A059893 -id:A059893 - cp's OEIS frontend

A005940 The Doudna sequence: write n-1 in binary; power of prime(k) in a(n) is # of 1's that are followed by k-1 0's.

1, 2, 3, 4, 5, 6, 9, 8, 7, 10, 15, 12, 25, 18, 27, 16, 11, 14, 21, 20, 35, 30, 45, 24, 49, 50, 75, 36, 125, 54, 81, 32, 13, 22, 33, 28, 55, 42, 63, 40, 77, 70, 105, 60, 175, 90, 135, 48, 121, 98, 147, 100, 245, 150, 225, 72, 343, 250, 375, 108, 625, 162, 243, 64, 17, 26, 39

Offset: 1

N. J. A. Sloane and J. H. Conway

A permutation of the natural numbers. - Robert G. Wilson v, Feb 22 2005
Fixed points: A029747. - Reinhard Zumkeller, Aug 23 2006
The even bisection, when halved, gives the sequence back. - Antti Karttunen, Jun 28 2014
From Antti Karttunen, Dec 21 2014: (Start)
This irregular table can be represented as a binary tree. Each child to the left is obtained by applying A003961 to the parent, and each child to the right is obtained by doubling the parent:
                                      1
                                      |
                   ...................2...................
                  3                                       4
        5......../ \........6                   9......../ \........8
       / \                 / \                 / \                 / \
      /   \               /   \               /   \               /   \
     /     \             /     \             /     \             /     \
    7       10         15       12         25       18         27       16
  11 14   21  20     35  30   45  24     49  50   75  36    125  54   81  32
etc.
Sequence A163511 is obtained by scanning the same tree level by level, from right to left. Also in binary trees A253563 and A253565 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) gives the distance of n from 1 in all these trees.
A252737(n) gives the sum and A252738(n) the product of terms on row n (where 1 is on row 0, 2 on row 1, 3 and 4 on row 2, etc.). A252745(n) gives the number of nodes on level n whose left child is larger than the right child, A252750 the difference between left and right child for each node from node 2 onward.
(End)
-A008836(a(1+n)) gives the corresponding numerator for A323505(n). - Antti Karttunen, Jan 19 2019
(a(2n+1)-1)/2 [= A244154(n)-1, for n >= 0] is a permutation of the natural numbers. - George Beck and Antti Karttunen, Dec 08 2019
From Peter Munn, Oct 04 2020: (Start)
Each term has the same even part (equivalently, the same 2-adic valuation) as its index.
Using the tree depicted in Antti Karttunen's 2014 comment:
Numbers are on the right branch (4 and descendants) if and only if divisible by the square of their largest prime factor (cf. A070003).
Numbers on the left branch, together with 2, are listed in A102750.
(End)
According to Kutz (1981), he learned of this sequence from American mathematician Byron Leon McAllister (1929-2017) who attributed the invention of the sequence to a graduate student by the name of Doudna (first name Paul?) in the mid-1950's at the University of Wisconsin. - Amiram Eldar, Jun 17 2021
From David James Sycamore, Sep 23 2022: (Start)
Alternative (recursive) definition: If n is a power of 2 then a(n)=n. Otherwise, if 2^j is the greatest power of 2 not exceeding n, and if k = n - 2^j, then a(n) is the least m*a(k) that has not occurred previously, where m is an odd prime.
Example: Use recursion with n = 77 = 2^6 + 13. a(13) = 25 and since 11 is the smallest odd prime m such that m*a(13) has not already occurred (see a(27), a(29),a(45)), then a(77) = 11*25 = 275. (End)
The odd bisection, when transformed by replacing all prime(k)^e in a(2*n - 1) with prime(k-1)^e, returns a(n), and thus gives the sequence back. - David James Sycamore, Sep 28 2022

			From _N. J. A. Sloane_, Aug 22 2022: (Start)
Let c_i = number of 1's in binary expansion of n-1 that have i 0's to their right, and let p(j) = j-th prime.  Then a(n) = Product_i p(i+1)^c_i.
If n=9, n-1 is 1000, c_3 = 1, a(9) = p(4)^1 = 7.
If n=10, n-1 = 1001, c_0 = 1, c_2 = 1, a(10) = p(1)*p(3) = 2*5 = 10.
If n=11, n-1 = 1010, c_1 = 1, c_2 = 1, a(11) = p(2)*p(3) = 15. (End)

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Antti Karttunen, Table of n, a(n) for n = 1..8192 (terms 1..1024 from Reinhard Zumkeller)
Michael De Vlieger, 6 row Doudna Tree diagram as mentioned in Comments.
Michael De Vlieger, Annotated fan-style binary tree showing 10 levels, with a color function where 2^m is shown in medium blue in row m, k < 2^m is darker blue, and k > 2^m is brighter green, with records in each row shown in red.
Ronald E. Kutz, Two unusual sequences, Two-Year College Mathematics Journal, Vol. 12, No. 5 (1981), pp. 316-319.
Index entries for sequences that are permutations of the natural numbers

Cf. A103969. Inverse is A005941 (A156552).
Cf. A125106. [From Franklin T. Adams-Watters, Mar 06 2010]
Cf. A252737 (gives row sums), A252738 (row products), A332979 (largest on row).
Related permutations of positive integers: A163511 (via A054429), A243353 (via A006068), A244154, A253563 (via A122111), A253565, A332977, A334866 (via A225546).
A000120, A003602, A003961, A006519, A053645, A070939, A246278, A250246, A252753, A253552 are used in a formula defining this sequence.
Formulas for f(a(n)) are given for f = A000265, A003963, A007949, A055396, A056239.
Numbers that occur at notable sets of positions in the binary tree representation of the sequence: A000040, A000079, A002110, A070003, A070826, A102750.
Cf. also A000142, A001511, A002450, A112798, A252463, A252464, A252745, A252750, A324054, A324106, A323505, A323508.
Cf. A106737, A290077, A323915, A324052, A324054, A324055, A324056, A324057, A324058, A324114, A324335, A324340, A324348, A324349 for various number-theoretical sequences applied to (i.e., permuted by) this sequence.
k-adic valuation: A007814 (k=2), A337821 (k=3).
Positions of multiples of 3: A091067.
Primorial deflation: A337376 / A337377.
Sum of prime indices of a(n) is A161511, reverse version A359043.
A048793 lists binary indices, ranked by A019565.
A066099 lists standard comps, partial sums A358134 (ranked by A358170).
Cf. A001222, A029837, A059893, A241916, A242628, A253566, A359042.

