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%I #14 Nov 09 2020 00:29:22
%S 3,6,9,13,15,17,21,25,28,29,33,37,38,41,45,49,53,54,57,61,65,66,69,70,
%T 73,77,81,85,86,89,93,97,101,102,105,109,113,117,118,120,121,125,129,
%U 133,134,137,141,145,149,150,153,156,157,161,165,166,169,173,177
%N 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.
%C 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.
%C 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).
%H Rémy Sigrist, <a href="/A338763/b338763.txt">Table of n, a(n) for n = 1..10000</a>
%H Rémy Sigrist, <a href="/A338763/a338763.gp.txt">PARI program for A338763</a>
%F a(n) <= a(n+1).
%e The first terms, alongside L_n, are:
%e n a(n) L_n
%e -- ---- ----------------------------------------------------------
%e 1 3 { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ... }
%e 2 6 { 3, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ... }
%e 3 9 { 6, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ... }
%e 4 13 { 6, 9, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ... }
%e 5 15 { 6, 9, 13, 8, 9, 10, 11, 12, 13, 14, 15, ... }
%e 6 17 { 15, 13, 8, 9, 10, 11, 12, 13, 14, 15, ... }
%e 7 21 { 15, 13, 17, 10, 11, 12, 13, 14, 15, ... }
%e 8 25 { 15, 13, 17, 21, 12, 13, 14, 15, ... }
%e 9 28 { 15, 13, 17, 21, 25, 14, 15, ... }
%e 10 29 { 28, 17, 21, 25, 14, 15, ... }
%o (PARI) See Links section.
%Y Cf. A326936, A338764.
%K nonn
%O 1,1
%A _Rémy Sigrist_, Nov 07 2020
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