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%I #18 Oct 05 2021 22:47:54
%S 1,1,1,1,2,3,2,6,6,8,13,18,21,35,45,61,90,121,162,241,323,450,638,865,
%T 1233,1698,2349,3315,4592,6382,8970,12421,17351,24320,33714,47218,
%U 65978,91784,128177,179807,249689,349549,489341,681468,953769,1334490,1860641,2606043,3643618,5086481,7124229,9960420
%N Number of positive numbers that can be built with n ones using +, -, and *, and require at least n ones.
%C a(n+1)/a(n) appears from the values through n=63 to be oscillating in a narrowing range around 7/5.
%H Glen Whitney, <a href="/A348083/b348083.txt">Table of n, a(n) for n = 1..64</a> (This file produced with the same C code that generated A348069, simply with division eliminated from the loop over operators.)
%F a(n) = |{k : A091333(k) = n}|.
%e For n=5, there are two numbers whose minimal expression using 1,+,-, and * uses five ones: 5 = 1+1+1+1+1 and 6 = (1+1)*(1+1+1), so a(5) = 2.
%e For n=10, there are eight numbers whose minimal expression uses ten ones: 22 = 3(2*3+1)+1, 23 = 2*2*2*3-1, 25 = 5*5, 26 = 3*3*3-1, 28 = 3*3*3+1, 30 = 2*3*5, 32 = 2*2*2*2*2, and 36 = 2*2*3*3. We use numbers k=1..5 in these expressions because each takes k ones to express. Note that n=10 is also the least n for which a(n) differs from A005421(n), which counts the solutions to A005245(k) = n.
%o (Python)
%o from functools import cache
%o @cache
%o def f(m):
%o if m == 0: return set()
%o if m == 1: return {1}
%o out = set()
%o for j in range(1, m//2+1):
%o for x in f(j):
%o for y in f(m-j):
%o out.update([x + y, x * y])
%o if x != y: out.add(abs(x-y))
%o return list(out)
%o def a(n): return len(f(n)) - len(f(n-1))
%o print([a(n) for n in range(1, 33)]) # _Michael S. Branicky_, Sep 27 2021
%Y Cf. A005421, A091333, A005245.
%K nonn
%O 1,5
%A _Glen Whitney_, Sep 27 2021
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