A002450 -id:A002450 - cp's OEIS frontend

A053646 Distance to nearest power of 2.

0, 0, 1, 0, 1, 2, 1, 0, 1, 2, 3, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25

Offset: 1

Henry Bottomley, Mar 22 2000

Sum_{j=1..2^(k+1)} a(j) = A002450(k) = (4^k - 1)/3. - Klaus Brockhaus, Mar 17 2003

			a(10)=2 since 8 is closest power of 2 to 10 and |8-10| = 2.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Klaus Brockhaus, Illustration for A053646, A081252, A081253 and A081254
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint 2016.
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
Index entries for sequences related to distance to nearest element of some set

Cf. A006165, A053188, A060973, A081134, A002450, A081252, A081253, A081254.

a:= n-> (h-> min(n-h, 2*h-n))(2^ilog2(n)):
seq(a(n), n=1..100);  # Alois P. Heinz, Mar 28 2021

np2[n_]:=Module[{min=Floor[Log[2,n]],max},max=min+1;If[2^max-nHarvey P. Dale, Feb 21 2012 *)

a(n)=vecmin(vector(n,i,abs(n-2^(i-1))))

for(n=1,89,p=2^floor(0.1^25+log(n)/log(2)); print1(min(n-p,2*p-n),","))

a(n) = my (p=#binary(n)); return (min(n-2^(p-1), 2^p-n)) \\ Rémy Sigrist, Mar 24 2018

def A053646(n): return min(n-(m := 2**(len(bin(n))-3)),2*m-n) # Chai Wah Wu, Mar 08 2022

a(2^k+i) = i for 1 <= i <= 2^(k-1); a(3*2^k+i) = 2^k-i for 1 <= i <= 2^k; (Sum_{k=1..n} a(k))/n^2 is bounded. - Benoit Cloitre, Aug 17 2002
a(n) = min(n-2^floor(log(n)/log(2)), 2*2^floor(log(n)/log(2))-n). - Klaus Brockhaus, Mar 08 2003
From Peter Bala, Aug 04 2022: (Start)
a(n) = a( 1 + floor((n-1)/2) ) + a( ceiling((n-1)/2) ).
a(2*n) = 2*a(n); a(2*n+1) = a(n) + a(n+1) for n >= 2. Cf. A006165. (End)
a(n) = 2*A006165(n) - n for n >= 2. - Peter Bala, Sep 25 2022

A018215 a(n) = n*4^n.

Original entry on oeis.org

0, 4, 32, 192, 1024, 5120, 24576, 114688, 524288, 2359296, 10485760, 46137344, 201326592, 872415232, 3758096384, 16106127360, 68719476736, 292057776128, 1236950581248, 5222680231936, 21990232555520, 92358976733184, 387028092977152, 1618481116086272

Offset: 0

N. J. A. Sloane, Peter Winkler (pw(AT)bell-labs.com)

Bisection of A001787. That is, a(n) = A001787(2*n). - Graeme McRae, Jul 12 2006

All numbers of the form n*4^n+(4^n-1)/3 have the property that they are sums of two squares and also their indices are the sum of two squares. This follows from the identity n*4^n+(4^n-1)/3 = 4*(4*(..(4*(4*n+1)+1)..)+1)+1. - Artur Jasinski, Nov 12 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Index entries for linear recurrences with constant coefficients, signature (8,-16).

Cf. A000302 (4^n), A001787, A002450, A083679.

Row n=4 of A258997.

[n*4^n: n in [0..25]]; // Vincenzo Librandi, Jun 01 2011

Table[n 4^n,{n,0,20}] (* or *) LinearRecurrence[{8,-16},{0,4},30] (* Harvey P. Dale, Apr 22 2018 *)

a(n) = n<<(2*n) \\ David A. Corneth, Apr 22 2018

G.f.: 4*x/(1-4*x)^2.
E.g.f.: 4*x*exp(4*x).
From Amiram Eldar, Jul 20 2020: (Start)
Sum_{n>=1} 1/a(n) = log(4/3) = A083679.
Sum_{n>=1} (-1)^(n+1)/a(n) = log(5/4). (End)

A036991 Numbers k with the property that in the binary expansion of k, reading from right to left, the number of 0's never exceeds the number of 1's.

