A133344 Complexity of the number n, counting 1's and built using +, *, ^ and # representing concatenation.

1, 2, 3, 4, 5, 5, 6, 5, 5, 6, 2, 3, 4, 5, 6, 6, 7, 6, 6, 7, 3, 4, 5, 5, 6, 6, 6, 7, 7, 8, 4, 5, 5, 6, 7, 6, 7, 8, 7, 8, 5, 5, 6, 6, 7, 7, 8, 7, 8, 8, 6, 7, 7, 8, 7, 8, 9, 9, 9, 8, 6, 6, 6, 7, 8, 7, 8, 8, 8, 9, 7, 8, 8, 8, 9, 10, 8, 9, 10, 10, 6, 7, 8, 7, 8, 8

Offset: 1

Author

Jonathan Vos Post, Oct 20 2007

Keywords

• base
• nonn

Comments

The complexity of an integer n is the least number of 1's needed to represent it using only additions, multiplications, exponentiation and parentheses. This allows juxtaposition of numbers to form larger integers, so for example, 2 = 1+1 has complexity 2, but unlike A003037, so does 11 = 1#1 (concatenating two numbers is an allowed operation). Similarly a(111) = 3. The complexity of a number has been defined in several different ways by different authors. See the Index to the OEIS for other definitions.

Examples

			An example (usually nonunique) of the derivation of the first 22 values.
a(1) = 1, the number of 1's in "1."
a(2) = 2, the number of 1's in "1+1 = 2."
a(3) = 3, the number of 1's in "1+1+1 = 3."
a(4) = 4, the number of 1's in "1+1+1+1 = 4."
a(5) = 5, the number of 1's in "1+1+1+1+1 = 5."
a(6) = 5, since there are 5 1's in "((1+1)*(1+1+1)) = 6."
a(7) = 6, since there are 6 1's in "1+(((1+1)*(1+1+1))) = 7."
a(8) = 5, since there are 5 1's in "(1+1)^(1+1+1) = 8."
a(9) = 5, since there are 5 1's in "(1+1+1)^(1+1) = 9."
a(10) = 6 since there are 6 1's in "1+((1+1+1)^(1+1)) = ten.
a(11) = 2 since there are 2 1's in "1#1 = eleven."
a(12) = 3 since there are 3 1's in "1+(1#1) = twelve."
a(13) = 4 since there are 4 1's in "1+1+(1#1) = thirteen."
a(14) = 5 since there are 5 1's in "1+1+1+(1#1) = fourteen."
a(16) = 6 since there are 6 1's in "(1+1+1+1)^(1+1)."
a(17) = 7 since there are 7 1's in "1+((1+1+1+1)^(1+1))."
a(18) = 6 since there are 6 1's in "1#((1+1)^(1+1+1))."
a(19) = 6 since there are 6 1's in "1#((1+1+1)^(1+1))."
a(20) = 7 since there are 7 1's in "(1#1)+((1+1+1)^(1+1))."
a(21) = 3 since there are 3 1's in "(1+1)#1."
a(22) = 4 since 22 = (1+1)*(1#1) = (1#1)+(1#1) = (1+1)#(1+1).
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000
• Index to sequences related to the complexity of n

Crossrefs

Cf. A003037, A025280, A005520, A005245, A005421, A117618.

Programs

Maple

with(numtheory):
a:= proc(n) option remember; local r; `if`(n=1, 1, min(
       seq(a(i)+a(n-i), i=1..n-1),
       seq(a(d)+a(n/d), d=divisors(n) minus {1, n}),
       seq(`if`(cat("", n)[i+1]<>"0", a(iquo(n, 10^(length(n)-i),
           'r'))+a(r), NULL), i=1..length(n)-1),
       seq(a(root(n, p))+a(p), p=divisors(igcd(seq(i[2],
           i=ifactors(n)[2]))) minus {0, 1})))
    end:
seq(a(n), n=1..120);  # Alois P. Heinz, Nov 06 2013