OFFSET
1,1
COMMENTS
a(n) > 10^floor(n/2).
All terms have last digit 1 or 9.
Squares of terms are listed in A085877.
Decimal representation of each term is formed by the reverse concatenation of initial terms of either A063006 or A091661.
Except for 3, there are no solutions for n>1 and m^2 == -1 (mod 10^n). See comment in A063006 under extensions. - Robert G. Wilson v, Jan 26 2013
If a(n)<(10^n)/2 then (10^n-a(n))^2 is also congruent (modulo 10^n), its just not the least. - Robert G. Wilson v, Jan 26 2013
LINKS
Max Alekseyev, Table of n, a(n) for n = 1..1000
FORMULA
Let b(n) = A224474(n) (or equivalently b(n) = A224473(n)), then for n >= 3, there are eight solutions in [0,10^n) to x^2 == 1 (mod 10^n), namely x = 1, 5*10^(n-1) - 1, 5*10^(n-1) + 1, 10^n - 1, b(n), 10^n - b(n), |b(n) - 5*10^(n-1)|, and 10^n - |b(n) - 5*10^(n-1)|, so a(n) = min{b(n), |b(n) - 5*10^(n-1)|, 10^n - b(n)} < 25*10^(n-2). - Jianing Song, Sep 23 2024
EXAMPLE
1249^2 = 1560001 == 1 (mod 10^4), and there is no smaller m > 1 such that m^2 == 1 (mod 10^4). Hence a(4) = 1249.
PROG
(PARI) install(Zn_quad_roots, GGG);
a181539(n) = vecsort(Zn_quad_roots(10^n, 0, -1)[2])[2]; \\ Max Alekseyev, Oct 13 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Kevin Batista (kevin762401(AT)yahoo.com), Oct 29 2010
EXTENSIONS
a(2) through a(4), a(7) through a(11) corrected, comment added, example replaced by Klaus Brockhaus, Nov 01 2010
Edited by N. J. A. Sloane, Oct 29 2010, Nov 09 2010
Definition to avoid the constant sequence a(n)=1 constrained by R. J. Mathar, Nov 18 2010
a(1) corrected, terms a(13) onward added by Max Alekseyev, Dec 10 2012
STATUS
approved