A090996 -id:A090996 - cp's OEIS frontend

A101211 Triangle read by rows: n-th row is length of run of leftmost 1's, followed by length of run of 0's, followed by length of run of 1's, etc., in the binary representation of n.

1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 3, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 3, 1, 4, 1, 4, 1, 3, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 3, 2, 2, 1, 2, 1, 1, 1, 2, 1, 2, 3, 2, 3, 1, 1, 4, 1, 5, 1, 5, 1, 4, 1, 1, 3, 1, 1, 1, 3, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 1

Offset: 1

Leroy Quet, Dec 13 2004

Row n has A005811(n) elements. In rows 2^(k-1)..2^k-1 we have all the compositions (ordered partitions) of k. Other orderings of compositions: A066099, A108244, and A124734. - Jason Kimberley, Feb 09 2013
A043276(n) = largest term in n-th row. - Reinhard Zumkeller, Dec 16 2013
From the first comment it follows that we have a bijection between the positive integers and the set of all compositions. - Emeric Deutsch, Jul 11 2017
From Robert Israel, Jan 23 2018: (Start)
If n is even, row 2*n is row n with its last element incremented by 1, and row 2*n+1 is row n with 1 appended.
If n is odd, row 2*n+1 is row n with its last element incremented by 1, and row 2*n is row n with 1 appended. (End)

			Since 9 is 1001 in binary, the 9th row is 1,2,1.
Since 11 is 1011 in binary, the 11th row is 1,1,2.
Triangle begins:
  1;
  1,1;
  2;
  1,2;
  1,1,1;
  2,1;
  3;
  1,3;

Antti Karttunen, The rows 1..1023 of the table, flattened

A070939(n) gives the sum of terms in row n, while A167489(n) gives the product of its terms. A090996 gives the first column. A227736 lists the terms of each row in reverse order.
Cf. also A227186.
Cf. A030308, A175911.
Cf. A318927 (concatenation of each row), A318926 (concatenations of reversed rows).
Cf. A382255 (Heinz numbers of the rows: Product_k prime(T(n,k))).

import Data.List (group)
a101211 n k = a101211_tabf !! (n-1) !! (k-1)
a101211_row n = a101211_tabf !! (n-1)
a101211_tabf = map (reverse . map length . group) $ tail a030308_tabf
-- Reinhard Zumkeller, Dec 16 2013

# Maple program due to W. Edwin Clark:
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: # Row n is obtained with the command c(n). - Emeric Deutsch, Jul 03 2017
# Maple program due to W. Edwin Clark, yielding the integer ind corresponding to a given composition (the index of the composition):
ind := proc (x) local X, j, i: X := NULL: for j to nops(x) do if type(j, odd) then X := X, seq(1, i = 1 .. x[j]) end if: if type(j, even) then X := X, seq(0, i = 1 .. x[j]) end if end do: X := [X]: add(X[i]*2^(nops(X)-i), i = 1 .. nops(X)) end proc; # Clearly, ind(c(n))= n. - Emeric Deutsch, Jan 23 2018

Table[Length /@ Split@ IntegerDigits[n, 2], {n, 38}] // Flatten (* Michael De Vlieger, Jul 11 2017 *)

apply( {A101211_row(n)=Vecrev((n=vecextract([-1..exponent(n)], bitxor(2*n, bitor(n,1))))[^1]-n[^-1])}, [1..19]) \\ replacing older code by M. F. Hasler, Mar 24 2025

from itertools import groupby
def arow(n): return [len(list(g)) for k, g in groupby(bin(n)[2:])]
def auptorow(rows):
    alst = []
    for i in range(1, rows+1): alst.extend(arow(i))
    return alst
print(auptorow(38)) # Michael S. Branicky, Oct 02 2021

a(n) = A227736(A227741(n)) = A227186(A056539(A227737(n)),A227740(n)) - Antti Karttunen, Jul 27 2013

More terms from Emeric Deutsch, Apr 12 2005

A360189 Triangle T(n,k), n>=0, 0<=k<=floor(log_2(n+1)), read by rows: T(n,k) = number of nonnegative integers <= n having binary weight k.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 1, 1, 3, 2, 1, 3, 3, 1, 3, 3, 1, 1, 4, 3, 1, 1, 4, 4, 1, 1, 4, 5, 1, 1, 4, 5, 2, 1, 4, 6, 2, 1, 4, 6, 3, 1, 4, 6, 4, 1, 4, 6, 4, 1, 1, 5, 6, 4, 1, 1, 5, 7, 4, 1, 1, 5, 8, 4, 1, 1, 5, 8, 5, 1, 1, 5, 9, 5, 1, 1, 5, 9, 6, 1, 1, 5, 9, 7, 1

Offset: 0

Alois P. Heinz, Mar 04 2023

T(n,k) is defined for all n >= 0 and k >= 0. Terms that are not in the triangle are zero.

