Matthew Malone - cp's OEIS frontend

A346707 "Look once to the left" sequence, omitting a(k) for each iteration k, starting with 1,2 (see example).

1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2

Offset: 1

Matthew Malone, Jul 29 2021

nonn

The ratio of number of 2's to number of 1's appears to converge to 1.3985918...

			Begin with [1, 2].
Iteration 1: append to self, omitting term 1: [1, 2] + [2] = [1, 2, 2].
Iteration 2: append to self, omitting term 2: [1, 2, 2] + [1, 2] = [1, 2, 2, 1, 2].
Iteration 3: append to self, omitting term 3: [1, 2, 2, 1, 2] + [1, 2, 1, 2] = [1, 2, 2, 1, 2, 1, 2, 1, 2].

Cf. A000051, A293838.

Block[{a = {1, 2}}, Do[a = Join[a, Delete[a, i]], {i, 7}]; a] (* Michael De Vlieger, Aug 04 2021 *)

a(n) = n-=2; while(n>0, my(k=logint(n,2)); n-=1<Kevin Ryde, Aug 03 2021

def sequence(iterations, start=[1,2]):
  a = start
  for k in range(0, iterations):
    a = a + a[:k] + a[k+1:]
  return a

A343453 The number of 3's minus the number of 2's among the first n terms of A342101.

Original entry on oeis.org

0, -1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 2, 2, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 0, 1, 1, 2, 2, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 3

Offset: 1

Matthew Malone, Apr 15 2021

sign

It appears that the sequence never goes below -1 and increases without bound.

It appears that if the first appearance of a number x occurs at index n and the first appearance of x+1 appears at index m then m/n approaches 4 as x increases.

			A342101 = [1, 2, 3, 1, 3, ...]. By the fifth term of A342101 we see 2 terms with value 3, and a single term with value 2, so a(5) = 2 - 1 = 1.

Kevin Ryde, PARI/GP Code and Notes.

Cf. A342101.

fun a(iter: Int): List = runningSum(twosVersusThrees(iter))
fun runningSum(a: List) = a.drop(1).fold(listOf(a[0])) { acc, cur ->
    acc + (acc.last() + cur)
}
fun twosVersusThrees(iter: Int): List = removeMiddle(listOf(0,-1,1), iter)
fun removeMiddle(initial: List, iter: Int): List {
    if (iter < 2) return initial
    val prev = removeMiddle(initial, iter-1)
    return prev + prev.subList(0, (prev.size - 1) / 2) + prev.subList((prev.size + 1) /2, prev.size)
}

Block[{a = {}, s = Nest[Join[#, Drop[#, {(Length[#] + 1)/2}]] &, Range[3], 6], c}, Array[Set[c[#], 0] &, 3]; Do[c[ s[[i]] ]++; AppendTo[a, c[3] - c[2]], {i, Min[Length@ s, 104]}]; a] (* Michael De Vlieger, May 01 2021 *)

PARI
```
\\ See links.
```

A338643 Cycle lengths arising from interpreting sequence prefixes as graph node offsets.

Original entry on oeis.org

1, 1, 2, 2, 3, 2, 3, 4, 3, 3, 4, 5, 3, 4, 4, 5, 6, 4, 4, 5, 5, 6, 7, 11, 5, 5, 6, 6, 7, 8, 5, 13, 6, 6, 7, 7, 8, 9, 6, 6, 16, 7, 7, 8, 8, 9, 10, 17, 7, 7, 10, 8, 8, 9, 9, 10, 11, 8, 10, 8, 8, 11, 9, 9, 10, 10, 11, 12, 8, 9, 11, 9, 9, 12, 10, 10, 11, 11, 12, 13

Offset: 1

Matthew Malone, Apr 21 2021

nonn

			Start with a(1) = 1. Treat a(1) = 1 as a node pointing 1 space ahead which, wrapping around the end of the length-1 prefix, points to itself. This graph has a cycle length of 1, so a(2) = 1.
Now a(1..2) = [1, 1]. a(1) = 1 points 1 space ahead to a(2) and a(2) = 1 points 1 space ahead and wraps around to a(1), completing a cycle of length 2 so a(3) = 2.
Skipping ahead a bit, a(1..5) = [1, 1, 2, 2, 3]. In this prefix, a(1) = 1 points to a(2) = 1, which points to a(3) = 2, which points to a(5) = 3, which wraps around and points to a(3), for a cycle length of 2, so a(6) = 2.

