A091067 -id:A091067 - cp's OEIS frontend

A255330 a(n) = total number of nodes in the finite subtrees branching from the node n in the infinite trunk of "number-of-runs beanstalk" (A255056).

1, 2, 0, 4, 1, 0, 7, 0, 3, 1, 0, 5, 2, 6, 0, 6, 0, 3, 1, 0, 5, 2, 12, 0, 2, 5, 0, 4, 2, 6, 0, 6, 0, 3, 1, 0, 5, 2, 12, 0, 2, 7, 1, 12, 4, 0, 2, 5, 0, 4, 2, 12, 0, 2, 5, 0, 4, 2, 6, 0, 6, 0, 3, 1, 0, 5, 2, 12, 0, 2, 7, 1, 12, 4, 0, 2, 7, 1, 10, 17, 0, 0, 1, 11, 4, 0, 2, 5, 0, 4, 2, 12, 0, 2, 7, 1, 12, 4, 0, 2, 5, 0, 4, 2, 12, 0, 2, 5, 0, 4, 2, 6, 0, 6, 0, 3, 1, 0, 5

Offset: 0

Antti Karttunen, Feb 21 2015

nonn

A255058 gives the number of branches (children) of the node n in the trunk, of which one is the next node of the infinite trunk itself. Thus, if A255058(n) = 1, then a(n) = 0.

			The edge-relation between nodes is given by A236840(child) = parent. Odd numbers are leaves, as there are no such k that A236840(k) were odd.
The node 11 in the infinite trunk is A255056(11) = 30. Apart from 32 [we have A236840(32) = 30] which is the next node (node 12) in the infinite trunk, it has a single leaf-child 31 [A236840(31) = 30] at the "left side" (less than 32), and a leaf-child 33 [A236840(33) = 30] (more than 32) at the "right side", and also at that side, a subtree of three nodes 34 <- 38 <- 43 [we have A236840(43) = 38, A236840(38) = 34 and A236840(34) = 30], thus in total there are 1+1+3 = 5 nodes in finite branches emanating from the node 11 of the infinite trunk, and a(11) = 5.

Antti Karttunen, Table of n, a(n) for n = 0..8590

Partial sums: A255333.

Cf. A091067, A236840, A255056, A255058, A255068, A255327, A255328, A255329, A255331.

(define (A255330 n) (if (zero? n) 1 (let ((k (A255057 n))) (add A255327 (A091067 k) (A255068 k)))))
;; Other code as in A255327.

a(0) = 1; a(n) = sum_{k = A091067(A255057(n)) .. A255068(A255057(n))} A255327(k).

a(n) = A255328(n) + A255329(n).

A055975 First differences of A003188 (decimal equivalent of the Gray Code).

Original entry on oeis.org

1, 2, -1, 4, 1, -2, -1, 8, 1, 2, -1, -4, 1, -2, -1, 16, 1, 2, -1, 4, 1, -2, -1, -8, 1, 2, -1, -4, 1, -2, -1, 32, 1, 2, -1, 4, 1, -2, -1, 8, 1, 2, -1, -4, 1, -2, -1, -16, 1, 2, -1, 4, 1, -2, -1, -8, 1, 2, -1, -4, 1, -2, -1, 64, 1, 2, -1, 4, 1, -2, -1, 8, 1, 2, -1, -4, 1, -2, -1, 16, 1, 2, -1, 4, 1, -2, -1, -8, 1, 2, -1, -4, 1, -2, -1, -32, 1, 2, -1, 4

Offset: 1

Alford Arnold, Jul 22 2000

Multiplicative with a(2^e) = 2^e, a(p^e) = (-1)^((p^e-1)/2) otherwise. - Mitch Harris, May 17 2005
a(A091072(n)) > 0; a(A091067(n)) < 0. - Reinhard Zumkeller, Apr 28 2012
In the binary representation of n, clear everything left of the least significant 1 bit, and negate if the bit left of it was set originally. - Ralf Stephan, Aug 23 2013
This sequence is the trace of n in the minimal alternating binary representation of n (defined at A256696). - Clark Kimberling, Apr 07 2015

			Since A003188 is 0, 1,  3, 2, 6,  7,  5, 4, 12, 13, 15, 14, 10, ...,
sequence begins  1, 2, -1, 4, 1, -2, -1, 8,  1,  2, -1,  4, ... .

