A005245 -id:A005245 - cp's OEIS frontend

A000792 a(n) = max{(n - i)*a(i) : i < n}; a(0) = 1.

1, 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 81, 108, 162, 243, 324, 486, 729, 972, 1458, 2187, 2916, 4374, 6561, 8748, 13122, 19683, 26244, 39366, 59049, 78732, 118098, 177147, 236196, 354294, 531441, 708588, 1062882, 1594323, 2125764, 3188646, 4782969, 6377292

Offset: 0

N. J. A. Sloane

Numbers of the form 3^k, 2*3^k, 4*3^k with a(0) = 1 prepended.
If a set of positive numbers has sum n, this is the largest value of their product.
In other words, maximum of products of partitions of n: maximal value of Product k_i for any way of writing n = Sum k_i. To find the answer, take as many of the k_i's as possible to be 3 and then use one or two 2's (see formula lines below).
a(n) is also the maximal size of an Abelian subgroup of the symmetric group S_n. For example, when n = 6, one of the Abelian subgroups with maximal size is the subgroup generated by (123) and (456), which has order 9. [Bercov and Moser] - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 19 2001
Also the maximum number of maximal cliques possible in a graph with n vertices (cf. Capobianco and Molluzzo). - Felix Goldberg (felixg(AT)tx.technion.ac.il), Jul 15 2001 [Corrected by Jim Nastos and Tanya Khovanova, Mar 11 2009]
Every triple of alternate terms {3*k, 3*k+2, 3*k+4} in the sequence forms a geometric progression with first term 3^k and common ratio 2. - Lekraj Beedassy, Mar 28 2002
For n > 4, a(n) is the least multiple m of 3 not divisible by 8 for which omega(m) <= 2 and sopfr(m) = n. - Lekraj Beedassy, Apr 24 2003
Maximal number of divisors that are possible among numbers m such that A080256(m) = n. - Lekraj Beedassy, Oct 13 2003
Or, numbers of the form 2^p*3^q with p <= 2, q >= 0 and 2p + 3q = n. Largest number obtained using only the operations +,* and () on the parts 1 and 2 of any partition of n into these two summands where the former exceeds the latter. - Lekraj Beedassy, Jan 07 2005
a(n) is the largest number of complexity n in the sense of A005520 (A005245). - David W. Wilson, Oct 03 2005
a(n) corresponds also to the ultimate occurrence of n in A001414 and thus stands for the highest number m such that sopfr(m) = n, for n >= 2. - Lekraj Beedassy, Apr 29 2002
a(n) for n >= 1 is a paradigm shift sequence with procedural length p = 0, in the sense of A193455. - Jonathan T. Rowell, Jul 26 2011
a(n) = largest term of n-th row in A212721. - Reinhard Zumkeller, Jun 14 2012
For n >= 2, a(n) is the largest number whose prime divisors (with multiplicity) add to n, whereas the smallest such number (resp. smallest composite number) is A056240(n) (resp. A288814(n)). - David James Sycamore, Nov 23 2017
For n >= 3, a(n+1) = a(n)*(1 + 1/s), where s is the smallest prime factor of a(n). - David James Sycamore, Apr 10 2018

			a{8} = 18 because we have 18 = (8-5)*a(5) = 3*6 and one can verify that this is the maximum.
a(5) = 6: the 7 partitions of 5 are (5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1) and the corresponding products are 5, 4, 6, 3, 4, 2 and 1; 6 is the largest.
G.f. = 1 + x + 2*x^2 + 3*x^3 + 4*x^4 + 6*x^5 + 9*x^6 + 12*x^7 + 18*x^8 + ...

