A117618 -id:A117618 - cp's OEIS frontend

A003037 Smallest number of complexity n: smallest number requiring n 1's to build using +, * and ^.

1, 2, 3, 4, 5, 7, 11, 13, 21, 23, 41, 43, 71, 94, 139, 211, 215, 431, 863, 1437, 1868, 2855, 5737, 8935, 15838, 15839, 54357, 95597, 139117, 233195, 470399, 1228247, 2183791, 4388063, 6945587, 13431919, 32329439, 46551023

Offset: 1

N. J. A. Sloane

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 does not allow juxtaposition of 1's to form larger integers, so for example, 2 = 1+1 has complexity 2, but 11 does not (concatenating two 1's is not an allowed operation). The complexity of a number has been defined in several different ways by different authors. See the Index to the OEIS for other definitions. - Jonathan Vos Post, Oct 20 2007

			An example (usually nonunique) of the derivation of the first 10 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) = 7, since there are 6 1's in "((1+1)*(1+1+1))+1 = 7."
a(7) = 11, since there are 7 1's in "((1+1+1)^(1+1))+1+1 = 11."
a(8) = 13, since there are 8 1's in "((1+1+1)*(1+1+1+1))+1 = 13."
a(9) = 21, since there are 9 1's in "(1+1+1)*(((1+1)*(1+1+1))+1) = 21."
a(10) = 23, since there are 10 1's in "1+((1+1)*(((1+1+1)^(1+1))+1+1)) = 23."

W. A. Beyer, M. L. Stein and S. M. Ulam, The Notion of Complexity. Report LA-4822, Los Alamos Scientific Laboratory of the University of California, Los Alamos, NM, December 1971.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

W. A. Beyer, Letter to N. J. A. Sloane, 1980
W. A. Beyer, M. L. Stein and S. M. Ulam, The Notion of Complexity. Report LA-4822, Los Alamos Scientific Laboratory of the University of California, Los Alamos, NM, December 1971. [Annotated scanned copy]
Index to sequences related to the complexity of n

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

xmax:= 5:  # get terms <= 10^xmax
C[1]:= {1}: A[1]:= 1: CU[1]:= {1}:
for n from 2 do
   C[n]:= {seq(seq(seq(op(select(`<=`,
[a+b,a*b,`if`(b*ilog10(a) <= xmax,a^b,NULL),`if`(a*ilog10(b) <= xmax,b^a,NULL)]
         ,10^xmax)),b=C[n-k]),a=C[k]),k=1..floor(n/2))}
         minus CU[n-1];
   if C[n] = {} then break fi;
   A[n]:= min(C[n]);
   CU[n]:= CU[n-1] union C[n];
od:
seq(A[i],i=1..n-1); # Robert Israel, Jan 08 2015

More terms from David W. Wilson, May 15 1997

More terms from Sean A. Irvine, Jan 07 2015

A182002 Smallest positive integer that cannot be computed using exactly n n's, the four basic arithmetic operations (+, -, *, /), and the parentheses.

Original entry on oeis.org

2, 2, 1, 10, 13, 22, 38, 91, 195, 443, 634, 1121, 3448, 6793, 17692

Offset: 1

Ali Dasdan, Apr 05 2012

			a(2) = 2 because two 2's can produce 0 = 2-2, 1 = 2/2, 4 = 2+2 = 2*2, so the smallest positive integer that cannot be computed is 2.
a(3) = 1 because no expression with three 3's gives 1.

Cf. A005245, A005520, A003313, A076142, A076091, A061373, A005421, A064097, A025280, A003037, A117618.

Cf. A171826, A171827, A171828, A171829, A258068, A258069, A258070, A258071.

f:= proc(n,b) option remember;
      `if`(n=1, {b}, {seq(seq(seq([k+m, k-m, k*m,
      `if`(m=0, NULL, k/m)][], m=f(n-i, b)), k=f(i, b)), i=1..n-1)})
    end:
a:= proc(n) local i, l;
      l:= sort([infinity, select(x-> is(x, integer) and x>0, f(n, n))[]]);
      for i do if l[i]<>i then return i fi od
    end:
seq(a(n), n=1..8); # Alois P. Heinz, Apr 13 2012

from fractions import Fraction
from functools import lru_cache
def a(n):
    @lru_cache()
    def f(m):
        if m == 1: return {Fraction(n, 1)}
        out = set()
        for j in range(1, m//2+1):
            for x in f(j):
                for y in f(m-j):
                    out.update([x + y, x - y, y - x, x * y])
                    if y: out.add(Fraction(x, y))
                    if x: out.add(Fraction(y, x))
        return out
    k, s = 1, f(n)
    while k in s: k += 1
    return k
print([a(n) for n in range(1, 10)]) # Michael S. Branicky, Jul 29 2022

a(11)-a(12) from Alois P. Heinz, Apr 22 2012
a(13)-a(14) from Michael S. Branicky, Jul 29 2022
a(15) from Michael S. Branicky, Jul 27 2023

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

Original entry on oeis.org

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

Jonathan Vos Post, Oct 20 2007

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.

			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).

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

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

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

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A003037 Smallest number of complexity n: smallest number requiring n 1's to build using +, * and ^.

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A182002 Smallest positive integer that cannot be computed using exactly n n's, the four basic arithmetic operations (+, -, *, /), and the parentheses.

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

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Maple