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a(1) = 1. a(n) = number of earlier terms which, when written in binary, are substrings of binary n.
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%I #19 Apr 12 2019 08:13:41

%S 1,1,2,3,3,5,4,4,5,5,8,7,8,6,5,7,7,5,7,10,9,11,15,10,9,13,13,12,17,11,

%T 9,7,7,8,11,8,13,11,16,16,15,10,17,18,18,21,22,15,15,14,12,19,18,19,

%U 24,22,20,24,25,21,25,17,14,11,11,11,13,12,17,16,21,15,11,19,26,17,23,22,24

%N a(1) = 1. a(n) = number of earlier terms which, when written in binary, are substrings of binary n.

%C First occurrence of k: 1, 3, 4, 7, 6, 14, 12, 11, 21, 20, 22, 28, 26, 50, 23, 39, 29, 44, 52, 57, 46, 47, 77, 55, 59, ...

%H Robert Israel, <a href="/A122954/b122954.txt">Table of n, a(n) for n = 1..10000</a>

%e Binary a(1) = 1, binary a(2) = 1, binary a(3) = 10, binary a(7) = 100 and binary a(8) = 100 are all substrings of binary 9 = 1001. So a(9) = 5.

%p A[1]:= 1:

%p S[1]:= "1":

%p for n from 2 to 100 do

%p B:= convert(convert(n,binary),string);

%p A[n]:= nops(select(t -> SearchText(S[t],B)>0, [$1..n-1]));

%p S[n]:= convert(convert(A[n],binary),string);

%p od:

%p seq(A[i],i=1..100); # _Robert Israel_, Apr 11 2019

%t f[ s_ ] := Append[ s, Length@ Select[ s, StringPosition[ ToString@ FromDigits@ IntegerDigits[ Length@s + 1, 2 ], ToString@ FromDigits@ IntegerDigits[ #, 2 ] ] != {} & ] ]; Nest[ f, {1}, 79 ] (* _Robert G. Wilson v_, Nov 01 2006 *)

%K nonn,look

%O 1,3

%A _Leroy Quet_, Oct 25 2006

%E More terms from _Robert G. Wilson v_, Nov 01 2006