A036991 -id:A036991 - cp's OEIS frontend

A036994 Numbers k with the property that reading from right to left in the binary expansion of k, the number of 1's always stays ahead of the number of 0's.

1, 3, 7, 11, 15, 23, 27, 31, 39, 43, 47, 55, 59, 63, 79, 87, 91, 95, 103, 107, 111, 119, 123, 127, 143, 151, 155, 159, 167, 171, 175, 183, 187, 191, 207, 215, 219, 223, 231, 235, 239, 247, 251, 255, 287, 303, 311, 315, 319, 335, 343, 347, 351, 359, 363, 367

Offset: 1

N. J. A. Sloane

Even numbers can't appear in this sequence. - Alonso del Arte, Sep 21 2011

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences related to binary expansion of n

Cf. A036990, A036991, A036992, A036993.

Cf. A030308.

a036994 n = a036994_list !! (n-1)
a036994_list = filter ((p 0) . a030308_row) [1, 3 ..] where
   p ones []    = ones > 0
   p ones (0:bs) = ones > 1 && p (ones - 1) bs
   p ones (1:bs) = p (ones + 1) bs
-- Reinhard Zumkeller, Aug 01 2013

aheadOnesRLQ[n_Integer] := Module[{digits, len, flag = True, iter = 1, ones = 0, zeros = 0}, digits = Reverse[IntegerDigits[n, 2]]; len = Length[digits]; While[flag && iter < len, If[digits[[iter]] == 1, ones++, zeros++]; flag = ones > zeros; iter++]; flag]; Select[Range[1, 201, 2], aheadOnesRLQ] (* Alonso del Arte, Sep 21 2011 *)
Select[Range[400],Min[Accumulate[Reverse[IntegerDigits[#,2]/. (0->-1)]]]> 0&] (* Harvey P. Dale, Apr 23 2016 *)

ok(x)={if(x<1,return(0));my(c=logint(x,2),c0=0,c1=0);for(i=0,c,if(bittest(x,i),c1++,c0++);if(c1<=c0,return(0)));1} \\ for(n=1,367,if(ok(n),print1(n,", "))) - Ruud H.G. van Tol, Sep 14 2022

from itertools import count, islice
def A036994_gen(startvalue=0): # generator of terms >= startvalue
    for n in count(max(startvalue,0)):
        s = bin(n)[2:]
        c, l = 0, len(s)
        for i in range(l):
            c += int(s[l-i-1])
            if 2*c <= i + 1:
                break
        else:
            yield n
A036994_list = list(islice(A036994_gen(),20)) # Chai Wah Wu, Dec 31 2021

More terms from Erich Friedman

A116385 Expansion of e.g.f. Bessel_I(2,2x) + 2*Bessel_I(3,2x) + Bessel_I(4,2x).

Original entry on oeis.org

0, 0, 1, 2, 5, 10, 21, 42, 84, 168, 330, 660, 1287, 2574, 5005, 10010, 19448, 38896, 75582, 151164, 293930, 587860, 1144066, 2288132, 4457400, 8914800, 17383860, 34767720, 67863915, 135727830, 265182525, 530365050, 1037158320, 2074316640

Offset: 0

Paul Barry, Feb 12 2006

Third column of the Riordan array A116382.

Apart from its root term -1: central terms of the triangle in A051631: a(n) = A051631(n+1, [(n+1)/2]). - Reinhard Zumkeller, Nov 13 2011

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Gennady Eremin, Dyck Numbers, II. Triplets and Rooted Trees in OEIS A036991, arXiv:2211.01135 [math.NT], 2022.

