A268811 Sequence of positive integers where each is chosen to be as small as possible subject to the condition that no three terms a(j), a(j+k), a(j+2k) (for any j and k) form a geometric progression.

1, 1, 2, 1, 1, 2, 2, 3, 3, 1, 1, 2, 1, 1, 2, 2, 3, 3, 2, 3, 3, 5, 5, 6, 5, 5, 6, 1, 1, 2, 1, 1, 2, 2, 3, 3, 1, 1, 2, 1, 1, 2, 2, 3, 3, 2, 3, 3, 5, 5, 6, 5, 5, 6, 2, 3, 3, 5, 5, 6, 5, 5, 6, 6, 7, 7, 6, 7, 7, 8, 8, 10, 6, 7, 7, 6, 7, 7, 8, 8, 10, 1, 1, 2, 1, 1, 2, 2, 3, 3, 1, 1, 2, 1, 1, 2, 2, 3, 3, 2, 3, 3, 5, 5, 6, 5, 5, 6, 1, 1, 2, 1, 1, 2, 2, 3, 3, 1, 1, 2, 1, 1, 2, 2, 3, 3, 2

Offset: 1

Author

Aaron David Fairbanks, Feb 13 2016

Keywords

• nonn
• easy

Comments

Apparently: all terms belong to A000452, and for any k > 0, the value A000452(k) first appears at index A265316(k+1). - Rémy Sigrist, May 13 2021

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000
• Rémy Sigrist, C program for A268811

Crossrefs

Cf. A000452, A229037, A265316.

Programs

C

// See Links section.

Python

A268811_list = []
for n in range(1000):
    i, j, b = 1, 1, set()
    while n-2*i >= 0:
        b.add(A268811_list[n-i]**2/A268811_list[n-2*i])
        i += 1
        while j in b:
            b.remove(j)
            j += 1
    A268811_list.append(j)