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A173419
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Length of shortest computation yielding n using addition, subtraction and multiplication (starting from 1).
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16
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0, 1, 2, 2, 3, 3, 4, 3, 3, 4, 4, 4, 5, 4, 4, 3, 4, 4, 5, 4, 5, 5, 5, 4, 4, 5, 4, 5, 5, 5, 5, 4, 5, 5, 5, 4, 5, 5, 5, 5, 6, 5, 6, 6, 5, 6, 6, 5, 5, 5, 6, 6, 6, 5, 6, 5, 6, 6, 6, 5, 6, 5, 5, 4, 5, 5, 6, 5, 6, 6, 6, 5, 6, 6, 5, 6, 6, 5, 5, 5, 4, 5, 5, 5, 6, 6, 6, 6, 6, 5, 6, 6, 6, 6, 6, 5, 6, 6, 5, 5, 6, 6, 6, 6, 6, 6, 6, 5
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OFFSET
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1,3
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COMMENTS
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Let x_0 = 1 and x_m = n, with each x_k = x_i + x_j, x_k = x_i * x_j, or x_k = x_i - x_j for some 0 <= i,j < k. a(n) is the least such m.
Shub & Smale ask if there is a c such that a(n!) <= (log n)^c for all n.
If for any sequence of nonzero integers (m_i) there is no constant c such that a(n! * m_n) <= (log n)^c, then "the Hilbert Nullstellensatz is intractable, and consequently the algebraic version of 'NP != P' is true" (Shub & Smale).
Conjecture: if n is prime then a(n) >= a(n-1). The conjecture is true for n < 1800. - Dmitry Kamenetsky, Dec 26 2019
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REFERENCES
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R. K. Guy, Unsolved Problems Number Theory, Sect. F26.
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LINKS
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FORMULA
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a(n) <= 2 log_2(n).
a(n) >= log_2(log_2(n)) + 1.
a(n) >= log_2(n)/log_2(log_2(n)) for almost all n, as proved by Moreira (improving DeMelo & Svaiter).
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EXAMPLE
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For n = 9, one sequence is (1, 1 + 1 = 2, 1 + 2 = 3, 3 * 3 = 9). Since no shorter sequence is possible, a(9) = 3.
For n = 96, one sequence is (1, 1 + 1 = 2, 2 + 2 = 4, 2 + 4 = 6, 4*4 = 16, 6*16 = 96); no shorter is possible so a(96) = 5.
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MAPLE
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g:= f->seq(f union {t}, t={seq(seq({i+j, i-j, i*j}[], j=f), i=f)} minus f):
F:= proc(n) F(n):= map(g, F(n-1)) end: F(0):= {{1}}:
S:= proc(n) S(n):= map(x->x[], F(n)) end:
a:= proc(n) local k; for k from 0 while not(n in S(k)) do od; k end:
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CROSSREFS
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KEYWORD
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nice,nonn
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AUTHOR
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STATUS
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approved
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