B. R. Barwell, Cutting String and Arranging Counters, J. Rec. Math., 4 (1971), 164-168.
B. R. Barwell, Journal of Recreational Mathematics, "Maximum Product": Solution to Prob. 2004;25(4) 1993, Baywood, NY.
M. Capobianco and J. C. Molluzzo, Examples and Counterexamples in Graph Theory, p. 207. North-Holland: 1978.
S. L. Greitzer, International Mathematical Olympiads 1959-1977, Prob. 1976/4 pp. 18;182-3 NML vol. 27 MAA 1978
J. L. Gross and J. Yellen, eds., Handbook of Graph Theory, CRC Press, 2004; p. 396.
P. R. Halmos, Problems for Mathematicians Young and Old, Math. Assoc. Amer., 1991, pp. 30-31 and 188.
L. C. Larson, Problem-Solving Through Problems. Problem 1.1.4 pp. 7. Springer-Verlag 1983.
D. J. Newman, A Problem Seminar. Problem 15 pp. 5;15. Springer-Verlag 1982.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..500
R. Bercov and L. Moser, On Abelian permutation groups, Canad. Math. Bull. 8 (1965) 627-630.
Walter Bridges and William Craig, On the distribution of the norm of partitions, arXiv:2308.00123 [math.CO], 2023.
J. Arias de Reyna and J. van de Lune, The question "How many 1's are needed?" revisited, arXiv preprint arXiv:1404.1850 [math.NT], 2014. See M_n.
J. Arias de Reyna and J. van de Lune, Algorithms for determining integer complexity, arXiv preprint arXiv:1404.2183 [math.NT], 2014.
Nigel Derby, 96.21 The MaxProduct partition, The Mathematical Gazette 96:535 (2012), pp. 148-151.
Tomislav Doslic, Maximum Product Over Partitions Into Distinct Parts, Journal of Integer Sequences, Vol. 8 (2005), Article 05.5.8.
Hans Havermann, Tables of sum-of-prime-factors sequences (overview with links to the first 50000 sums).
J. Iraids, K. Balodis, J. Cernenoks, M. Opmanis, R. Opmanis and K. Podnieks, Integer Complexity: Experimental and Analytical Results, arXiv preprint arXiv:1203.6462 [math.NT], 2012.
Andrew Kenney and Caroline Shapcott, Maximum Part-Products of Odd Palindromic Compositions, Journal of Integer Sequences, Vol. 18 (2015), Article 15.2.6.
E. F. Krause, Maximizing The Product of Summands, Mathematics Magazine, MAA Oct 1996, Vol. 69, no. 5 pp. 270-271.
MathPro, 20000 Problems Under the Sea, Problem 14856.Putnam 1979/A1.
J. W. Moon and L. Moser, On cliques in graphs, Israel J. Math. 3 (1965), 23-28.
Natasha Morrison and Alex Scott, Maximizing the number of induced cycles in a graph, Preprint, 2016. See f_2(n).
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
F. Pluvinage, Developing problem solving experiences in practical action projects, The Mathematics Enthusiast, ISSN 1551-3440, Vol. 10, nos. 1 & 2, pp. 219-244.
D. A. Rawsthorne, How many 1's are needed?, Fib. Quart. 27 (1989), 14-17.
J. T. Rowell, Solution Sequences for the Keyboard Problem and its Generalizations, Journal of Integer Sequences, 18 (2015), #15.10.7.
Robert Schneider and Andrew V. Sills, The Product of Parts or "Norm" of a Partition, Integers (2020) Vol. 20A, Article #A13.
J. Scholes, 40th Putnam 1979 Problem A1.
J. Scholes, 18th IMO 1976 Problem 4.
Andrew V. Sills and Robert Schneider, The product of parts or "norm" of a partition, arXiv:1904.08004 [math.NT], 2019.
V. Vatter, Maximal independent sets and separating covers, Amer. Math. Monthly, 118 (2011), 418-423.
Robert G. Wilson v, Letter to N. J. Sloane, circa 1991.
A. C.-C. Yao, On a problem of Katona on minimal separating systems, Discrete Math., 15 (1976), 193-199.
Index to sequences related to the complexity of n
Index entries for linear recurrences with constant coefficients, signature (0,0,3).

