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

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

Author

Matthew Malone, Aug 02 2019

Keywords

• nonn
• easy

Examples

			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,...}
		

Crossrefs

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

Programs

Kotlin

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)
}

Mathematica

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 *)

PARI

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

Formula

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.