Original entry on oeis.org

0, 1, 3, 5, 7, 11, 13, 15, 19, 21, 23, 27, 29, 31, 39, 43, 45, 47, 51, 53, 55, 59, 61, 63, 71, 75, 77, 79, 83, 85, 87, 91, 93, 95, 103, 107, 109, 111, 115, 117, 119, 123, 125, 127, 143, 151, 155, 157, 159, 167, 171, 173, 175, 179, 181, 183, 187, 189, 191, 199, 203

Offset: 1

N. J. A. Sloane

List of binary words that correspond to a valid pairing of parentheses. - Joerg Arndt, Nov 27 2004
This sequence includes as subsequences A000225, A002450, A007583, A036994, A052940, A112627, A113836, A113841, A290114; and also A015521 (without 0), A083713 (without 0), A086224 (without 6), A182512 (without 0). - Gennady Eremin, Nov 27 2021 and Aug 26 2023
Partial differences are powers of 2 (cf. A367626, A367627). - Gennady Eremin, Dec 23 2021
This is the sequence A030101(A014486(n)), n >= 0, sorted into ascending order. See A014486 for more references, illustrations, etc., concerning Dyck paths and other associated structures enumerated by the Catalan numbers. - Antti Karttunen, Sep 25 2023
The terms in this sequence with a given length in base 2 are counted by A001405. For example, the number of terms of bit length k=5 (these are 19, 21, 23, 27, 29, and 31) is equal to A001405(k-1) = A001405(4) = 6. - Gennady Eremin, Nov 07 2023

			From _Joerg Arndt_, Dec 05 2021: (Start)
List of binary words with parentheses for those in the sequence (indicated by P). The binary words are scanned starting from the least significant bit, while the parentheses words are written left to right:
     Binary   Parentheses (if the value is in the sequence)
00:  ..... P  [empty string]
01:  ....1 P   ()
02:  ...1.
03:  ...11 P   (())
04:  ..1..
05:  ..1.1 P   ()()
06:  ..11.
07:  ..111 P   ((()))
08:  .1...
09:  .1..1
10:  .1.1.
11:  .1.11 P   (()())
12:  .11..
13:  .11.1 P   ()(())
14:  .111.
15:  .1111 P   (((())))
16:  1....
17:  1...1
18:  1..1.
19:  1..11 P   (())()
(End)

Alois P. Heinz, Table of n, a(n) for n = 1..13496 (all terms < 2^16; first 1000 terms from Reinhard Zumkeller)
Joerg Arndt, Matters Computational (The Fxtbook), section 1.28, pp. 78-80.
Gennady Eremin, Dyck Numbers, IV. Nested patterns in OEIS A036991, arXiv:2306.10318, 2023. See pp. 1-2, 4, 7-11, 13-14.
Gennady Eremin, Partitioning the set of natural numbers into Mersenne trees and into arithmetic progressions; Natural Matrix and Linnik's constant, arXiv:2405.16143 [math.CO], 2024. See pp 2-5, 14.
Gennady Eremin, Infinite matrix of odd natural numbers. A bit about Sophie Germain prime numbers, arXiv:2501.17090 [math.GM], 2025. See pp. 3, 11.
Index entries for sequences related to binary expansion of n

Cf. A350577 (primes subsequence).
See also A014486, A030101, A036988, A036990, A036992. A036994 is a subset (requires the count of zeros to be strictly less than the count of 1's).
See also A030308, A000225, A002450, A007583, A350346, A367625, A367626 & A367627 (first differences).

a036991 n = a036991_list !! (n-1)
a036991_list = filter ((p 1) . a030308_row) [0..] where
   p     []    = True
   p ones (0:bs) = ones > 1 && p (ones - 1) bs
   p ones (1:bs) = p (ones + 1) bs
-- Reinhard Zumkeller, Jul 31 2013

q:= proc(n) local l, t, i; l:= Bits[Split](n); t:=0;
      for i to nops(l) do t:= t-1+2*l[i];
        if t<0 then return false fi
      od: true
    end:
select(q, [$0..300])[];  # Alois P. Heinz, Oct 09 2019

moreOnesRLQ[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[0, 203], moreOnesRLQ] (* Alonso del Arte, Sep 21 2011 *)
Join[{0},Select[Range[210],Min[Accumulate[Reverse[IntegerDigits[#,2]]/.{0->-1}]]>-1&]] (* Harvey P. Dale, Apr 18 2014 *)