			T(6,2) = 3: 3, 5, 6, or in binary: 11_2, 101_2, 110_2.
T(15,3) = 4: 7, 11, 13, 14, or in binary: 111_2, 1011_2, 1101_2, 1110_2.
Triangle T(n,k) begins:
  1;
  1, 1;
  1, 2;
  1, 2, 1;
  1, 3, 1;
  1, 3, 2;
  1, 3, 3;
  1, 3, 3, 1;
  1, 4, 3, 1;
  1, 4, 4, 1;
  1, 4, 5, 1;
  1, 4, 5, 2;
  1, 4, 6, 2;
  1, 4, 6, 3;
  1, 4, 6, 4;
  1, 4, 6, 4, 1;
  ...

Alois P. Heinz, Rows n = 0..2^12-1, flattened
Peter J. Taylor, Closed form for Sum_{k=0..n} [wt(k) = m] where wt(n) is the binary weight of n, answer to question on MathOverflow (2024).
Wikipedia, Iverson bracket

Columns k=0-2 give: A000012, A029837(n+1) = A113473(n) for n>0, A340068(n+1).
Last elements of rows give A090996(n+1).
Cf. A000027, A000120, A000225, A000788, A006046, A007318, A116520, A116522, A116525, A116526, A130665, A130667, A161342, A231500, A231501, A231502, A272020, A361257.

b:= proc(n) option remember; `if`(n<0, 0,
      b(n-1)+x^add(i, i=Bits[Split](n)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n)):
seq(T(n), n=0..23);

T(n,k) = my(v1); v1 = Vecrev(binary(n+1)); v1 = Vecrev(select(x->(x>0),v1,1)); sum(j=0, min(k,#v1-1), binomial(v1[j+1]-1,k-j)) \\ Mikhail Kurkov, Nov 27 2024

T(n,k) = T(n-1,k) + [A000120(n) = k] where [] is the Iverson bracket and T(n,k) = 0 for n<0.
T(2^n-1,k) = A007318(n,k) = binomial(n,k).
T(n,floor(log_2(n+1))) = A090996(n+1).
Sum_{k>=0} T(n,k) = n+1.
Sum_{k>=0} k * T(n,k) = A000788(n).
Sum_{k>=0} k^2 * T(n,k) = A231500(n).
Sum_{k>=0} k^3 * T(n,k) = A231501(n).
Sum_{k>=0} k^4 * T(n,k) = A231502(n).
Sum_{k>=0} 2^k * T(n,k) = A006046(n+1).
Sum_{k>=0} 3^k * T(n,k) = A130665(n).
Sum_{k>=0} 4^k * T(n,k) = A116520(n+1).
Sum_{k>=0} 5^k * T(n,k) = A130667(n+1).
Sum_{k>=0} 6^k * T(n,k) = A116522(n+1).
Sum_{k>=0} 7^k * T(n,k) = A161342(n+1).
Sum_{k>=0} 8^k * T(n,k) = A116526(n+1).
Sum_{k>=0} 10^k * T(n,k) = A116525(n+1).
Sum_{k>=0} n^k * T(n,k) = A361257(n).
T(n,k) = Sum_{j=0..min(k, A000120(n+1)-1)} binomial(A272020(n+1,j+1)-1,k-j) for n >= 0, k >= 0 (see Peter J. Taylor link). - Mikhail Kurkov, Nov 27 2024

A125985 Signature-permutation of Vaillé's 1997 bijection on 'bridges' (Dyck paths).

