A020988 -id:A020988 - cp's OEIS frontend

A258996 Permutation of the positive integers: this permutation transforms the enumeration system of positive irreducible fractions A002487/A002487' (Calkin-Wilf) into the enumeration system A162911/A162912 (Drib), and vice versa.

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

Offset: 1

Yosu Yurramendi, Jun 16 2015

As A258746 the permutation is self-inverse. Except for fixed points 1, 2, 3 it consists completely of 2-cycles: (4,6), (5,7), (8,10), (9,11), (12,14), (13,15), (16,26), (17,27), ..., (21,31), ..., (32,42), ... . - Yosu Yurramendi, Mar 31 2016

When terms of sequence |n - a(n)|/2 (n > 3) are considered only once, and they are sorted in increasing order, A147992 is obtained. - Yosu Yurramendi, Apr 05 2016

Yosu Yurramendi, Table of n, a(n) for n = 1..16383
Index entries for sequences that are permutations of the natural numbers

Cf. A092569, A117120, A258746. Similar R-programs: A332769, A284447.

maxlevel <- 5 # by choice
a <- 1
for(m in 0:maxlevel) for(k in 0:(2^m-1)){
  a[2^(m+1) + 2*k    ] = 2*a[2^(m+1) - 1 - k]
  a[2^(m+1) + 2*k + 1] = 2*a[2^(m+1) - 1 - k] + 1}
a

# Given n, compute a(n) by taking into account the binary representation of n
maxblock <- 7 # by choice
a <- 1:3
for(n in 4:2^maxblock){
  ones <- which(as.integer(intToBits(n)) == 1)
  nbit <- as.integer(intToBits(n))[1:tail(ones, n = 1)]
  anbit <- nbit
  anbit[seq(2, length(anbit) - 1, 2)] <- 1 - anbit[seq(2, length(anbit) - 1, 2)]
  a <- c(a, sum(anbit*2^(0:(length(anbit) - 1))))
}
a
# Yosu Yurramendi, Mar 30 2021

a(1) = 1, a(2) = 2, a(3) = 3. For n = 2^m + k, m > 1, 0 <= k < 2^m. If m is even, then a(2^(m+1)+k) = a(2^m + k) + 2^m and a(2^(m+1) + 2^m+k) = a(2^m+k) + 2^(m+1). If m is odd, then a(2^(m+1) + k) = a(2^m+k) + 2^(m+1) and a(2^(m+1) + 2^m+k) = a(2^m+k) + 2^m.
From Yosu Yurramendi, Mar 23 2017: (Start)
A258746(a(n)) = a(A258746(n)), n > 0.
A092569(a(n)) = a(A092569(n)), n > 0.
A117120(a(n)) = a(A117120(n)), n > 0;
A065190(a(n)) = a(A065190(n)), n > 0;
A054429(a(n)) = a(A054429(n)), n > 0;
A063946(a(n)) = a(A063946(n)), n > 0. (End)
a(1) = 1, for m >= 0  and 0 <= k < 2^m, a(2^(m+1) + 2*k) = 2*a(2^(m+1) - 1 - k), a(2^(m+1) + 2*k + 1) = 2*a(2^(m+1) - 1 - k) + 1. - Yosu Yurramendi, May 23 2020
a(n) = A020988(A102572(n)) XOR n. - Alan Michael Gómez Calderón, Mar 11 2025

A056830 Alternate digits 1 and 0.

Original entry on oeis.org

0, 1, 10, 101, 1010, 10101, 101010, 1010101, 10101010, 101010101, 1010101010, 10101010101, 101010101010, 1010101010101, 10101010101010, 101010101010101, 1010101010101010, 10101010101010101, 101010101010101010

