A138651 - OEIS

%I #13 Jan 15 2022 10:03:07

%S 1,2,3,7,14,26,52,104,201,379,751,1422,2679,5068,9383,17249,31285,

%T 57171,103476,186921,333991,594921,1044076,1837839,3198889,5561453,

%U 9689270,16763804,28909679,49142265,83418913

%N Number of distinct values obtained when each of the operators # in the expression 1#2#3#...#n is replaced by + (add) or x (multiply) in all possible ways, for n=1,2,3,...

%C a(n) can be found incrementally by storing pairs (v, p) where v is the value of the expression and p is the product of all terms appearing after the last + operator.  The process starts with (1, 1) and at stage n, each (v, p) maps to (v+n, n), (v-p+p*n, p*n) for operator + and *, respectively. See second Python program. - _Michael S. Branicky_, Jan 14 2022

%e For n=4, the eight expressions {1+2+3+4,1+2+3x4,1+2x3+4,1+2x3x4,1x2+3+4,1x2+3x4, 1x2x3+4,1x2x3x4} are obtained, with the eight values {10,15,11,25,9,14,10,24} respectively, seven of which are distinct, so a(4)=7.

%o (Python)

%o from itertools import product

%o def a(n): return len(set(eval("1" + "".join(op+str(i) for op, i in zip(ops, range(2, n+1)))) for ops in product("+*", repeat=n-1)))

%o print([a(n) for n in range(1, 16)]) # _Michael S. Branicky_, Jan 14 2022

%o (Python) # faster version for initial segment of sequence (see Comments)

%o def afindn(terms):

%o     vset, vpset = {1}, {(1, 1)}

%o     print(len(vset), end=", ")

%o     for n in range(2, terms+1):

%o         newvset, newvpset = set(), set()

%o         for v, p in vpset:

%o             newvs = [v+n, v+p*(n-1)]

%o             newvset.update(newvs)

%o             if n != terms: # saves memory for that last term

%o                 newvpset.update([(newvs[0], n), (newvs[1], p*n)])

%o         vset, vpset = newvset, newvpset

%o         print(len(vset), end=", ")

%o afindn(26) # _Michael S. Branicky_, Jan 14 2022

%Y Cf. A069765, A078389.

%K nonn,more

%O 1,2

%A _John W. Layman_, May 15 2008

%E a(17)-a(26) from _Wojciech Florek_, Feb 27 2018

%E a(27)-a(31) from _Michael S. Branicky_, Jan 14 2022