A257264 -id:A257264 - cp's OEIS frontend

A257507 Row 2 of A257264: a(n) = A011371(A055938(n)).

1, 3, 4, 7, 10, 10, 11, 15, 18, 18, 22, 23, 25, 25, 26, 31, 34, 34, 38, 39, 41, 41, 46, 47, 49, 50, 54, 54, 56, 56, 57, 63, 66, 66, 70, 71, 73, 73, 78, 79, 81, 82, 86, 86, 88, 88, 94, 95, 97, 98, 102, 102, 104, 105, 110, 110, 113, 116, 117, 117, 119, 119, 120, 127, 130, 130, 134, 135, 137, 137, 142, 143, 145, 146

Offset: 1

Antti Karttunen, May 03 2015

nonn

The sequence gives the parent node of each leaf-vertex (A055938) in binary beanstalk.

			Terms of A055938 are the leaf-nodes in Paul Tek's illustration. This sequence gives the corresponding parent-node (in that illustration a node immediately below where the arrow points), for each term of A055938[1..]: 2, 5, 6, 9, 12, 13, 14, ...
As A055938(4) = 9, and 9's parent node is 7 (because A011371(9) = 7), a(4) = 7.
As A055938(5) = 12, and 12's parent node is 10, a(5) = 10.
As A055938(6) = 13, and 13's parent node is 10, a(6) = 10.

Antti Karttunen, Table of n, a(n) for n = 1..16384
Paul Tek, Illustration of how natural numbers in range 0 .. 133 are organized as a binary tree in the binary beanstalk

Row 2 of A257264.
Cf. A011371, A055938.
Cf. A257508 (same sequence with duplicates removed), A257512 (the terms which occur twice).

(define (A257507 n) (A011371 (A055938 n)))

a(n) = A011371(A055938(n)).

A055938 Integers not generated by b(n) = b(floor(n/2)) + n (complement of A005187).

Original entry on oeis.org

2, 5, 6, 9, 12, 13, 14, 17, 20, 21, 24, 27, 28, 29, 30, 33, 36, 37, 40, 43, 44, 45, 48, 51, 52, 55, 58, 59, 60, 61, 62, 65, 68, 69, 72, 75, 76, 77, 80, 83, 84, 87, 90, 91, 92, 93, 96, 99, 100, 103, 106, 107, 108, 111, 114, 115, 118, 121, 122, 123, 124, 125, 126, 129

Offset: 1

Alford Arnold, Jul 21 2000

Note that the lengths of the consecutive runs in a(n) form sequence A001511.
Integers that are not a sum of distinct integers of the form 2^k-1. - Vladeta Jovovic, Jan 24 2003
Also n! never ends in this many 0's in base 2 - Carl R. White, Jan 21 2008
A079559(a(n)) = 0. - Reinhard Zumkeller, Mar 18 2009
These numbers are dead-end points when trying to apply the iterated process depicted in A071542 in reverse, i.e. these are positive integers i such that there does not exist k with A000120(i+k)=k. See also comments at A179016. - Antti Karttunen, Oct 26 2012
Conjecture: a(n)=b(n) defined as b(1)=2, for n>1, b(n+1)=b(n)+1 if n is already in the sequence, b(n+1)=b(n)+3 otherwise. If so, then see Cloitre comment in A080578. - Ralf Stephan, Dec 27 2013
Numbers n for which A257265(m) = 0. - Reinhard Zumkeller, May 06 2015. Typo corrected by Antti Karttunen, Aug 08 2015
Numbers which have a 2 in their skew-binary representation (cf. A169683). - Allan C. Wechsler, Feb 28 2025

			Since A005187 begins 0 1 3 4 7 8 10 11 15 16 18 19 22 23 25 26 31... this sequence begins 2 5 6 9 12 13 14 17 20 21

Complement of A005187. Setwise difference of A213713 and A213717.
Row 1 of arrays A257264, A256997 and also of A255557 (when prepended with 1). Equally: column 1 of A256995 and A255555.
Cf. also arrays A254105, A254107 and permutations A233276, A233278.
Left inverses: A234017, A256992.
Cf. A001511, A046699, A079559, A080578, A086343, A227359, A227408, A234016.
Gives positions of zeros in A213714, A213723, A213724, A213731, A257265, positions of ones in A213725-A213727 and A256989, positions of nonzeros in A254110.
Cf. also A010061 (integers that are not a sum of distinct integers of the form 2^k+1).
Analogous sequence for factorial base number system: A219658, for Fibonacci number system: A219638, for base-3: A096346. Cf. also A136767-A136774.
Cf. A257508, A257509, A257126.