select( {is_A036991(n,c=1)=!n||!until(!n>>=1,(c-=(-1)^bittest(n,0))||return)}, [0..99]) \\ M. F. Hasler, Nov 26 2021

def ok(n):
    if n == 0: return True # by definition
    count = {"0": 0, "1": 0}
    for bit in bin(n)[:1:-1]:
        count[bit] += 1
        if count["0"] > count["1"]: return False
    return True
print([k for k in range(204) if ok(k)]) # Michael S. Branicky, Nov 25 2021

from itertools import count, islice
def A036991_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 n
A036991_list = list(islice(A036991_gen(),20)) # Chai Wah Wu, Dec 30 2021

If a(n) = A000225(k) for some k, then a(n+1) = a(n) + A060546(k). - Gennady Eremin, Nov 07 2023

More terms from Erich Friedman
Edited by N. J. A. Sloane, Sep 14 2008 at the suggestion of R. J. Mathar
Offset corrected and example adjusted accordingly by Reinhard Zumkeller, Jul 31 2013

A062052 Numbers with exactly 2 odd integers in their Collatz (or 3x+1) trajectory.

Original entry on oeis.org

5, 10, 20, 21, 40, 42, 80, 84, 85, 160, 168, 170, 320, 336, 340, 341, 640, 672, 680, 682, 1280, 1344, 1360, 1364, 1365, 2560, 2688, 2720, 2728, 2730, 5120, 5376, 5440, 5456, 5460, 5461, 10240, 10752, 10880, 10912, 10920, 10922, 20480, 21504, 21760, 21824

Offset: 1

David W. Wilson

nonn

The Collatz (or 3x+1) function is f(x) = x/2 if x is even, 3x+1 if x is odd.
The Collatz trajectory of n is obtained by applying f repeatedly to n until 1 is reached.
The sequence consists of terms of A002450 and their 2^k multiples. The first odd integer in the trajectory is one of the terms of A002450 and the second odd one is the terminal 1. - Antti Karttunen, Feb 21 2006
This sequence looks to appear first in the literature on page 1285 in R. E. Crandall.

			The Collatz trajectory of 5 is (5,16,8,4,2,1), which contains 2 odd integers.

Reinhard Zumkeller and T. D. Noe, Table of n, a(n) for n = 1..1000 (first 100 terms from Reinhard Zumkeller)
R. E. Crandall, On the 3x+1 problem, Math. Comp., 32 (1978) 1281-1292.
J. Shallit and D. Wilson, The "3x+1" Problem and Finite Automata, Bulletin of the EATCS #46 (1992) pp. 182-185.
Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for 2-automatic sequences.

Cf. A062053, A062054, A062055, A062056, A062057, A062058, A062059, A062060, A122196, A122197.
Is this a subset of A115774?
Column k=2 of A354236.

import Data.List (elemIndices)
a062052 n = a062052_list !! (n-1)
a062052_list = map (+ 1) $ elemIndices 2 a078719_list
-- Reinhard Zumkeller, Oct 08 2011

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; countOdd[lst_] := Length[Select[lst, OddQ]]; Select[Range[22000], countOdd[Collatz[#]] == 2 &] (* T. D. Noe, Dec 03 2012 *)

for(n=2,100000,s=n; t=0; while(s!=1,if(s%2==0,s=s/2,s=3*s+1; t++); if(s*t==1,print1(n,","); ); ))

def a(n):
    l=[n, ]
    while True:
        if n%2==0: n//=2
        else: n = 3*n + 1
        if n not in l:
            l.append(n)
            if n<2: break
        else: break
    return len([i for i in l if i % 2])
print([n for n in range(1, 22001) if a(n)==2]) # Indranil Ghosh, Apr 14 2017

A078719(a(n)) = 2; A006667(a(n)) = 1.

a(n) = 2^x * (4^y - 1)/3 where x = A122196(n) - 1 and y = A122197(n) + 1. - Alan Michael Gómez Calderón, Jan 16 2025 after Antti Karttunen

A096773 a(n) = 4*a(n-2) + 1 with a(1) = 0, a(2) = 3.

