A307371 Numbers k such that the digits of sqrt(k) begin with k.

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

Offset: 1

Author

Dmitry Kamenetsky, Apr 17 2019

Keywords

• nonn
• base
• easy

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)

Examples

			sqrt(9998) = 99.989..., which begins with "9998", so 9998 is in the sequence.
		

Links

• Dmitry Kamenetsky, Java program to compute terms

Crossrefs

Cf. A307588, A307600.

Programs

Python

A307371_list = [0, 1, 98, 99, 100, 9998]
for _ in range(100):
    A307371_list.append(101*A307371_list[-3]-100*A307371_list[-6]) # Chai Wah Wu, Jan 18 2020

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)

Extensions

a(12)-a(25) from Jon E. Schoenfield, May 01 2019