a055938 n = a055938_list !! (n-1)
a055938_list = concat $
   zipWith (\u v -> [u+1..v-1]) a005187_list $ tail a005187_list
-- Reinhard Zumkeller, Nov 07 2011

a[0] = 0; a[1] = 1; a[n_Integer] := a[Floor[n/2]] + n; b = {}; Do[ b = Append[b, a[n]], {n, 0, 105}]; c =Table[n, {n, 0, 200}]; Complement[c, b]
(* Second program: *)
t = Table[IntegerExponent[(2n)!, 2], {n, 0, 100}]; Complement[Range[t // Last], t] (* Jean-François Alcover, Nov 15 2016 *)

L=listcreate();for(n=1,1000,for(k=2*n-hammingweight(n)+1,2*n+1-hammingweight(n+1),listput(L,k)));Vec(L) \\ Ralf Stephan, Dec 27 2013

def a053644(n): return 0 if n==0 else 2**(len(bin(n)[2:]) - 1)
def a043545(n):
    x=bin(n)[2:]
    return int(max(x)) - int(min(x))
def a079559(n): return 1 if n==0 else a043545(n + 1)*a079559(n + 1 - a053644(n + 1))
print([n for n in range(1, 201) if a079559(n)==0]) # Indranil Ghosh, Jun 11 2017, after the comment by Reinhard Zumkeller

;; utilizing COMPLEMENT-macro from Antti Karttunen's IntSeq-library)
(define A055938 (COMPLEMENT 1 A005187))
;; Antti Karttunen, Aug 08 2015

a(n) = A080578(n+1) - 2 = A080468(n+1) + 2*n (conjectured). - Ralf Stephan, Dec 27 2013
From Antti Karttunen, Aug 08 2015: (Start)
Other identities. For all n >= 1:
A234017(a(n)) = n.
A256992(a(n)) = n.
A257126(n) = a(n) - A005187(n).
(End)

More terms from Robert G. Wilson v, Jul 24 2000

A256993 a(1) = 0; for n > 1, a(n) = 1 + a(A256992(n)).

Original entry on oeis.org

0, 1, 2, 3, 2, 3, 4, 3, 4, 4, 5, 3, 4, 5, 4, 5, 4, 5, 6, 5, 5, 4, 5, 6, 6, 5, 4, 5, 6, 5, 6, 5, 6, 6, 7, 5, 6, 6, 6, 7, 5, 6, 6, 6, 5, 7, 7, 6, 6, 5, 7, 7, 6, 7, 6, 6, 7, 5, 6, 7, 6, 7, 6, 7, 6, 7, 8, 7, 7, 6, 7, 8, 7, 7, 6, 7, 7, 8, 6, 7, 7, 7, 8, 6, 7, 6, 7, 8, 8, 7, 7, 6, 8, 7, 7, 8, 6, 8, 7, 7, 8, 7, 6, 8, 7, 8, 8, 7, 7, 8, 8, 6, 7, 7, 7, 8, 7, 8, 8, 7, 6, 7, 8, 7, 8, 7, 8, 7

Offset: 1

Antti Karttunen, Apr 15 2015

nonn

Number of iterations of A256992 needed to reach one when starting from n.

Antti Karttunen, Table of n, a(n) for n = 1..8192

Cf. A070939, A071542, A254110, A255559, A256991, A256992, A257264, A257265, A279341, A279343, A279345, A279346.

a(1) = 0; for n > 1, a(n) = 1 + a(A256992(n)).
Other observations. For all n >= 1 it holds that:
a(n) >= A254110(n).
a(n) >= A256989(n).
a(n) >= A255559(n)-1.
Also it seems that a(n) - A070939(n) = -1, 0 or +1 for all n >= 1. [Compare A256991 and A256992 to see the connection.]
It is also very likely that a(n) <= A071542(n) for all n.
From Antti Karttunen, Dec 10 2016: (Start)
For all n >= 2, a(n) = A070939(A279341(n)) = A070939(A279343(n)).
For all n >= 2, a(n) = A279345(n) + A279346(n) - 1.
(End)

A257265 Distance to n from a leaf nearest to n in the binary beanstalk.