Offset: 0

Henry Bottomley, Aug 30 2000

Fibonacci bit-representations of numbers for which there is only one possible representation and for which the maximal and minimal bit-representations (A104326 and A014417) are equal. The numbers represented are equal to the numbers in A000071 (subtract the first term of that sequence). For example, 10101 = 12 because 8+5+1. - Casey Mongoven, Mar 19 2006
Sequence A000975 written in base 2. - Jaroslav Krizek, Aug 05 2009
The absolute value of alternating sum of the first n repunits: a(n) = abs(Sum_{k=0..n} (-1)^k*A002275(n)). - Ilya Gutkovskiy, Dec 02 2015
Binary representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 131", based on the 5-celled von Neumann neighborhood. See A279053 for references and links. - Robert Price, Dec 05 2016

			n  a(n)             A000975(n)   (If a(n) is interpreted in base 2.)
------------------------------
0  0 ....................... 0
1  1 ....................... 1
2  10 ...................... 2 = 2^1
3  101 ..................... 5
4  1010 ................... 10 = 2^1 + 2^3
5  10101 .................. 21
6  101010 ................. 42 = 2^1 + 2^3 + 2^5
7  1010101 ................ 85
8  10101010 .............. 170 = 2^1 + 2^3 + 2^5 + 2^7
9  101010101 ............. 341
10 1010101010 ............ 682 = 2^1 + 2^3 + 2^5 + 2^7 + 2^9
11 10101010101 .......... 1365
12 101010101010 ......... 2730 = 2^1 + 2^3 + 2^5 + 2^7 + 2^9 + 2^11, etc.
- _Bruno Berselli_, Dec 02 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..800
Index entries for linear recurrences with constant coefficients, signature (10,1,-10).
Index entries for 2-automatic sequences.

Partial sums of A015585.

Cf. A000975, A033113, A033114, A033115, A033116, A033117, A033118, A033119, A059848, A062864.

List([0..30], n-> Int(10^(n+1)/99) ); # G. C. Greubel, Jul 14 2019

[Round((20*10^n-11)/198) : n in [0..30]]; // Vincenzo Librandi, Jun 25 2011

A056830 := proc(n) floor(10^(n+1)/99) ; end proc:

CoefficientList[Series[x/((1-x^2)*(1-10*x)), {x,0,30}], x] (* G. C. Greubel, Sep 26 2017 *)

Vec(x/((1-x)*(1+x)*(1-10*x))+O(x^30)) \\ Charles R Greathouse IV, Feb 13 2017

[floor(10^(n+1)/99) for n in (0..30)] # G. C. Greubel, Jul 14 2019

a(n) = +10*a(n-1) + a(n-2) - 10*a(n-3).
a(n) = floor(10^(n+1)/(10^2-1)) = a(n-2)+10^(n-1) = 10*a(n-1) + (1 - (-1)^n)/2.
From Paul Barry, Nov 12 2003: (Start)
a(n+1) = Sum_{k=0..floor(n/2)} 10^(n-2*k).
a(n+1) = Sum_{k=0..n} Sum_{j=0..k} (-1)^(j+k)*10^j.
G.f.: x/((1-x)*(1+x)*(1-10*x)).
a(n) = 9*a(n-1) + 10*a(n-2) + 1.
a(n) = 10^(n+1)/99 - (-1)^n/22 - 1/18. (End)
a(n) = A007088(A107909(A104161(n))) = A007088(A000975(n)). - Reinhard Zumkeller, May 28 2005
a(n) = round((20*10^n-11)/198) = floor((10*10^n-1)/99) = ceiling((10*10^n-10)/99) = round((10*10^n-10)/99). - Mircea Merca, Dec 27 2010
From Daniel Forgues, Sep 20 2018: (Start)
If a(n) is interpreted in base 2:
a(2n) = Sum_{k=1..n} 2^(2n-1), n >= 0; a(2n-1) = a(2n)/2, n >= 1.
a(2n) = A020988(n), n >= 0.
a(0) = 0; a(2n) = 4*a(2n-2) + 2, n >= 1. (End)

More terms from Casey Mongoven, Mar 19 2006

A080674 a(n) = (4/3)*(4^n - 1).

