

A307371


Numbers k such that the digits of sqrt(k) begin with k.


9



0, 1, 98, 99, 100, 9998, 9999, 10000, 999998, 999999, 1000000, 99999998, 99999999, 100000000, 9999999998, 9999999999, 10000000000, 999999999998, 999999999999, 1000000000000, 99999999999998, 99999999999999, 100000000000000, 9999999999999998, 9999999999999999
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OFFSET

1,3


COMMENTS

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)21/k < k <= 10^(2m). Thus the only solutions for k are 10^(2m), 10^(2m)1 and 10^(2m)2. (End)


LINKS



FORMULA

a(n) = 101*a(n3)  100*a(n6) 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

(Python)
A307371_list = [0, 1, 98, 99, 100, 9998]
for _ in range(100):


CROSSREFS



KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



