf. A002113, A168576, A056810, A002778, A002779.

Sequence in context: A017284 A017392 A017656 * A013861 A291145 A256839

Adjacent sequences: A186077 A186078 A186079 * A186081 A186082 A186083

Keyword

nonn,base

Author

Matevz Markovic, Feb 11 2011

Extensions

a(11)-a(13) using extensions of A056810 from Hugo Pfoertner, Oct 22 2021

Status

approved