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%I #19 Apr 01 2016 10:39:45
%S 4,8,363,484,5445,46464,252,2138312,12321,0,44,23232,31213,686,
%T 53187678135,44944,272,24642,171,0,525,88,575,46464,5221225,62426,
%U 36963,252,464,0,1783799973871,291080192,2112,4114,53235,69696,333,20102,93639,0,656,858858
%N a(n) = smallest palindrome k > n such that k/n is a square; a(n) = 0 if no solution exists.
%C If n is a multiple of 10 then a(n) = 0, since no palindrome ends in 0.
%C 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
%C 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
%H Giovanni Resta, <a href="/A260726/a260726.txt">Known values up to a(200)</a>
%e a(3) = 363, because 363/3 = 11^2. 363 * 3 = 1089, which is also a square.
%e a(15) = 53187678135, because 53187678135/15 = 59547^2 and 53187678135 * 15 = 893205^2.
%p ispali:= proc(n) local L; L:= convert(n,base,10); ListTools:-Reverse(L)=L end proc:
%p f:= proc(n) local m;
%p if n mod 10 = 0 then return 0 fi;
%p for m from 2 to 10^6 do if ispali(m^2*n) then return m^2*n fi od:
%p -1 # signals time-out
%p end proc:
%p seq(f(n), n=1..50); # _Robert Israel_, Aug 21 2015
%t 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 *)
%o (Python)
%o for k in range(1,150):
%o ....c=2
%o ....while c < 10**8:
%o ........kk=k*c**2
%o ........if kk==int(str(kk)[::-1])
%o ............print (k,kk)
%o ............c=10**9
%o ........c=c+1
%Y Cf. A002113, A000290, A023108, A061563.
%K nonn,base
%O 1,1
%A _Pieter Post_, Jul 30 2015
%E Missing a(13) from _Giovanni Resta_, Aug 05 2015
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