OFFSET
1,1
COMMENTS
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.
For each k >= 1, 10^(4*k)-2*10^(3*k)+10^(2*k)-2*10^k+3 is a term.
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)
LINKS
Robert Israel, Table of n, a(n) for n = 1..1312
Robert Israel, Proof that A115431, A116117 and A116135 are the same
EXAMPLE
8083_8081 = 8991^2.
98009803_98009800 = 98999900 * 98999902, where _ denotes
concatenation
MAPLE
f:= proc(n) local S;
S:= map(t -> rhs(op(t))^2 mod 10^n+2, [msolve(x^2+2, 10^n+1)]);
op(sort(select(t -> t-2 >= 10^(n-1) and t-2 < 10^n and issqr(t-2 + t*10^n), S)))
end proc:
seq(f(n), n=1..20); # Robert Israel, Feb 20 2019
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Giovanni Resta, Jan 24 2006
STATUS
approved