OFFSET
1,1
COMMENTS
If n is a multiple of 10 then a(n) = 0, since no palindrome ends in 0.
Up to 200 only 3 terms are currently unknown, a(125) > 5.2*10^28, a(177) > 3.5*10^27 and a(185) > 4.5*10^27. See Links for a table of known values. - Giovanni Resta, Aug 05 2015
If a(125) > 0, then the first 3 digits of a(125) are 521 and the last 3 digits of a(125) are 125. Proof: Let m^2 = a(125)/125. Then m is odd as otherwise 125*m^2 is a multiple of 10 which is not a palindrome. Since m is odd, m^2 == 1 mod 8 and thus 125*m^2 == 125 mod 1000. - Chai Wah Wu, Mar 31 2016
LINKS
Giovanni Resta, Known values up to a(200)
EXAMPLE
a(3) = 363, because 363/3 = 11^2. 363 * 3 = 1089, which is also a square.
a(15) = 53187678135, because 53187678135/15 = 59547^2 and 53187678135 * 15 = 893205^2.
MAPLE
ispali:= proc(n) local L; L:= convert(n, base, 10); ListTools:-Reverse(L)=L end proc:
f:= proc(n) local m;
if n mod 10 = 0 then return 0 fi;
for m from 2 to 10^6 do if ispali(m^2*n) then return m^2*n fi od:
-1 # signals time-out
end proc:
seq(f(n), n=1..50); # Robert Israel, Aug 21 2015
MATHEMATICA
palQ[n_] := Block[{d = IntegerDigits@ n}, d == Reverse@ d]; a[n_] := If[ Mod[n, 10] == 0, 0, Block[{q = 2}, While[! palQ[q^2 * n], q++]; q^2 * n]]; Array[a, 42] (* Giovanni Resta, Aug 18 2015 *)
PROG
(Python)
for k in range(1, 150):
....c=2
....while c < 10**8:
........kk=k*c**2
........if kk==int(str(kk)[::-1])
............print (k, kk)
............c=10**9
........c=c+1
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Pieter Post, Jul 30 2015
EXTENSIONS
Missing a(13) from Giovanni Resta, Aug 05 2015
STATUS
approved