A352969 Start with {1}, then at each step replace it with the set of all pairwise products and sums of its elements (an element can be paired with itself). a(n) gives the number of elements after n-th step.

1, 2, 4, 11, 52, 678, 67144, 357306081

Offset

0,2

Links

Table of n, a(n) for n=0..7.

Example

After the 1st step the set becomes {1*1, 1+1} = {1, 2}. It has 2 distinct elements so a(1) = 2.
After the 2nd step the set becomes {1*1, 1+1, 1*2, 1+2, 2*2, 2+2} = {1, 2, 2, 3, 4, 4} = {1, 2, 3, 4}. It has 4 distinct elements so a(2) = 4.

Mathematica

Length /@ NestList[DeleteDuplicates[Flatten[Table[With[{a = #[[k]], b = #[[;; k]]}, {a b, a + b}], {k, Length[#]}]]] &, {1}, 6]

Prog

(PARI) lista(nn) = {my(v = [1]); print1(#v, ", "); for (n=1, nn, v = setunion(setbinop((x, y)->(x+y), v), setbinop((x, y)->(x*y), v)); print1(#v, ", "); ); } \\ Michel Marcus, Apr 13 2022
(Python)
from math import prod
from itertools import combinations_with_replacement
from functools import lru_cache
@lru_cache(maxsize=None)
def A352969_set(n):
    if n == 0:
        return {1}
    return set(sum(x) for x in combinations_with_replacement(A352969_set(n-1), 2)) | set(prod(x) for x in combinations_with_replacement(A352969_set(n-1), 2))
def A353969(n):
    return len(A352969_set(n)) # Chai Wah Wu, Apr 15 2022

Crossrefs

Cf. A263995, A273525, A353535, A353536.

Sequence in context: A267013 A318532 A057857 * A091233 A173312 A351184

Adjacent sequences: A352966 A352967 A352968 * A352970 A352971 A352972

Keyword

nonn,more,hard

Author

Vladimir Reshetnikov, Apr 12 2022

Extensions

a(7) from Thomas Scheuerle, Apr 13 2022

a(7) corrected by Chai Wah Wu, Apr 15 2022

Status

approved