Original entry on oeis.org

0, 1, 3, 2, 8, 7, 5, 6, 4, 22, 21, 18, 20, 17, 13, 12, 19, 15, 16, 10, 11, 14, 9, 64, 63, 59, 62, 58, 50, 49, 61, 55, 57, 46, 48, 54, 45, 36, 35, 32, 34, 31, 60, 56, 41, 52, 40, 47, 53, 43, 44, 27, 26, 33, 29, 30, 51, 38, 39, 42, 24, 25, 28, 37, 23, 196, 195, 190, 194, 189

Offset: 0

Antti Karttunen, Jan 02 2007

nonn

Vaillé shows in 1997 paper that this automorphism transforms a 'derivation' of a Dyck path to its 'compression', i.e., in OEIS terms, A125985(A126310(n)) = A126309(A125985(n)) holds for all n. He also proves that A057515(A125985(n)) = A126307(n) and A057514(A125985(n)) = A072643(n) - A057514(n) + 1 (the latter identity for all n >= 1).

A. Karttunen, Table of n, a(n) for n = 0..2055
J. Vaillé, Une Bijection Explicative de Plusieurs Propriétés Remarquables des Ponts, European J. Combin. 18 (1997), no. 1, 117-124.
Index entries for signature-permutations of Catalan automorphisms

Inverse: A125986. The number of cycles, maximum cycle sizes and LCM's of all cycle sizes in range [A014137(n-1)..A014138(n-1)] of this permutation are given by A126291, A126292 and A126293. The fixed points are given by A126300/A126301.

(define (A125985 n) (A080300 (rising-list->binexp (A125985-aux2 (A014486 n)))))
(define (A125985-aux2 n) (let loop ((lists (A125985-aux1 n)) (z (list)) (m 1)) (if (null? lists) z (loop (cdr lists) (m-join z (car lists) m) (+ m 1)))))
(define (A125985-aux1 n) (if (zero? n) (list) (let ((begin_from (<< 1 (- (- (A000523 n) (A090996 n)) 1)))) (let loop ((s (A090996 n)) (t 0) (nth_list 1) (p begin_from) (b (if (= 0 (A004198bi n begin_from)) 0 1)) (lists (list (list)))) (cond ((< s 1) (cond ((< p 1) (reverse! lists)) (else (loop (- t (- 1 b)) b (+ 1 nth_list) (>> p 1) (if (= 0 (A004198bi n (>> p 1))) 0 1) (cons (list (+ b 1 nth_list)) lists))))) (else (loop (- s (- 1 b)) (+ t b) nth_list (>> p 1) (if (= 0 (A004198bi n (>> p 1))) 0 1) (cons (cons (+ b nth_list) (car lists)) (cdr lists)))))))))
(define (A125985-aux2 n) (let loop ((lists (A125985-aux1 n)) (z (list)) (m 1)) (if (null? lists) z (loop (cdr lists) (m-join z (car lists) m) (+ m 1)))))
(define (m-join a b m) (let loop ((a a) (b b) (c (list))) (cond ((and (not (pair? a)) (not (pair? b))) (reverse! c)) ((not (pair? a)) (loop a (cdr b) (cons (car b) c))) ((not (pair? b)) (loop (cdr a) b (cons (car a) c))) ((equal? (car a) (car b)) (loop (cdr a) (cdr b) (cons (car a) c))) ((> (car b) m) (loop a (cdr b) (cons (car b) c))) (else (loop (cdr a) b (cons (car a) c))))))
(define (rising-list->binexp rising-list) (let loop ((s 0) (i 0) (h 0) (fs rising-list)) (cond ((null? fs) (+ s (<< (- (<< 1 h) 1) i))) ((> (car fs) h) (loop s (+ i 1) (car fs) (cdr fs))) (else (loop (+ s (<< (- (<< 1 (+ 1 (- h (car fs)))) 1) i)) (+ i 2 (- h (car fs))) (car fs) (cdr fs))))))
(define (>> n i) (if (zero? i) n (>> (floor->exact (/ n 2)) (- i 1))))
(define (<< n i) (if (<= i 0) (>> n (- i)) (<< (+ n n) (- i 1))))

A083368 a(n) is the position of the highest one in A003754(n).