a005940 n = f (n - 1) 1 1 where
   f 0 y _          = y
   f x y i | m == 0 = f x' y (i + 1)
           | m == 1 = f x' (y * a000040 i) i
           where (x',m) = divMod x 2
-- Reinhard Zumkeller, Oct 03 2012
(Scheme, with memoization-macro definec from Antti Karttunen's IntSeq-library)
(define (A005940 n) (A005940off0 (- n 1))) ;; The off=1 version, utilizing any one of three different offset-0 implementations:
(definec (A005940off0 n) (cond ((< n 2) (+ 1 n)) (else (* (A000040 (- (A070939 n) (- (A000120 n) 1))) (A005940off0 (A053645 n))))))
(definec (A005940off0 n) (cond ((<= n 2) (+ 1 n)) ((even? n) (A003961 (A005940off0 (/ n 2)))) (else (* 2 (A005940off0 (/ (- n 1) 2))))))
(define (A005940off0 n) (let loop ((n n) (i 1) (x 1)) (cond ((zero? n) x) ((even? n) (loop (/ n 2) (+ i 1) x)) (else (loop (/ (- n 1) 2) i (* x (A000040 i)))))))
;; Antti Karttunen, Jun 26 2014

f := proc(n,i,x) option remember ; if n = 0 then x; elif type(n,'even') then procname(n/2,i+1,x) ; else procname((n-1)/2,i,x*ithprime(i)) ; end if; end proc:
A005940 := proc(n) f(n-1,1,1) ; end proc: # R. J. Mathar, Mar 06 2010

f[n_] := Block[{p = Partition[ Split[ Join[ IntegerDigits[n - 1, 2], {2}]], 2]}, Times @@ Flatten[ Table[q = Take[p, -i]; Prime[ Count[ Flatten[q], 0] + 1]^q[[1, 1]], {i, Length[p]}] ]]; Table[ f[n], {n, 67}] (* Robert G. Wilson v, Feb 22 2005 *)
Table[Times@@Prime/@(Join@@Position[Reverse[IntegerDigits[n,2]],1]-Range[DigitCount[n,2,1]]+1),{n,0,100}] (* Gus Wiseman, Dec 28 2022 *)

A005940(n) = { my(p=2, t=1); n--; until(!n\=2, n%2 && (t*=p) || p=nextprime(p+1)); t } \\ M. F. Hasler, Mar 07 2010; update Aug 29 2014

a(n)=my(p=2, t=1); for(i=0,exponent(n), if(bittest(n,i), t*=p, p=nextprime(p+1))); t \\ Charles R Greathouse IV, Nov 11 2021

from sympy import prime
import math
def A(n): return n - 2**int(math.floor(math.log(n, 2)))
def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))
print([b(n - 1) for n in range(1, 101)]) # Indranil Ghosh, Apr 10 2017

from math import prod
from itertools import accumulate
from collections import Counter
from sympy import prime
def A005940(n): return prod(prime(len(a)+1)**b for a, b in Counter(accumulate(bin(n-1)[2:].split('1')[:0:-1])).items()) # Chai Wah Wu, Mar 10 2023

From Reinhard Zumkeller, Aug 23 2006, R. J. Mathar, Mar 06 2010: (Start)
a(n) = f(n-1, 1, 1)
  where f(n, i, x) = x if n = 0,
                   = f(n/2, i+1, x) if n > 0 is even
                   = f((n-1)/2, i, x*prime(i)) otherwise. (End)
From Antti Karttunen, Jun 26 2014: (Start)
Define a starting-offset 0 version of this sequence as:
  b(0)=1, b(1)=2, [base cases]
and then compute the rest either with recurrence:
  b(n) = A000040(1+(A070939(n)-A000120(n))) * b(A053645(n)).
or
  b(2n) = A003961(b(n)), b(2n+1) = 2 * b(n). [Compare this to the similar recurrence given for A163511.]
Then define a(n) = b(n-1), where a(n) gives this sequence A005940 with the starting offset 1.
Can be also defined as a composition of related permutations:
a(n+1) = A243353(A006068(n)).
a(n+1) = A163511(A054429(n)). [Compare the scatter plots of this sequence and A163511 to each other.]
This permutation also maps between the partitions as enumerated in the lists A125106 and A112798, providing identities between:
A161511(n) = A056239(a(n+1)). [The corresponding sums ...]
A243499(n) = A003963(a(n+1)). [... and the products of parts of those partitions.]
(End)
From Antti Karttunen, Dec 21 2014  - Jan 04 2015: (Start)
A002110(n) =  a(1+A002450(n)). [Primorials occur at (4^n - 1)/3 in the offset-0 version of the sequence.]
a(n) = A250246(A252753(n-1)).
a(n) = A122111(A253563(n-1)).
For n >= 1, A055396(a(n+1)) = A001511(n).
For n >= 2, a(n) = A246278(1+A253552(n)).
(End)
From Peter Munn, Oct 04 2020: (Start)
A000265(a(n)) = a(A000265(n)) = A003961(a(A003602(n))).
A006519(a(n)) = a(A006519(n)) = A006519(n).
a(n) = A003961(a(A003602(n))) * A006519(n).
A007814(a(n)) = A007814(n).
A007949(a(n)) = A337821(n) = A007814(A003602(n)).
a(n) = A225546(A334866(n-1)).
(End)
a(2n) = 2*a(n), or generally a(2^k*n) = 2^k*a(n). - Amiram Eldar, Oct 03 2022
If n-1 = Sum_{i} 2^(q_i-1), then a(n) = Product_{i} prime(q_i-i+1). These are the Heinz numbers of the rows of A125106. If the offset is changed to 0, the inverse is A156552. - Gus Wiseman, Dec 28 2022

More terms from Robert G. Wilson v, Feb 22 2005
Sign in a formula switched and Maple program added by R. J. Mathar, Mar 06 2010
Binary tree illustration and keyword tabf added by Antti Karttunen, Dec 21 2014

A054429 Simple self-inverse permutation of natural numbers: List each block of 2^n numbers (from 2^n to 2^(n+1) - 1) in reverse order.

