A335292 To calculate a(n), iterate the function f(x) = x * ceiling(a(n-1) / 2) + 1 on the value x = 1 until the result has not appeared in the sequence previously. Take a(1) = 1.

1, 2, 3, 7, 5, 4, 15, 9, 6, 13, 8, 21, 12, 43, 23, 157, 80, 41, 22, 133, 68, 35, 19, 11, 259, 131, 67, 1191, 597, 300, 151, 77, 40, 421, 212, 107, 55, 29, 16, 73, 38, 20, 111, 57, 30, 241, 122, 62, 32, 17, 10, 31, 273, 138, 70, 36, 343, 173, 88, 45, 24, 1885

Offset

1,2

Comments

By definition, this sequence never repeats its values.

Links

Omer Demir, Table of n, a(n) for n = 1..10000

Example

a(6) = 4. Therefore, ceiling(a(6) / 2) is equal to 2, and a(7) should be 1 * 2 + 1 = 3. However, since 3 can already be found in the sequence (coincidentally, at position n = 3), one must iterate the function again to get a(7) = 3 * 2 + 1 = 7. But 7 is also already in the sequence (at position n = 4), so the function must be iterated once more to find that a(7) is indeed equal to 7 * 2 + 1 = 15.

Prog

(Python)
from math import ceil
sequence = [1]
terms = 1000
for n in range(2, terms + 1):
    nextnum = 1
    while nextnum in sequence:
        nextnum = nextnum * ceil(sequence[-1] / 2) + 1
    sequence.append(nextnum)
print(sequence)

Crossrefs

Sequence in context: A021425 A355278 A329412 * A060940 A318739 A205299

Adjacent sequences: A335289 A335290 A335291 * A335293 A335294 A335295

Keyword

nonn

Author

Omer Demir, May 30 2020

Status

approved