A348532 a(n) is the number of multisets of integers that are possible to reach by starting with n occurrences of 0 and by splitting and reverse splitting.

1, 1, 2, 2, 7, 9, 43, 59, 338, 490, 3097, 4639, 31283, 48107, 338553, 531469, 3857036, 6157068, 45713546, 73996100

Offset: 0

Author

Tejo Vrush, Oct 21 2021

Keywords

• nonn
• more

Comments

Splitting is taking 2 occurrences of the same integer and incrementing one of them by 1 and decrementing the other occurrence by 1.

Reverse splitting is taking two elements with a difference of 2 and incrementing the smaller one by 1 and decrementing the larger one by 1. It is the opposite of splitting.

Examples

			For n = 5, the multisets are as follows:
  {{0,0,0,0,0}}   {{-1,0,0,0,1}}   {{-1,-1,0,1,1}}
  {{-1,-1,0,0,2}} {{-1,-1,-1,1,2}} {{-2,0,0,1,1}}
  {{-2,0,0,0,2}}  {{-2,-1,1,1,1}}  {{-2,-1,0,1,2}}.
  Therefore, a(5) = 9.
For n = 6, the multisets are as follows:
  {{0,0,0,0,0,0}}     {{-1,0,0,0,0,1}}     {{-1,-1,0,0,1,1}}
  {{-1,-1,0,0,0,2}}   {{-1,-1,-1,1,1,1}}   {{-1,-1,-1,0,1,2}}
  {{-1,-1,-1,0,0,3}}* {{-1,-1,-1,-1,2,2}}* {{-1,-1,-1,-1,1,3}}*
  {{-2,0,0,0,1,1}}    {{-2,0,0,0,0,2}}     {{-2,-1,0,1,1,1}}
  {{-2,-1,0,0,1,2}}   {{-2,-1,0,0,0,3}}*   {{-2,-1,-1,1,1,2}}
  {{-2,-1,-1,0,2,2}}  {{-2,-1,-1,0,1,3}}   {{-2,-1,-1,-1,2,3}}*
  {{-2,-2,1,1,1,1}}*  {{-2,-2,0,1,1,2}}    {{-2,-2,0,0,2,2}}
  {{-2,-2,0,0,1,3}}   {{-2,-2,-1,1,2,2}}   {{-2,-2,-1,1,1,3}}
  {{-2,-2,-1,0,2,3}}  {{-2,-2,-2,2,2,2}}*  {{-2,-2,-2,1,2,3}}*
  {{-3,0,0,0,0,3}}*   {{-3,0,0,0,1,2}}*    {{-3,0,0,1,1,1}}*
  {{-3,-1,1,1,1,1}}*  {{-3,-1,0,1,1,2}}    {{-3,-1,0,0,2,2}}
  {{-3,-1,0,0,1,3}}   {{-3,-1,-1,1,2,2}}   {{-3,-1,-1,1,1,3}}
  {{-3,-1,-1,0,2,3}}  {{-3,-2,1,1,1,2}}*   {{-3,-2,0,1,2,2}}
  {{-3,-2,0,1,1,3}}   {{-3,-2,0,0,2,3}}    {{-3,-2,-1,2,2,2}}*
  {{-3,-2,-1,1,2,3}}.
  Therefore, a(6) = 43.
The ones marked with an asterisk are the ones that need reverse splitting
to be reached. They are not produced using the rules of A347913.
		

Crossrefs

Cf. A000571, A347913.

Programs

Python

def nextq(q):
    used, used2 = set(), set()
    for i in range(len(q)-1):
        for j in range(i+1, len(q)):
            if q[i] == q[j]:
                if q[i] in used: continue
                used.add(q[i])
                qc = list(q); qc[i] -= 1; qc[j] += 1
                yield tuple(sorted(qc))
            elif q[j] - q[i] == 2:  # assumes q is sorted
                if q[i] in used2: continue
                used2.add(q[i])
                qc = list(q); qc[i] += 1; qc[j] -= 1
                yield tuple(sorted(qc))
def a(n):
    s = tuple(0 for i in range(n)); reach = {s}; expand = list(reach)
    while len(expand) > 0:
        q = expand.pop()
        for qq in nextq(q):
            if qq not in reach:
                reach.add(qq)
                expand.append(qq)
    return len(reach)
print([a(n) for n in range(13)]) # Michael S. Branicky, Oct 21 2021

Formula

It appears that a(n) = A000571(n) for odd n.

Extensions

a(6)-a(19) from Michael S. Branicky, Oct 21 2021