Original entry on oeis.org

1, 3, 2, 7, 6, 5, 4, 15, 14, 13, 12, 11, 10, 9, 8, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 127, 126, 125, 124, 123, 122, 121

Offset: 1

Antti Karttunen

a(n) gives the position of the inverse of the n-th term in the full Stern-Brocot tree: A007305(a(n)+2) = A047679(n) and A047679(a(n)) = A007305(n+2). - Reinhard Zumkeller, Dec 22 2008
From Gary W. Adamson, Jun 21 2012: (Start)
The mapping and conversion rules are as follows:
By rows, we have ...
   1;
   3,  2;
   7,  6,  5,  4;
  15, 14, 13, 12, 11, 10, 9, 8;
  ... onto which we are to map one-half of the Stern-Brocot infinite Farey Tree:
  1/2
  1/3, 2/3
  1/4, 2/5, 3/5, 3/4
  1/5, 2/7, 3/8, 3/7, 4/7, 5/8, 5/7, 4/5
  ...
The conversion rules are: Convert the decimal to binary, adding a duplicate of the rightmost binary term to its right. For example, 10 = 1010, which becomes 10100. Then, from the left, record the number of runs = [1,1,1,2], the continued fraction representation of 5/8. Check: 10 decimal corresponds to 5/8 as shown in the overlaid mapping. Take decimal 9 = 1001 which becomes 10011, with a continued fraction representation of [1,2,2] = 5/7. Check: 9 decimal corresponds to 5/7 in the Farey Tree map. (End)
From Indranil Ghosh, Jan 19 2017: (Start)
a(n) is the value generated when n is converted into its Elias gamma code, the 1's and 0's are interchanged and the resultant is converted back to its decimal value for all values of n > 1. For n = 1, A054429(n) = 1 but after converting 1 to Elias gamma code, interchanging the 1's and 0's and converting it back to decimal, the result produced is 0.
For example, let n = 10. The Elias gamma code for 10 is '1110010'. After interchanging the 1's and 0's it becomes "0001101" and 1101_2 = 13_10. So a(10) = 13. (End)
From Yosu Yurramendi, Mar 09 2017 (similar to Zumkeller's comment): (Start)
A002487(a(n)) = A002487(n+1), A002487(a(n)+1) = A002487(n), n > 0.
A162909(a(n)) = A162910(n), A162910(a(n)) = A162909(n), n > 0.
A162911(a(n)) = A162912(n), A162912(a(n)) = A162911(n), n > 0.
A071766(a(n)) = A245326(n), A245326(a(n)) = A071766(n), n > 0.
A229742(a(n)) = A245325(n), A245325(a(n)) = A229742(n), n > 0.
A020651(a(n)) = A245327(n), A245327(a(n)) = A020651(n), n > 0.
A020650(a(n)) = A245328(n), A245328(a(n)) = A020650(n), n > 0. (End)
From Yosu Yurramendi, Mar 29 2017: (Start)
A063946(a(n)) = a(A063946(n)) = A117120(n), n > 0.
A065190(a(n)) = a(A065190(n)) = A092569(n), n > 0.
A258746(a(n)) = a(A258746(n)) = A165199(n), n > 0.
A258996(a(n)) = a(A258996(n)), n > 0.
A117120(a(n)) = a(A117120(n)), n > 0.
A092569(a(n)) = a(A092569(n)), n > 0. (End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Index entries for sequences that are permutations of the natural numbers
Index entries for sequences related to binary expansion of n

See also A054424, A054430.
{A000027, A054429, A059893, A059894} form a 4-group.
Cf. A053644, A115303, A115304, A115305, A115306, A115307, A115308, A115309, A106649.
This is Guy Steele's sequence GS(6, 5) (see A135416).

a054429 n = a054429_list !! (n-1)
a054429_list = f [1..] where
   f xs@(x:_) = reverse us ++ f vs where (us, vs) = splitAt x xs
-- Reinhard Zumkeller, Jun 01 2015, Feb 21 2014

A054429 := n -> 3*2^ilog2(n) - n - 1:
seq(A054429(n), n = 1..70); # [Updated by Peter Luschny, Apr 24 2024]

Flatten[Table[Range[2^(n+1)-1,2^n,-1],{n,0,6}]] (* Harvey P. Dale, Dec 17 2013 *)

A054429(n)= 3<<#binary(n\2)-n-1 \\ M. F. Hasler, Aug 18 2014

from itertools import count, islice
def A054429_gen(): # generator of terms
    return (m for n in count(0) for m in range((1<A054429_list = list(islice(A054429_gen(),30)) # Chai Wah Wu, Jul 27 2023

maxblock <- 10 # by choice
a <- NULL
for(m in 0:maxblock) a <- c(a, rev(2^m:(2^(m+1)-1)))
a
# Yosu Yurramendi, Mar 10 2017

a(n) = ReflectBinTreePermutation(n).
a(n) = if n=1 then 1 else 2*a(floor(n/2)) + 1 - n mod 2. - Reinhard Zumkeller, Feb 18 2003
G.f.: 1/(1-x) * ((x-2x^2)/(1-x) + Sum_{k>=0} 3*2^k*x^2^k). - Ralf Stephan, Sep 15 2003
A000120(a(n)) = A000120(A059894(n)) = A023416(n) + 1. - Ralf Stephan, Oct 05 2003
A115310(n, 1) = a(n). - Reinhard Zumkeller, Jan 20 2006
a(1) = 1, a(2^(m+1) + k) = a(2^m+k) + 2^(m+1),
a(2^(m+1) + 2^m+k) = a(2^m+k) + 2^m, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Apr 06 2017
a(n) = A117120(A063946(n)) = A063946(A117120(n)) = A092569(A065190(n)) = A065190(A092569(n)), n > 0. - Yosu Yurramendi, Apr 10 2017
a(n) = 3*A053644(n) - n - 1. - Alan Michael Gómez Calderón, Feb 28 2025

A333381 Number of maximal anti-runs of the n-th composition in standard order.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Mar 24 2020

nonn

Anti-runs are sequences without any adjacent equal terms.
A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again.
For n > 0, also one plus the number of adjacent equal pairs in the n-th composition in standard order.