See A007600 for a left inverse.
Cf. A000793, A009490, A034891, A062943, A007601, A062723, A069188, A087902.
Cf. array A064364, rightmost (nonvanishing) numbers in row n >= 2.
See A056240 and A288814 for the minimal numbers whose prime factors sums up to n.
A000792, A178715, A193286, A193455, A193456, and A193457 are closely related as paradigm shift sequences for (p = 0, ..., 5 respectively).
Cf. A202337 (subsequence).
Cf. A038754, A005245, A005520.

a000792 n = a000792_list !! n
a000792_list = 1 : f [1] where
   f xs = y : f (y:xs) where y = maximum $ zipWith (*) [1..] xs
-- Reinhard Zumkeller, Dec 17 2011

I:=[1,1,2,3,4]; [n le 5 select I[n] else 3*Self(n-3): n in [1..45]]; // Vincenzo Librandi, Apr 14 2015

A000792 := proc(n)
    m := floor(n/3) ;
    if n mod 3 = 0 then
        3^m ;
    elif n mod 3 = 1 then
        4*3^(m-1) ;
    else
        2*3^m ;
    end if;
    floor(%) ;
end proc: # R. J. Mathar, May 26 2013

a[1] = 1; a[n_] := 4* 3^(1/3 *(n - 1) - 1) /; (Mod[n, 3] == 1 && n > 1); a[n_] := 2*3^(1/3*(n - 2)) /; Mod[n, 3] == 2; a[n_] := 3^(n/3) /; Mod[n, 3] == 0; Table[a[n], {n, 0, 40}]
CoefficientList[Series[(1 + x + 2x^2 + x^4)/(1 - 3x^3), {x, 0, 50}], x] (* Harvey P. Dale, May 01 2011 *)
f[n_] := Max[ Times @@@ IntegerPartitions[n, All, Prime@ Range@ PrimePi@ n]]; f[1] = 1; Array[f, 43, 0] (* Robert G. Wilson v, Jul 31 2012 *)
a[ n_] := If[ n < 2, Boole[ n > -1], 2^Mod[-n, 3] 3^(Quotient[ n - 1, 3] + Mod[n - 1, 3] - 1)]; (* Michael Somos, Jan 23 2014 *)
Join[{1, 1}, LinearRecurrence[{0, 0, 3}, {2, 3, 4}, 50]] (* Jean-François Alcover, Jan 08 2019 *)
Join[{1,1},NestList[#+Divisors[#][[-2]]&,2,41]] (* James C. McMahon, Aug 09 2024 *)

{a(n) = floor( 3^(n - 4 - (n - 4) \ 3 * 2) * 2^( -n%3))}; /* Michael Somos, Jul 23 2002 */

lista(nn) = {print1("1, 1, "); print1(a=2, ", "); for (n=1, nn, a += a/divisors(a)[2]; print1(a, ", "););} \\ Michel Marcus, Apr 14 2015

A000792(n)=if(n>1,3^((n-2)\3)*(2+(n-2)%3),1) \\ M. F. Hasler, Jan 19 2019

G.f.: (1 + x + 2*x^2 + x^4)/(1 - 3*x^3). - Simon Plouffe in his 1992 dissertation.
a(3n) = 3^n; a(3*n+1) = 4*3^(n-1) for n > 0; a(3*n+2) = 2*3^n.
a(n) = 3*a(n-3) if n > 4. - Henry Bottomley, Nov 29 2001
a(n) = n if n <= 2, otherwise a(n-1) + Max{gcd(a(i), a(j)) | 0 < i < j < n}. - Reinhard Zumkeller, Feb 08 2002
A007600(a(n)) = n; Andrew Chi-Chih Yao attributes this observation to D. E. Muller. - Vincent Vatter, Apr 24 2006
a(n) = 3^(n - 2 - 2*floor((n - 1)/3))*2^(2 - (n - 1) mod 3) for n > 1. - Hieronymus Fischer, Nov 11 2007
From Kiyoshi Akima (k_akima(AT)hotmail.com), Aug 31 2009: (Start)
a(n) = 3^floor(n/3)/(1 - (n mod 3)/4), n > 1.
a(n) = 3^(floor((n - 2)/3))*(2 + ((n - 2) mod 3)), n > 1. (End)
a(n) = (2^b)*3^(C - (b + d))*(4^d), n > 1, where C = floor((n + 1)/3), b = max(0, ((n + 1) mod 3) - 1), d = max(0, 1 - ((n + 1) mod 3)). - Jonathan T. Rowell, Jul 26 2011
G.f.: 1 / (1 - x / (1 - x / (1 + x / (1 - x / (1 + x / (1 + x^2 / (1 + x))))))). - Michael Somos, May 12 2012
3*a(n) = 2*a(n+1) if n > 1 and n is not divisible by 3. - Michael Somos, Jan 23 2014
a(n) = a(n-1) + largest proper divisor of a(n-1), n > 2. - Ivan Neretin, Apr 13 2015
a(n) = max{a(i)*a(n-i) : 0 < i < n} for n >= 4. - Jianing Song, Feb 15 2020
a(n+1) = a(n) + A038754(floor( (2*(n-1) + 1)/3 )), for n > 1. - Thomas Scheuerle, Oct 27 2022

