%I #13 Feb 20 2019 15:01:34

%S 6,5346,8083,10578,45531,58626,2392902,2609443,7272838,51248898,

%T 98009803,159728062051,360408196038,523637103531,770378933826,

%U 998000998003,1214959556998,1434212848998,3860012299771,4243705560771

%N Numbers k such that the concatenation of k with k-2 gives a square.

%C From _Robert Israel_, Feb 20 2019: (Start) The same as A116117 and A116135 (see link).

%C So there are two equivalent definitions: numbers k such that k concatenated with k-6 gives the product of two numbers which differ by 4; and numbers k such that k concatenated with k-3 gives the product of two numbers which differ by 2.

%C For each k >= 1, 10^(4*k)-2*10^(3*k)+10^(2*k)-2*10^k+3 is a term.

%C If k is a term and k-2 has length m, then all prime factors of 10^m+1 must be congruent to 1 or 3 (mod 8). In particular, we can't have m == 2 (mod 4) or m == 3 (mod 6), as in those cases 10^m+1 would be divisible by 101 or 7 respectively. (End)

%H Robert Israel, <a href="/A115431/b115431.txt">Table of n, a(n) for n = 1..1312</a>

%H Robert Israel, <a href="/A115431/a115431.pdf">Proof that A115431, A116117 and A116135 are the same</a>

%e 8083_8081 = 8991^2.

%e 98009803_98009800 = 98999900 * 98999902, where _ denotes

%e concatenation

%p f:= proc(n) local S;

%p S:= map(t -> rhs(op(t))^2 mod 10^n+2, [msolve(x^2+2,10^n+1)]);

%p op(sort(select(t -> t-2 >= 10^(n-1) and t-2 < 10^n and issqr(t-2 + t*10^n), S)))

%p end proc:

%p seq(f(n),n=1..20); # _Robert Israel_, Feb 20 2019

%Y Cf. A030465, A102567, A115426, A115437, A115428, A115429, A115430, A115432, A115433, A115434, A115435, A115436, A115442.

%K base,nonn

%O 1,1

%A _Giovanni Resta_, Jan 24 2006