			The 46th composition in standard order is (2,1,1,2), with maximal anti-runs ((2,1),(1,2)), so a(46) = 2.

Wikipedia, Longest increasing subsequence

Anti-runs summing to n are counted by A003242(n).
A triangle counting maximal anti-runs of compositions is A106356.
A triangle counting maximal runs of compositions is A238279.
Partitions whose first differences are an anti-run are A238424.
All of the following pertain to compositions in standard order (A066099):
- Adjacent equal pairs are counted by A124762.
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767.
- Strictly increasing runs are counted by A124768.
- Strictly decreasing runs are counted by A124769.
- Strict compositions are ranked by A233564.
- Constant compositions are ranked by A272919.
- Normal compositions are ranked by A333217.
- Adjacent unequal pairs are counted by A333382.
- Anti-runs are ranked by A333489.
Cf. A000120, A029931, A048793, A059893, A070939, A114994, A225620, A228351.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Length[Split[stc[n],UnsameQ]],{n,0,100}]

For n > 0, a(n) = A124762(n) + 1.

A272020 Irregular triangle read by rows: strictly decreasing sequences of positive numbers given in lexicographic order.

Original entry on oeis.org

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

Offset: 0

Peter Kagey, Apr 17 2016

Length of n-th row given by A000120(n);
Min of n-th row given by A001511(n);
Sum of n-th row given by A029931(n);
Product of n-th row given by A096111(n);
Max of n-th row given by A113473(n);
Numerator of sum of reciprocals of n-th row given by A116416(n);
Denominator of sum of reciprocals of n-th row given by A116417(n);
LCM of n-th row given by A271410(n).
The first appearance of n is at A001787(n - 1).
n-th row begins at index A000788(n - 1) for n > 0.
Also the reversed positions of 1's in the reversed binary expansion of n. Also the reversed partial sums of the n-th composition in standard order (row n of A066099). Reversing rows gives A048793. - Gus Wiseman, Jan 17 2023

			Row n is given by the exponents in the binary expansion of 2*n. For example, row 5 = [3, 1] because 2*5 = 2^3 + 2^1.
Row 0: []
Row 1: [1]
Row 2: [2]
Row 3: [2, 1]
Row 4: [3]
Row 5: [3, 1]
Row 6: [3, 2]
Row 7: [3, 2, 1]

Peter Kagey, Table of n, a(n) for n = 0..10000

Cf. A000120, A096111, A116416, A116417, A271410, A001787.
Cf. A048793 gives the rows in reverse order.
Cf. A272011.
Lasts are A001511.
Heinz numbers of the rows are A019565.
Firsts are A029837 or A070939 or A113473.
Row sums are A029931.
A066099 lists standard comps, partial sums A358134, weighted sum A359042.
Cf. A005940, A059893, A125106, A161511, A242628.

T:= proc(n) local i, l, m; l:= NULL; m:= n;
      if n=0 then return [][] fi; for i while m>0 do
      if irem(m, 2, 'm')=1 then l:=i, l fi od; l
    end:
seq(T(n), n=0..35);  # Alois P. Heinz, Nov 27 2024

Table[Reverse[Join@@Position[Reverse[IntegerDigits[n,2]],1]],{n,0,100}] (* Gus Wiseman, Jan 17 2023 *)

A007306 Denominators of Farey tree fractions (i.e., the Stern-Brocot subtree in the range [0,1]).

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 5, 5, 4, 5, 7, 8, 7, 7, 8, 7, 5, 6, 9, 11, 10, 11, 13, 12, 9, 9, 12, 13, 11, 10, 11, 9, 6, 7, 11, 14, 13, 15, 18, 17, 13, 14, 19, 21, 18, 17, 19, 16, 11, 11, 16, 19, 17, 18, 21, 19, 14, 13, 17, 18, 15, 13, 14, 11, 7, 8, 13, 17, 16, 19, 23, 22, 17, 19, 26, 29, 25, 24

Offset: 0

N. J. A. Sloane

Also number of odd entries in n-th row of triangle of Stirling numbers of the second kind (A008277). - Benoit Cloitre, Feb 28 2004
Apparently (except for the first term) the number of odd entries in the alternated diagonals of Pascal's triangle at 45 degrees slope. - Javier Torres (adaycalledzero(AT)hotmail.com), Jul 26 2009
The Kn3 and Kn4 triangle sums, see A180662 for their definitions, of Sierpiński's triangle A047999 equal a(n+1). - Johannes W. Meijer, Jun 05 2011
From Yosu Yurramendi, Jun 23 2014: (Start)
If the terms (n>1) are written as an array:
  2,
  3,  3,
  4,  5,  5,  4,
  5,  7,  8,  7,  7,  8,  7,  5,
  6,  9, 11, 10, 11, 13, 12,  9,  9, 12, 13, 11, 10, 11,  9,  6,
  7, 11, 14, 13, 15, 18, 17, 13, 14, 19, 21, 18, 17, 19, 16, 11, 11, 16, 19,17,18,
then the sum of the k-th row is 2*3^(k-2), each column is an arithmetic progression. The differences of the arithmetic progressions give the sequence itself (a(2^(m+1)+1+k) - a(2^m+1+k) = a(k+1), m >= 1, 1 <= k <= 2^m), because a(n) = A002487(2*n-1) and A002487 has these properties. A071585 also has these properties. Each row is a palindrome: a(2^(m+1)+1-k) = a(2^m+k), m >= 0, 1 <= k <= 2^m.
If the terms (n>0) are written in this way:
  1,
  2, 3,
  3, 4,  5,  5,
  4, 5,  7,  8,  7,  7,  8,  7,
  5, 6,  9, 11, 10, 11, 13, 12,  9,  9, 12, 13, 11, 10, 11,  9,
  6, 7, 11, 14, 13, 15, 18, 17, 13, 14, 19, 21, 18, 17, 19, 16, 11, 11, 16, 19,
each column is an arithmetic progression and the steps also give the sequence itself (a(2^(m+1)+k) - a(2^m+k) = a(k), m >= 0, 0 <= k < 2^m). Moreover, by removing the first term of each column:
a(2^(m+1)+k) = A049448(2^m+k+1), m >= 0, 0 <= k < 2^m.
(End)
k > 1 occurs in this sequence phi(k) = A000010(k) times. - Franklin T. Adams-Watters, May 25 2015
Except for the initial 1, this is the odd bisection of A002487. The even bisection of A002487 is A002487 itself. - Franklin T. Adams-Watters, May 25 2015
For all m >= 0, max_{k=1..2^m} a(2^m+k) = A000045(m+3) (Fibonacci sequence). - Yosu Yurramendi, Jun 05 2016
For all n >= 2, max(m: a(2^m+k) = n, 1<=k<=2^m) = n-2. - Yosu Yurramendi, Jun 05 2016
a(2^m+1) = m+2, m >= 0; a(2^m+2) = 2m+1, m>=1; min_{m>=0, k=1..2^m} a(2^m+k) = m+2; min_{m>=2, k=2..2^m-1} a(2^m+k) = 2m+1. - Yosu Yurramendi, Jun 06 2016
a(2^(m+2) + 2^(m+1) - k) - a(2^(m+1) + 2^m-k) = 2*a(k+1), m >= 0, 0 <= k <= 2^m. - Yosu Yurramendi, Jun 09 2016
If the initial 1 is omitted, this is the number of nonzero entries in row n of the generalized Pascal triangle P_2, see A282714 [Leroy et al., 2017]. - N. J. A. Sloane, Mar 02 2017
Apparently, this sequence was introduced by Johann Gustav Hermes in 1894. His paper gives a strong connection between this sequence and the so-called "Gaussian brackets" ("Gauss'schen Klammer"). For an independent discussion about Gaussian brackets, see the relevant MathWorld article and the article by Herzberger (1943). Srinivasan (1958) gave another, more modern, explanation of the connection between this sequence and the Gaussian brackets. (Parenthetically, J. G. Hermes is the mathematician who completed or constructed the regular polygon with 65537 sides.) - Petros Hadjicostas, Sep 18 2019

