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A057027
Triangle T read by rows: row n consists of the numbers C(n,2)+1 to C(n+1,2); numbers in odd-numbered places form an increasing sequence and the others a decreasing sequence.
10
1, 2, 3, 4, 6, 5, 7, 10, 8, 9, 11, 15, 12, 14, 13, 16, 21, 17, 20, 18, 19, 22, 28, 23, 27, 24, 26, 25, 29, 36, 30, 35, 31, 34, 32, 33, 37, 45, 38, 44, 39, 43, 40, 42, 41, 46, 55, 47, 54, 48, 53, 49, 52, 50, 51, 56, 66, 57, 65, 58, 64, 59, 63, 60, 62, 61, 67, 78, 68, 77, 69, 76
OFFSET
1,2
COMMENTS
Arrange the quotients F(i)/F(j) of Fibonacci numbers, for 2<=i<j<=n, in increasing order. Then the positions of the F(i)/F(n-k) are the first n-k-2 terms of the diagonal T(i,i-k), for k=0,1,2,...,n-3.
FORMULA
From Werner Schulte, Sep 09 2024: (Start)
T(n, k) = (n^2 + (-1)^k * (n - k) + (3 + (-1)^k) / 2) / 2.
T(n, 1) = (n^2 - n + 2) / 2 = A000124(n).
T(n, 2) = (n^2 + n) / 2 = A000217(n) for n >= 2.
T(n, k) = T(n, k-2) - (-1)^k for 3 <= k <= n. (End)
G.f.: x*y*(1 + x*(y - 1) - x^4*(y - 1)*y^2 + x^5*y^3 + x^3*y*(y^2 - y - 1) - x^2*(y^2 + y - 1))/((1 - x)^3*(1 - x*y)^3*(1 + x*y)). - Stefano Spezia, Sep 10 2024
EXAMPLE
For n=6, the ordered quotients are 1/8, 1/5, 2/8, 1/3, 3/8, 2/5, 1/2, 3/5, 5/8, 2/3; the positions of 1/5, 2/5, 3/5 are 2, 6, 8 (first terms of diagonal T(i, i-1)).
Triangle starts:
1;
2, 3;
4, 6, 5;
7,10, 8, 9;
...
MATHEMATICA
nn= 12; t = Table[Range[Binomial[n, 2] + 1, Binomial[n + 1, 2]], {n, nn}]; Table[t[[n, If[OddQ@ k, Ceiling[k/2], -k/2] ]], {n, nn}, {k, n}] // Flatten (* Michael De Vlieger, Jul 02 2016 *)
CROSSREFS
Reflection of the array in A057028 about its central column, a permutation of the natural numbers.
Inverse permutation to A064578. Central column: A057029.
Column 1 is A000124, column 2 is A000217.
Row sums are A006003.
Sequence in context: A361251 A194969 A194981 * A371246 A090894 A371360
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, Jul 28 2000
EXTENSIONS
Corrected and extended by Vladeta Jovovic, Oct 18 2001
STATUS
approved