OFFSET
1,1
COMMENTS
All terms == 75 mod 90. It appears that all terms == 8625 mod 9000.
All members of the sequence are obtained as follows.
Let f(t) = 25 t - 7/2 + 1/(16 t), and 0 < t_1 < t_2 < t_3 < t_4 where f(t_1) = f(t_4) = 1 and f(t_2) = f(t_3) = 1/10.
If -log(t_2)/log(25) -1 <= frac(d*log(10)/log(25)) < -log(t_1)/log(25) - 1, then
f(t)*10^d is in the sequence, where a = floor(d*log(10)/log(25)) - 1 and t = 25^a/10^d.
If -log(t_4)/log(25) < frac(d*log(10)/log(25)) <= -log(t_3)/log(25), then
f(t)*10^d is in the sequence, where a = floor(d*log(10)/log(25)) and t = 25^a/10^d.
By the equidistribution theorem and irrationality of log(10)/log(25), there are infinitely many terms of each type.
LINKS
Robert Israel, Table of n, a(n) for n = 1..177
EXAMPLE
For n=1, 180625 = 425^2 and 680625 = 825^2 are squares, so 80625 is in the sequence.
MAPLE
f:= t -> 25*t - 7/2 + 1/(16*t):
t1, t2, t3, t4:= 9/100-(1/50)*14^(1/2), 9/125-(1/500)*671^(1/2), 9/125+(1/500)*671^(1/2), 9/100+(1/50)*14^(1/2):
v1, v2, v3, v4:= -log(t1)/log(25)-1, -log(t2)/log(25)-1, -log(t3)/log(25), -log(t4)/log(25):
A:= NULL:
for d from 1 to 100 do
v:= frac(d*log(10)/log(25));
if is(v >= v2 and v < v1) then
a:= d*log(10)/log(25) - v - 1;
A:= A, 10^d*f(25^a/10^d);
elif is(v > v4 and v <= v3) then
a:= d*log(10)/log(25) - v;
A:= A, 10^d*f(25^a/10^d);
fi
od:
A;
PROG
(PARI) is_A264805(n)=issquare(10^#digits(n)+n)&&issquare(6*10^#digits(n)+n) \\ M. F. Hasler, Nov 25 2015
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Robert Israel, Nov 25 2015
STATUS
approved