cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A005245 The (Mahler-Popken) complexity of n: minimal number of 1's required to build n using + and *.

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, 11, 9, 10, 10, 9, 10, 11, 10, 11, 10, 11, 11, 11, 10, 11, 11, 11, 11, 12, 11, 12, 12, 11, 12, 13, 11, 12, 12, 12, 12, 13, 11, 12, 12, 12, 13, 14, 12, 13, 13, 12, 12, 13, 13, 14, 13, 14, 13, 14, 12, 13, 13, 13, 13, 14, 13, 14
Offset: 1

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Author

Keywords

Comments

The complexity of an integer n is the least number of 1's needed to represent it using only additions, multiplications 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 ("pasting together" 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.
Guy asks if a(p) = a(p-1) + 1 for prime p. Martin Fuller found the least counterexample p = 353942783 in 2008, see Fuller link. - Charles R Greathouse IV, Oct 04 2012
It appears that this sequence is lower than A348262 {1,+,^} only a finite number of times. - Gordon Hamilton and Brad Ballinger, May 23 2022
The second Altman links proves that {a(n) - 3*log_3(n)} is a well-ordered subset of the reals whose intersection with [0,k) has order type omega^k for each positive integer k, so this set itself has order type omega^omega. - Jianing Song, Apr 13 2024

Examples

			From _Lekraj Beedassy_, Jul 04 2009: (Start)
   n.........minimal expression........ a(n) = number of 1's
   1..................1...................1
   2.................1+1..................2
   3................1+1+1.................3
   4.............(1+1)*(1+1)..............4
   5............(1+1)*(1+1)+1.............5
   6............(1+1)*(1+1+1).............5
   7...........(1+1)*(1+1+1)+1............6
   8..........(1+1)*(1+1)*(1+1)...........6
   9...........(1+1+1)*(1+1+1)............6
  10..........(1+1+1)*(1+1+1)+1...........7
  11.........(1+1+1)*(1+1+1)+1+1..........8
  12.........(1+1)*(1+1)*(1+1+1)..........7
  13........(1+1)*(1+1)*(1+1+1)+1.........8
  14.......{(1+1)*(1+1+1)+1}*(1+1)........8
  15.......{(1+1)*(1+1)+1}*(1+1+1)........8
  16.......(1+1)*(1+1)*(1+1)*(1+1)........8
  17......(1+1)*(1+1)*(1+1)*(1+1)+1.......9
  18........(1+1)*(1+1+1)*(1+1+1).........8
  19.......(1+1)*(1+1+1)*(1+1+1)+1........9
  20......{(1+1+1)*(1+1+1)+1}*(1+1).......9
  21......{(1+1)*(1+1+1)+1}*(1+1+1).......9
  22.....{(1+1)*(1+1+1)+1}*(1+1+1)+1.....10
  23....{(1+1)*(1+1+1)+1}*(1+1+1)+1+1....11
  24......(1+1)*(1+1)*(1+1)*(1+1+1).......9
  25.....(1+1)*(1+1)*(1+1)*(1+1+1)+1.....10
  26....{(1+1)*(1+1)*(1+1+1)+1}*(1+1)....10
  27.......(1+1+1)*(1+1+1)*(1+1+1)........9
  28......(1+1+1)*(1+1+1)*(1+1+1)+1......10
  29.....(1+1+1)*(1+1+1)*(1+1+1)+1+1.....11
  30.....{(1+1+1)*(1+1+1)+1}*(1+1+1).....10
  31....{(1+1+1)*(1+1+1)+1}*(1+1+1)+1....11
  32....(1+1)*(1+1)*(1+1)*(1+1)*(1+1)....10
  33...(1+1)*(1+1)*(1+1)*(1+1)*(1+1)+1...11
  34..{(1+1)*(1+1)*(1+1)*(1+1)+1}*(1+1)..11
  .........................................
(End)
		