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 7, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 8, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 7, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 9, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2

Offset: 1

Gary W. Adamson, Jun 04 2003

Previous name was: A Fibbinary system represents a number as a sum of distinct Fibonacci numbers (instead of distinct powers of two). Using representations without adjacent zeros, a(n) = the highest bit-position which changes going from n-1 to n.
A003754(n), when written in binary, is the representation of n.
Often one uses Fibbinary representations without adjacent ones (the Zeckendorf expansion).
a(A000071(n+3)) = n. - Reinhard Zumkeller, Aug 10 2014
From Gus Wiseman, Jul 24 2025: (Start)
Conjecture: To obtain this sequence, start with A245563 (maximal run lengths of binary indices), then remove empty and duplicate rows (giving A385817), then take the first term of each remaining row. Some variations:
- For sum instead of first term we appear to have A200648.
- For length instead of first term we appear to have A200650+1.
- For last instead of first term we have A385892.
(End)

			27 is represented 110111, 28 is 111010; the fourth position changes, so a(28)=4.

Jay Kappraff, Beyond Measure: A Guided Tour Through Nature, Myth and Number, World Scientific, 2002, page 460.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

A035612 is the analogous sequence for Zeckendorf representations.
A001511 is the analogous sequence for power-of-two representations.
Cf. A003714, A003754.
Cf. A000045, A000071.
Appears to be the first element of each row of A385817, see A083368, A200648, A200650, A385818, A385892.
A000120 counts ones in binary expansion.
A245563 gives run lengths of binary indices, see A089309, A090996, A245562, A246029, A328592.
A384877 gives anti-run lengths of binary indices, ranks A385816.
Cf. A001629, A037800, A044813, A048793, A069010, A200649, A384878, A385886.

a083368 n = a083368_list !! (n-1)
a083368_list = concat $ h $ drop 2 a000071_list where
   h (a:fs@(a':_)) = (map (a035612 . (a' -)) [a .. a' - 1]) : h fs
-- Reinhard Zumkeller, Aug 10 2014

For n = F(a)-1 to F(a+1)-2, a(n) = A035612(F(a+1)-1-n).

a(n) = a(k)+1 if n = ceiling(phi*k) where phi is the golden ratio; otherwise a(n) = 1. - Tom Edgar, Aug 25 2015

Edited by Don Reble, Nov 12 2005

Shorter name from Joerg Arndt, Jul 27 2025

A126307 a(n) is the length of the leftmost ascent (i.e., height of the first peak) in the n-th Dyck path encoded by A014486(n).

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 2, 2, 3, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Antti Karttunen, Jan 02 2007

nonn

In other words, this sequence gives the number of leading 1's in the terms of A063171.

			A014486(20) = 228 (11100100 in binary), encodes the following Dyck path:
    /\
   /  \/\
  /      \
and the first rising (left-hand side) slope has length 3, thus a(20)=3.

a(n) = A099563(A071156(n)) = A057515(A125985(n)) = A080237(A057164(n)) = A057515(A057504(A057164(n))).

a(n) = A090996(A014486(n)).

A342126 The binary expansion of a(n) corresponds to that of n where all the 1's have been replaced by 0's except in the first run of 1's.

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 6, 7, 8, 8, 8, 8, 12, 12, 14, 15, 16, 16, 16, 16, 16, 16, 16, 16, 24, 24, 24, 24, 28, 28, 30, 31, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 48, 48, 48, 48, 48, 48, 48, 48, 56, 56, 56, 56, 60, 60, 62, 63, 64, 64, 64, 64

Offset: 0

Rémy Sigrist, Apr 25 2021

In other words, this sequence gives the first run of 1's in the binary expansion of a number.

A023758(n) appears A057728(n) times.

			The first terms, alongside their binary expansion, are:
  n   a(n)  bin(n)  bin(a(n))
  --  ----  ------  ---------
   0     0       0          0
   1     1       1          1
   2     2      10         10
   3     3      11         11
   4     4     100        100
   5     4     101        100
   6     6     110        110
   7     7     111        111
   8     8    1000       1000
   9     8    1001       1000
  10     8    1010       1000
  11     8    1011       1000
  12    12    1100       1100
  13    12    1101       1100
  14    14    1110       1110
  15    15    1111       1111

Rémy Sigrist, Table of n, a(n) for n = 0..8192
Index entries for sequences related to binary expansion of n

Cf. A023758, A057728, A087734, A090996, A342410.

a(n) = { my (b=binary(n), p=1); for (k=1, #b, b[k] = p*=b[k]); fromdigits(b, 2) }

def A342126(n):
    s = bin(n)[2:]
    i = s.find('0')
    return n if i == -1 else (2**i-1)*2**(len(s)-i) # Chai Wah Wu, Apr 29 2021

a(n) = n - A087734(n).
a(2*n) = 2*a(n).
a(a(n)) = a(n).
a(n) <= n with equality iff n belongs to A023758.