Original entry on oeis.org

0, 3, 1, 13, 5, 53, 21, 213, 85, 853, 341, 3413, 1365, 13653, 5461, 54613, 21845, 218453, 87381, 873813, 349525, 3495253, 1398101, 13981013, 5592405, 55924053, 22369621, 223696213, 89478485, 894784853, 357913941, 3579139413, 1431655765

Offset: 1

Gottfried Helms, Aug 15 2004

Remainders for classes m of integers n (mod 2^(m+1)). After applying one Collatz (3x+1)-transformation to the so-classified integers the result can be written in two classes (mod 6) only.
This classifying scheme covers all positive integers.
With one 3x+1-transformation T(x;p) := x' = (3x+1)/2^p, all numbers x, described in the form, with the free parameter i >= 0, x = i*2^N + a(N) result in x', describable by the two classes with the same parameter i:
x' = i*6 + 1 (for odd N>2), or x' = i*6 + 5 (for even N). Thus
x =  4*i +  3 -> x' = 6*i + 5, x =   8*i +  1 -> x' = 6*i + 1,
x = 16*i + 13 -> x' = 6*i + 5, x =  32*i +  5 -> x' = 6*i + 1,
x = 64*i + 53 -> x' = 6*i + 5, x = 128*i + 21 -> x' = 6*i + 1,
....
all with "i" as a free parameter >= 0 covering all positive integers.

			a(1) = (2^0-1)/3 =  0, a(2) = (5*2^1 - 1) / 3 =  3,
a(3) = (2^2-1)/3 =  1, a(4) = (5*2^3 - 1) / 3 = 13,
a(5) = (2^4-1)/3 =  5, a(6) = (5*2^5 - 1) / 3 = 53,
a(7) = (2^6-1)/3 = 21.
....

Michael De Vlieger, Table of n, a(n) for n = 1..3323
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for linear recurrences with constant coefficients, signature (1,4,-4).

Bisections are A002450 & A072197.
After the initial 0, column 1 of A257852.
Cf. A176965.

[(2^(n-1)*(3 + 2*(-1)^n) - 1)/3: n in [1..40]]; // Vincenzo Librandi, Jul 12 2015

a[1] = 0; a[2] = 3; a[n_] := a[n] = 4a[n - 2] + 1; Table[ a[n], {n, 35}] (* Robert G. Wilson v, Aug 20 2004 *)
Table[(2^(n - 1)*(3 + 2*(-1)^(n)) - 1)/3, {n, 10}] (* L. Edson Jeffery, Jul 12 2015 *)
nxt[{a_,b_}]:={b,4a+1}; NestList[nxt,{0,3},40][[;;,1]] (* or *) LinearRecurrence[{1,4,-4},{0,3,1},40] (* Harvey P. Dale, Mar 19 2025 *)

apply( {A096773(n) = if(n%2, 1, 5)<<(n-1)\3}, [1..55]) \\ M. F. Hasler, May 28 2024

# To map any (odd) v to its (r,c):
use bigint; $v=149; $r=$c=0; while(1){ $b=($v&1); $v>>=1; if ($b==($v&1)){ $c=($v>>1); last} $r++} $r&=1; # this splits the binary representation into two parts, at the first repeated digit from the right: the number of bits on the right is the row value, and the binary value on the left is the column value. Example: 149 => 1.00.10101 => (r,c)=(5,1). Ruud H.G. van Tol, Sep 23 2021

A096773=lambda n:((1 if n&1 else 5)<M. F. Hasler, May 28 2024

a(2m) = (5*2^(2m-1) - 1)/3, a(2m-1) = (2^(2m-2)-1)/3.
From Paul Curtz, Jul 01 2008; corrected by Bob Selcoe, Jul 28 2018: (Start)
a(2n) = 10*a(2n-1) + 3.
a(n+1) - 2*a(n) = A001045(n+2), signed. (End)
a(n) = (2^(n-1)*(3 + 2*(-1)^n) - 1)/3. - L. Edson Jeffery, Jul 12 2015
a(2n) = A086893(2n), a(2n+1) = A086893(2n-1), n > 0. - Yosu Yurramendi, Jan 17 2017
G.f.: -x^2*(-3+2*x) / ( (x-1)*(2*x+1)*(2*x-1) ). - R. J. Mathar, Mar 07 2017
a(2n) = A072197(n-1), n > 0; a(2n+1) = A002450(n), n >= 0. - Yosu Yurramendi, Mar 07 2017
a(2n) = (A266753(n) + A004171(n-1))/2, a(2n+1) = (A266753(n) - A004171(n-1))/2, n > 0. - Yosu Yurramendi, Mar 07 2017
a(n) = least residue 2*3^(2^(n-4)-1) - 1 (mod 2^n), n >= 5. - Bob Selcoe, Jul 26 2018
a(n) = 2*A176965(n-1) + 1 for n > 1. - Loren M. Pearson, Dec 06 2024

A125118 Triangle read by rows: T(n,k) = value of the n-th repunit in base (k+1) representation, 1<=k<=n.

