

A068995


Integer parts of the square roots of the schizophrenic numbers (A014824).


3



1, 3, 11, 35, 111, 351, 1111, 3513, 11111, 35136, 111111, 351364, 1111111, 3513641, 11111111, 35136418, 111111111, 351364184, 1111111111, 3513641844, 11111111111, 35136418446, 111111111111, 351364184463, 1111111111111
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OFFSET

1,2


COMMENTS

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 selfsimilar.


LINKS

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


FORMULA

From Christopher Hohl, Jun 27 2019: (Start)
a(2n1) = A014824(n)  A014824(n1), for n>=1;
a(2n2) = floor(a(2n1) / sqrt(10)), for n>=2. (End)


EXAMPLE

123 is the third schizophrenic number; its square root has integer part 11.


MATHEMATICA

h[n_ /; n == 0] := 0; h[n_ /; n > 0] := 10*h[n  1] + n; t = Table[Floor[Sqrt[h[i]]], {i, 1, 40}]


CROSSREFS

Cf. A014824.
Sequence in context: A026154 A025181 A004054 * A109196 A032637 A034576
Adjacent sequences: A068992 A068993 A068994 * A068996 A068997 A068998


KEYWORD

base,nonn


AUTHOR

Joseph L. Pe, Mar 14 2002


STATUS

approved