A279209 Length of first run of 0's in binary expansion of n.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 1, 0, 3, 2, 1, 1, 2, 1, 1, 0, 4, 3, 2, 2, 1, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 0, 5, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 4, 3, 2, 2, 1, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 0, 6, 5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

N. J. A. Sloane, Dec 22 2016

			4 = 100_2 so a(4) = 2.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000

Cf. A090996, A279210.

Table[First[Map[Length, DeleteCases[Split@IntegerDigits[n, 2], w_/;Times@@w>0]]/.{}->{0}], {n, 0, 200}] (* Vincenzo Librandi, Dec 23 2016 *) (* after Michael De Vlieger *)

A181631 Triangle by rows, number of leading 1's in the maximal Fibonacci representation (A104326).

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 1, 1, 2, 3, 4, 1, 1, 1, 2, 2, 3, 4, 5, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5, 6, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 6, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 7, 8

Offset: 1

Gary W. Adamson, Nov 02 2010

Row sums = A001911: (1, 3, 6, 11, 19, 32, ...).
A112310 = number of 1's in the maximal Fibonacci representation, which has headings of (..., 8, 5, 3, 2, 1) filling entries from the right to left; as opposed to the minimal Fibonacci representation (A014417) which starts from the left. For example, 8 in maximal = 1011 = (5 + 2 + 1) whereas in minimal = (10000) = (8).
Rows have (1, 2, 3, 5, 8, ...) terms.

			First few rows of the triangle:
  1;
  1, 2;
  1, 2, 3;
  1, 1, 2, 3, 4;
  1, 1, 1, 2, 2, 3, 4, 5;
  1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5, 6;
  ...
Example: a(14) = 1 since 14 in the maximal Fibonacci representation is 10111.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000045 (row lengths), A003754, A001911 (row sums), A014417, A090996, A104326, A112310.

f[s_] := Module[{i = FirstPosition[s, 0]}, If[MissingQ[i], Length[s], i[[1]] - 1]]; f /@ Select[IntegerDigits[#, 2] & /@ Range[300], SequencePosition[#, {0, 0}] == {} &] (* Amiram Eldar, May 31 2025 *)

a(n) = A090996(A003754(n+1)). - Amiram Eldar, May 31 2025

More terms from Amiram Eldar, May 31 2025

A228349 Triangle read by rows: T(j,k) is the k-th part in nondecreasing order of the j-th region of the set of compositions (ordered partitions) of n in colexicographic order, if 1<=j<=2^(n-1) and 1<=k<=A006519(j).

Original entry on oeis.org

1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 3, 4, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 5, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 3, 4, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Omar E. Pol, Aug 26 2013

Triangle read by rows in which row n lists the A006519(n) elements of the row A001511(n) of triangle A090996, n >= 1.

The equivalent sequence for partitions is A220482.

			----------------------------------------------------------
.             Diagram                Triangle
Compositions    of            of compositions (rows)
of 5          regions          and regions (columns)
----------------------------------------------------------
.            _ _ _ _ _
5           |_        |                                 5
1+4         |_|_      |                               1 4
2+3         |_  |     |                             2   3
1+1+3       |_|_|_    |                           1 1   3
3+2         |_    |   |                         3       2
1+2+2       |_|_  |   |                       1 2       2
2+1+2       |_  | |   |                     2   1       2
1+1+1+2     |_|_|_|_  |                   1 1   1       2
4+1         |_      | |                 4               1
1+3+1       |_|_    | |               1 3               1
2+2+1       |_  |   | |             2   2               1
1+1+2+1     |_|_|_  | |           1 1   2               1
3+1+1       |_    | | |         3       1               1
1+2+1+1     |_|_  | | |       1 2       1               1
2+1+1+1     |_  | | | |     2   1       1               1
1+1+1+1+1   |_|_|_|_|_|   1 1   1       1               1
.
Written as an irregular triangle in which row n lists the parts of the n-th region the sequence begins:
1;
1,2;
1;
1,1,2,3;
1;
1,2;
1;
1,1,1,1,2,2,3,4;
1;
1,2;
1;
1,1,2,3;
1;
1,2;
1;
1,1,1,1,1,1,1,1,2,2,2,2,3,3,4,5;
...
Alternative interpretation of this sequence:
Triangle read by rows in which row r lists the regions of the last section of the set of compositions of r:
[1];
[1,2];
[1],[1,1,2,3];
[1],[1,2],[1],[1,1,1,1,2,2,3,4];
[1],[1,2],[1],[1,1,2,3],[1],[1,2],[1],[1,1,1,1,1,1,1,1,2,2,2,2,3,3,4,5];