Rémy Sigrist, Table of n, a(n) for n = 1..50000
Rémy Sigrist, C program for A338643

C
```
See Links section.
```

fun graph_sequence(terms: Int): List {
    val a = mutableListOf(1)
    repeat(terms-1) { a += a.loopLength() }
    return a
}
fun List.loopLength(): Int {
    val visitedIndices = mutableMapOf()
    var idx = 0
    var counter = 0
    while (idx !in visitedIndices) {
        visitedIndices[idx] = counter++
        idx = (idx + this[idx]) % this.size
    }
    return counter - (visitedIndices[idx] ?: 0)
}

A342101 Remove middle term and append, starting with [1, 2, 3].

Original entry on oeis.org

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

Offset: 1

Matthew Malone, Feb 28 2021

nonn

			Start with [1, 2, 3], take that sequence, remove the middle term, 2, and append to the original sequence, yielding [1, 2, 3, 1, 3]. Then repeat this process to give [1, 2, 3, 1, 3, 1, 2, 1, 3], and so on.

Alois P. Heinz, Table of n, a(n) for n = 1..65537
Kevin Ryde, PARI/GP Code and Notes.

Cf. A000051.

fun A342101(iter: Int): List = removeMiddle(listOf(1,2,3), iter)
fun removeMiddle(initial: List, iter: Int): List {
  if (iter < 2) return initial
  val prev = removeMiddle(initial, iter-1)
  return prev + prev.subList(0, (prev.size - 1) / 2) + prev.subList((prev.size + 1) /2, prev.size)
}

T:= proc(n) option remember; `if`(n=1, [$1..3][],
      subsop(2^(n-2)+1=[][], [seq(T(i), i=1..n-1)])[])
    end:
seq(T(n), n=1..8);  # Alois P. Heinz, Apr 12 2021

Nest[Join[#, Drop[#, {(Length[#] + 1)/2}]] &, Range[3], 6] (* Michael De Vlieger, May 01 2021 *)

first(n) = { my(v = [1,2,3]); for(i = 1, logint(n-1, 2), cv = v[^(#v + 1)\2]; v = concat(v, cv) ); v } \\ David A. Corneth, Apr 14 2021

PARI
```
\\ Also see links.
```

def aupton(terms):
  alst = [1, 2, 3]
  while len(alst) < terms:
    alst += alst[:len(alst)//2] + alst[(len(alst)+1)//2:]
  return alst[:terms]
print(aupton(87)) # Michael S. Branicky, Mar 26 2021

A309435 Iteratively replace the product of two sequentially chosen consecutive integers with those integers.

Original entry on oeis.org

1, 2, 3, 4, 5, 2, 3, 7, 8, 9, 5, 2, 11, 3, 4, 13, 14, 15, 16, 17, 18, 19, 4, 5, 3, 7, 2, 11, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 11, 3, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 9, 5, 46, 47, 48, 49, 50, 51, 4, 13, 53, 54, 55, 7, 8, 57, 58, 59, 60

Offset: 1

Matthew Malone, Aug 02 2019

			A_0 = {1,2,3,4,5,...}
A_0(0) * A_0(1) = 1 * 2 = 2, which is not found after A_0(1), so A_1 = A_0.
A_1(1) * A_1(2) = 2 * 3 = 6, which *is* found after A_1(2), so A_2 = {1,2,3,4,5,2,3,7,8,...}
A_2(2) * A_2(3) = 3 * 4 = 12, A_3 = {1,2,3,4,5,2,3,7,8,9,10,11,3,4,13,...}
A_3(3) * A_3(4) = 4 * 5 = 20, A_4 = {1,2,3,4,5,2,3,7,8,9,10,11,3,4,13,...,19,4,5,21,...}
A_4(4) * A_4(5) = 5 * 2 = 10, A_5 = {1,2,3,4,5,2,3,7,8,9,5,2,11,3,4,13,...,19,4,5,21,...}

This sequence is similar to A309503 except it uses multiplication instead of addition.

fun generate(len: Int): List {
    fun gen_inner(len: Int, level: Int): List {
        if (level < 1) return (1..len).toList()
        val prev = gen_inner(len, level - 1)
        if (level == len) return prev.take(len)
        val (a,b) = prev[level - 1] to prev[level]
        return if (prev.drop(level + 1).contains(a*b)) {
            prev.indexOfFirst { it == a*b }.let { idx ->
                prev.take(idx) + a + b + prev.drop(idx + 1)
            }
        } else prev
    }
    return gen_inner(len,len)
}