Original entry on oeis.org

0, 4, 20, 84, 340, 1364, 5460, 21844, 87380, 349524, 1398100, 5592404, 22369620, 89478484, 357913940, 1431655764, 5726623060, 22906492244, 91625968980, 366503875924, 1466015503700, 5864062014804, 23456248059220, 93824992236884, 375299968947540, 1501199875790164

Offset: 0

N. J. A. Sloane, Mar 02 2003

a(n) is the number of steps which are made when generating all n-step random walks that begin in a given point P on a two-dimensional square lattice. To make one step means to move along one edge on the lattice. - Pawel P. Mazur (Pawel.Mazur(AT)pwr.wroc.pl), Mar 10 2005
Conjectured to be the number of integers from 0 to (10^n)-1 that lack 0, 1, 2, 3, 4 and 5 as a digit. - Alexandre Wajnberg, Apr 25 2005
Gives the values of m such that binomial(4*m + 4,m) is odd. Cf. A002450, A020988 and A263132. - Peter Bala, Oct 11 2015
Also the partial sums of 4^n for n>0, cf. A000302. - Robert G. Wilson v, Sep 18 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..170
Peter Bala, A characterization of A002450, A020988 and A080674.
Mattia Fregola, Cellular Automata RULE13 generating OEIS sequence A080674
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Row n = 4 of A228275.

Cf. A000301, A000302, A002450, A020988, A024036, A080674, A263132.

[(4/3)*(4^n-1): n in [0..40] ]; // Vincenzo Librandi, Apr 28 2011

Table[4*(4^n-1)/3,{n,0,100}]  (* Vladimir Joseph Stephan Orlovsky, Jan 30 2012 *)
LinearRecurrence[{5,-4},{0,4},40] (* Harvey P. Dale, May 05 2018 *)

vector(100, n, n--; (4/3)*(4^n-1)) \\ Altug Alkan, Oct 11 2015

Vec(4*x/((x-1)*(4*x-1)) + O(x^40)) \\ Colin Barker, Oct 12 2015

a(n) = 2*A020988(n) = A002450(n+1) - 1 = 4*A002450(n).
a(n) = Sum_{i = 1..n} 4^i. - Adam McDougall (mcdougal(AT)stolaf.edu), Sep 29 2004
a(n) = 4*a(n-1) + 4. - Alexandre Wajnberg, Apr 25 2005
a(n) = 4^n + a(n-1) (with a(0) = 0). - Vincenzo Librandi, Aug 08 2010
From Colin Barker, Oct 12 2015: (Start)
a(n) = 5*a(n-1) - 4*a(n-2).
G.f.: 4*x / ((x-1)*(4*x-1)). (End)
E.g.f.: 4*exp(x)*(exp(3*x) - 1)/3. - Elmo R. Oliveira, Dec 17 2023

A359194 Binary complement of 3*n.

Original entry on oeis.org

1, 0, 1, 6, 3, 0, 13, 10, 7, 4, 1, 30, 27, 24, 21, 18, 15, 12, 9, 6, 3, 0, 61, 58, 55, 52, 49, 46, 43, 40, 37, 34, 31, 28, 25, 22, 19, 16, 13, 10, 7, 4, 1, 126, 123, 120, 117, 114, 111, 108, 105, 102, 99, 96, 93, 90, 87, 84, 81, 78, 75, 72, 69, 66, 63, 60, 57

Offset: 0

Joshua Searle, Dec 19 2022

The binary complement takes the binary value of a number and turns any 1s to 0s and vice versa. This is equivalent to subtracting from the next larger Mersenne number.
It is currently unknown whether every starting positive integer, upon iteration, reaches 0.
From M. F. Hasler, Dec 26 2022: (Start)
This map enjoys the following properties:
(P1) a(2*n) = a(n)*2 + 1 (since 3*(2*n) is 3*n shifted one binary digit to the left, and the one's complement yields that of 3*n with a '1' appended).
(P2) As an immediate consequence of (P1), all even-indexed values are odd.
(P3) Also from (P1), by immediate induction we have a(2^n) = 2^n-1 for all n >= 0.
(P4) Also from (P1), a(4*n) = a(n)*4 + 3.
(P5) Similarly, a(4*n+1) = a(n)*4 (because the 1's complement of 3 is 0).
(P6) From (P5), a(n) = 0 for all n in A002450 (= (4^k-1)/3). [For the initial value at n = 0 the discrepancy is explained by the fact that the number 0 should be considered to have zero digits, but here the result is computed with 0 considered to have one binary digit.] (End)

			a(7) = 10 because 3*7 = 21 = 10101_2, whose binary complement is 01010_2 = 10.
a(42) = 1 because 3*42 = 126 = 1111110_2, whose binary complement is 0000001_2 = 1.
a(52) = 99 by
  3*n        = binary 10011100
  complement = binary 01100011 = 99.

