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a(n) = ceiling((n-1-sqrt(n+1))/2).
1

%I #28 Sep 08 2022 08:46:25

%S 0,0,0,1,1,2,2,2,3,3,4,4,5,5,5,6,6,7,7,8,8,9,9,9,10,10,11,11,12,12,13,

%T 13,14,14,14,15,15,16,16,17,17,18,18,19,19,20,20,20,21,21,22,22,23,23,

%U 24,24,25,25,26,26,27,27,27,28,28,29,29,30,30,31,31,32

%N a(n) = ceiling((n-1-sqrt(n+1))/2).

%C a(n) is a lower bound for the number of items outside the instance of n-1 at one end of a Colombian variant Langford pairing (A336747). For example, one of the most lop-sided pairings for n=7 is 4 1 6 1 7 4 3 5 2 6 3 2 7 5, and there are a(n)=2 items to the left of the first '6'. This bound is tight until at least n=184.

%H Edward Moody, <a href="https://github.com/EdwardMGraphite/colombian-langford">Java program for enumerating Colombian Langford pairings</a>

%t Table[Ceiling[(n - 1 - Sqrt[n + 1])/2], {n, 1, 100}] (* _Amiram Eldar_, Aug 21 2020 *)

%o (Magma) [Ceiling((n-1-Sqrt(n+1))/2) : n in [1..100]]; // _Wesley Ivan Hurt_, Aug 21 2020

%o (PARI) a(n) = ceil((n-1-sqrt(n+1))/2); \\ _Michel Marcus_, Aug 19 2020

%Y Cf. A336747.

%K nonn,easy

%O 1,6

%A _Edward Moody_, Aug 16 2020