A354858 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that shares a factor with a(n-1) and both the sum a(n) + a(n-1) is distinct from all previous sums, a(i) + a(i-1), i=2..n-1, and the product a(n) * a(n-1) is distinct from all previous products, a(i) * a(i-1), i=2..n-1.

1, 2, 2, 4, 4, 6, 3, 9, 6, 8, 8, 10, 10, 12, 9, 15, 10, 16, 12, 15, 15, 18, 14, 20, 15, 21, 18, 20, 20, 22, 22, 24, 21, 27, 24, 26, 26, 28, 21, 35, 20, 38, 19, 57, 3, 63, 7, 56, 6, 58, 10, 55, 20, 52, 22, 55, 25, 60, 9, 69, 12, 70, 14, 72, 15, 75, 18, 70, 21, 75, 24, 68, 26, 72, 28, 74, 30, 65

Offset: 1

Author

Scott R. Shannon, Jun 09 2022

Keywords

• nonn
• look

Comments

This sequence uses a combination of the term selection rules of A354755 and A354753. The first forty-five terms are the same as A354755 beyond which they differ; see the examples below. In the first 500000 terms only six terms are prime, 2,3,7,19, with 2 and 3 occurring twice, the last being a(47) = 7. It is unknown if more appear. The only fixed points are 1,2,4,6, and it is likely no more exist.

Examples

			a(7) = 3 as a(6) = 6, and 3 is the smallest number that shares a factor with 6 and whose sum and product with the previous term, 6 + 3 = 9 and 6 * 3 = 18, have not previously appeared. Note 2 shares a factor with 6 but 6 + 2 = 8, and a sum of 8 has already occurred with a(4) + a(5) = 4 + 4 = 8, so 2 cannot be chosen.
a(46) = 63 as a(45) = 3, and 63 is the smallest number that shares a factor with 3 and whose sum and product with the previous term, 3 + 63 = 66 and 3 * 63 = 189, have not previously appeared. Note 60 shares a factor with 3 but the product 3 * 60 = 180 has already occurred with a(19) * a(20) = 12 * 15 = 180, so 60 cannot be chosen. This is the first term to differ from A354755.
		

Links

• Scott R. Shannon, Image of the first 500000 terms. The green line is y = n.

Crossrefs

Cf. A354755, A354753, A088177, A064413, A354803.

Programs

PARI

lista(nn) = my(va = vector(nn), vp = vector(nn-2), vs = vector(nn-2)); va[1] = 1; va[2] = 2; for (n=3, nn, my(k=2); while ((gcd(k, va[n-1]) == 1) || #select(x->(x==k*va[n-1]), vp) || #select(x->(x==k+va[n-1]), vs), k++); va[n] = k; vp[n-2] = k*va[n-1]; vs[n-2] = k+va[n-1];); va; \\ Michel Marcus, Jun 17 2022