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%I #44 Jan 18 2020 11:07:30
%S 0,1,98,99,100,9998,9999,10000,999998,999999,1000000,99999998,
%T 99999999,100000000,9999999998,9999999999,10000000000,999999999998,
%U 999999999999,1000000000000,99999999999998,99999999999999,100000000000000,9999999999999998,9999999999999999
%N Numbers k such that the digits of sqrt(k) begin with k.
%C From _Chai Wah Wu_, Jan 17 2020: (Start)
%C 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 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 Dmitry Kamenetsky, <a href="/A307371/a307371.java.txt">Java program to compute terms</a>
%F From _Chai Wah Wu_, Jan 17 2020: (Start)
%F a(n) = 101*a(n-3) - 100*a(n-6) for n > 6.
%F G.f.: x^2*(100*x^4 - x^3 + 99*x^2 + 98*x + 1)/(100*x^6 - 101*x^3 + 1). (End)
%e sqrt(9998) = 99.989..., which begins with "9998", so 9998 is in the sequence.
%o (Python)
%o A307371_list = [0, 1, 98, 99, 100, 9998]
%o for _ in range(100):
%o A307371_list.append(101*A307371_list[-3]-100*A307371_list[-6]) # _Chai Wah Wu_, Jan 18 2020
%Y Cf. A307588, A307600.
%K nonn,base
%O 1,3
%A _Dmitry Kamenetsky_, Apr 17 2019
%E a(12)-a(25) from _Jon E. Schoenfield_, May 01 2019
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