A255331 -id:A255331 - cp's OEIS frontend

A255332 Partial sums of A255331.

-1, -1, -1, -5, -4, -4, -11, -11, -14, -13, -13, -10, -10, -16, -16, -22, -22, -25, -24, -24, -21, -21, -33, -33, -33, -38, -38, -34, -34, -40, -40, -46, -46, -49, -48, -48, -45, -45, -57, -57, -57, -50, -49, -61, -59, -59, -59, -64, -64, -60, -60, -72, -72, -72, -77, -77, -73, -73, -79, -79, -85, -85, -88, -87, -87, -84

Offset: 0

Antti Karttunen, Feb 21 2015

a(805) = 54 is the first positive term.
Is a(836) = -32 the last negative term?
The conspicuous "noncontinuity" which occurs in the scatter plot for the first time at n=5790 is caused by a sudden negative record at A255331(5790) = -708. Note that A255328(5790) = 708.

Antti Karttunen, Table of n, a(n) for n = 0..16140
A. Karttunen, Scatter plot of sequence plotted up to n=8590 (Drawn with the help of OEIS server's graph-utility).
A. Karttunen, Ratio A255332(n)/a(n) plotted with the help of OEIS-server

Cf. A255125, A255328, A255329, A255331, A255332, A255333.

Analogous sequences: A218789, A230409.

a(0) = -1; for n >= 1: a(n) = a(n-1) + A255331(n).

A255056 Trunk of number-of-runs beanstalk: The unique infinite sequence such that a(n-1) = a(n) - number of runs in binary representation of a(n).

Original entry on oeis.org

0, 2, 4, 6, 10, 12, 14, 18, 22, 26, 28, 30, 32, 36, 42, 46, 50, 54, 58, 60, 62, 64, 68, 74, 78, 84, 90, 94, 96, 100, 106, 110, 114, 118, 122, 124, 126, 128, 132, 138, 142, 148, 152, 156, 162, 168, 174, 180, 186, 190, 192, 196, 202, 206, 212, 218, 222, 224, 228, 234, 238, 242, 246, 250, 252, 254

Offset: 0

Antti Karttunen, Feb 14 2015

All numbers of the form (2^n)-2 are present, which guarantees the uniqueness and also provides a well-defined method to compute the sequence, for example, via a partially reversed version A255066.

The sequence was inspired by a similar "binary weight beanstalk", A179016, sharing some general properties with it (like its partly self-copying behavior, see A255071), but also differing in some aspects. For example, here the branching degree is not the constant 2, but can vary from 1 to 4. (Cf. A255058.)

Antti Karttunen, Table of n, a(n) for n = 0..16142

First differences: A255336.
Terms halved: A255057.
Cf. A255053 & A255055 (the lower & upper bound for a(n)) and also A255123, A255124 (distances to those limits).
Cf. A255327, A255058 (branching degree for node n), A255330 (number of nodes in the finite subtrees branching from the node n), A255331, A255332
Subsequence: A000918 (except for -1).
Cf. A255061, A255062, A255071, A255072, A255066, A255122.
Cf. A254113, A254114.
Cf. A255063, A255064, A255125, A255126.
Similar "beanstalk's trunk" sequences using some other subtracting map than A236840: A179016, A219648, A219666.

(define (A255056 n) (A255066 (A255122 n)))

a(n) = A255066(A255122(n)).
Other identities and observations. For all n >= 0:
a(n) = 2*A255057(n).
A255072(a(n)) = n.
A255053(n) <= a(n) <= A255055(n).

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).

Original entry on oeis.org

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).

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).

A255329 a(n) = total number of nodes in the finite subtrees branching "right" (to the "larger side") from the node n in the infinite trunk of "number-of-runs beanstalk" (A255056).

Original entry on oeis.org

0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 4, 1, 0, 0, 0, 0, 0, 1, 0, 4, 1, 0, 0, 1, 0, 0, 4, 1, 0, 0, 0, 0, 0, 1, 0, 4, 1, 0, 0, 1, 7, 1, 0, 3, 0, 1, 0, 0, 4, 1, 0, 0, 1, 0, 0, 4, 1, 0, 0, 0, 0, 0, 1, 0, 4, 1, 0, 0, 1, 7, 1, 0, 3, 0, 1, 7, 1, 0, 16, 0, 0, 1, 0, 3, 0, 1, 0, 0, 4, 1, 0, 0, 1, 7, 1, 0, 3, 0, 1, 0, 0, 4, 1, 0, 0, 1, 0, 0, 4, 1, 0, 0, 0, 0, 0, 1, 0, 4

Offset: 0

Antti Karttunen, Feb 21 2015

nonn

			See example in A255330. Here we count only the nodes at the right side, thus a(11) = 1+3 = 4.

