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A186080
Fourth powers that are palindromic in base 10.
4
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
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
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