Cf. A001405.

a116385 n = a051631 (n+1) $ (n+1) `div` 2
-- Reinhard Zumkeller, Nov 13 2011

With[{nn=40},CoefficientList[Series[BesselI[2,2x]+2BesselI[3,2x]+ BesselI[ 4,2x],{x,0,nn}],x]Range[0,nn]!] (* Harvey P. Dale, Sep 14 2011 *)

a(n)= binomial(n+3, (n+3)\2) - 3*binomial(n+1, (n+1)\2) \\ Bill McEachen, Dec 12 2022

E.g.f.: (d/dx)(Bessel_I(3,2x),x) + 2*Bessel_I(3,2x).
a(n) = C(n+1,floor((n-2)/2))*(1+(-1)^n)/2 + C(n,floor((n-3)/2))*(1-(-1)^n).
Conjecture: (n+4)*a(n) -2*a(n-1) +(-7*n-8)*a(n-2) +6*a(n-3) +12*(n-2)*a(n-4)=0. - R. J. Mathar, Jun 13 2014
a(n) = A001405(n+3) - 3*A001405(n+1) (from Eremin link). - Bill McEachen, Dec 12 2022
G.f.: (-1 - x + x^2 + B(x) - 3*x^2*B(x))/x^3, where B(x) is the g.f. of A001405. - Gennady Eremin, Oct 09 2023

A350346 Binary numbers such that when read from right to left, the number of 0's never exceeds the number of 1's.

Original entry on oeis.org

0, 1, 11, 101, 111, 1011, 1101, 1111, 10011, 10101, 10111, 11011, 11101, 11111, 100111, 101011, 101101, 101111, 110011, 110101, 110111, 111011, 111101, 111111, 1000111, 1001011, 1001101, 1001111, 1010011, 1010101, 1010111, 1011011, 1011101, 1011111, 1100111

Offset: 1

Gennady Eremin, Dec 26 2021

Binary expansion of A036991 terms.
Dyck language interpreted as binary numbers in ascending order (inverse encode of A063171).
The first term a(1)=0 corresponds to an empty string, denote it NULL.
Restoring the leading 0's (need the same number of 0's and 1's) and then replacing "0" by the left parenthesis "(" and "1" by the right parenthesis ")" give well-formed parenthesis strings: 0 -> NULL, 1=01 -> (), 11=0011 -> (()), 101=0101 -> ()(), 111=000111 -> ((())), 1011=001011 -> (()()), 1101=001101 -> (())() and so on.
Chomsky-2 grammar with axiom s, terminal alphabet {0, 1} and three rules s -> ss, s -> 0s1, s -> 01 (compare A063171).

			s -> ss -> 0s1s -> 00s11s -> 000111s -> 00011101 = 11101.

Gennady Eremin, Table of n, a(n) for n = 1..5000
Gennady Eremin, Dyck Numbers, III. Enumeration and bijection with symmetric Dyck paths, arXiv:2302.02765 [math.CO], 2023.
Gennady Eremin, Dyck Numbers, IV. Nested patterns in OEIS A036991, arXiv:2306.10318, 2023.

Cf. A007088, A036991, A014486, A063171.

Join[{0},Select[Table[FromDigits[IntegerDigits[n,2]],{n,120}],Min[Accumulate[ Reverse[ IntegerDigits[#]]/.(0->-1)]]>=0&]] (* Harvey P. Dale, Apr 29 2022 *)

def ok(n):
    if n == 0: return True
    count = {"0": 0, "1": 0}
    for bit in bin(n)[:1:-1]:
        count[bit] += 1
        if count["0"] > count["1"]: return False
    return True # A036991
nn = 1; print(1, 0)
for n in range(1, 23230):  # printing b-file
    if ok(n) == False: continue
    nn += 1; print(nn, bin(n)[2:])

from itertools import count, islice
def A350346_gen(): # generator of terms
    yield 0
    for n in count(1):
        s = bin(n)[2:]
        c, l = 0, len(s)
        for i in range(l):
            c += int(s[l-i-1])
            if 2*c <= i:
                break
        else:
            yield int(s)
A350346_list = list(islice(A350346_gen(),20)) # Chai Wah Wu, Dec 30 2021

a(n) = A007088(A036991(n)). - Michel Marcus, Dec 26 2021

A114567 a(n) is the number k such that the binary expansion of n mod 2^k is the postorder traversal of a binary tree, where 1 indicates a node and 0 indicates there are no children on that side.