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

Cf. A255056, A255057, A255327, A255328, A255330, A255331.

(define (A255329 n) (add A255327 (A255056 (+ n 1)) (A255068 (A255057 n))))
;; Other code as in A255327.

a(n) = sum_{k = A255056(n+1) .. A255068(A255057(n))} A255327(k).

a(n) = A255330(n) - A255328(n).

A255058 Branching degree of node n in the trunk of number-of-runs beanstalk: a(n) = A106836(1+A255057(n)).

Original entry on oeis.org

3, 3, 1, 4, 2, 1, 4, 1, 3, 2, 1, 4, 3, 4, 1, 3, 1, 3, 2, 1, 4, 3, 4, 1, 3, 3, 1, 3, 3, 4, 1, 3, 1, 3, 2, 1, 4, 3, 4, 1, 3, 3, 2, 4, 4, 1, 3, 3, 1, 3, 3, 4, 1, 3, 3, 1, 3, 3, 4, 1, 3, 1, 3, 2, 1, 4, 3, 4, 1, 3, 3, 2, 4, 4, 1, 3, 3, 2, 4, 4, 1, 1, 2, 3, 4, 1, 3, 3, 1, 3, 3, 4, 1, 3, 3, 2, 4, 4, 1, 3, 3, 1, 3, 3, 4, 1, 3, 3, 1, 3, 3, 4, 1, 3, 1, 3, 2, 1, 4

Offset: 0

Antti Karttunen, Feb 20 2015

nonn

Iff a(n) = 1, then A255330(n) = 0.

If a(n) = 1, then A255331(n) = 0.

			The node 11 in the infinite trunk is A255056(11) = 30. Apart from 32, which is the next node (node 12) in the infinite trunk, it has one leaf-child 31 at the "left side" (less than 32), and one leaf-child 33 (more than 32) at the "right side", and also at that side a subtree of three nodes (34 <- 38 <- 43), starting from 34, so in total there are four branches emanating from 30, [i.e., four different k such that A236840(k) = 30], thus a(11) = 4.
Note that a(0) = 3, as for node zero, we count among its children the following cases A236840(2) = 0, A236840(1) = 0, and also A236840(0) = 0, with 0 being exceptionally its own child.

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

Cf. A005811, A106836, A236840, A255330, A255331, A255056, A255057.

(define (A255058 n) (A106836 (+ 1 (A255057 n))))

a(n) = A106836(1+A255057(n)).

A262894 a(n) = A262888(n) - A262889(n).

Original entry on oeis.org

6, 0, 41, 0, 0, 5, 0, 16, 0, 2, 0, -1, 1, 22, 4, 0, 0, -3, 0, -1, -13, 0, 105, 2, -1, -1, -2, 1, 18, 7, 0, 0, 0, 1, -1, -3, 0, 0, -5, 0, -4, 13, 0, -7, 0, 0, -7, 6, 1, 0, 0, 0, 41, 0, 0, 0, 90, -1, 0, 5, 0, -2, 0, 1, -1, 0, 12, -1, 0, 3, 61, 0, 0, 0, 0, 0, 0, 0, 117, 7, 0, 2, 10, 0, 0, 1, 23, -1, 1, 1, 0, 0, 1, 0, -5, -1, 0, 1, 2, 2, 568, -1, 1, 1, 4, -1, 5, 9, -3, 0, -22, -1, 0

Offset: 0

Antti Karttunen, Oct 04 2015

sign

Antti Karttunen, Table of n, a(n) for n = 0..8107

Cf. A262888, A262889, A262895 (partial sums).

Cf. also A255331.

(define (A262894 n) (- (A262888 n) (A262889 n)))

a(n) = A262888(n) - A262889(n).

cp's OEIS Frontend

A255332 Partial sums of A255331.

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A255056 Trunk of number-of-runs beanstalk: The unique infinite sequence such that a(n-1) = a(n) - number of runs in binary representation of a(n).

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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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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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A255329 a(n) = total number of nodes in the finite subtrees branching "right" (to the "larger side") from the node n in the infinite trunk of "number-of-runs beanstalk" (A255056).

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A255058 Branching degree of node n in the trunk of number-of-runs beanstalk: a(n) = A106836(1+A255057(n)).

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A262894 a(n) = A262888(n) - A262889(n).

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