Original entry on oeis.org

1, 3, 4, 7, 13, 21, 15, 40, 85, 156, 31, 121, 341, 781, 1555, 63, 364, 1365, 3906, 9331, 19608, 127, 1093, 5461, 19531, 55987, 137257, 299593, 255, 3280, 21845, 97656, 335923, 960800, 2396745, 5380840, 511, 9841, 87381, 488281, 2015539, 6725601, 19173961, 48427561, 111111111

Offset: 1

Reinhard Zumkeller, Nov 21 2006

			First 4 rows:
1: [1]_2
2: [11]_2 ........ [11]_3
3: [111]_2 ....... [111]_3 ....... [111]_4
4: [1111]_2 ...... [1111]_3 ...... [1111]_4 ...... [1111]_5
_
1: 1
2: 2+1 ........... 3+1
3: (2+1)*2+1 ..... (3+1)*3+1 ..... (4+1)*4+1
4: ((2+1)*2+1)*2+1 ((3+1)*3+1)*3+1 ((4+1)*4+1)*4+1 ((5+1)*5+1)*5+1.

G. C. Greubel, Rows n = 1..50 of the triangle, flattened
Eric Weisstein's World of Mathematics, Repunit

This triangle shares some features with triangle A104878.
This triangle is a portion of rectangle A055129.
Each term of A110737 comes from the corresponding row of this triangle.
Diagonals (adjusting offset as necessary): A060072, A023037, A031973, A173468.
Columns (adjusting offset as necessary): A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218724, A218725, A218726, A218727, A218728, A218729, A218730, A218731, A218732, A218733, A218734, A132469, A218736, A218737, A218738, A218739, A218740, A218741, A218742, A218743, A218744, A218745, A218746, A218747, A218748, A218749, A218750, A218751, A218753, A218752.
Cf. A023037, A031973, A125119, A125120 (row sums).

[((k+1)^n -1)/k : k in [1..n], n in [1..12]]; // G. C. Greubel, Aug 15 2022

Table[((k+1)^n -1)/k, {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Aug 15 2022 *)

def A125118(n,k): return ((k+1)^n -1)/k
flatten([[A125118(n,k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Aug 15 2022

T(n, k) = Sum_{i=0..n-1} (k+1)^i.
T(n+1, k) = (k+1)*T(n, k) + 1.
Sum_{k=1..n} T(n, k) = A125120(n).
T(2*n-1, n) = A125119(n).
T(n, 1) = A000225(n).
T(n, 2) = A003462(n) for n>1.
T(n, 3) = A002450(n) for n>2.
T(n, 4) = A003463(n) for n>3.
T(n, 5) = A003464(n) for n>4.
T(n, 9) = A002275(n) for n>8.
T(n, n) = A060072(n+1).
T(n, n-1) = A023037(n) for n>1.
T(n, n-2) = A031973(n) for n>2.
T(n, k) = A055129(n, k+1) = A104878(n+k, k+1), 1<=k<=n. - Mathew Englander, Dec 19 2020

A295235 Numbers k such that the positions of the ones in the binary representation of k are in arithmetic progression.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 17, 18, 20, 21, 24, 28, 30, 31, 32, 33, 34, 36, 40, 42, 48, 56, 60, 62, 63, 64, 65, 66, 68, 72, 73, 80, 84, 85, 96, 112, 120, 124, 126, 127, 128, 129, 130, 132, 136, 144, 146, 160, 168, 170, 192, 224, 240, 248

