OFFSET
1,3
COMMENTS
Each next row is the last row concatenated with the last row reversed with elements incremented. A000120 is produced by a similar principle, omitting the reversal step. [Edited by Andrey Zabolotskiy, Mar 27 2020]
From Gary W. Adamson, Aug 08 2009: (Start)
The row with 8 terms: (1, 2, 3, 2, 3, 4, 3, 2); can be used to generate the numbers of hydrogen bonds per codon/anti-codon; superimposed on the DNA codon array of A147995 as follows: top row and left column of an 8 X 8 array is composed of the 8 terms (1, 2, 3, 2, 3, 4, 3, 2). If rows and columns have an offset of "1", then odd rows circulate downward starting from the position (n,n). Even rows circulate in the opposite direction starting from position (n,n).
This produces the array:
1 2 3 2 3 4 3 2
2 1 2 3 4 3 2 3
3 2 1 2 3 2 3 4
2 3 2 1 2 3 2 3
3 4 3 2 1 2 3 2
4 3 2 3 2 1 4 3
3 2 3 4 3 2 1 2
2 3 4 3 2 3 2 1
...
This produces a semi-magic square with a diagonal of (1,1,1,...). Using the simple replacement rule ("complement to 10"): (1->9); (2->8); (3->7); (4->6) we obtain the chart of DNA hydrogen bonds per codon/anti-codon shown in A147995. Top row of the hydrogen bond array as well as left column = (9, 8, 7, 8, 7, 6, 7, 8).
Alternatively, using the circulant rule for alternate rows and putting all 9's along the diagonal, we obtain the chart of hydrogen bonds. (End)
Rows tend to A088748 (which can also be generated from the dragon curve, A014577). - Gary W. Adamson, Aug 30 2009
Conjecture: positions of records are A081254. - Andrey Zabolotskiy, Mar 27 2020
REFERENCES
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, pp. 116-117.
LINKS
Andrey Zabolotskiy, Table of n, a(n) for n = 1..8191
EXAMPLE
Fractions in the left branch of the infinite Stern-Brocot tree (the fractions between 0 and 1), are:
1/2;
1/3, 2/3;
1/4, 2/5, 3/5, 3/4;
1/5, 2/7, 3/8, 3/7, 4/7, 5/8, 5/7, 4/5;
...
and their corresponding continued fraction representations are:
[2]
[3] [1,2]
[4] [2,2] [1,1,2] [1,3]
[5] [3,2] [2,1,2] [2,3] [1,1,3] [1,1,1,2] [1,2,2] [1,4]
...
with the number of terms in each continued fraction representation generating the present triangle:
1
1 2
1 2 3 2
1 2 3 2 3 4 3 2
...
MATHEMATICA
sb[n_List] := Block[{k = l = Length[n], a = n}, While[k > 1, a = Insert[ a, (Numerator[ a[[k]]] + Numerator[ a[[k - 1]]]) / (Denominator[ a[[k]]] + Denominator[ a[[k - 1]]]), k]; k-- ]; a]; sbn[n_] := Complement[ Nest[ sb, {0, 1}, n], Nest[ sb, {0, 1}, n - 1]]; f[n_] := Length /@ (ContinuedFraction /@ sbn[n]) - 1; Flatten[ Table[ f[n], {n, 7}]] (* Robert G. Wilson v, Jun 09 2004 *)
PROG
(Haskell)
a088696 n = a088696_list !! (n-1)
a088696_list = f [1] where
f (x:xs) = x : f (xs ++ [x + 1 - x `mod` 2, x + x `mod` 2])
-- Reinhard Zumkeller, Mar 07 2011
(Python)
a = [[1]]
for n in range(6):
a.append(a[-1] + [x+1 for x in a[-1][::-1]])
print(sum(a, []))
# Andrey Zabolotskiy, Mar 27 2020, after Jon Maiga
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Gary W. Adamson, Oct 07 2003
EXTENSIONS
Edited and extended by Robert G. Wilson v, Jun 09 2004
STATUS
approved