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A343856
Irregular table read by rows; the first row is [1]; to obtain the next row, replace each odd-indexed term u with (u, u), and each even-indexed term v with (2*v).
1
1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 2, 2, 4, 2, 2, 1, 1, 2, 2, 2, 4, 2, 2, 8, 2, 2, 4, 1, 1, 2, 2, 2, 4, 2, 2, 8, 2, 2, 4, 8, 8, 4, 2, 2, 8, 1, 1, 2, 2, 2, 4, 2, 2, 8, 2, 2, 4, 8, 8, 4, 2, 2, 8, 8, 8, 16, 4, 4, 4, 2, 2, 16, 1, 1, 2, 2, 2, 4, 2, 2, 8, 2, 2, 4, 8, 8, 4, 2, 2, 8, 8, 8, 16, 4, 4, 4, 2, 2, 16, 8, 8, 16, 16, 16, 8, 4, 4, 8, 2, 2, 4, 16, 16
OFFSET
1,6
COMMENTS
Sequence A061419 and A343857 gives row lengths and partial sums, respectively.
The n-th row sums to 2^(n-1).
This sequence has fractal features.
As with Jim Conant's iterative dissection of a square (A328078), at each iteration, we split in two odd-indexed elements.
This sequence has similarities with A205592: in A205592:
- we start with A205592(1) = 1,
- for k = 1, 2, ...:
if k is odd: append two copies of A205592(k),
if k is even: append 2*A205592(k).
EXAMPLE
Table begins:
1: [1]
2: [1, 1]
3: [1, 1, 2]
4: [1, 1, 2, 2, 2]
5: [1, 1, 2, 2, 2, 4, 2, 2]
6: [1, 1, 2, 2, 2, 4, 2, 2, 8, 2, 2, 4]
7: [1, 1, 2, 2, 2, 4, 2, 2, 8, 2, 2, 4, 8, 8, 4, 2, 2, 8]
PROG
(PARI) { a = r = [1]; for (n=1, 8, i = 0; a=concat(a, r = concat(apply (v -> if (i++%2, [v, v], [2*v]), r)))); print (a) }
(Python)
def auptorow(rows):
alst, row, newrow = [1], [1], []
for r in range(2, rows+1):
for i, v in enumerate(row, start=1): newrow += [v, v] if i%2 else [2*v]
alst, row, newrow = alst + newrow, newrow, []
return alst
print(auptorow(9)) # Michael S. Branicky, May 04 2021
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Rémy Sigrist, May 01 2021
STATUS
approved