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A140361 Number of additions, subtractions, or multiplications necessary to reach n starting from 1 and 2. 2
0, 0, 0, 1, 1, 2, 2, 3, 2, 2, 3, 3, 3, 4, 3, 3, 2, 3, 3, 4, 3, 4, 4, 4, 3, 3, 4, 3, 4, 4, 4, 4, 3, 4, 4, 4, 3, 4, 4, 4, 4, 5, 4, 5, 5, 4, 5, 5, 4, 4, 4, 5, 5, 5, 4, 5, 4, 5, 5, 5, 4, 5, 4, 4, 3, 4, 4, 5, 4, 5, 5, 5, 4, 5, 5, 4, 5, 5, 4, 4, 4, 3, 4, 4, 4, 5, 5, 5, 5, 5, 4, 5, 5, 5, 5, 5, 4, 5, 5, 4 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,6
COMMENTS
Koiran calls this function tau(n). - Leonid Broukhis, Aug 04 2008
In the model used here a computation of length h of an integer n is a sequence of integers (n_{-1}=1, n_0=2, n_1, ..., n_h=n) such that for each i >= 1 there exist j,k < i and o in {+,-,*} with n_i = n_j o n_k. a(0)=a(1)=a(2)=0 and for n >= 3, a(n) is equal to the length of a shortest computation of n. - Alois P. Heinz, Sep 20 2012
LINKS
P. Koiran, Valiant's model and the cost of computing integers, Comput. Complex. 13 (2004), pp. 131-146.
FORMULA
a(n) = A173419(n) - 1 for n > 1. Hence log_2(log_2(n)) <= a(n) <= 2*log_2(n) - 1. - Charles R Greathouse IV, Sep 20 2012
EXAMPLE
a(7) = 3 because we have 7 = (1+2)+(2*2), or 7 = 2*(2+2)-1 and there is no shorter way; the sequences are (1,2,3,4,7) or (1,2,4,8,7), respectively.
MAPLE
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, 2}}:
S:= proc(n) S(n):= map(x->x[], F(n)) minus S(n-1) end: S(0):= {0, 1, 2}:
a:= proc(n) local k; for k from 0 while not(n in S(k)) do od; k end:
seq(a(n), n=0..100); # Alois P. Heinz, Sep 24 2012
CROSSREFS
Cf. A173419.
Sequence in context: A136032 A135975 A334796 * A237769 A187182 A362816
KEYWORD
nonn
AUTHOR
Leonid Broukhis, Jul 21 2008
EXTENSIONS
Corrected, from 6 to 5, a(59) = ((2+2)*2)*8-1-4 and a(94) = (((2+2)+2)+4)*10-6, by Leonid Broukhis, Aug 04 2008
a(24) and a(96) corrected by Charles R Greathouse IV, Sep 20 2012
STATUS
approved

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Last modified April 25 11:16 EDT 2024. Contains 371967 sequences. (Running on oeis4.)