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

Cf. A003188, A006519 (unsigned), A007814.
MASKTRANSi transform of A053644 (conjectural).
Cf. A119972, A119974, A220466.

a055975 n = a003188 n - a003188 (n-1)
a055975_list = zipWith (-) (tail a003188_list) a003188_list
-- Reinhard Zumkeller, Apr 28 2012

nmax:=100: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := (-1)^(n+1)*2^p od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Jan 27 2013

f[n_]:=BitXor[n,Floor[n/2]];Differences[Array[f,120,0]] (* Harvey P. Dale, Jul 18 2011, applying Robert G. Wilson v's program from A003188 *)

a(n)=((-1)^((n/2^valuation(n,2)-1)/2)*2^valuation(n,2)) \\ Ralf Stephan

def A055975(n): return (n^(n>>1))-((n-1)^(n-1>>1)) # Chai Wah Wu, Jun 29 2022

a(2n) = 2a(n), a(2n+1) = (-1)^n. G.f. sum(k>=0, 2^k*t/(1+t^2), t=x^2^k). a(n) = 2^A007814(n) * (-1)^((n/2^A007814(n)-1)/2). - Ralf Stephan, Oct 29 2003

a((2*n-1)*2^p) = (-1)^(n+1)*2^p, p >= 0. - Johannes W. Meijer, Jan 27 2013

More terms from Larry Reeves (larryr(AT)acm.org), Sep 05 2000

A255068 a(n) is the largest k such that A255070(k) = n.

Original entry on oeis.org

2, 5, 6, 10, 11, 13, 14, 18, 21, 22, 23, 26, 27, 29, 30, 34, 37, 38, 42, 43, 45, 46, 47, 50, 53, 54, 55, 58, 59, 61, 62, 66, 69, 70, 74, 75, 77, 78, 82, 85, 86, 87, 90, 91, 93, 94, 95, 98, 101, 102, 106, 107, 109, 110, 111, 114, 117, 118, 119, 122, 123, 125, 126, 130, 133, 134, 138, 139, 141, 142, 146, 149, 150

Offset: 0

Antti Karttunen, Feb 21 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..8192
Kevin Ryde, Iterations of the Dragon Curve, see index TurnRight, with a(n) = TurnRight(n) - 1.

First differences are A106836 (from its second term onward).
Cf. A091067, A255070, A255327.
Sequence A341522 sorted into ascending order.

a(n) = my(t=1); n=2*n+2; forstep(i=logint(n,2),0,-1, if(bittest(n,i)==t, n++;t=!t)); n-1; \\ Kevin Ryde, Mar 21 2021

(define (A255068 n) (- (A091067 (+ n 1)) 1))

a(n) = A091067(n+1) - 1.

A255328 a(n) = total number of nodes in the finite subtrees branching "left" (to the "smaller side") from the node n in the infinite trunk of "number-of-runs beanstalk" (A255056).

Original entry on oeis.org

1, 1, 0, 4, 0, 0, 7, 0, 3, 0, 0, 1, 1, 6, 0, 6, 0, 3, 0, 0, 1, 1, 12, 0, 1, 5, 0, 0, 1, 6, 0, 6, 0, 3, 0, 0, 1, 1, 12, 0, 1, 0, 0, 12, 1, 0, 1, 5, 0, 0, 1, 12, 0, 1, 5, 0, 0, 1, 6, 0, 6, 0, 3, 0, 0, 1, 1, 12, 0, 1, 0, 0, 12, 1, 0, 1, 0, 0, 10, 1, 0, 0, 0, 11, 1, 0, 1, 5, 0, 0, 1, 12, 0, 1, 0, 0, 12, 1, 0, 1, 5, 0, 0, 1, 12, 0, 1, 5, 0, 0, 1, 6, 0, 6, 0, 3, 0, 0, 1

Offset: 0

Antti Karttunen, Feb 21 2015

nonn

			See example in A255330. Here we count only the nodes at the left side, thus a(11) = 1.

Antti Karttunen, Table of n, a(n) for n = 0..8590

Cf. A255056, A255057, A255327, A255329, A255330, A255331.

(define (A255328 n) (if (zero? n) 1 (add A255327 (A091067 (A255057 n)) (A255056 (+ n 1)))))
;; Other code as in A255327.

a(0) = 1; a(n) = sum_{k = A091067(A255057(n)) .. A255056(n+1)} A255327(k).

a(n) = A255330(n) - A255329(n).

A266121 Lexicographically first injection of natural numbers beginning with a(1)=1 such that 1+(a(n)*a(n+1)) is a fibbinary number (A003714), i.e., has no adjacent 1's in its base-2 representation.