Michael S. Branicky, Table of n, a(n) for n = 0..10000
Joshua Searle, Collatz-inspired sequences.

Trisection of A035327.
Cf. A002450, A020988 (indices of 1's).
Cf. A256078.

a(n)=if(n, bitneg(3*n, exponent(3*n)+1), 1) \\ Rémy Sigrist, Dec 22 2022

def a(n): return 1 if n == 0 else (m:=3*n)^((1 << m.bit_length()) - 1)
print([a(n) for n in range(67)]) # Michael S. Branicky, Dec 20 2022

a(n) = A035327(3*n).

a(n) = 0 iff n belongs to A002450 \ {0}. - Rémy Sigrist, Dec 22 2022

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)).

A039301 Number of distinct quadratic residues mod 4^n.

Original entry on oeis.org

1, 2, 4, 12, 44, 172, 684, 2732, 10924, 43692, 174764, 699052, 2796204, 11184812, 44739244, 178956972, 715827884, 2863311532, 11453246124, 45812984492, 183251937964, 733007751852, 2932031007404, 11728124029612, 46912496118444, 187649984473772, 750599937895084

Offset: 0

David W. Wilson

Number of distinct n-digit suffixes of base 4 squares.

Vincenzo Librandi, Table of n, a(n) for n = 0..170
Index entries for linear recurrences with constant coefficients, signature (5,-4).

[Floor((4^n+10)/6): n in [0..40] ]; // Vincenzo Librandi, Apr 28 2011

Floor[(4^Range[0, 30] + 10)/6] (* Paolo Xausa, Jan 29 2025 *)

a(n) = floor((4^n+10)/6).
a(n) = A007583(n-1)+1 = A020988(n-2)+2 = A083584(n-2)+3. - Ralf Stephan, Jun 14 2003
Also, a(0)=1 and, for n>0, a(n) = (4^n+8)/6. - Bruno Berselli, Jul 27 2010
G.f.: (1-3*x-2*x^2)/((1-x)*(1-4*x)). - Bruno Berselli, Jul 27 2010
a(n)-5*a(n-1)+4*a(n-2) = 0 for n>1. - Bruno Berselli, Jul 27 2010

A083584 a(n) = (8*4^n - 5)/3.

Original entry on oeis.org

1, 9, 41, 169, 681, 2729, 10921, 43689, 174761, 699049, 2796201, 11184809, 44739241, 178956969, 715827881, 2863311529, 11453246121, 45812984489, 183251937961, 733007751849, 2932031007401, 11728124029609, 46912496118441

Offset: 0

Paul Barry, May 01 2003

a(n) = A007583(n+1) - 2 = A020988(n) - 1 = A039301(n+2) - 3. - Ralf Stephan, Jun 14 2003

Sum of n-th row of triangle of powers of 4: 1; 4 1 4; 16 4 1 4 16; 64 16 4 1 4 16 64; .... - Philippe Deléham, Feb 24 2014

			a(0) = 1;
a(1) = 4 + 1 + 4 = 9;
a(2) = 16 + 4 + 1 + 4 + 16 = 41;
a(3) = 64 + 16 + 4 + 1 + 4 + 16 + 64 = 169; etc. - _Philippe Deléham_, Feb 24 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..170
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A083855.

[(8*4^n-5)/3: n in [0..40] ]; // Vincenzo Librandi, Apr 28 2011

(8 4^Range[0,30]-5)/3 (* or *) LinearRecurrence[{5,-4},{1,9},30] (* Harvey P. Dale, Oct 23 2011 *)

a(n)=(8*4^n-5)/3 \\ Charles R Greathouse IV, Oct 07 2015

print([8*4**n//3 - 1 for n in range(50)]) # Karl V. Keller, Jr., May 21 2022

a(n) = (8*4^n - 5)/3.
G.f.: (1+4*x)/((1-x)*(1-4*x)).
E.g.f.: (8*exp(4*x) - exp(x))/3.
a(0)=1, a(1)=9, a(n) = 5*a(n-1) - 4*a(n-2). - Harvey P. Dale, Oct 23 2011
a(n) = 4*a(n-1) + 5, a(0) = 1. - Philippe Deléham, Feb 24 2014
a(n+1) = 2^(2^n+1) + a(n), a(1)=1. - Ben Paul Thurston, Dec 27 2015

A087230 a(n) is the 2-adic valuation of 6*n + 4.