Michael De Vlieger, Table of n, a(n) for n = 1..13312 (rows 1 <= n <= 2^11 = 2048).

Main triangle: Right border gives A001511. Row j has length A006519(j). Row sums give A038712.

Cf. A001787, A001792, A011782, A029837, A045623, A065120, A070939, A090996, A186114, A187816, A187818, A206437, A220482, A228347, A228348, A228350, A228351, A228366, A228367, A228370, A228371, A228525, A228526.

Table[Map[Length@ TakeWhile[IntegerDigits[#, 2], # == 1 &] &, Range[2^(# - 1), 2^# - 1]] &@ IntegerExponent[2 n, 2], {n, 32}] // Flatten (* Michael De Vlieger, May 23 2017 *)

A153036 Integer parts of the full Stern-Brocot tree.

Original entry on oeis.org

0, 1, 0, 2, 0, 0, 1, 3, 0, 0, 0, 0, 1, 1, 2, 4, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 3, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Reinhard Zumkeller, Dec 22 2008

			a(1): 1;
a(2..3): 1x0, 2;
a(4..7): 2x0, 1x1, 3;
a(8..15): 4x0, 2x1, 1x2, 4;
a(16..31): 8x0, 4x1, 2x2, 1x3, 5;
a(32..63): 16x0, 8x1, 4x2, 2x3, 1x4, 6;
a(64..127): 32x0, 16x1, 8x2, 4x3, 2x4, 1x5, 7;
a(128..255): 64x0, 32x1, 16x2, 8x3, 4x4, 2x5, 1x6, 8;
a(256..511): 128x0, 64x1, 32x2, 16x3, 8x4, 4x5, 2x6, 1x7, 9.

Jon Maiga, Computer-generated formulas for A153036, Sequence Machine.
N. J. A. Sloane, Stern-Brocot or Farey Tree
Index entries for sequences related to Stern's sequences

Cf. A130321.
If every block of terms of length 2^k is reversed, we get A290256; other permutations within these blocks give A007814 and A272729-1.
Cf. A090996, A043545, A007814, A145342.

a(n+1) = floor(A007305(n+2)/A047679(n)). [Corrected by Andrey Zabolotskiy, Jul 23 2020]
a(n) = if n=2^k-1 then k else Log2(n)-1-Log2(2^(Log2(n)+1)-(n+1)), where Log2=A000523.
From Andrey Zabolotskiy, Oct 07 2021: (Start)
Formulas discovered by Sequence Machine (and also essentially by Kevin Ryde):
a(n) = A090996(n) - A043545(n).
a(n) = A007814(A145342(n+1)). (End)

a(0) = 0 added by Andrey Zabolotskiy, Jul 23 2020

cp's OEIS Frontend

A101211 Triangle read by rows: n-th row is length of run of leftmost 1's, followed by length of run of 0's, followed by length of run of 1's, etc., in the binary representation of n.

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A360189 Triangle T(n,k), n>=0, 0<=k<=floor(log_2(n+1)), read by rows: T(n,k) = number of nonnegative integers <= n having binary weight k.

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A125985 Signature-permutation of Vaillé's 1997 bijection on 'bridges' (Dyck paths).

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A083368 a(n) is the position of the highest one in A003754(n).

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A126307 a(n) is the length of the leftmost ascent (i.e., height of the first peak) in the n-th Dyck path encoded by A014486(n).

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A342126 The binary expansion of a(n) corresponds to that of n where all the 1's have been replaced by 0's except in the first run of 1's.

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A279209 Length of first run of 0's in binary expansion of n.

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