A307707 Lexicographically earliest sequence of nonnegative integers in which, for all k >= 0, there are exactly k pairs of consecutive terms whose sum is k.
0, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 3, 3, 3, 3, 4, 3, 4, 3, 4, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8
Offset: 1
Links
- Jean-Marc Falcoz, Table of n, a(n) for n = 1..11326
Crossrefs
Programs
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Mathematica
m = 107; a[1]=0; a24[n_] := Ceiling[(Sqrt[8n+1]-1)/2]; Array[a, m] /. Solve[Table[a[n] + a[n+1] == a24[n], {n, 1, m-1}]][[1]] (* Jean-François Alcover, Jun 02 2019, after Rémy Sigrist's formula *)
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PARI
v=0; rem=wanted=1; for (n=1, 107, print1 (v", "); v=wanted-v; if (rem--==0, rem=wanted++)) \\ Rémy Sigrist, Apr 23 2019
Formula
a(n) + a(n+1) = A002024(n). - Rémy Sigrist, Apr 24 2019
Let t_m = m*(m+1)/2. Write n = t_m - i with m >= 1 and 0 <= i < m. Then a(n) = m/2 if m is even, or if m is odd, a(n) = (m-1)/2 + (i-1 mod 2). - N. J. A. Sloane, Nov 16 2024
Extensions
Definition clarified by Rémy Sigrist and N. J. A. Sloane, Nov 17 2024
Comments