More terms and better description from Therese Biedl (biedl(AT)uwaterloo.ca), Jan 19 2000

A064097 A quasi-logarithm defined inductively by a(1) = 0 and a(p) = 1 + a(p-1) if p is prime and a(n*m) = a(n) + a(m) if m,n > 1.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 4, 5, 5, 5, 4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 6, 6, 7, 6, 7, 5, 7, 6, 7, 6, 7, 7, 7, 6, 7, 7, 8, 7, 7, 8, 9, 6, 8, 7, 7, 7, 8, 7, 8, 7, 8, 8, 9, 7, 8, 8, 8, 6, 8, 8, 9, 7, 9, 8, 9, 7, 8, 8, 8, 8, 9, 8, 9, 7, 8, 8, 9, 8, 8, 9, 9, 8, 9, 8, 9, 9, 9, 10, 9, 7, 8, 9, 9, 8, 9, 8, 9, 8

Offset: 1

Thomas Schulze (jazariel(AT)tiscalenet.it), Sep 16 2001

nonn

Note that this is the logarithm of a completely multiplicative function. - Michael Somos
Number of iterations of r -> r - (largest divisor d < r) needed to reach 1 starting at r = n. a(n) = a(n - A032742(n)) + 1 for n >= 2. - Jaroslav Krizek, Jan 28 2010
From Antti Karttunen, Apr 04 2020: (Start)
Krizek's comment above stems from the fact that n - n/p = (p-1)*(n/p), where p is the least prime dividing n [= A020639(n), thus n/p = A032742(n)] and because this is fully additive sequence we can write a(n) = a(p) + a(n/p) = (1+a(p-1)) + a(n/p) = 1 + a((p-1)*(n/p)) = 1 + a(n - n/p), for any composite n.
Note that in above formula p can be any prime factor of n, not only the smallest, which proves Robert G. Wilson v's comment in A333123 that all such iteration paths from n to 1 are of the same length, regardless of the route taken.
(End)
From Antti Karttunen, May 11 2020: (Start)
Moreover, those paths form the chains of a graded poset, which is also a lattice. See the Mathematics Stack Exchange link.
Keeping the formula otherwise same, but changing it for primes either as a(p) = 1 + a(A064989(p)), a(p) = 1 + a(floor(p/2)) or a(p) = 1 + a(A048673(p)) gives sequences A056239, A064415 and A334200 respectively.
(End)
a(n) is the number of iterations r->A060681(r) to reach 1 starting at r=n. - R. J. Mathar, Nov 06 2023

			a(19) = 6: 19 - 1 = 18; 18 - 9 = 9; 9 - 3 = 6; 6 - 3 = 3; 3 - 1 = 2; 2 - 1 = 1. That is a total of 6 iterations. - _Jaroslav Krizek_, Jan 28 2010
From _Antti Karttunen_, Apr 04 2020: (Start)
We can follow also alternative routes, where we do not always select the largest proper divisor to subtract, for example, from 19 to 1, we could go as:
19-1 = 18; 18-(18/3) = 12; 12-(12/2) = 6; 6-(6/3) = 4; 4-(4/2) = 2; 2-(2/2) = 1, or as
19-1 = 18; 18-(18/3) = 12; 12-(12/3) = 8; 8-(8/2) = 4; 4-(4/2) = 2; 2-(2/2) = 1,
both of which also have exactly 6 iterations.
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Passawan Noppakaew and Prapanpong Pongsriiam, Product of Some Polynomials and Arithmetic Functions, J. Int. Seq. (2023) Vol. 26, Art. 23.9.1.
Hugo Pfoertner, Addition chains
Mathematics Stack Exchange, Does a graded poset on the positive integers generated from subtracting factors define a lattice?
Wikipedia, Addition chain

