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%I #13 Jun 29 2019 02:20:10
%S 1,3,11,35,111,351,1111,3513,11111,35136,111111,351364,1111111,
%T 3513641,11111111,35136418,111111111,351364184,1111111111,3513641844,
%U 11111111111,35136418446,111111111111,351364184463,1111111111111
%N Integer parts of the square roots of the schizophrenic numbers (A014824).
%C a(n) appears to result from (alternately) intermeshing two subsequences, one of the form 11, 111, 1111, ..., the other of the form 35, 351, 3513, .... In both subsequences, the current term is an initial segment of the next term. If the first k (k an even number) terms are deleted from a(n), a(n) can be reconstructed from the resulting sequence by deleting appropriate digits from the end of terms. In this sense, a(n) is self-similar.
%F From _Christopher Hohl_, Jun 27 2019: (Start)
%F a(2n-1) = A014824(n) - A014824(n-1), for n>=1;
%F a(2n-2) = floor(a(2n-1) / sqrt(10)), for n>=2. (End)
%e 123 is the third schizophrenic number; its square root has integer part 11.
%t h[n_ /; n == 0] := 0; h[n_ /; n > 0] := 10*h[n - 1] + n; t = Table[Floor[Sqrt[h[i]]], {i, 1, 40}]
%Y Cf. A014824.
%K base,nonn
%O 1,2
%A _Joseph L. Pe_, Mar 14 2002
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