A198069 Table read by rows, T(0,0) = 1 and for n>0, 0<=k<=2^(n-1) T(n,k) = gcd(k,2^(n-1)).
1, 1, 1, 2, 1, 2, 4, 1, 2, 1, 4, 8, 1, 2, 1, 4, 1, 2, 1, 8, 16, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, 32, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 32, 64, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2
Offset: 0
Examples
1 1, 1 2, 1, 2 4, 1, 2, 1, 4 8, 1, 2, 1, 4, 1, 2, 1, 8 16, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16
Links
- Reinhard Zumkeller, Rows n = 0..13 of triangle, flattened
Programs
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Haskell
a198069 n k = a198069_tabf !! n !! k a198069_row n = a198069_tabf !! n a198069_tabf = [0] : iterate f [1, 1] where f (x:xs) = ys ++ tail (reverse ys) where ys = (2 * x) : xs -- Reinhard Zumkeller, May 26 2013
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Maple
# In triangular form: seq(print(seq(gcd(k,2^(n-1)),k=0..2^(n-1))),n=0..6);
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Mathematica
Join[{1},Flatten[Table[GCD[k,2^(n-1)],{n,10},{k,0,2^(n-1)}]]] (* Harvey P. Dale, Oct 30 2021 *)
Formula
For n > 0: Let S be the n-th row, S' = replace the initial term by its double, then row (n+1) = concatenation of S' and the reverse of S' without the initial term. - Reinhard Zumkeller, May 26 2013