Similar to A061373 which uses the same recurrence relation but a(1) = 1.
Cf. A003313, A005245, A020639, A032742, A052126, A054725, A060681, A064415, A073933, A076142, A076091, A175125, A256653, A307742, A323076, A323077, A329697, A333123, A334200, A334202, A334203, and array A334111.
Cf. A000079 (position of last occurrence), A105017 (position of records), A334197 (positions of record jumps upward).
Partial sums of A334090.
Cf. also A056239.

import Data.List (genericIndex)
a064097 n = genericIndex a064097_list (n-1)
a064097_list = 0 : f 2 where
   f x | x == spf  = 1 + a064097 (spf - 1) : f (x + 1)
       | otherwise = a064097 spf + a064097 (x `div` spf) : f (x + 1)
       where spf = a020639 x
-- Reinhard Zumkeller, Mar 08 2013

a:= proc(n) option remember;
      add((1+a(i[1]-1))*i[2], i=ifactors(n)[2])
    end:
seq(a(n), n=1..120);  # Alois P. Heinz, Apr 26 2019
# alternative which can be even used outside this entry
A064097 := proc(n)
        option remember ;
        add((1+procname(i[1]-1))*i[2], i=ifactors(n)[2]) ;
end proc:
seq(A064097(n),n=1..100) ; # R. J. Mathar, Aug 07 2022

quasiLog := (Length@NestWhileList[# - Divisors[#][[-2]] &, #, # > 1 &] - 1) &;
quasiLog /@ Range[1024]
(* Terentyev Oleg, Jul 17 2011 *)
fi[n_] := Flatten[ Table[#[[1]], {#[[2]]}] & /@ FactorInteger@ n]; a[1] = 0; a[n_] := If[ PrimeQ@ n, a[n - 1] + 1, Plus @@ (a@# & /@ fi@ n)]; Array[a, 105] (* Robert G. Wilson v, Jul 17 2013 *)
a[n_] := Length@ NestWhileList[# - #/FactorInteger[#][[1, 1]] &, n, # != 1 &] - 1; Array[a, 100] (* or *)
a[n_] := a[n - n/FactorInteger[n][[1, 1]]] +1; a[1] = 0; Array[a, 100]  (* Robert G. Wilson v, Mar 03 2020 *)

NN=200; an=vector(NN);
a(n)=an[n];
for(n=2,NN,an[n]=if(isprime(n),1+a(n-1), sumdiv(n,p, if(isprime(p),a(p)*valuation(n,p)))));
for(n=1,100,print1(a(n)", "))

a(n)=if(isprime(n), return(a(n-1)+1)); if(n==1, return(0)); my(f=factor(n)); apply(a,f[,1])~ * f[,2] \\ Charles R Greathouse IV, May 10 2016

(define (A064097 n) (if (= 1 n) 0 (+ 1 (A064097 (A060681 n))))) ;; After Jaroslav Krizek's Jan 28 2010 formula.
(define (A060681 n) (- n (A032742 n))) ;; See also code under A032742.
;; Antti Karttunen, Aug 23 2017

Conjectures: for n>1, log(n) < a(n) < (5/2)*log(n); lim n ->infinity sum(k=1, n, a(k))/(n*log(n)-n) = C = 1.8(4)... - Benoit Cloitre, Oct 30 2002
Conjecture: for n>1, floor(log_2(n)) <= a(n) < (5/2)*log(n). - Robert G. Wilson v, Aug 10 2013
a(n) = Sum_{k=1..n} a(p_k)*e_k if n is composite with factorization p_1^e_1 * ... * p_k^e_k. - Orson R. L. Peters, May 10 2016
From Antti Karttunen, Aug 23 2017: (Start)
a(1) = 0; for n > 1, a(n) = 1 + a(A060681(n)). [From Jaroslav Krizek's Jan 28 2010 formula in comments.]
a(n) = A073933(n) - 1. (End)
a(n) = A064415(n) + A329697(n) [= A054725(n) + A329697(n), for n > 1]. - Antti Karttunen, Apr 16 2020
a(n) = A323077(n) + A334202(n) = a(A052126(n)) + a(A006530(n)). - Antti Karttunen, May 12 2020

