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%I #17 Oct 23 2022 01:44:32
%S 11,111,1100,1111,11000,11100,11111,110000,110011,111000,111100,
%T 111111,1100000,1100011,1100111,1110000,1110011,1111000,1111100,
%U 1111111,11000000,11000011,11000111,11001100,11001111,11100000,11100011,11100111,11110000,11110011,11111000,11111100,11111111
%N Binary numbers (digits in {0, 1}) with no run of digits with length < 2.
%C This is the binary representation of the terms in A033015.
%C The sequence can be seen as a table where row r contains the terms with r digits. Then row r+1 is obtained by from the terms of row r by duplicating their last digit, and from those of row r-1 by appending twice the 1's complement of their last digit. This yields the row lengths given in FORMULA.
%F a(n) = A007088(A033015(n)).
%F The number of terms with n digits is Fibonacci(n-1); the largest such term is A000042(n) = A002275(n).
%e There can't be a terms with only 1 digit, so the smallest term is a(1) = 11.
%e The only 3-digit term is a(2) = 111, since in 100 the digit 1 is alone, and in 101 and 110 the digit 0 is alone.
%e With four digits we must have either no or two digits 0 and they must be at the end (to avoid isolated '1's), i.e., a(3) = 1100 and a(4) = 1111.
%o (PARI) {is_A355280(n,d=digits(n))=vecmax(d)==1 && is_A033015(fromdigits(d,2))}
%o A355280(n)=A007088(A033015(n))
%o concat(apply( {A355280_row(n)=if(n>2, setunion([x*10+x%10|x<-A355280_row(n-1)],[x*100+11*(1-x%10)|x<-A355280_row(n-2)]), n>1, [11],[])}, [1..8])) \\ "Row" of n-digit terms. For (very) large n one should implement memoization instead of this naive recursion.
%o (Python)
%o def A355280_row(n): return [] if n<2 else [11] if n==2 else sorted(
%o [x*10+x%10 for x in A355280_row(n-1)] +
%o [x*100+11-x%10*11 for x in A355280_row(n-2)]) # _M. F. Hasler_, Oct 17 2022
%Y Cf. A033015 (the same terms converted from base 2 to base 10),
%Y Subsequence of A007088 (the binary numbers); A000042 (numbers in base 1) = A002275 \ {0} (repunits) are subsequences; A061851 is the subsequence of palindromes.
%K nonn,base
%O 1,1
%A _M. F. Hasler_, Oct 17 2022
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