Original entry on oeis.org

1, 3, 1, 5, 1, 5, 1, 7, 1, 3, 1, 7, 1, 7, 1, 9, 1, 3, 1, 7, 1, 7, 1, 9, 1, 3, 1, 9, 1, 9, 1, 11, 1, 3, 1, 5, 1, 5, 1, 9, 1, 3, 1, 9, 1, 9, 1, 11, 1, 3, 1, 9, 1, 9, 1, 11, 1, 3, 1, 11, 1, 11, 1, 13, 1, 3, 1, 5, 1, 5, 1, 9, 1, 3, 1, 9, 1, 9, 1, 11, 1, 3, 1, 9, 1, 9, 1, 11, 1, 3, 1, 11, 1, 11, 1, 13, 1, 3, 1, 5, 1

Offset: 0

Mike Stay, Feb 15 2006

Postorder traversals of a binary tree form an instantaneous code; any integer has a unique decomposition into codewords. To get the first codeword, find a(n). Then set n' = floor(n/2^(a(n))), find a(n'), and so on.

Equivalently, the number of bits of n, starting from the least significant, in which the number of 0's first exceeds the number of 1's (including 0's above the most significant bit of n if necessary). - Kevin Ryde, Sep 30 2024

			a(37) = 1 + a(floor(37/2)) + a(floor(37/2^(a(floor(37/2)) + 1)))
  = 1 + a(18) + a(floor(37/2^(a(18) + 1)))
  = 1 + 1 + a(floor(37/2^(1 + 1)))
  = 2 + a(9)
  = 2 + 1 + a(floor(9/2)) + a(floor(9/2^(a(floor(9/2)) + 1)))
  = 3 + a(4) + a(floor(9/2^(a(4) + 1)))
  = 3 + 1 + a(floor(9/4))
  = 4 + a(2)
  = 5.
37 mod 2^5 = 5 = 00101, which is the postorder traversal of the binary tree with a root node and a single left child.

Antti Karttunen, Table of n, a(n) for n = 0..65537
Index entries for sequences related to binary expansion of n

Cf. A036991.

a := proc(n) option remember; if 0 = n mod 2 then 1; else 1 + a(floor(n/2)) + a(floor(n/2^(a(floor(n/2)) + 1))); end if; end proc; # Petros Hadjicostas, Nov 20 2019

a[n_] := a[n] = If[EvenQ[n], 1, 1 + a[Floor[n/2]] + a[ Floor[ n/2^( a[Floor[n/2]] + 1)]]]; a /@ Range[0, 100] (* Giovanni Resta, Nov 21 2019 *)

A114567(n) = if(!(n%2),1,1+A114567(n\2) + A114567(n>>(1+A114567(n\2)))); \\ Antti Karttunen, Mar 30 2021, after the Maple-program

a(n) = my(delta=1); for(i=0,oo, if(bittest(n,i), delta++, delta--||return(i+1))); \\ Kevin Ryde, Sep 30 2024

a(n) = 1, if n is even, and a(n) = 1 + a(floor(n/2)) + a(floor(n/2^{a(floor(n/2)) + 1})), if n is odd.

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A036994 Numbers k with the property that reading from right to left in the binary expansion of k, the number of 1's always stays ahead of the number of 0's.

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A116385 Expansion of e.g.f. Bessel_I(2,2x) + 2*Bessel_I(3,2x) + Bessel_I(4,2x).

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A350346 Binary numbers such that when read from right to left, the number of 0's never exceeds the number of 1's.

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A114567 a(n) is the number k such that the binary expansion of n mod 2^k is the postorder traversal of a binary tree, where 1 indicates a node and 0 indicates there are no children on that side.

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Comments

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Crossrefs

Programs

Maple

Mathematica

PARI

PARI

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