A338763 - OEIS

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A338763

Let L_1 = (1, 2, ...); for any n > 0, let M_n = Min_{k > 0} L_n(k) + L_n(k+1) and K_n = Min_{ k | L_n(k) + L_n(k+1) = M_n }, L_{n+1} is obtained by replacing the two terms L_n(K_n) and L_n(K_n+1) by M_n in L_n; a(n) = M_n.

3, 6, 9, 13, 15, 17, 21, 25, 28, 29, 33, 37, 38, 41, 45, 49, 53, 54, 57, 61, 65, 66, 69, 70, 73, 77, 81, 85, 86, 89, 93, 97, 101, 102, 105, 109, 113, 117, 118, 120, 121, 125, 129, 133, 134, 137, 141, 145, 149, 150, 153, 156, 157, 161, 165, 166, 169, 173, 177

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

In other words, we start with the natural numbers and repeatedly replace the leftmost pair of consecutive terms with minimal sum by its sum; a(n) corresponds to the sum at n-th step.

This sequence is weakly increasing, and tends to infinity (as for any m > 0, there are only finitely many runs of two or more consecutive integers with a sum < m).

LINKS

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Rémy Sigrist, PARI program for A338763

FORMULA

a(n) <= a(n+1).

EXAMPLE

The first terms, alongside L_n, are:

n a(n) L_n

-- ---- ----------------------------------------------------------

1 3 { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ... }

2 6 { 3, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ... }

3 9 { 6, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ... }