			[ 0/1; 1/1; ] 1/2; 1/3, 2/3; 1/4, 2/5, 3/5, 3/4; 1/5, 2/7, 3/8, 3/7, 4/7, 5/8, 5/7, 4/5; ...

P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 61.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 158.
J. C. Lagarias, Number Theory and Dynamical Systems, pp. 35-72 of S. A. Burr, ed., The Unreasonable Effectiveness of Number Theory, Proc. Sympos. Appl. Math., 46 (1992). Amer. Math. Soc.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Alois P. Heinz, Table of n, a(n) for n = 0..10000
Suayb S. Arslan, Asymptotically MDS Array BP-XOR Codes, arXiv:1709.07949 [cs.IT], 2017.
Alexander Bogomolny, Stern-Brocot tree.
J. Hermes, Anzahl der Zerlegungen einer ganzen, rationalen Zahl in Summanden (The number of partitions of a rational integer, Part I), Mathematischen Annalen 45(3) (1894), 371-380.
J. Hermes, Anzahl der Zerlegungen einer ganzen, rationalen Zahl in Summanden (The number of partitions of a rational integer, Part I), Mathematischen Annalen 45(3) (1894), 371-380.
J. Hermes, Anzahl der Zerlegungen einer ganzen, rationalen Zahl in Summanden, II (The number of partitions of a rational integer, Part II), Mathematischen Annalen 47(2-3) (1896), 281-297.
M. Herzberger, Gaussian optics and Gaussian brackets, Journal of the Optical Society of America 33(12) (1943), 651-655. [This paper gives a clear description of Gaussian brackets that are related to this sequence as explained by Hermes (1894).]
Jennifer Lansing, On the Stern sequence and a related sequence, Ph.D. dissertation in Mathematics, University of Illinois at Urbana-Champaign, 2014. [This doctoral dissertation discusses the so-called Stern sequence on which Hermes' papers are based (according to Srinivasan (1958)).]
Julien Leroy, Michel Rigo, and Manon Stipulanti, Counting the number of non-zero coefficients in rows of generalized Pascal triangles, Discrete Mathematics 340 (2017), 862-881.
Julien Leroy, Michel Rigo, and Manon Stipulanti, Counting subword occurrences in base-b expansions, Integers (2018) 18A, Article #A13.
G. Melançon, Lyndon factorization of sturmian words, Discr. Math. 210 (2000), 137-149.
N. J. A. Sloane, Stern-Brocot or Farey Tree.
M. S. Srinivasan, The enumeration of positive rational numbers, Proc. Indian Acad. Sci. Sect. A 47 (1958), 12-24.
M. Stern, Über eine zahlentheoretische Function, Journal für die reine und angewandte Mathematik 55 (1858), 193-220. [According to Srinivasan (1958), Hermes's (1894) paper, where this sequence is introduced, is based on Stern's sequence.]
Manon Stipulanti, Convergence of Pascal-Like Triangles in Parry-Bertrand Numeration Systems, arXiv:1801.03287 [math.CO], 2018.
Javier Torres Suarez, Number theory - geometric connection (part 2) (YouTube video that mentions this sequence - link sent by Pacha Nambi, Aug 26 2009).
Eric Weisstein's World of Mathematics, Gaussian brackets; they are related to this sequence.
Wikipedia, Johann Gustav Hermes. [He is the person who introduced this sequence and the person who completed or constructed a regular polygon with 65537 sides.]
Index entries for fraction trees
Index entries for sequences related to Stern's sequences

For numerators see A007305.
Cf. A001222, A002487, A006842, A006843, A047679, A054424, A065674-A065675, A065810, A260443, A277324, A277328, A283986, A283987, A283988, A284009, A284265, A284266, A284267, A284268, A284565, A284566, A285106, A285107, A285108, A287731, A287732.
See also A282714.