More terms from Michael Somos, Sep 25 2001

A005520 Smallest number of complexity n: smallest number requiring n 1's to build using + and *.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 10, 11, 17, 22, 23, 41, 47, 59, 89, 107, 167, 179, 263, 347, 467, 683, 719, 1223, 1438, 1439, 2879, 3767, 4283, 6299, 10079, 11807, 15287, 21599, 33599, 45197, 56039, 81647, 98999, 163259, 203999, 241883, 371447, 540539, 590399, 907199

Offset: 1

N. J. A. Sloane

Ed Pegg Jr, www.mathpuzzle.com, Apr 10 2001, notes that all the new terms are -1 mod 120. - That is, this holds at least for all terms from a(45) = 590399 up to the highest known term a(89) given in the b-file at the end of 2015. Comment clarified by Antti Karttunen, Dec 14 2015
Largest number of complexity n is given by A000792. - David W. Wilson, Oct 03 2005
After 1438 = 2 * 719, all elements through 8206559 are primes. Equivalently, except for a(4) = 4, a(7) = 10, a(10) = 22 and a(25) = 1438, we have a(1) through a(53) are all primes. - Jonathan Vos Post, Apr 07 2006
a(54)-a(89) are all primes. - Janis Iraids, Apr 21 2011
Previous observations (primes with property -1 mod 120) still hold. - Martins Opmanis, Oct 16 2009
The prime 353942783 = A189125(1) = 2*3 + (1 + 2^2*3^2)*(2 + 3^4(1 + 2*3^10)) is the smallest counterexample to Guy's first hypothesis on integer complexity. - Jonathan Vos Post, Mar 30 2012
The sequence A265360 (second smallest number of complexity n) seems to have similar properties. - Janis Iraids via Antti Karttunen, Dec 15 2015

			Examples from Piotr Fabian:
1 = 1, 1 "one": first 1, a(1) = 1
2 = 1+1, 2 "ones": first 2, a(2) = 2
3 = 1+1+1, 3 "ones": first 3, a(3) = 3
4 = 1+1+1+1, 4 "ones": first 4, a(4) = 4
5 = 1+1+1+1+1, 5 "ones": first 5, a(5) = 5
6 = (1+1)*(1+1+1), 5 "ones"
7 = 1+((1+1)*(1+1+1)), 6 "ones": first 6, a(6) = 7
8 = (1+1)*(1+1+1+1), 6 "ones"
9 = (1+1+1)*(1+1+1), 6 "ones"
10 = 1+((1+1+1)*(1+1+1)), 7 "ones": first 7, a(7) = 10
11 = 1+(1+(1+1+1)*(1+1+1)), 8 "ones": first 8, a(8) = 11
12 = (1+1)*((1+1)*(1+1+1)), 7 "ones"

R. K. Guy, Unsolved Problems in Number Theory, Sec. F26 (related material).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Janis Iraids, Table of n, a(n) for n = 1..89
Kazuyuki Amano, Integer Complexity and Mixed Binary-Ternary Representation, 33rd Int'l Symp. Alg. Comp. (ISAAC 2022) Leibniz Int'l Proc. Informatics (LIPIcs) Vol. 248.
J. Arias de Reyna, Complejidad de los números naturales, Gaceta R. Soc. Mat. Esp., 3 (2000), 230-250.
J. Arias de Reyna, Complejidad de los números naturales, Gaceta R. Soc. Mat. Esp., 3 (2000), 230-250. [Cached copy, with permission]
J. Arias de Reyna, Complexity of natural numbers, arXiv:2111.03345 [math.NT], 2021. See also Integers (2024) Vol. 24, A19, p. 6.
J. Arias de Reyna and J. van de Lune, The question "How many 1's are needed?" revisited, arXiv preprint arXiv:1404.1850 [math.NT], 2014.
Jānis Iraids, Kaspars Balodis, Juris Čerņenoks, Mārtiņš Opmanis, Rihards Opmanis, and Kārlis Podnieks, Integer complexity: experimental and analytical results, arXiv:1203.6462 [math.NT], 2012.
Ed Pegg Jr., Math Games: Integer Complexity, www.mathpuzzle.com, April 12, 2004.
D. A. Rawsthorne, How many 1's are needed?, Fib. Quart. 27 (1989), 14-17.
Eric Weisstein's World of Mathematics, Integer Complexity
Index to sequences related to the complexity of n

Cf. A005245, A025280, A003037, A005421, A048183, A181898, A182061, A265360.

