cp's OEIS Frontend

We may start either with multiplication or summation. After summation the next step will be multiplication or vice versa.

From Thomas Scheuerle, Oct 30 2022: (Start)

The only zero in this sequence is at a(7). Proof: Let k be any number greater than 1026. If k is even subtract 2, if k is odd subtract 3, then divide by two. Repeat this process until k < 1024. Obviously we will get some number between 511 and 1024. By computation it is known that all these numbers can be reached. They can be reached if we start with multiplication and if we start with addition we can reach all these numbers too.

Conjecture: All numbers greater than 145 can be reached in at least 3 different ways. Numbers greater than 145 which can be reached in only three ways are all of the form 27*2^k - 5 (A304387). This is conjectured to arise from the relation: 27*2^(k+2) - 5 = 3*(27*2^(k+1) - 5) - 2*(27*2^k - 5) which is in some sense here the worst case in between *2 and *3. (End)

Proof of the conjecture: We use the identities in A358095. For n > 145, if a(n) <= 2, A358095(n) <= 2. Therefore, n must be a*3*2^k - 2, where a is in the set {3, 7, 26, 30, 32, 34, 40, 49} and k is a positive integer. However, A358096(a*3*2^k - 2) = A358095(a*3*2^(k-1) - 1) >= 3, a contradiction. Hence all numbers greater than 145 can be reached in at least 3 different ways. If a(n) = 3, the only possibilities are that A358095(n) = 0 and A358096(n) = 3 or A358095(n) = 3 and A358096(n) = 0. For the first one, n must be 9*2^k - 2, so A358096(n) = A358095(9*2^(k-1) - 1) >= 4. For the second one, if n is in the form a*3*2^k - 2, A358096(n) = A358095(a*3*2^(k-1) - 1) >= 4; if n = 108*2^k - 3, A358096(n) = A358095(36*2^k - 1) >= 4; if n is in the form 108*2^k - 5, A358096(n) = 0. Hence a(n) = 3 iff n = 27*2^k - 5 for n > 145. - Yifan Xie, Jan 07 2025

From Yifan Xie, Jan 07 2025: (Start)

The following identities can be proved by induction (k, t are nonnegative integers):

a(n) = 0 iff n = 2 or n = 9*2^k - 2.

a(n) = 1 iff n = 1, 3, 6, 8, 9 or n = 21*2^k - 2.

If n > 71, a(n) = 2 iff n = m*3*2^k - 2, where m is in the set {26, 30, 32, 34, 40, 49}.

If n > 85, a(n) = 3 iff n = 108*2^k - 2, 108*2^k - 3, 108*2^k - 4, 108*2^k - 5, 123*2^k - 2, 126*2^k - 2, 132*2^k - 2, 150*2^k - 2, 174*2^k - 2, 210*2^k - 2, (108*2^t - 3)*2^k - 2.

If n > 60, a(n) = 4 iff n = 114*2^k - 2. (End)