[1] cat [&+[Binomial(n+k,2*k) mod 2: k in [0..n]]: n in [0..80]]; // Vincenzo Librandi, Jun 10 2019

A007306 := proc(n): if n=0 then 1 else A002487(2*n-1) fi: end: A002487 := proc(m) option remember: local a, b, n; a := 1; b := 0; n := m; while n>0 do if type(n, odd) then b := a + b else a := a + b end if; n := floor(n/2); end do; b; end proc: seq(A007306(n),n=0..77); # Johannes W. Meijer, Jun 05 2011

a[0] = 1; a[n_] := Sum[ Mod[ Binomial[n+k-1, 2k] , 2], {k, 0, n}]; Table[a[n], {n, 0, 77}] (* Jean-François Alcover, Dec 16 2011, after Paul Barry *)
a[0] = 0; a[1] = 1;
Flatten[{1,Table[a[2*n] = a[n]; a[2*n + 1] = a[n] + a[n + 1], {n, 0, 50}]}] (* Horst H. Manninger, Jun 09 2021 *)

{a(n) = if( n<1, n==0, n--; sum( k=0, n, binomial( n+k, n-k)%2))};

{a(n) = my(m); if( n<2, n>=0, m = 2^length( binary( n-1)); a(n - m/2) + a(m-n+1))}; /* Michael Somos, May 30 2005 */

from sympy import binomial
def a(n):
    return 1 if n<1 else sum(binomial(n + k - 1, 2*k) % 2 for k in range(n + 1))
print([a(n) for n in range(101)]) # Indranil Ghosh, Mar 22 2017

from functools import reduce
def A007306(n): return sum(reduce(lambda x,y:(x[0],sum(x)) if int(y) else (sum(x),x[1]),bin((n<<1)-1)[-1:2:-1],(1,0))) if n else 1 # Chai Wah Wu, May 18 2023

maxrow <- 6 # by choice
a <- c(1,2)
for(m in 0:maxrow) for(k in 1:2^m){
  a[2^(m+1)+k  ] <- a[2^m+k] + a[k]
  a[2^(m+1)-k+1] <- a[2^m+k]
}
a
# Yosu Yurramendi, Jan 05 2015

# Given n, compute directly a(n)
# by taking into account the binary representation of n-1
# aa <- function(n){
  b <- as.numeric(intToBits(n))
  l <- sum(b)
  m <- which(b == 1)-1
  d <- 1
  if(l > 1) for(j in 1:(l-1)) d[j] <- m[j+1]-m[j]+1
  f <- c(1,m[1]+2) # In A002487: f <- c(0,1)
  if(l > 1) for(j in 3:(l+1)) f[j] <- d[j-2]*f[j-1]-f[j-2]
  return(f[l+1])
}
# a(0) = 1, a(1) = 1, a(n) = aa(n-1)   n > 1
#
# Example
n <- 73
aa(n-1)
#
# Yosu Yurramendi, Dec 15 2016

@CachedFunction
def a(n):
    return a((odd_part(n-1)+1)/2)+a((odd_part(n)+1)/2) if n>1 else 1
[a(n) for n in (0..77)] # after Alessandro De Luca, Peter Luschny, May 20 2014

def A007306(n):
    if n == 0: return 1
    M = [1, 1]
    for b in (n-1).bits():
        M[b] = M[0] + M[1]
    return M[1]
print([A007306(n) for n in (0..77)]) # Peter Luschny, Nov 28 2017

(define (A007306 n) (if (zero? n) 1 (A002487 (+ n n -1)))) ;; Code for A002487 given in that entry. - Antti Karttunen, Mar 21 2017

Recurrence: a(0) to a(8) are 1, 1, 2, 3, 3, 4, 5, 5, 4; thereafter a(n) = a(n-2^p) + a(2^(p+1)-n+1), where 2^p < n <= 2^(p+1). [J. Hermes, Math. Ann., 1894; quoted by Dickson, Vol. 1, p. 158] - N. J. A. Sloane, Mar 24 2019
a(4*n) = -a(n)+2*a(2*n); a(4*n+1) = -a(n)+a(2*n)+a(2*n+1); a(4*n+2)=a(n)-a(2*n)+2*a(2*n+1); a(4*n+3) = 4*a(n)-4*a(2*n)+3*a(2*n+1). Thus a(n) is a 2-regular sequence. - Jeffrey Shallit, Dec 26 2024
For n > 0, a(n) = A002487(n-1) + A002487(n) = A002487(2*n-1).
a(0) = 1; a(n) = Sum_{k=0..n-1} C(n-1+k, n-1-k) mod 2, n > 0. - Benoit Cloitre, Jun 20 2003
a(n+1) = Sum_{k=0..n} binomial(2*n-k, k) mod 2; a(n) = 0^n + Sum_{k=0..n-1} binomial(2(n-1)-k, k) mod 2. - Paul Barry, Dec 11 2004
a(n) = Sum_{k=0..n} C(n+k,2*k) mod 2. - Paul Barry, Jun 12 2006
a(0) = a(1) = 1; a(n) = a(A003602(n-1)) + a(A003602(n)), n > 1. - Alessandro De Luca, May 08 2014
a(n) = A007305(n+(2^m-1)), m=A029837(n), n=1,2,3,... . - Yosu Yurramendi, Jul 04 2014
a(n) = A007305(2^(m+1)-n) - A007305(2^m-n), m >= (A029837(n)+1), n=1,2,3,... - Yosu Yurramendi, Jul 05 2014
a(2^m) = m+1, a(2^m+1) = m+2 for m >= 0. - Yosu Yurramendi, Jan 01 2015
a(n+2) = A007305(n+2) + A047679(n) n >= 0. - Yosu Yurramendi, Mar 30 2016
a(2^m+2^r+k) = a(2^r+k)(m-r+1) - a(k), m >= 2, 0 <= r <= m-1, 0 <= k < 2^r. Example: a(73) = a(2^6+2^3+1) = a(2^3+1)*(6-3+1) - a(1) = 5*4 - 1 = 19 . - Yosu Yurramendi, Jul 19 2016
From Antti Karttunen, Mar 21 2017 & Apr 12 2017: (Start)
For n > 0, a(n) = A001222(A277324(n-1)) = A001222(A260443(n-1)*A260443(n)).
The following decompositions hold for all n > 0:
a(n) = A277328(n-1) + A284009(n-1).
a(n) = A283986(n) + A283988(n) = A283987(n) + 2*A283988(n).
a(n) = 2*A284265(n-1) + A284266(n-1).
a(n) = A284267(n-1) + A284268(n-1).
a(n) = A284565(n-1) + A284566(n-1).
a(n) = A285106(n-1) + A285108(n-1) = A285107(n-1) + 2*A285108(n-1). (End)
a(A059893(n)) = a(n+1) for n > 0. - Yosu Yurramendi, May 30 2017
a(n) = A287731(n) + A287732(n) for n > 0. - I. V. Serov, Jun 09 2017
a(n) = A287896(n) + A288002(n) for n > 1.
a(n) = A287896(n-1) + A002487(n-1) - A288002(n-1) for n > 1.
a(n) = a(n-1) + A002487(n-1) - 2*A288002(n-1) for n > 1. - I. V. Serov, Jun 14 2017
From Yosu Yurramendi, May 14 2019: (Start)
For m >= 0, M >= m, 0 <= k < 2^m,
a((2^(m+1) + A119608(2^m+k+1))*2^(M-m) - A000035(2^m+k)) =
a((2^(m+2) - A119608(2^m+k+1))*2^(M-m) - A000035(2^m+k)-1) =
a(2^(M+2) - (2^m+k)) = a(2^(M+1) + (2^m+k) + 1) =
a(2^m+k+1)*(M-m) + a(2^(m+1)+2^m+k+1). (End)
a(k) = sqrt(A007305(2^(m+1)+k)*A047679(2^(m+1)+k-2) - A007305(2^m+k)*A047679(2^m+k-2)), m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Jun 09 2019
G.f.: 1 + x * (1 + x) * Product_{k>=0} (1 + x^(2^k) + x^(2^(k+1))). - Ilya Gutkovskiy, Jul 19 2019
Conjecture: a(n) = a(n-1) + b(n-1) - 2*(a(n-1) mod b(n-1)) for n > 1 with a(0) = a(1) = 1 where b(n) = a(n) - b(n-1) for n > 1 with b(1) = 1. - Mikhail Kurkov, Mar 13 2022