T = Range[100]; Do[p = T[[i]] T[[i + 1]]; Do[If[T[[j]] == p, T = Join[ T[[;; j-1]], {T[[i]], T[[i+1]]}, T[[j+1 ;;]]]; Break[]], {j, i+2, Length@ T}], {i, Length@T}]; T (* Giovanni Resta, Sep 20 2019 *)

a = vector(92, k, k); for (n=1, #a, print1 (a[n] ", "); s=a[n]*a[n+1]; for (k=n+2, #a, if (a[k]==s, a=concat([a[1..k-1], a[n..n+1], a[k+1..#a]]); break))) \\ Rémy Sigrist, Aug 03 2019

To generate the sequence, start with the integers, A_0={1,2,3,4,5,...}. To generate A_{n+1} calculate x = A_n(n) * A_n(n+1). Replace the next instance of x in A_n (after A_n(n+1)) with A_n(n), A_n(n+1). The limit of this process gives the sequence.

A309503 Iteratively replace the sum of two sequentially chosen consecutive integers with those integers.

Original entry on oeis.org

1, 2, 1, 2, 4, 5, 2, 4, 5, 2, 8, 4, 5, 2, 8, 11, 8, 4, 13, 14, 15, 16, 4, 13, 4, 14, 8, 11, 16, 4, 8, 13, 22, 23, 24, 4, 8, 13, 4, 14, 8, 13, 14, 24, 4, 14, 15, 30, 15, 16, 24, 8, 33, 34, 13, 22, 36, 37, 14, 24, 8, 13, 4, 14, 24, 16, 41, 42, 8, 13, 22, 44

Offset: 1

Matthew Malone, Aug 05 2019

			A_0 = {1,2,3,4,5,...}.
A_0(0) + A_0(1) = 1 + 2 = 3, which is found after A_0(1), so A_1 = {1,2,1,2,4,5,...}.
A_1(1) + A_1(2) = 2 + 1 = 3, which is *not* found after A_1(2), so A_2 = A_1.
A_2(2) + A_2(3) = 1 + 2 = 3, A_3 = A_2.
A_3(3) + A_3(4) = 2 + 4 = 6, A_4 = {1,2,1,2,4,5,2,4,7,8,9,10,...}.
A_4(4) + A_4(5) = 4 + 5 = 9, A_5 = {1,2,1,2,4,5,2,4,7,8,4,5,10,...}.

This sequence is similar to A309435 except it uses addition instead of multiplication.

fun generate(len: Int): List {
    fun gen_inner(len: Int, level: Int): List {
        if (level < 1) return (1..len).toList()
        val prev = gen_inner(len, level - 1)
        if (level == len) return prev.take(len)
        val (a, b) = prev[level - 1] to prev[level]
        return if (prev.drop(level + 1).contains(a+b)) {
            prev.indexOfFirst { it == a+b }.let { idx ->
                prev.take(idx) + a + b + prev.drop(idx + 1)
            }
        } else prev
    }
    return gen_inner(len, len)
}

T = Range[100]; Do[s = T[[i]] + T[[i + 1]]; Do[If[T[[j]] == s, T = Join[ T[[;; j-1]], {T[[i]], T[[i+1]]}, T[[j+1 ;;]]]; Break[]], {j, i+2, Length@ T}], {i, Length@T}]; T (* Giovanni Resta, Sep 20 2019 *)

a = vector(92, k, k);
for (n=1, #a-1, s=a[n]+a[n+1]; print1 (a[n] ", "); for (k=n+2, #a - 1, if (a[k]==s, a=concat([a[1..k-1], a[n..n+1], a[k+1..#a]]); break)))

To generate the sequence, start with the integers, A_0={1,2,3,4,5,...}. To generate A_{n+1} calculate x = A_n(n) + A_n(n+1). Replace the next instance of x in A_n (after A_n(n+1)) with A_n(n), A_n(n+1). The limit of this process gives the sequence.

cp's OEIS Frontend

User: Matthew Malone

A346707 "Look once to the left" sequence, omitting a(k) for each iteration k, starting with 1,2 (see example).

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PARI

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A343453 The number of 3's minus the number of 2's among the first n terms of A342101.

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Kotlin

Mathematica

PARI

A338643 Cycle lengths arising from interpreting sequence prefixes as graph node offsets.

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C

Kotlin

A342101 Remove middle term and append, starting with [1, 2, 3].

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Kotlin

Maple

Mathematica

PARI

PARI

Python

A309435 Iteratively replace the product of two sequentially chosen consecutive integers with those integers.

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Programs

Kotlin

Mathematica

PARI

Formula

A309503 Iteratively replace the sum of two sequentially chosen consecutive integers with those integers.

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Author

Keywords

Examples

Crossrefs

Programs

Kotlin

Mathematica

PARI

Formula