A309503 Iteratively replace the sum of two sequentially chosen consecutive integers with those integers.
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
Examples
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,...}.
Crossrefs
This sequence is similar to A309435 except it uses addition instead of multiplication.
Programs
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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[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 *) -
PARI
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)))
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.