A025280 Complexity of n: number of 1's required to build n using +, *, and ^.

1, 2, 3, 4, 5, 5, 6, 5, 5, 6, 7, 7, 8, 8, 8, 6, 7, 7, 8, 8, 9, 9, 10, 8, 7, 8, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 10, 11, 11, 12, 11, 10, 11, 12, 9, 8, 9, 10, 10, 11, 8, 9, 9, 10, 10, 11, 11, 12, 12, 11, 7, 8, 9, 10, 11, 12, 12, 13, 9, 10, 10, 10, 11, 12, 11, 12, 11, 7, 8, 9, 10, 11, 12, 11

Offset: 1

Author

N. J. A. Sloane, David W. Wilson

Keywords

• nonn

References

• R. K. Guy, Unsolved Problems Number Theory, Sect. F26.

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000
• R. K. Guy, Some suspiciously simple sequences, Amer. Math. Monthly 93 (1986), 186-190; 94 (1987), 965; 96 (1989), 905.
• J. Iraids, K. Balodis, J. Cernenoks, M. Opmanis, R. Opmanis and K. Podnieks, Integer Complexity: Experimental and Analytical Results. arXiv preprint arXiv:1203.6462, 2012. - From _N. J. A. Sloane_, Sep 22 2012
• Index to sequences related to the complexity of n

Crossrefs

Cf. A005245 (variant using + and *), A091333 (using +, -, and *), A091334 (using +, -, *, and ^), A348089 (using +, -, *, /, and ^), A348262 (using + and ^).

Cf. A003037, A005520, A005208.

Programs

Maple

with(numtheory):
a:= proc(n) option remember; `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(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..100);  # Alois P. Heinz, Mar 08 2013

Mathematica

root[x_, n_] := With[{f = FactorInteger[x]}, Times @@ (f[[All, 1]]^(f[[All, 2]]/n))]; Clear[a]; a[n_] := a[n] = If[n == 1, 1, Min[Table[a[i] + a[n-i], {i, 1, n-1}], Table[a[d] + a[n/d], {d, Divisors[n][[2 ;; -2]]}], Table[a[root[n, p]] + a[p], {p, DeleteCases[Divisors[GCD @@ FactorInteger[n][[All, 2]]], 0|1]}]]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Mar 12 2014, after Alois P. Heinz *)

Python

from math import gcd
from sympy import divisors, factorint, integer_nthroot
from functools import cache
@cache
def a(n):
    if n == 1: return 1
    p = min(a(i)+a(n-1) for i in range(1, n//2+1))
    divs, m = divisors(n), n
    if len(divs) > 2:
        m = min(a(d)+a(n//d) for d in divs[1:len(divs)//2+1])
    f = factorint(n)
    edivs, e = divisors(gcd(*f.values())), n
    if len(edivs) > 1:
        e = min(a(integer_nthroot(n, r)[0])+a(r) for r in edivs[1:])
    return min(p, m, e)
print([a(n) for n in range(1, 88)]) # Michael S. Branicky, Mar 24 2024 after Alois P. Heinz

Formula

a(n) = A005208(n) + 1.