N = 10^6;
fact = cell(1,N);
for n = 2:sqrt(N)
  for m = [n^2:n:N]
    fact{m} = [fact{m},n];
  end
end
R = zeros(1,N);
R(1) = 1;
A(1) = 1;
mmax = 1;
for n = 2:N
  m = min(R(1:floor(n/2)) + R([n-1:-1:ceil(n/2)]));
  if numel(fact{n}) > 0
    m = min(m, min(R(fact{n}) + R(n ./ fact{n})));
  end
  R(n) = m;
  if m > mmax
    A(m) = n;
    mmax = m;
  elseif A(m) == 0
    A(m) = n;
  end
end
A % Robert Israel, Dec 14 2015

N:= 100000: # to get all terms <= N
R:= Vector(N):
R[1]:= 1: A[1]:= 1:
for n from 2 to N do
  inds:= [seq(n-i,i=1..floor(n/2))];
  m:= min(R[1..floor(n/2)] + R[inds]);
  for d in select(`<=`,numtheory:-divisors(n),floor(sqrt(n))) minus {1} do
    m:= min(m, R[d]+R[n/d])
  od;
  R[n]:= m;
  if not assigned(A[m]) then A[m]:= n fi;
od:
seq(A[m],m=1..max(R)); # Robert Israel, Dec 14 2015

nn = 10000;
R = Table[0, nn];
R[[1]] = 1; Clear[A]; A[1] = 1;
For[n = 2, n <= nn, n++,
  inds = Table[n-i, {i, 1, n/2}];
  m = Min[R[[1 ;; Floor[n/2]]] + R[[inds]]];
  Do[
    m = Min[m, R[[d]] + R[[n/d]]], {d,
    Select[Rest[Divisors[n]], # <= Sqrt[n]&]}
  ];
  R[[n]] = m;
  If[!IntegerQ[A[m]], A[m] = n];
];
Table[A[m], {m, 1, Max[R]}] (* Jean-François Alcover, Aug 05 2018, after Robert Israel *)

def aupton(nn):
  alst, R = [1], {0: {1}} # R[n] is set reachable using n+1 1's (n ops)
  for n in range(1, nn):
    R[n]  = set(a+b for i in range(n//2+1) for a in R[i] for b in R[n-1-i])
    R[n] |= set(a*b for i in range(n//2+1) for a in R[i] for b in R[n-1-i])
    alst.append(min(R[n] - R[n-1]))
  return alst
print(aupton(35)) # Michael S. Branicky, Jun 08 2021

Corrected and extended by David W. Wilson, May 1997
Extended to terms a(40)=163259 and a(41)=203999 by John W. Layman, Nov 03 1999
Further terms from Piotr Fabian (PCF(AT)who.net), Mar 30 2001
a(68)-a(89) from Janis Iraids, Apr 20 2011

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

Original entry on oeis.org

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

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

nonn

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

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

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

Cf. A003037, A005520, A005208.

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

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 *)

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

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

A091333 Number of 1's required to build n using +, -, *, and parentheses.

Original entry on oeis.org

1, 2, 3, 4, 5, 5, 6, 6, 6, 7, 8, 7, 8, 8, 8, 8, 9, 8, 9, 9, 9, 10, 10, 9, 10, 10, 9, 10, 11, 10, 11, 10, 11, 11, 11, 10, 11, 11, 11, 11, 12, 11, 12, 12, 11, 12, 12, 11, 12, 12, 12, 12, 12, 11, 12, 12, 12, 13, 13, 12, 13, 13, 12, 12, 13, 13, 14, 13, 13, 13, 13, 12, 13, 13, 13, 13, 14

Offset: 1

Jens Voß, Dec 30 2003

nonn

Consider an alternate complexity measure b(n) which gives the minimum number of 1's necessary to build n using +, -, *, and / (where this additional operation is strict integer division, defined only for n/d where d|n). It turns out that b(n) coincides with a(n) for all n up to 50221174, see A348069. - Glen Whitney, Sep 23 2021

In respect of the previous comment: when creating A362471, where repunits are allowed, we found a difference if we allowed n/d with noninteger (intermediate) results. So, see also A362626. - Peter Munn, Apr 29 2023

			A091333(23) = 10 because 23 = (1+1+1+1) * (1+1+1) * (1+1) - 1. (Note that 23 is also the smallest index at which A091333 differs from A005245.)