Original entry on oeis.org

1, 3, 5, 4, 2, 8, 9, 7, 12, 6, 14, 24, 11, 13, 20, 16, 10, 26, 40, 17, 15, 39, 28, 19, 27, 25, 23, 48, 22, 30, 44, 31, 33, 32, 18, 36, 29, 47, 45, 52, 21, 55, 49, 84, 61, 43, 51, 53, 80, 34, 64, 37, 35, 59, 75, 117, 93, 91, 57, 41, 100, 82, 50, 104, 42, 98, 106, 90, 114, 72, 58, 144, 65, 63, 151, 56, 38, 54, 76, 71, 60

Offset: 1

Antti Karttunen, Dec 23 2015

It is conjectured that this sequence is not only injective, but also surjective on N, i.e., that it is a true permutation of natural numbers.
A similar sequence, but with condition that "(a(n)*a(n+1)) must be a member of A003714" yields a sequence: 1, 2, 4, 5, 8, 9, 16, 10, 13, 20, ... (A269361) which certainly is not a bijection, because it contains only terms of A091072.
Also, with above condition and the initial value a(1) = 3 the algorithm generates A269363 which contains only terms of A091067. See also A266191.

			After the initial a(1) = 1, for obtaining the value of a(2) we try the first unused number, which is 2, but (1*2)+1 = 3, which in binary is "11", thus 2 is not qualified at this point of time. So next we try 3, and (1*3)+1 = 4, which in binary is "100", and that satisfies our criterion (no adjacent 1-bits), thus we set a(2) = 3.
For a(3), we test with the least unused numbers 2, 4, 5, etc., yielding products (3*2)+1 = 7 = "111", (3*4)+1 = 13 = "1101" and (3*5)+1 = 16 = "10000" in binary, and only 5 satisfies the criterion, thus we set a(3) = 5.

Antti Karttunen, Table of n, a(n) for n = 1..16384
Index entries for sequences that are permutations of the natural numbers

Left inverse: A266122 (also the right inverse if this sequence is a permutation of natural numbers).
Cf. A003714, A003987, A091067, A091072, A269361, A269363, A269365, A269366, A269367.
Cf. also A266191 and A266117 for similar permutations.

Minor typo in the description corrected by Antti Karttunen, Feb 25 2016

A255070 (1/2)*(n minus number of runs in the binary expansion of n): a(n) = (n - A005811(n)) / 2 = A236840(n)/2.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 2, 3, 3, 3, 3, 4, 5, 5, 6, 7, 7, 7, 7, 8, 8, 8, 9, 10, 11, 11, 11, 12, 13, 13, 14, 15, 15, 15, 15, 16, 16, 16, 17, 18, 18, 18, 18, 19, 20, 20, 21, 22, 23, 23, 23, 24, 24, 24, 25, 26, 27, 27, 27, 28, 29, 29, 30, 31, 31, 31, 31, 32, 32, 32, 33, 34, 34, 34, 34, 35

Offset: 0

Antti Karttunen, Feb 14 2015

Antti Karttunen, Table of n, a(n) for n = 0..8192
Helmut Prodinger and Friedrich J. Urbanek, Infinite 0-1-Sequences Without Long Adjacent Identical Blocks, Discrete Mathematics, Volume 28, Number 3, 1979, pages 277-289. Also first author's copy. See theorem 3.6, n_1(k) = a(k).

Least inverse: A091067 (also the positions of records).
Greatest inverse: A255068.
Run lengths: A106836.
Cf. A005811, A060833, A236840, A255054, A255056, A255072, A269363, A269367.

a[n_] := (n - Length@ Split[IntegerDigits[n, 2]])/2; a[0] = 0; Array[a, 100, 0] (* Amiram Eldar, Jul 16 2023 *)

Scheme
```
(define (A255070 n) (/ (A236840 n) 2))
```

a(n) = A236840(n) / 2 = (n - A005811(n)) / 2.
Other identities:
a(A091067(n)) = n for all n >= 1.
a(A255068(n)) = n for all n >= 0.
a(A269363(n)) = A269367(n). - Antti Karttunen, Aug 12 2019

A256696 R(k), the minimal alternating binary representation of k, concatenated for k = 0, 1, 2,....

