A377136 Triangle read by rows. Each row is a permutation of a block of consecutive numbers; the blocks are disjoint and every positive number belongs to some block. The length of row n is the (n+1)-st Fibonacci number for n > 0; see Comments.
1, 2, 3, 4, 5, 7, 6, 8, 10, 12, 11, 9, 13, 15, 17, 19, 20, 18, 16, 14, 21, 23, 25, 27, 29, 31, 33, 32, 30, 28, 26, 24, 22, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 53, 51, 49, 47, 45, 43, 41, 39, 37, 35, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 88, 86, 84, 82, 80, 78, 76, 74, 72, 70, 68, 66, 64, 62, 60, 58, 56, 89
Offset: 1
Examples
Triangle begins: k = 1 2 3 4 5 6 7 8 n=1: 1; n=2: 2; n=3: 3, 4; n=4: 5, 7, 6; n=5: 8, 10, 12, 11, 9; n=6: 13, 15, 17, 19, 20, 18, 16, 14; Subtracting F(n)-1 from each term in row n produces a permutation of 1 .. F(n): 1; 1; 1,2; 1,3,2; 1,3,5,4,2; 1,3,5,7,8,6,4,2; ...
Links
- Boris Putievskiy, Table of n, a(n) for n = 1..6764
- Boris Putievskiy, Integer Sequences: Irregular Arrays and Intra-Block Permutations, arXiv:2310.18466 [math.CO], 2023.
- Index entries for sequences that are permutations of the natural numbers
Programs
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Mathematica
T[n_,k_]:=Module[{P,Result}, P= If[2*k-1 <=Fibonacci[n],2*k-1,2*(Fibonacci[n]+1-k)]; Result=P+Fibonacci[n+1]-1; Result]; Nmax=6; Table[T[n,k],{n,1,Nmax},{k,1,Fibonacci[n]}]//Flatten
Formula
T(n,k) for 1 <= k <= F(n) (see Example):
T(n,k) = P(n,k) + F(n+1)-1, T(n,k) = P(n,k) + A000045(n+1)-1, where P(n,k) = 2*k-1 if 2*k-1 <= F(n), P(n,k) = 2*(F(n)+1-k) if 2*k-1 > F(n).
Comments