Original entry on oeis.org

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

Offset: 0

Labos Elemer, Aug 28 2003

In Collatz-algorithm if initiated with m=odd value, the first 3x+1 step is followed by a(n) step corresponding to division by 2. Compare to A085058 and A087229. Each 2nd term is either =1 or equals corresponding term of A087229, depending on whether the odd number congruent to 1 or 3 modulo 4.
From K. G. Stier, Aug 19 2014: (Start)
Sequence exhibits a "pseudo" ruler function (A001511) behavior. It is similar to the latter in repeating equal terms m>0 after each 2^m steps. However, the first occurrence of m in the mentioned ruler function is simply at n=log_2(m), while in the given sequence this property develops two distinct (odd and even) strands:
First occurrence of
m=1 at a(1); m=2 at a(0)
m=3 at a(6); m=4 at a(2)
m=5 at a(26); m=6 at a(10)
m=7 at a(106); m=8 at a(42)
m=9 at a(426); m=10 at a(170)
...
where values n in the odd strand (1,6,26,106,426,...) equal sequence A020989, and values n in the even strand (0,2,10,42,170,...) equal sequence A020988. (End)

			n=85: m = 6*85+4 = 514 and Collatz-iteration goes on by one dividing step, a(85)=1.

Kenny Lau, Table of n, a(n) for n = 0..10000

Cf. A007814, A016957, A085058, A087229, A020988, A020989, A061547.

a:= n-> padic[ordp](6*n+4, 2):
seq(a(n), n=0..120);  # Alois P. Heinz, Mar 16 2021

Table[Part[Part[FactorInteger[6*w+4], 1], 2], {w, 0, 100}]
Table[IntegerExponent[6*n + 4, 2], {n, 0, 100}] (* Amiram Eldar, Jan 27 2022 *)

forstep(n=0, 1000, 1, m=6*n+4; print1(valuation(m, 2), ", ") ) \\ K. G. Stier, Aug 19 2014

sub a {
  my $nv= ((shift() << 1) | 1);
  my $bp= 1;
  while (($nv & 1) xor ($nv & 2)) {
    $nv>>= 1;
    $bp++;
  }
  return $bp;
} # Ruud H.G. van Tol, Nov 16 2021

n=100; N=3*n+2; val=[1]*(N+1); exp=2
while exp <= N:
    for j in range(exp,N+1,exp): val[j] += 1
    exp *= 2
for i in range(n+1): print(i,val[3*i+2])
# Kenny Lau, Jun 09 2018

def A087230(n): return (~(m:=6*n+4) & m-1).bit_length() # Chai Wah Wu, Jul 02 2022

a(n) = A007814(A016957(n)). - Michel Marcus, Jan 27 2022

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 2. - Amiram Eldar, Sep 10 2024

a(0) = 2 prepended by Andrey Zabolotskiy, Jan 27 2022, based on Ihar Senkevich's contribution

A263132 Positive values of m such that binomial(4*m - 1, m) is odd.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 11, 12, 16, 22, 24, 32, 43, 44, 48, 64, 86, 88, 96, 128, 171, 172, 176, 192, 256, 342, 344, 352, 384, 512, 683, 684, 688, 704, 768, 1024, 1366, 1368, 1376, 1408, 1536, 2048, 2731, 2732, 2736, 2752, 2816, 3072, 4096, 5462, 5464, 5472, 5504

Offset: 1

Peter Bala, Oct 10 2015

This sequence, when viewed as a set, equals the set of numbers of the form 4^n * ceiling(2^k/3) for n >= 0, k >= 1, i.e., the product subset in Z of A000302 and A005578 regarded as sets. See the example below.
Equivalently, this sequence, when viewed as a set, equals the set of numbers of the form 2^n * (2^(2*k + 1) + 1)/3 for n,k >= 0, i.e., the product subset in Z of A000079 and A007583 regarded as sets. See the example below.
2*a(n) gives the values of m such that binomial(4*m - 2,m) is odd. 4*a(n) gives the values of m such that binomial(4*m - 3,m) is odd (other than m = 1) and also the values of m such that binomial(4*m - 4,m) is odd.

