A044813 Positive integers having distinct base-2 run lengths.

1, 3, 4, 6, 7, 8, 14, 15, 16, 24, 28, 30, 31, 32, 35, 39, 48, 49, 55, 57, 59, 60, 62, 63, 64, 67, 79, 96, 97, 111, 112, 120, 121, 123, 124, 126, 127, 128, 131, 135, 143, 159, 192, 193, 223, 224, 225, 239, 241, 247, 248, 249, 251, 252, 254, 255, 256, 259, 263

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• base

Comments

A005811(a(n)) = A165413(a(n)). - Reinhard Zumkeller, Mar 02 2013

From Emeric Deutsch, Jan 25 2018: (Start)

Also, the indices of the compositions that have distinct parts. For the definition of the index of a composition see A298644. For example, 223 is in the sequence since its binary form is 11011111 and the composition [2,1,5] has distinct parts. 100 is not in the sequence since its binary form is 1100100 and the parts of the composition [2,2,1,2] are not distinct.

The command c(n) from the Maple program yields the composition having index n. (End)

Links

• Reinhard Zumkeller and Gheorghe Coserea, Table of n, a(n) for n = 1..20000 (first 5000 terms from Reinhard Zumkeller)

Crossrefs

Cf. A030308, A298644.

Programs

Haskell

import Data.List (group, nub)
a044813 n = a044813_list !! (n-1)
a044813_list = filter p [1..] where
   p x = nub xs == xs where
         xs = map length $ group $ a030308_row x
-- Reinhard Zumkeller, Mar 02 2013

Maple

Runs := proc (L) local j, r, i, k: j := 1: r[j] := L[1]: for i from 2 to nops(L) do if L[i] = L[i-1] then r[j] := r[j], L[i] else j := j+1: r[j] := L[i] end if end do: [seq([r[k]], k = 1 .. j)] end proc: RunLengths := proc (L) map(nops, Runs(L)) end proc: c := proc (n) ListTools:-Reverse(convert(n, base, 2)): RunLengths(%) end proc: A := {}: for n to 300 do if nops(convert(c(n), set)) = nops(c(n)) then A := `union`(A, {n}) else end if end do: A; # most of the Maple program is due to W. Edwin Clark. - Emeric Deutsch, Jan 25 2018

Mathematica

f[n_] := Unequal@@Length/@Split[IntegerDigits[n,2]]; Select[Range[300],f] (* Ray Chandler, Oct 21 2011 *)

PARI

is(n) = {
  my(v = 0, hist = vector(1 + logint(n+1, 2)));
  while(n != 0,
        v = valuation(n, 2); n >>= v; n++;
        hist[v+1]++; if (hist[v+1] >= 2, return(0));
        v = valuation(n, 2); n >>= v; n--;
        hist[v+1]++; if (hist[v+1] >= 2, return(0)));
  return(1);
};
seq(n) = {
  my(k = 1, top = 0, v = vector(n));
  while(top < n, if (is(k), v[top++] = k); k++);
  return(v);
};
seq(59) \\ Gheorghe Coserea, Nov 02 2015

Python

from itertools import groupby
def ok(n):
  runlengths = [len(list(g)) for k, g in groupby(bin(n)[2:])]
  return len(runlengths) == len(set(runlengths))
print([i for i in range(1, 264) if ok(i)]) # Michael S. Branicky, Jan 04 2021

Formula

a(Sum_{k=0..n} A032020(k)) = 2^n, for n>1. - Gheorghe Coserea, May 30 2017

Extensions

Extended by Ray Chandler, Oct 21 2011