Formula fixed and extended by Franklin T. Adams-Watters, Jul 07 2009

Incorrect Maple program removed by Johannes W. Meijer, Jun 05 2011

A007931 Numbers that contain only 1's and 2's. Nonempty binary strings of length n in lexicographic order.

Original entry on oeis.org

1, 2, 11, 12, 21, 22, 111, 112, 121, 122, 211, 212, 221, 222, 1111, 1112, 1121, 1122, 1211, 1212, 1221, 1222, 2111, 2112, 2121, 2122, 2211, 2212, 2221, 2222, 11111, 11112, 11121, 11122, 11211, 11212, 11221, 11222, 12111, 12112, 12121, 12122

Offset: 1

R. Muller

Numbers written in the dyadic system [Smullyan, Stillwell]. - N. J. A. Sloane, Feb 13 2019
Logic-binary sequence: prefix it by the empty word to have all binary words on the alphabet {1,2}.
The least binary word of length k is a(2^k - 1).
See Mathematica program for logic-binary sequence using (0,1) in place of (1,2); the sequence starts with 0,1,00,01,10. - Clark Kimberling, Feb 09 2012
A007953(a(n)) = A014701(n+1); A007954(a(n)) = A048896(n). - Reinhard Zumkeller, Oct 26 2012
a(n) is n written in base 2 where zeros are not allowed but twos are. The two distinct digits used are 1, 2 instead of 0, 1. To obtain this sequence from the "canonical" base 2 sequence with zeros allowed, just replace any 0 with a 2 and then subtract one from the group of digits situated on the left: (10-->2; 100-->12; 110-->22; 1000-->112; 1010-->122). - Robin Garcia, Jan 31 2014
For numbers made of only two different digits, see also A007088 (digits 0 & 1), A032810 (digits 2 & 3), A032834 (digits 3 & 4), A256290 (digits 4 & 5), A256291 (digits 5 & 6), A256292 (digits 6 & 7), A256340(digits 7 & 8), A256341 (digits 8 & 9), and A032804-A032816 (in other bases). Numbers with exactly two distinct (but unspecified) digits in base 10 are listed in A031955, for other bases in A031948-A031954. - M. F. Hasler, Apr 04 2015
The variant with digits {0, 1} instead of {1, 2} is obtained by deleting all initial digits in sequence A007088 (numbers written in base 2). - M. F. Hasler, Nov 03 2020

			Positive numbers may not start with 0 in the OEIS, otherwise this sequence would have been written as: 0, 1, 00, 01, 10, 11, 000, 001, 010, 011, 100, 101, 110, 111, 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 00000, 00001, 00010, 00011, 00100, 00101, 00110, 00111, 01000, 01001, 01010, 01011, ...
From _Hieronymus Fischer_, Jun 06 2012: (Start)
a(10)   = 122.
a(100)  = 211212.
a(10^3) = 222212112.
a(10^4) = 1122211121112.
a(10^5) = 2111122121211112.
a(10^6) = 2221211112112111112.
a(10^7) = 11221112112122121111112.
a(10^8) = 12222212122221111211111112.
a(10^9) = 22122211221212211212111111112. (End)

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 2. - From N. J. A. Sloane, Jul 26 2012
K. Atanassov, On the 97th, 98th and the 99th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 3, 89-93.
R. M. Smullyan, Theory of Formal Systems, Princeton, 1961.
John Stillwell, Reverse Mathematics, Princeton, 2018. See p. 90.

Hieronymus Fischer, Table of n, a(n) for n = 1..10000 (terms up to 2^10-2 from T. D. Noe, corrected by Sean A. Irvine, April 18 2019)
K. Atanassov, On Some of Smarandache's Problems, American Research Press, 1999, 16-21.
EMIS, Mirror site for Southwest Journal of Pure and Applied Mathematics
R. R. Forslund, A logical alternative to the existing positional number system, Southwest Journal of Pure and Applied Mathematics, Vol. 1, 1995.
R. R. Forslund, Positive Integer Pages
James E. Foster, A Number System without a Zero-Symbol, Mathematics Magazine, Vol. 21, No. 1. (1947), pp. 39-41.
F. Smarandache, Only Problems, Not Solutions!.
Index entries for 10-automatic sequences.