Offset: 1

Rémy Sigrist, Nov 18 2017

Also numbers k of the form Sum_{b=0..h-1} 2^(i+j*b) for some h >= 0, i >= 0, j > 0 (in fact, h = A000120(k), and if k > 0, i = A007814(k)).
There is a simple bijection between the finite sets of nonnegative integers in arithmetic progression and the terms of this sequence: s -> Sum_{i in s} 2^i; the term 0 corresponds to the empty set.
For any n > 0, A054519(n) gives the numbers of terms with n+1 digits in binary representation.
For any n >= 0, n is in the sequence iff 2*n is in the sequence.
For any n > 0, A000695(a(n)) is in the sequence.
The first prime numbers in the sequence are: 2, 3, 5, 7, 17, 31, 73, 127, 257, 8191, 65537, 131071, 262657, 524287, ...
This sequence contains the following sequences: A000051, A000079, A000225, A000668, A002450, A019434, A023001, A048645.
For any k > 0, 2^k - 2, 2^k - 1, 2^k, 2^k + 1 and 2^k + 2 are in the sequence (e.g., 14, 15, 16, 17, and 18).
Every odd term is a binary palindrome (and thus belongs to A006995).
Odd terms are A064896. - Robert Israel, Nov 20 2017

			The binary representation of the number 42 is "101010" and has ones evenly spaced, hence 42 appears in the sequence.
The first terms, alongside their binary representations, are:
   n  a(n)  a(n) in binary
  --  ----  --------------
   1    0           0
   2    1           1
   3    2          10
   4    3          11
   5    4         100
   6    5         101
   7    6         110
   8    7         111
   9    8        1000
  10    9        1001
  11   10        1010
  12   12        1100
  13   14        1110
  14   15        1111
  15   16       10000
  16   17       10001
  17   18       10010
  18   20       10100
  19   21       10101
  20   24       11000

Rémy Sigrist, Table of n, a(n) for n = 1..1000

Cf. A000051, A000079, A000120, A000225, A000668, A000695, A002450, A006995, A007814, A019434, A023001, A048645, A054519, A064896.

Cf. A029931, A048793 (binary indices triangle), A070939, A291166, A325328 (prime indices rather than binary indices), A326669, A326675.

f:= proc(d) local i,j,k;
  op(sort([seq(seq(add(2^(d-j*k),k=0..m),m=1..d/j),j=1..d),2^(d+1)]))
end proc:
0,1,seq(f(d),d=0..10); # Robert Israel, Nov 20 2017

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Select[Range[100],SameQ@@Differences[bpe[#]]&] (* Gus Wiseman, Jul 22 2019 *)

is(n) = my(h=hammingweight(n)); if(h<3, return(1), my(i=valuation(n,2),w=#binary(n)); if((w-i-1)%(h-1)==0, my(j=(w-i-1)/(h-1)); return(sum(k=0,h-1,2^(i+j*k))==n), return(0)))

A016153 a(n) = (9^n-4^n)/5.

Original entry on oeis.org

0, 1, 13, 133, 1261, 11605, 105469, 953317, 8596237, 77431669, 697147165, 6275373061, 56482551853, 508359743893, 4575304803901, 41178011670565, 370603178776909, 3335432903959477, 30018913315504477, 270170288559017029

Offset: 0

N. J. A. Sloane

a(n) is also the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x,y)=xF(n-1)(x,y)+yF(n-2)(x,y),F(0)(x,y)=0,F(1)(x,y)=1, when we write 13x for x and -36x^2 for y. - Mario Catalani (mario.catalani(AT)unito.it), Dec 09 2002

John Elias, Illustration: Union of Cantor Square and Koch Snowflake fractals
R. Flórez, R. A. Higuita and A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
Index entries for linear recurrences with constant coefficients, signature (13,-36).

Cf. A015441, A002450.

Join[{a=0,b=1},Table[c=13*b-36*a;a=b;b=c,{n,60}]](*and/or*)f[n_]:=(9^n-4^n)/5;f[Range[0,60]] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2011 *)

PARI
```
a(n)=(9^n-4^n)/5
```

G.f.: x/((1-4*x)*(1-9*x)). a(n)=13*a(n-1)-36*a(n-2).
a(n) = A015441(2*n).
From Peter Bala, Jul 23 2025: (Start)
The following products telescope. Cf. A002450:
Product_{k >= 1} 1 + 6^k/a(k+1) = 3; Product_{k >= 1} 1 - 6^k/a(k+1) = 3/13.
Product_{k >= 1} 1 + (-6)^k/a(k+1) = 3/5; Product_{k >= 1} 1 - (-6)^k/a(k+1) = 15/13.
(End)

A162911 Numerators of drib tree fractions, where drib is the bit-reversal permutation tree of the Bird tree.

