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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
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
1,2
COMMENTS
Row n consists of permutation of the integers from F(n+2) to F(n+3) - 1, where F(n) = A000045(n). The permutation is generated using Kevin Ryde's formula from A194959.
The sequence is an intra-block permutation of the positive integers.
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).
EXAMPLE
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;
...
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
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Boris Putievskiy, Oct 17 2024
STATUS
approved