This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A307371 #47 Dec 05 2024 18:58:32 %S A307371 0,1,98,99,100,9998,9999,10000,999998,999999,1000000,99999998, %T A307371 99999999,100000000,9999999998,9999999999,10000000000,999999999998, %U A307371 999999999999,1000000000000,99999999999998,99999999999999,100000000000000,9999999999999998,9999999999999999 %N A307371 Numbers k such that the digits of sqrt(k) begin with k. %C A307371 From _Chai Wah Wu_, Jan 17 2020: (Start) %C A307371 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. %C A307371 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) %H A307371 Dmitry Kamenetsky, <a href="/A307371/a307371.java.txt">Java program to compute terms</a> %F A307371 From _Chai Wah Wu_, Jan 17 2020: (Start) %F A307371 a(n) = 101*a(n-3) - 100*a(n-6) for n > 6. %F A307371 G.f.: x^2*(100*x^4 - x^3 + 99*x^2 + 98*x + 1)/(100*x^6 - 101*x^3 + 1). (End) %e A307371 sqrt(9998) = 99.989..., which begins with "9998", so 9998 is in the sequence. %o A307371 (Python) %o A307371 A307371_list = [0, 1, 98, 99, 100, 9998] %o A307371 for _ in range(100): %o A307371 A307371_list.append(101*A307371_list[-3]-100*A307371_list[-6]) # _Chai Wah Wu_, Jan 18 2020 %Y A307371 Cf. A307588, A307600. %K A307371 nonn,base,easy %O A307371 1,3 %A A307371 _Dmitry Kamenetsky_, Apr 17 2019 %E A307371 a(12)-a(25) from _Jon E. Schoenfield_, May 01 2019