Original entry on oeis.org

0, 1, 2, 4, -1, 4, 8, -4, 1, 8, -2, 8, -1, 8, 16, -8, 1, 16, -8, 2, 16, -8, 4, -1, 16, -4, 16, -4, 1, 16, -2, 16, -1, 16, 32, -16, 1, 32, -16, 2, 32, -16, 4, -1, 32, -16, 4, 32, -16, 8, -4, 1, 32, -16, 8, -2, 32, -16, 8, -1, 32, -8, 32, -8, 1, 32, -8, 2, 32

Offset: 0

Clark Kimberling, Apr 09 2015

Suppose that b = (b(0), b(1), ... ) is an increasing sequence of positive integers satisfying b(0) = 1 and b(n+1) <= 2*b(n) for n >= 0.  Let B(n) be the least b(m) >= n.  Let R(0) = 1, and for n > 0, let R(n) = B(n) - R(B(n) - n).  The resulting sum of the form R(n) = B(n) - B(m(1)) + B(m(2)) - ... + ((-1)^k)*B(k) is the minimal alternating b-representation of n.  The sum B(n) + B(m(2)) + ... is the positive part of R(n), and the sum B(m(1)) + B(m(3)) + ... , the nonpositive part of R(n).  The number ((-1)^k)*B(k) is the trace of n.
If b(n) = 2^n, the sum R(n) is the minimal alternating binary representation of n.
A055975 = trace of n, for n >= 1.
A091072 gives the numbers having positive trace.
A091067 gives the numbers having negative trace.
A072339 = number of terms in R(n).
A073122 = sum of absolute values of the terms in R(n).

			R(0) = 0
R(1) = 1
R(2) = 2
R(3) = 4 - 1
R(4) = 4
R(9) = 8 - 4 + 1
R(11) = 16 - 8 + 4 - 1

D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1981, Vol. 2 (2nd ed.), p. 196, Exercise 27.

Clark Kimberling, Table of n, a(n) for n = 0..1000

Cf. A000079, A256655 (Fibonacci based), A055975, A091072, A091067, A072339, A073122, A256701, A256702.

z = 100; b[n_] := 2^n; bb = Table[b[n], {n, 0, 40}];
s[n_] := Table[b[n + 1], {k, 1, b[n]}];
h[0] = {1}; h[n_] := Join[h[n - 1], s[n - 1]];
g = h[10]; r[0] = {0};
r[n_] := If[MemberQ[bb, n], {n}, Join[{g[[n]]}, -r[g[[n]] - n]]]
u = Flatten[Table[r[n], {n, 0, z}]]

A269363 Lexicographically first injection of natural numbers beginning with a(1)=3 such that for all n >= 1, a(n)*a(n+1) is a fibbinary number (A003714), i.e., has no adjacent 1's in its base-2 representation.

Original entry on oeis.org

3, 6, 7, 12, 11, 15, 22, 24, 14, 19, 27, 38, 28, 23, 46, 48, 43, 30, 39, 35, 59, 44, 31, 75, 62, 87, 51, 83, 56, 47, 88, 54, 76, 55, 96, 86, 60, 71, 67, 70, 78, 112, 79, 107, 102, 91, 120, 139, 118, 140, 119, 142, 131, 134, 155, 240, 156, 135, 152, 108, 95, 92, 103, 179, 184, 115, 147, 224, 94, 175, 123, 150, 111, 158, 214, 163, 203

Offset: 1

Antti Karttunen, Feb 25 2016

The sequence is conjectured to be a permutation of A091067.

The scatter plot is quite interesting (essentially the same as A269367). Compare also to the graph of A269361.

Antti Karttunen, Table of n, a(n) for n = 1..65537

Cf. A003714, A003987, A091067, A266121, A266191, A269361, A269365, A269366.

Cf. A269367 (the terms ranked with A255070).

fibbinaryQ[n_] := BitAnd[n, 2 n]==0; a[1]=3; a[n_] := a[n] = For[k=1, True, k++, If[Mod[k, 4] != 1, If[fibbinaryQ[a[n-1] k], If[FreeQ[Array[a, n-1], k], Return[k]]]]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Mar 02 2016 *)

A269367 Suspected permutation of natural numbers: a(n) = A255070(A269363(n)).

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 9, 11, 6, 8, 12, 17, 13, 10, 21, 23, 19, 14, 18, 16, 28, 20, 15, 35, 30, 41, 24, 39, 27, 22, 42, 25, 36, 26, 47, 40, 29, 34, 32, 33, 37, 55, 38, 51, 49, 43, 59, 67, 57, 68, 58, 69, 64, 65, 75, 119, 76, 66, 74, 52, 46, 44, 50, 87, 90, 56, 71, 111, 45, 85, 60, 72, 54, 77, 104, 79, 99, 88, 95, 48, 53, 78, 102, 82, 31

Offset: 1

Antti Karttunen, Feb 27 2016

Lexicographically first injection of natural numbers beginning with a(1)=1 such that for all n >= 1, A091067(a(n))*A091067(a(n+1)) is a fibbinary number (A003714), i.e., has no adjacent 1's in its base-2 representation.

