OFFSET
1,3
COMMENTS
From Chai Wah Wu, Jan 17 2020: (Start)
Theorem: A number n is a term if and only if n is 0, 1, 10^(2m), 10^{2m)-1 or 10^{2m}-2 for some m >= 1.
Proof: k <= sqrt(k)*10^m < k+1. For m = 0, the only solutions are 0 and 1. For m > 0, k^2 <= k*10^(2m) < (k+1)^2. This is equivalent to k <= 10^2m < k + 2 + 1/k, i.e., 10^(2m)-2-1/k < k <= 10^(2m). Thus the only solutions for k are 10^(2m), 10^(2m)-1 and 10^(2m)-2. (End)
LINKS
Dmitry Kamenetsky, Java program to compute terms
FORMULA
From Chai Wah Wu, Jan 17 2020: (Start)
a(n) = 101*a(n-3) - 100*a(n-6) for n > 6.
G.f.: x^2*(100*x^4 - x^3 + 99*x^2 + 98*x + 1)/(100*x^6 - 101*x^3 + 1). (End)
EXAMPLE
sqrt(9998) = 99.989..., which begins with "9998", so 9998 is in the sequence.
PROG
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Dmitry Kamenetsky, Apr 17 2019
EXTENSIONS
a(12)-a(25) from Jon E. Schoenfield, May 01 2019
STATUS
approved