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

1,3

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

Table of n, a(n) for n=1..25.

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

(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

Crossrefs

Cf. A307588, A307600.

Sequence in context: A120310 A129887 A205067 * A273460 A095605 A095589

Adjacent sequences: A307368 A307369 A307370 * A307372 A307373 A307374

Keyword

nonn,base,easy

Author

Dmitry Kamenetsky, Apr 17 2019

Extensions

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

Status

approved