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%I #20 Dec 17 2017 08:02:57
%S 9,12,17,19,24,25,28,33,40,49,51,52,57,60,64,67,72,73,79,81,84,88,89,
%T 96,97,99,103,105,108,112,115,116,121,124,129,134,136,144,145,148,156,
%U 161,163,168,169,172,177,180,184,192,193,199
%N Bases N in which ab and ba are different squares, for some a and b.
%C From _Robert Israel_, Mar 14 2016: (Start)
%C Leading 0's are not allowed.
%C Conjecture: all odd squares (A016754) except 1 are terms of the sequence. (End)
%C N=(2n+1)^2, a=n^2, b=4n^2+2n+1 shows that (2n+1)^2 is a term, so this sequence is infinite. - _Michael R Peake_, Mar 21 2017
%H Robert Israel, <a href="/A174525/b174525.txt">Table of n, a(n) for n = 1..10000</a>
%e 17_9 and 71_9 are squares. 14_12 and 41_12 are squares.
%p filter:= proc(n) local x,a,b,R;
%p for x from ceil(sqrt(n)) to n-1 do
%p a:= x^2 mod n;
%p if a=0 then next fi;
%p b:= (x^2-a)/n;
%p if assigned(R[b,a]) then return true fi;
%p R[a,b]:= 1;
%p od;
%p false
%p end proc:
%p select(filter, [$1..1000]); # _Robert Israel_, Mar 14 2016
%o (MATLAB)
%o Match = zeros(1,100);
%o for N=2:200, Tens=zeros(1,N-1);Units=zeros(1,N-1); for a=N-1:-1:sqrt(N),c=a^2;Tens(a)=floor(c/N);Units(a)=rem(c,N);end; for a=N-1:-1:sqrt(N),h=find((Units==Tens(a))&([1:N-1]~=a)); if length(h),Match=any(Units(a)==Tens(h)); if Match,Sol(N)=Sol(N)+1;end;end;end;end;
%o find(Match > 0)
%Y Cf. A016754.
%K base,easy,nonn
%O 1,1
%A _Michael R Peake_, Mar 21 2010
%E MATLAB program corrected by _Robert Israel_, Mar 14 2016
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