Original entry on oeis.org

1, 1, 2, 2, 3, 1, 3, 3, 5, 1, 4, 3, 4, 2, 5, 5, 8, 2, 7, 4, 5, 3, 7, 4, 7, 1, 5, 5, 7, 3, 8, 8, 13, 3, 11, 7, 9, 5, 12, 5, 9, 1, 6, 7, 10, 4, 11, 7, 11, 3, 10, 5, 6, 4, 9, 7, 12, 2, 9, 8, 11, 5, 13, 13, 21, 5, 18, 11, 14, 8, 19, 9, 16, 2, 11, 12, 17, 7, 19, 9, 14, 4, 13, 6, 7, 5, 11, 10, 17, 3, 13

Offset: 1

Ralf Hinze (ralf.hinze(AT)comlab.ox.ac.uk), Aug 05 2009

The drib tree is an infinite binary tree labeled with rational numbers. It is generated by the following iterative process: start with the rational 1; for the left subtree increment and then reciprocalize the current rational; for the right subtree interchange the order of the two steps: the rational is first reciprocalized and then incremented. Like the Stern-Brocot and the Bird tree, the drib tree enumerates all the positive rationals (A162911(n)/A162912(n)).
From Yosu Yurramendi, Jul 11 2014: (Start)
If the terms (n>0) are written as an array (left-aligned fashion) with rows of length 2^m, m = 0,1,2,3,...
   1,
   1, 2,
   2, 3,1, 3,
   3, 5,1, 4, 3, 4,2, 5,
   5, 8,2, 7, 4, 5,3, 7,4, 7,1, 5, 5, 7,3, 8,
   ...
then the sum of the m-th row is 3^m (m = 0,1,2,), each column k is a Fibonacci-type sequence.
If the rows are written in a right-aligned fashion:
                                                                           1
                                                                        1, 2
                                                                   2, 3,1, 3
                                                        3, 5,1, 4, 3, 4,2, 5
                                   5, 8,2, 7,4, 5,3, 7, 4, 7,1, 5, 5, 7,3, 8
                                                                         ...
then each column k also is a Fibonacci-type sequence.
If the sequence is considered by blocks of length 2^m, m = 0,1,2,..., the blocks of this sequence are the reverses of blocks of A162912 (a(2^m+k) = A162912(2^(m+1)-1-k), m = 0,1,2,..., k = 0..2^m-1).
(End)
From Yosu Yurramendi, Jan 12 2017: (Start)
a(2^(m+2m'  )     + A020988(m'))   = A000045(m+1), m>=0, m'>=0
a(2^(m+2m'+1)     + A020989(m'))   = A000045(m+3), m>=0, m'>=0
a(2^(m+2m'  ) - 1 - A002450(m'))   = A000045(m+1), m>=0, m'>=0
a(2^(m+2m'+1) - 1 - A072197(m'-1)) = A000045(m+3), m>=0, m'>0
a(2^(m+1) -1) = A000045(m+2), m>=0. (End)

			The first four levels of the drib tree:
  [1/1],
  [1/2, 2/1],
  [2/3, 3/1, 1/3, 3/2],
  [3/5, 5/2, 1/4, 4/3, 3/4, 4/1, 2/5, 5/3].

Ralf Hinze, Functional pearls: the bird tree, J. Funct. Programming 19 (2009), no. 5, 491-508.
Index entries for fraction trees

This sequence is the composition of A162909 and A059893: a(n) = A162909(A059893(n)). This sequence is a permutation of A002487(n+1).

import Ratio; drib :: [Rational]; drib = 1 : map (recip . succ) drib \/ map (succ . recip) drib; (a : as) \/ bs = a : (bs \/ as); a162911 = map numerator drib; a162912 = map denominator drib

a(n) = my(x = 0, y = 1); forstep(i = logint(n, 2), 0, -1, [x, y] = if(bittest(n, i), [y, x + y], [x + y, x])); y \\ Mikhail Kurkov, Oct 12 2023

blocklevel <- 6 # arbitrary
a <- 1
for(m in 0:blocklevel) for(k in 0:(2^m-1)){
  a[2^(m+1)+2*k  ] <- a[2^(m+1)-1-k]
  a[2^(m+1)+2*k+1] <- a[2^(m+1)-1-k] + a[2^m+k]
}
a
# Yosu Yurramendi, Jul 11 2014