Janis Iraids, Table of n, a(n) for n = 1..10000
Juris Cernenoks, Janis Iraids, Martins Opmanis, Rihards Opmanis and Karlis Podnieks, Integer Complexity: Experimental and Analytical Results II, arXiv:1409.0446 [math.NT], 2014.
Janis Iraids, A C program to compute the sequence
J. Iraids, K. Balodis, J. Cernenoks, M. Opmanis, R. Opmanis and K. Podnieks, Integer Complexity: Experimental and Analytical Results. arXiv preprint arXiv:1203.6462 [math.NT], 2012. - From _N. J. A. Sloane_, Sep 22 2012
Index to sequences related to the complexity of n

Cf. A005245 (variant using + and *), A025280 (using +, *, and ^), A091334 (using +, -, *, and ^), A348089 (using +, -, *, /, and ^), A348262 (using + and ^).

Cf. A348069, A362471, A362626.

from functools import cache
@cache
def f(m):
    if m == 0: return set()
    if m == 1: return {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])
                if x != y: out.add(abs(x-y))
    return out
def aupton(terms):
    tocover, alst, n = set(range(1, terms+1)), [0 for i in range(terms)], 1
    while len(tocover) > 0:
        for k in f(n) - f(n-1):
            if k <= terms:
                alst[k-1] = n
                tocover.discard(k)
        n += 1
    return alst
print(aupton(77)) # Michael S. Branicky, Sep 28 2021

A173419 Length of shortest computation yielding n using addition, subtraction and multiplication (starting from 1).

Original entry on oeis.org

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

Offset: 1

Charles R Greathouse IV, Feb 17 2010, Apr 22 2010

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

			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.

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

Alois P. Heinz, Table of n, a(n) for n = 1..1800
Peter Borwein and Joe Hobart, The extraordinary power of division in straight line programs, American Mathematical Monthly 119:7 (2012), pp. 584-592.
F. Cucker, M. Shub, and S. Smale, Separation of Complexity classes in Koiran's weak model, Theoretical Computer Science 133:1 (1994), pp. 3-14.
W. DeMelo and B. F. Svaiter, The cost of computing integers, Proc. Amer. Math. Soc. 124 (1996), pp. 1377-1378.
P. Koiran, Valiant's model and the cost of computing integers, Comput. Complex. 13 (2004), pp. 131-146.
Carlos Gustavo T. de A. Moreira, On asymptotic estimates for arithmetic cost functions, Proceedings of the American Mathematical Society 125:2 (1997), pp. 347-353.
Richard J. Mathar, Extended list of chain examples
Michael Shub and Steve Smale, On the intractability of Hilbert's Nullstellensatz and an algebraic version of "NP = P", Duke Mathematical Journal 81:1 (1995), pp. 47-54.
Al Zimmermann's Programming Contests, Factorials
Index to sequences related to the complexity of n

Records are essentially A141414.

Cf. A003313 (shortest chain using just addition), A005245 (number of 1s using just addition and multiplication), A217032(n):=A173419(n!).

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:
seq(a(n), n=1..110);  # Alois P. Heinz, Sep 24 2012

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).
a(n) <= A005245(n) <= A003313(n) <= A014701(n) <= 2*A000523(n). - Charles R Greathouse IV, Feb 07 2022

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

Original entry on oeis.org

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

cp's OEIS Frontend

A133374 Difference between largest number of complexity n in the sense of A005245 and smallest number of complexity n in the sense of A005245.

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A380464 Integers k such that A005245(m*k) < A005245(k) for some m.

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A000792 a(n) = max{(n - i)*a(i) : i < n}; a(0) = 1.

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A064097 A quasi-logarithm defined inductively by a(1) = 0 and a(p) = 1 + a(p-1) if p is prime and a(n*m) = a(n) + a(m) if m,n > 1.

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A005520 Smallest number of complexity n: smallest number requiring n 1's to build using + and *.

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

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