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%I #25 Nov 07 2024 22:02:13
%S 1,1,11,11,11,11,111,111,111,111,111,111,111,111,1111,1111,1111,1111,
%T 1111,1111,1111,1111,1111,1111,1111,1111,1111,1111,1111,1111,11111,
%U 11111,11111,11111,11111,11111,11111,11111,11111,11111,11111,11111
%N Repeat 2^d times the repunit A002275(d); d = 1, 2, 3...
%C Yields the length of the n-th (nonempty) binary word (or word over any 2-letter alphabet, like A007931 or A032810 or A032834) in tally mark notation (A000042).
%H Felix Fröhlich, <a href="/A256077/b256077.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A002275(A000523(n+1)) = A032810(n)-A007931(n) = A032834(n)-A032810(n), etc.
%t lim = 5; lst = Table[(10^n - 1)/9, {n, 0, lim}]; Reap@ For[i = 1, i <= lim, i++, Sow@ Table[lst[[i + 1]], {d, 2^i}]] // Flatten // Rest (* _Michael De Vlieger_, Apr 01 2015 *)
%o (PARI) a(n)=10^#binary(n+1)\90
%o (Python)
%o def A256077(n): return (10**((n+1).bit_length()-1)-1)//9 # _Chai Wah Wu_, Nov 07 2024
%K nonn
%O 1,3
%A _M. F. Hasler_, Mar 21 2015