			1) Notice how this sequence can be read from Table 1 below by moving through the table in a sequence of 'knight moves' (1 down and 2 to the left) starting from the first row. For example, starting at 11 on the top row we move in a series of knights moves 11 -> 12 -> 16, then return to the top row at 22 and move 22 -> 24 -> 32, return to the top row at 43 and move 43 -> 44 -> 48 -> 64, then return to top row at 86 and so on.
........................................................
.   Table 1: 4^n * ceiling(2^k/3) for n >= 0, k >= 1   .
........................................................
n\k|   1    2    3    4     5     6    7    8     9
---+----------------------------------------------------
0  |   1    2    3    6    11    22   43   86   171 ...
1  |   4    8   12   24    44    88  172  ...
2  |  16   32   48   96   176    ...
3  |  64  128  192  ...
4  | 256  ...
...
2) Notice how this sequence can be read from Table 2 below in a sequence of 'knight moves' (2 down and 1 to the left) starting from the first two rows. For example, starting at 43 in the first row we jump 43 -> 44 -> 48 -> 64, then return to the second row at 86 and jump 86 -> 88 -> 96 -> 128, followed by 171 -> 172 -> 176 -> 192 -> 256, and so on.
....................................................
.   Table 2: 2^n * (2^(2*k + 1) + 1)/3, n,k >= 0   .
....................................................
n\k|   0    1     2     3      4      5
---+----------------------------------------------
0  |   1    3    11    43    171    683  ...
1  |   2    6    22    86    342   1366  ...
2  |   4   12    44   172    684   2732  ...
3  |   8   24    88   344   1368   5464  ...
4  |  16   48   176   688   2736  10928  ...
5  |  32   96   352  1376   5472  21856  ...
6  |  64  192   704  2752  10944  43712  ...
7  | 128  384  1408  5504  21888  87424  ...
8  | 256 ...

Cf. A000079, A000267, A000302, A005578, A007583, A048716, A384688.

Other odd binomials: A002450 (4*m+1,m), A020988 (4*m+2,m), A263133 (4*m+3,m), A080674 (4*m+4,m), A118113 (3*m-2,m), A003714 (3*m,m).

[n: n in [1..6000] | Binomial(4*n-1, n) mod 2 eq 1]; // Vincenzo Librandi, Oct 12 2015

for n from 1 to 5000 do if mod(binomial(4*n-1, n), 2) = 1 then print(n) end if end do;

Select[Range[6000],OddQ[Binomial[4#-1,#]]&] (* Harvey P. Dale, Dec 26 2015 *)

for(n=1, 1e4, if (binomial(4*n-1, n) % 2 == 1, print1(n", "))) \\ Altug Alkan, Oct 11 2015

a(n) = my(r,s=sqrtint(4*n-3,&r)); (1<Kevin Ryde, Jun 14 2025

Python

A263132_list = [m for m in range(1,10**6) if not ~(4*m-1) & m] # Chai Wah Wu, Feb 07 2016

Formula

a(n) = A263133(n) + 1.

m is a term if and only if m AND NOT (4*m-1) = 0 where AND and NOT are bitwise operators. - Chai Wah Wu, Feb 07 2016

a(n) = (2^A000267(n-1) + 2^A384688(n-1)) / 3. - Kevin Ryde, Jun 14 2025

Extensions

More terms from Vincenzo Librandi, Oct 12 2015

cp's OEIS Frontend

A258996 Permutation of the positive integers: this permutation transforms the enumeration system of positive irreducible fractions A002487/A002487' (Calkin-Wilf) into the enumeration system A162911/A162912 (Drib), and vice versa.

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A056830 Alternate digits 1 and 0.

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A080674 a(n) = (4/3)*(4^n - 1).

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A359194 Binary complement of 3*n.

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

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A265747 Numbers written in Jacobsthal greedy base.

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