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A085305
Numbers such that first reversing digits and then squaring equals the result of first squaring and then reversing.
8
0, 1, 2, 3, 11, 12, 13, 21, 22, 31, 101, 102, 103, 111, 112, 113, 121, 122, 201, 202, 211, 212, 221, 301, 311, 1001, 1002, 1003, 1011, 1012, 1013, 1021, 1022, 1031, 1101, 1102, 1103, 1111, 1112, 1113, 1121, 1122, 1201, 1202, 1211, 1212, 1301, 2001, 2002, 2011
OFFSET
1,3
COMMENTS
Only digits {0, 1, 2, 3} seem to arise.
Numbers (other than 0) that end in zero are excluded. - N. J. A. Sloane, Mar 20 2010
REFERENCES
David Wells, The Dictionary of Curious and Interesting Numbers. London: Penguin Books (1997): p. 124.
FORMULA
Solutions to rev(x^2) = rev(x)^2.
EXAMPLE
n = 13 is a term because 31^2 = 961 = rev(169) = rev(13^2) = rev(rev(31)^2).
MATHEMATICA
rt[x_] := tn[Reverse[IntegerDigits[x]]] Do[s = rt[n^2]; s1=rt[n]^2; If[Equal[s, s1]&&!Equal[Mod[n, 10], 0], Print[{n, s, rt[s1]}]], {n, 0, 1000000}]
(* Second program: *)
Select[Range[0, 1999], Mod[#, 10] != 0 && FromDigits[Reverse[IntegerDigits[#^2]]] == FromDigits[Reverse[IntegerDigits[#]]]^2 &] (* Alonso del Arte, Oct 08 2012; corrected by Jean-François Alcover, Jan 11 2021 *)
PROG
a085305 n = a085305_list !! (n-1)
a085305_list = 0 : filter (\x -> x `mod` 10 > 0
&& a004086 (x^2) == (a004086 x)^2) [1..]
-- Reinhard Zumkeller, Jul 08 2011
(Magma) [0] cat [ m: n in [1..1810] | Reverse(Intseq(m^2)) eq Intseq(Seqint(Reverse(Intseq(m)))^2) where m is n+Floor((n-1)/9) ]; // Bruno Berselli, Jul 08 2011
(PARI) isok(x) = (x==0) || ((x%10) && fromdigits(Vecrev(digits(x^2))) == fromdigits(Vecrev(digits(x)))^2); \\ Michel Marcus, Jan 11 2021
CROSSREFS
Cf. A085306. See A061909 for another version.
Sequence in context: A007932 A334054 A035122 * A189818 A116032 A116029
KEYWORD
base,nonn
AUTHOR
Labos Elemer, Jun 27 2003
STATUS
approved