Original entry on oeis.org

2, 1, 0, 1, 1, 0, 0, 1, 2, 0, 1, 1, 0, 0, 0, 1, 2, 0, 1, 2, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 2, 0, 1, 2, 0, 0, 1, 1, 0, 1, 2, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 2, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 2, 0, 1, 2, 0, 0, 1, 1, 0, 1, 2, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 2, 1, 0, 1, 2, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 2, 1, 0, 1, 1, 0, 0, 0, 2, 1, 0, 2, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 2

Offset: 0

Antti Karttunen, Apr 29 2015

nonn

If n is one of the terms of A055938 (that is, one of the leaves of the binary beanstalk), then a(n) is zero. Otherwise, a(n) is the least number of iterations of A011371 required to reach n, when we start from any term of A055938.
Compare to the definition of A213725, which gives 1 + distance to the farthest leaf in the binary beanstalk (giving 0 for the nodes which reside in A179016, the infinite stem of the beanstalk, as there the maximum distance would not have a finite value).
Note that although the recursive formula given here mirrors the one given at A213725, it cannot be implemented in a naive way, precisely because of that infinite stem, as it would lead to recursion without end. Instead, with ordinary eagerly evaluating programming languages we have to employ, for example, a breadth-first search, as in the given Scheme-program.
Question: Will there ever appear a term larger than 2? (Only terms 0 - 2 occur in range 0 .. 2097151).

			For 0, the nearest leaf is 2, as when we start from 2, and always subtract the binary weight, A000120, we have: 2 - A000120(2) = A011371(2) = 1, and A011371(1) = 0, thus it takes two steps to get to 0, and there are no other terms of A055938 from which it would take fewer steps), so a(0) = 2, and also a(1) = 1, because it's one step nearer to 2.
a(2) = 0, because 2 is one of the terms of A055938.
a(8) = 2, because 12, 13 and 14 are the three nearest leaves to 8, and A011371(12) = A011371(13) = 10, A011371(14) = 11, A011371(10) = A011371(11) = 8 (thus it takes two iterations of A011371 to reach 8 from any of those three leaves) and there are no leaves nearer.
Please see also Paul Tek's illustration.

Antti Karttunen, Table of n, a(n) for n = 0..16384
Ela, peano
Haskell Wiki, Peano numbers
Paul Tek, Illustration of how natural numbers in range 0 .. 133 are organized as a binary tree in the binary beanstalk

Cf. A055938 (positions of 0's), A257508 (of 1's), A257509 (of 2's).
Cf. also A000120, A011371, A079559, A213723, A213724.
Cf. also A179016, A213725, A257264.

a257265 = length . us where
   us n = if a079559 n == 0
             then [] else () : zipWith (const $ const ())
                               (us $ a213723 n) (us $ a213724 n)
-- Reinhard Zumkeller, May 03 2015

;; Do a breadth-first search over the descendants, which are at each step of iteration sorted by their distance from the starting node.
(define (A257265 n) (let loop ((descendants (list (cons 0 n)))) (let ((dist (caar descendants)) (node (cdar descendants))) (cond ((zero? (A079559 node)) dist) (else (loop (sort (append (list (cons (+ 1 dist) (A213724 node)) (cons (+ 1 dist) (A213723 node))) (cdr descendants)) (lambda (a b) (< (car a) (car b))))))))))

If A079559(n) = 0, then a(n) = 0, otherwise a(n) = 1 + min(a(A213723(n)), a(A213724(n))). [But please see the comments above.]

A262898 Square array A(row,col) read by antidiagonals: A(1,col) = A045765(col); for row > 1, if A(row-1,col) = 0 then A(row,col) = 0, otherwise A(row,col) = A049820(A(row-1,col)).

Original entry on oeis.org

7, 8, 5, 13, 4, 3, 19, 11, 1, 1, 20, 17, 9, 0, 0, 24, 14, 15, 6, 0, 0, 25, 16, 10, 11, 2, 0, 0, 28, 22, 11, 6, 9, 0, 0, 0, 33, 22, 18, 9, 2, 6, 0, 0, 0, 36, 29, 18, 12, 6, 0, 2, 0, 0, 0, 37, 27, 27, 12, 6, 2, 0, 0, 0, 0, 0, 40, 35, 23, 23, 6, 2, 0, 0, 0, 0, 0, 0, 43, 32, 31, 21, 21, 2, 0, 0, 0, 0, 0, 0, 0, 49, 41, 26, 29, 17, 17, 0, 0, 0, 0, 0, 0, 0, 0, 50, 46, 39, 22, 27, 15, 15, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Antti Karttunen, Oct 06 2015

The array is read by downwards antidiagonals: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), etc.
Column n gives the trajectory of iterates of A049820, when starting from A045765(n), thus stepping through successive parent-nodes when starting from the n-th leaf in the tree generated by edge-relation A049820(child) = parent, until finally reaching the fixed point 0, which is the root of the whole tree.
A portion of the hanging tail of each column (upward from the first encountered zero) converges towards A259934, although not in monotone fashion.

