OFFSET
1,1
COMMENTS
From Robert Israel, Mar 14 2016: (Start)
Leading 0's are not allowed.
Conjecture: all odd squares (A016754) except 1 are terms of the sequence. (End)
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
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
EXAMPLE
17_9 and 71_9 are squares. 14_12 and 41_12 are squares.
MAPLE
filter:= proc(n) local x, a, b, R;
for x from ceil(sqrt(n)) to n-1 do
a:= x^2 mod n;
if a=0 then next fi;
b:= (x^2-a)/n;
if assigned(R[b, a]) then return true fi;
R[a, b]:= 1;
od;
false
end proc:
select(filter, [$1..1000]); # Robert Israel, Mar 14 2016
PROG
(MATLAB)
Match = zeros(1, 100);
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;
find(Match > 0)
CROSSREFS
KEYWORD
base,easy,nonn
AUTHOR
Michael R Peake, Mar 21 2010
EXTENSIONS
MATLAB program corrected by Robert Israel, Mar 14 2016
STATUS
approved