A349472 a(1)=1; for n > 1, a(n) is the smallest unused positive number such that gcd(a(n-1)+n,a(n)) > 1.

1, 3, 2, 4, 6, 8, 5, 13, 10, 12, 23, 7, 14, 16, 31, 47, 18, 9, 20, 15, 21, 43, 11, 25, 22, 24, 17, 27, 26, 28, 59, 35, 30, 32, 67, 103, 34, 33, 36, 19, 38, 40, 83, 127, 42, 44, 39, 29, 45, 50, 101, 48, 202, 46, 303, 359, 52, 54, 113, 173, 51, 226, 68, 55, 56, 58, 60, 62, 131, 57, 64, 66

Offset: 1

Author

Scott R. Shannon, Nov 19 2021

Keywords

• nonn

Comments

After 200000 terms the smallest unused number is 28349, although like similar sequences this is almost certainly a permutation of the positive integers. In the same range the fixed points are 4, 21, 50, 1269; it is likely no more exist. See the linked image.

Examples

			a(2) = 3 as a(1) + 2 = 3, 3 has not previously appeared, and gcd(3,3) > 1.
a(3) = 2 as a(2) + 3 = 6, 2 has not previously appeared, and gcd(6,2) > 1.
a(12) = 7 as a(11) + 12 = 35, 7 has not previously appeared, and gcd(35,7) > 1.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Scott R. Shannon, Image of the first 200000 terms. The green line is y = n.

Crossrefs

Cf. A064413, A347113, A098550, A336957.

Programs

Mathematica

nn = 72; c[] = False; u = 2; a[1] = j = 1; c[1] = True; Do[Set[{k, m}, {u, n + j}]; While[Or[c[k], CoprimeQ[k, m]], k++]; Set[{a[n], c[k], j}, {k, True, k}]; If[k == u, While[c[u], u++]], {n, 2, nn}]; Array[a, nn] (* _Michael De Vlieger, Oct 05 2022 *)

Python

from math import gcd
terms, appears = [1], {}
for n in range(2, 100):
    t = 2
    while True:
        if appears.get(t) is None and gcd(terms[-1]+n, t) > 1  :
            appears[t] = True; terms.append(t); break
        t += 1
print(terms) # Gleb Ivanov, Nov 19 2021