			The top left corner of the array:
7, 8, 13, 19, 20, 24, 25, 28, 33, 36, 37, 40, 43, 49, 50, 52, 55, 56
5, 4, 11, 17, 14, 16, 22, 22, 29, 27, 35, 32, 41, 46, 44, 46, 51, 48
3, 1,  9, 15, 10, 11, 18, 18, 27, 23, 31, 26, 39, 42, 38, 42, 47, 38
1, 0,  6, 11,  6,  9, 12, 12, 23, 21, 29, 22, 35, 34, 34, 34, 45, 34
0, 0,  2,  9,  2,  6,  6,  6, 21, 17, 27, 18, 31, 30, 30, 30, 39, 30
0, 0,  0,  6,  0,  2,  2,  2, 17, 15, 23, 12, 29, 22, 22, 22, 35, 22
0, 0,  0,  2,  0,  0,  0,  0, 15, 11, 21,  6, 27, 18, 18, 18, 31, 18
0, 0,  0,  0,  0,  0,  0,  0, 11,  9, 17,  2, 23, 12, 12, 12, 29, 12
0, 0,  0,  0,  0,  0,  0,  0,  9,  6, 15,  0, 21,  6,  6,  6, 27,  6
0, 0,  0,  0,  0,  0,  0,  0,  6,  2, 11,  0, 17,  2,  2,  2, 23,  2
0, 0,  0,  0,  0,  0,  0,  0,  2,  0,  9,  0, 15,  0,  0,  0, 21,  0
0, 0,  0,  0,  0,  0,  0,  0,  0,  0,  6,  0, 11,  0,  0,  0, 17,  0
0, 0,  0,  0,  0,  0,  0,  0,  0,  0,  2,  0,  9,  0,  0,  0, 15,  0
0, 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  6,  0,  0,  0, 11,  0
0, 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  2,  0,  0,  0,  9,  0
0, 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  6,  0
...

Antti Karttunen, Table of n, a(n) for n = 1..5050; the first 100 antidiagonals of the array

Transpose: A262899.
Cf. A045765 (row 1), A262902 (row 2).
Cf. A049820, A259934.
Cf. also A257264.

(define (A262898 n) (A262898bi (A002260 n) (A004736 n)))
(define (A262898bi row col) (if (= 1 row) (A045765 col) (if (zero? (A262898bi (- row 1) col)) 0 (A049820 (A262898bi (- row 1) col)))))

A(1,col) = A045765(col), and for row > 1, if A(row-1,col) = 0 then A(row,col) = 0, otherwise A(row,col) = A049820(A(row-1,col)).

A257509 Numbers n for which A257265(n) = 2; numbers for which the nearest descendant leaf in the binary beanstalk is two edges away.

Original entry on oeis.org

0, 8, 16, 19, 32, 35, 42, 53, 64, 67, 74, 85, 89, 101, 109, 112, 128, 131, 138, 149, 153, 165, 173, 176, 184, 197, 205, 208, 221, 224, 231, 240, 256, 259, 266, 277, 281, 293, 301, 304, 312, 325, 333, 336, 349, 352, 359, 368, 375, 389, 397, 400, 413, 416, 423, 432, 445, 448, 455, 464, 470, 480, 487, 492, 512

Offset: 1

Antti Karttunen, May 03 2015

nonn

Numbers n for which A257265(n) = 2.

			8 is present, because 12, 13 and 14 are the three leaves (terms of A055938) nearest to 8, and A011371(12) = A011371(13) = 10, A011371(14) = 11, A011371(10) = A011371(11) = 8 (thus it takes two iterations of A011371 to reach 8 from any of those three leaves). See also Paul Tek's illustration.

Antti Karttunen, Table of n, a(n) for n = 1..16385
Paul Tek, Illustration of how natural numbers in range 0 .. 133 are organized as a binary tree in the binary beanstalk

First differences: A256489.
Positions of 2's in A257265.
Subsequence of A005187.
Cf. A011371, A055938, A257508, A257264.

a257509 n = a257509_list !! (n-1)
a257509_list = filter ((== 2) . a257265) [0..]
-- Reinhard Zumkeller, May 06 2015

cp's OEIS Frontend

A257507 Row 2 of A257264: a(n) = A011371(A055938(n)).

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A055938 Integers not generated by b(n) = b(floor(n/2)) + n (complement of A005187).

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A256993 a(1) = 0; for n > 1, a(n) = 1 + a(A256992(n)).

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A257265 Distance to n from a leaf nearest to n in the binary beanstalk.

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A262898 Square array A(row,col) read by antidiagonals: A(1,col) = A045765(col); for row > 1, if A(row-1,col) = 0 then A(row,col) = 0, otherwise A(row,col) = A049820(A(row-1,col)).

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A257509 Numbers n for which A257265(n) = 2; numbers for which the nearest descendant leaf in the binary beanstalk is two edges away.

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