Cf. A007932 (digits 1-3), A059893, A045670, A052382 (digits 1-9), A059939, A059941, A059943, A032924, A084544, A084545, A046034 (prime digits 2,3,5,7), A089581, A084984 (no prime digits); A001742, A001743, A001744: loops; A202267 (digits 0, 1 and primes), A202268 (digits 1,4,6,8,9), A014261 (odd digits), A014263 (even digits).
Cf. A007088 (digits 0 & 1), A032810 (digits 2 & 3), A032834 (digits 3 & 4), A256290 (digits 4 & 5), A256291 (digits 5 & 6), A256292 (digits 6 & 7), A256340 (digits 7 & 8), A256341 (digits 8 & 9), and A032804-A032816 (in other bases).
Cf. A020450 (primes).

a007931 n = f (n + 1) where
   f x = if x < 2 then 0 else (10 * f x') + m + 1
     where (x', m) = divMod x 2
-- Reinhard Zumkeller, Oct 26 2012

[n: n in [1..100000] | Set(Intseq(n)) subset {1,2}]; // Vincenzo Librandi, Aug 19 2016

# Maple program to produce the sequence:
a:= proc(n) local m, r, d; m, r:= n, 0;
      while m>0 do d:= irem(m, 2, 'm');
        if d=0 then d:=2; m:= m-1 fi;
        r:= d, r
      od; parse(cat(r))/10
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Aug 26 2016
# Maple program to invert this sequence: given a(n), it returns n. - N. J. A. Sloane, Jul 09 2012
invert7931:=proc(u)
local t1,t2,i;
t1:=convert(u,base,10);
[seq(t1[i]-1,i=1..nops(t1))];
[op(%),1];
t2:=convert(%,base,2,10);
add(t2[i]*10^(i-1),i=1..nops(t2))-1;
end;

f[n_] := FromDigits[Rest@IntegerDigits[n + 1, 2] + 1]; Array[f, 42] (* Robert G. Wilson v Sep 14 2006 *)
(* Next, A007931 using (0,1) instead of (1,2) *)
d[n_] := FromDigits[Rest@IntegerDigits[n + 1, 2] + 1]; Array[FromCharacterCode[ToCharacterCode[ToString[d[#]]] - 1] &, 100] (* Peter J. C. Moses, at request of Clark Kimberling, Feb 09 2012 *)
Flatten[Table[FromDigits/@Tuples[{1,2},n],{n,5}]] (* Harvey P. Dale, Sep 13 2014 *)

apply( {A007931(n)=fromdigits([d+1|d<-binary(n+1)[^1]])}, [1..44]) \\ M. F. Hasler, Nov 03 2020, replacing older code from Mar 26 2015

/* inverse function */ apply( {A007931_inv(N)=fromdigits([d-1|d<-digits(N)],2)+2<M. F. Hasler, Nov 09 2020

def a(n): return int(bin(n+1)[3:].replace('1', '2').replace('0', '1'))
print([a(n) for n in range(1, 45)]) # Michael S. Branicky, May 13 2021

def A007931(n): return int(s:=bin(n+1)[3:])+(10**(len(s))-1)//9 # Chai Wah Wu, Jun 13 2025

To get a(n), write n+1 in base 2, remove initial 1, add 1 to all remaining digits: e.g., eleven (11) in base 2 is 1011; remove initial 1 and add 1 to remaining digits: a(10)=122. - Clark Kimberling, Mar 11 2003
Conversely, given a(n), to get n: subtract 1 from all digits, prefix with an initial 1, convert this binary number to base 10, subtract 1. E.g., a(6)=22 -> 11 -> 111 -> 7 -> 6. - N. J. A. Sloane, Jul 09 2012
a(n) = A053645(n+1)+A002275(A000523(n)) = a(n-2^b(n))+10^b(n) where b(n) = A059939(n) = floor(log_2(n+1)-1). - Henry Bottomley, Feb 14 2001
From Hieronymus Fischer, Jun 06 2012 and Jun 08 2012: (Start)
The formulas are designed to calculate base-10 numbers only using the digits 1 and 2.
a(n) = Sum_{j=0..m-1} (1 + b(j) mod 2)*10^j, where m = floor(log_2(n+1)), b(j) = floor((n+1-2^m)/(2^j)).
Special values:
a(k*(2^n-1)) = k*(10^n-1)/9, k= 1,2.
a(3*2^n-2) = (11*10^n-2)/9 = 10^n+2*(10^n-1)/9.
a(2^n-2) = 2*(10^(n-1)-1)/9, n>1.
Inequalities:
a(n) <= (10^log_2(n+1)-1)/9, equality holds for n=2^k-1, k>0.
a(n) > (2/10)*(10^log_2(n+1)-1)/9.
Lower and upper limits:
lim inf a(n)/10^log_2(n) = 1/45, for n --> infinity.
lim sup a(n)/10^log_2(n) = 1/9, for n --> infinity.
G.f.: g(x) = (1/(x(1-x)))*sum_{j=0..infinity} 10^j* x^(2*2^j)*(1 + 2 x^2^j)/(1 + x^2^j).
Also: g(x) = (1/(1-x))*(h_(2,0)(x) + h_(2,1)(x) - 2*h_(2,2)(x)), where h_(2,k)(x) = sum_{j>=0} 10^j*x^(2^(j+1)-1)*x^(k*2^j)/(1-x^2^(j+1)).
Also: g(x) = (1/(1-x)) sum_{j>=0} (1 - 3(x^2^j)^2 + 2(x^2^j)^3)*x^2^j*f_j(x)/(1-x^2^j), where f_j(x) = 10^j*x^(2^j-1)/(1-(x^2^j)^2). The f_j obey the recurrence f_0(x) = 1/(1-x^2), f_(j+1)(x) = 10x*f_j(x^2). (End)

Some crossrefs added by Hieronymus Fischer, Jun 06 2012

Edited by M. F. Hasler, Mar 26 2015

cp's OEIS Frontend

A065289 A065287 conjugated with A059893, inverse of A065290.

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A258209 Numbers k for which A256999(A059893(k)) = k.

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A153150 Self-inverse permutation of natural numbers: A059893-conjugate of A056539.

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A274048 a(n) = A116640(A018900(n)) = A116623(A059893(A018900(n))).

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A005940 The Doudna sequence: write n-1 in binary; power of prime(k) in a(n) is # of 1's that are followed by k-1 0's.

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A054429 Simple self-inverse permutation of natural numbers: List each block of 2^n numbers (from 2^n to 2^(n+1) - 1) in reverse order.

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A333381 Number of maximal anti-runs of the n-th composition in standard order.

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A272020 Irregular triangle read by rows: strictly decreasing sequences of positive numbers given in lexicographic order.

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