A352790 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not appeared that is a divisor of either a(n-2)+a(n-1) or a(n-2)*a(n-1).

1, 2, 3, 5, 4, 9, 6, 15, 7, 11, 18, 22, 8, 10, 16, 13, 26, 39, 65, 52, 20, 12, 24, 32, 14, 23, 37, 30, 67, 97, 41, 46, 29, 25, 27, 45, 36, 54, 72, 21, 28, 42, 35, 49, 84, 19, 38, 57, 95, 76, 171, 114, 285, 90, 50, 60, 40, 48, 44, 33, 66, 99, 55, 77, 121, 198, 242, 88, 110, 80, 100, 64, 82, 73

Offset: 1

Author

Scott R. Shannon, Apr 03 2022

Keywords

• nonn

Comments

The sequence shows large variations in its values due to some terms being required to equal the product of two large primes as all other divisors of both the sum and product have been used, e.g., a(25893) = 68485691. In the first 50000 terms there are 28233 occasions where the chosen term is a divisor of the sum and 22777 occasions where it is a divisor of the product. In the same range the fixed points are 1, 2, 3, 42, 3674, 11520, 18515, 39501 - it is likely more exist. The smallest number not to appear is 761, while the primes do not occur in their natural order. It is possible the sequence is finite as two terms could appear whose sum and product divisors have all been used. If not then the sequence is likely a permutation of the positive integers.

From Michael De Vlieger, Apr 03 2022: (Start)

Let S = a(n-1)+a(n+2) and let P = a(n-1)*a(n-2). Let u be the smallest missing number in a(1..n-1).

It is possible that S is prime, but P is prime iff either a(n-1)=1 or a(n-2)=1; since a(1) = 1 is given and followed by 2, for n > 2, P is always composite.

The axiom a(n) = (k | S) or (k | P) implies (k <= S) or (k <= P). Consequently, u <= k <= max(S,P). Let D contain divisors {d : d | S and d >= u} and let E contain factors {d : d | P and d >= u}. A solution k must appear in T = D U E.

For sufficiently large n, S is large, but P is large and composite.

In this sequence we may have equality of (one) input and output, since input S or P does not necessarily already exist in a.

(End)

Examples

			a(4) = 5 as a(2)+a(3) = 5 and a(2)*a(3) = 6. The combined divisors of these two numbers are 1,2,3,5,6, and of those 5 is the smallest that has not appeared.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Michael De Vlieger, Annotated log-log scatterplot of a(n), n = 1..2^16, showing records in red, local minima in blue, fixed points in gold, and accentuating primes in green.
• Scott R. Shannon, Line graph of the first 50000 terms.

Crossrefs

Cf. A352768, A332301, A351001, A349493, A251622.

Programs

Mathematica

nn = 74; c[] = 0; u = a[1] = c[1] = 1; a[2] = c[2] = 2; While[c[u] > 0, u++]; Do[k = 1; t = TakeWhile[Reverse@ Union[Divisors[#1 + #2], Divisors[#1*#2]], # >= u &] & @@ {a[i - 2], a[i - 1]}; While[c[t[[-k]]] > 0, k++]; Set[{a[i], c[t[[-k]]]}, {t[[-k]], i}]; If[a[i] == u, While[c[u] > 1, u++]], {i, Length[s] + 1, nn}]; Array[a, nn] (* _Michael De Vlieger, Apr 03 2022 *)