Antti Karttunen, Table of n, a(n) for n = 1..13217
Index entries for sequences that are permutations of the natural numbers

Left inverse: A269368 (also the right inverse if this is a permutation of natural numbers).

Cf. A003714, A091067, A266121, A269361, A269363, A255070.

a(n) = A255070(A269363(n)).

A060833 Separate the natural numbers into disjoint sets A, B with 1 in A, such that the sum of any 2 distinct elements of the same set never equals 2^k + 2. Sequence gives elements of set A.

Original entry on oeis.org

1, 4, 7, 8, 12, 13, 15, 16, 20, 23, 24, 25, 28, 29, 31, 32, 36, 39, 40, 44, 45, 47, 48, 49, 52, 55, 56, 57, 60, 61, 63, 64, 68, 71, 72, 76, 77, 79, 80, 84, 87, 88, 89, 92, 93, 95, 96, 97, 100, 103, 104, 108, 109, 111, 112, 113, 116, 119, 120, 121, 124, 125, 127, 128

Offset: 1

Sen-Peng Eu, May 01 2001

Can be constructed as follows: place of terms of (2^k+1,2^k+2,...,2^k) are the reflection from (2,3,4,...,2^k,1). [Comment not clear to me - N. J. A. Sloane]
If n == 0 mod 4, then n is in the sequence.  If n == 2 mod 4, then n is not in the sequence.  The number 2n - 1 is in the sequence if and only if n is in the sequence. For n > 1, n is in the sequence if and only if A038189(n-1) = 1. - N. Sato, Feb 12 2013
The set B contains all numbers 2^(k-1)+1 = (2^k+2)/2 (half of the "forbidden sums"), (2, 3, 5, 9, 17, 33, 65,...) = 1/2 * (4, 6, 10, 18, 34, 66, 130, 258,...). - M. F. Hasler, Feb 12 2013

Alois P. Heinz, Table of n, a(n) for n = 1..1000
Kevin Ryde, Iterations of the Dragon Curve, see index TurnRight, with a(n) = TurnRight(n-2) + 1 for n>=2.

Essentially one more than A091067.
First differences: A106836.
A082410(a(n)) = 0.

a:= proc(n) option remember; local k, t;
      if n=1 then 1
    else for k from 1+a(n-1) do t:= k-1;
           while irem(t, 2, 'r')=0 do t:=r od;
           if irem(t, 4)=3 then return k fi
         od
      fi
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Feb 12 2013

a[n_] := a[n] = Module[{k, t, q, r}, If[n == 1, 1, For[k = 1+a[n-1], True, k++, t = k-1; While[{q, r} = QuotientRemainder[t, 2]; r == 0, t = q]; If[Mod[t, 4] == 3, Return[k]]]]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Jan 30 2017, after Alois P. Heinz *)

a(n) = if(n=2*n-2, my(t=1); forstep(i=logint(n,2),0,-1, if(bittest(n,i)==t, n++;t=!t))); n+1; \\ Kevin Ryde, Mar 21 2021

a(1) = 1; and for n > 1: a(n) = A091067(n-1)+1. - Antti Karttunen, Feb 20 2015, based on N. Sato's Feb 12 2013 comment above.

More terms from Larry Reeves (larryr(AT)acm.org), May 10 2001

cp's OEIS Frontend

A255330 a(n) = total number of nodes in the finite subtrees branching from the node n in the infinite trunk of "number-of-runs beanstalk" (A255056).

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A055975 First differences of A003188 (decimal equivalent of the Gray Code).

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A255068 a(n) is the largest k such that A255070(k) = n.

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A255328 a(n) = total number of nodes in the finite subtrees branching "left" (to the "smaller side") from the node n in the infinite trunk of "number-of-runs beanstalk" (A255056).

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A266121 Lexicographically first injection of natural numbers beginning with a(1)=1 such that 1+(a(n)*a(n+1)) is a fibbinary number (A003714), i.e., has no adjacent 1's in its base-2 representation.

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A255070 (1/2)*(n minus number of runs in the binary expansion of n): a(n) = (n - A005811(n)) / 2 = A236840(n)/2.

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A256696 R(k), the minimal alternating binary representation of k, concatenated for k = 0, 1, 2,....

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A269363 Lexicographically first injection of natural numbers beginning with a(1)=3 such that for all n >= 1, a(n)*a(n+1) is a fibbinary number (A003714), i.e., has no adjacent 1's in its base-2 representation.

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