A351017 Number of binary words of length n with all distinct run-lengths.
1, 2, 2, 6, 6, 10, 22, 26, 38, 54, 114, 130, 202, 266, 386, 702, 870, 1234, 1702, 2354, 3110, 5502, 6594, 9514, 12586, 17522, 22610, 31206, 48630, 60922, 83734, 111482, 149750, 196086, 261618, 336850, 514810, 631946, 862130, 1116654, 1502982, 1916530, 2555734, 3242546
Offset: 0
Keywords
Examples
The a(0) = 1 through a(6) = 22 words: {} 0 00 000 0000 00000 000000 1 11 001 0001 00001 000001 011 0111 00011 000011 100 1000 00111 000100 110 1110 01111 000110 111 1111 10000 001000 11000 001110 11100 001111 11110 011000 11111 011100 011111 100000 100011 100111 110000 110001 110111 111001 111011 111100 111110 111111
Crossrefs
The version for partitions is A098859.
The complement is counted by twice A261982.
A000120 counts binary weight.
A005811 counts runs in binary expansion.
A011782 counts integer compositions.
A242882 counts compositions with distinct multiplicities.
A297770 counts distinct runs in binary expansion.
A325545 counts compositions with distinct differences.
A329767 counts binary words by runs-resistance.
A351014 counts distinct runs in standard compositions.
A351204 counts partitions where every permutation has all distinct runs.
A351290 ranks compositions with all distinct runs.
Programs
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Mathematica
Table[Length[Select[Tuples[{0,1},n],UnsameQ@@Length/@Split[#]&]],{n,0,10}]
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Python
from itertools import groupby, product def adrl(s): runlens = [len(list(g)) for k, g in groupby(s)] return len(runlens) == len(set(runlens)) def a(n): if n == 0: return 1 return 2*sum(adrl("1"+"".join(w)) for w in product("01", repeat=n-1)) print([a(n) for n in range(20)]) # Michael S. Branicky, Feb 08 2022
Formula
a(n>0) = 2 * A032020(n).
Extensions
a(25)-a(32) from Michael S. Branicky, Feb 08 2022
More terms from David A. Corneth, Feb 08 2022 using data from A032020