a(n) where a(1) = 1; a(2n) = b(n); a(2n+1) = a(n) + b(n); and b(1) = 1; b(2n) = a(n) + b(n); b(2n+1) = a(n).
a(2^(m+1)+2*k) = a(2^(m+1)-k-1), a(2^(m+1)+2*k+1) = a(2^(m+1)-k-1) + a(2^m+k), a(1) = 1, m>=0, k=0..2^m-1. - Yosu Yurramendi, Jul 11 2014
a(2^(m+1) + 2*k) = A162912(2^m + k), m >= 0, 0 <= k < 2^m.
a(2^(m+1) + 2*k + 1) = a(2^m + k) + A162912(2^m + k), m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Mar 30 2016
a(n*2^m + A176965(m)) = A268087(n), n > 0, m > 0. - Yosu Yurramendi, Feb 20 2017
a(n) = A002487(A258996(n)), n > 0. - Yosu Yurramendi, Jun 23 2021

A265747 Numbers written in Jacobsthal greedy base.

Original entry on oeis.org

0, 1, 2, 10, 11, 100, 101, 102, 110, 111, 200, 1000, 1001, 1002, 1010, 1011, 1100, 1101, 1102, 1110, 1111, 10000, 10001, 10002, 10010, 10011, 10100, 10101, 10102, 10110, 10111, 10200, 11000, 11001, 11002, 11010, 11011, 11100, 11101, 11102, 11110, 11111, 20000, 100000, 100001, 100002, 100010, 100011, 100100

Offset: 0

Antti Karttunen, Dec 17 2015

These are called "Jacobsthal Representation Numbers" in Horadam's 1996 paper.
Sum_{i=0..} digit(i)*A001045(2+digit(i)) recovers n from such representation a(n), where digit(0) stands for the least significant digit (at the right), and A001045(k) gives the k-th Jacobsthal number.
No larger digits than 2 will occur, which allows representing the same sequence in a more compact form by base-3 coding in A265746.
Sequence A197911 gives the terms with no digit "2" in their representation, while its complement A003158 gives the terms where "2" occurs at least once.
Numbers beginning with digit "2" in this representation are given by A020988(n) [= 2*A002450(n) = 2*A001045(2n)].

			For n=7, when selecting the terms of A001045 with the greedy algorithm, we need terms A001045(4) + A001045(2) + A001045(2) = 5 + 1 + 1, thus a(7) = "102".
For n=10, we need A001045(4) + A001045(4) = 5+5, thus a(10) = "200".

Antti Karttunen, Table of n, a(n) for n = 0..10923
A. F. Horadam, Jacobsthal Representation Numbers, Fib Quart. 34 (1996), 40-54. (See especially page 50, which is page 11 in PDF.)

Cf. A001045, A002450, A020988, A007089, A197911, A003158.
Cf. A265745 (sum of digits).
Cf. A265746 (same numbers interpreted in base-3, then shown in decimal).
Cf. A084639 (positions of repunits).
Cf. A007961, A014417, A014418, A244159 for analogous sequences.

jacob[n_] := (2^n - (-1)^n)/3; maxInd[n_] := Floor[Log2[3*n + 1]]; A265747[n_] := A265747[n] = 10^(maxInd[n] - 2) + A265747[n - jacob[maxInd[n]]]; A265747[0] = 0; Array[A265747, 100, 0] (* Amiram Eldar, Jul 21 2023 *)

A130249(n) = floor(log(3*n + 1) / log(2));
A001045(n) = (2^n - (-1)^n) / 3;
A265747(n) = {if(n==0, 0, my(d=n - A001045(A130249(n))); 10^(A130249(n)-2) + if(d == 0, 0, A265747(d)));} \\ Amiram Eldar, Jul 21 2023

def greedyJ(n): m = (3*n+1).bit_length() - 1; return (m, (2**m-(-1)**m)//3)
def a(n):
    if n == 0: return 0
    place, value = greedyJ(n)
    return 10**(place-2) + a(n - value)
print([a(n) for n in range(49)]) # Michael S. Branicky, Jul 11 2021

a(0) = 0; for n >= 1, a(n) = 10^(A130249(n)-2) + a(n-A001045(A130249(n))).

a(n) = A007089(A265746(n)).

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