References

  • M. Criton, "Les uns de Germain", Problem No. 4, pp. 13 ; 68 in '7 x 7 Enigmes Et Défis Mathématiques pour tous', vol. 25, Editions POLE, Paris 2001.
  • R. K. Guy, Unsolved Problems in Number Theory, Sect. F26.
  • K. Mahler and J. Popken, Over een Maximumprobleem uit de Rekenkunde (Dutch: On a maximum problem in arithmetic), Nieuw Arch. Wiskunde, (3) 1 (1953), pp. 1-15.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A025280 (variant using +, *, and ^), A091333 (using +, -, and *), A091334 (using +, -, *, and ^), A348089 (using +, -, *, / and ^), A348262 (using + and ^).
Cf. A000792 (largest integer of given complexity), A003313, A076142, A076091, A061373, A005421, A064097, A005520, A003037, A161906, A161908, A244743.

Programs

  • Haskell
    import Data.List (genericIndex)
    a005245 n = a005245_list `genericIndex` (n-1)
    a005245_list = 1 : f 2 [1] where
       f x ys = y : f (x + 1) (y : ys) where
         y = minimum $
             (zipWith (+) (take (x `div` 2) ys) (reverse ys)) ++
             (zipWith (+) (map a005245 $ tail $ a161906_row x)
                          (map a005245 $ reverse $ init $ a161908_row x))
    -- Reinhard Zumkeller, Mar 08 2013
    
  • Maple
    with(numtheory):
    a:= proc(n) option remember;
          `if`(n=1, 1, min(seq(a(i)+a(n-i), i=1..n/2),
           seq(a(d)+a(n/d), d=divisors(n) minus {1, n})))
        end:
    seq(a(n), n=1..100); # Alois P. Heinz, Apr 18 2012
  • Mathematica
    a[n_] := a[n] = If[n == 1, 1,
       Min[Table[a[i] + a[n-i], {i, 1, n/2}] ~Join~
       Table[a[d] + a[n/d], {d, Divisors[n] ~Complement~ {1, n}}]]];
    Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Dec 12 2013, after Alois P. Heinz *)
  • PARI
    A005245(n /* start by calling this with the largest needed n */, lim/* see below */) = { local(d); n<6 && return(n);
    if(n<=#A005245, A005245[n]&return(A005245[n]) /* return memoized result if available */,
    A005245=vector(n) /* create vector if needed - should better reuse existing data if available */);
    for(i=1, n-1, A005245[i] || A005245[i]=A005245(i,lim)); /* compute all previous elements */
    A005245[n]=min( vecmin(vector(min(n\2,if(lim>0,lim,n)), k, A005245[k]+A005245[n-k])) /* additive possibilities - if lim>0 is given, consider a(k)+a(n-k) only for k<=lim - we know it is save to use lim=1 up to n=2e7 */, if( isprime(n), n, vecmin(vector((-1+#d=divisors(n))\2, i, A005245[d[i+1]]+A005245[d[ #d-i]]))/* multiplicative possibilities */))}
    \\ See also the Python program by Tim Peters at A005421.
    \\ M. F. Hasler, Jan 30 2008
    
  • Python
    from functools import lru_cache
    from itertools import takewhile
    from sympy import divisors
    @lru_cache(maxsize=None)
    def A005245(n): return min(min(A005245(a)+A005245(n-a) for a in range(1,(n>>1)+1)),min((A005245(d)+A005245(n//d) for d in takewhile(lambda d:d*d<=n,divisors(n)) if d>1),default=n)) if n>1 else 1 # Chai Wah Wu, Apr 29 2023

Formula

It is known that a(n) <= A061373(n) but I think 0 <= A061373(n) - a(n) <= 1 also holds. - Benoit Cloitre, Nov 23 2003 [That's false: the numbers {46, 235, 649, 1081, 7849, 31669, 61993} require {1, 2, 3, 4, 5, 6, 7} fewer 1's in A005245 than in A061373. - Ed Pegg Jr, Apr 13 2004]
It is known from the work of Selfridge and Coppersmith that 3 log_3 n <= a(n) <= 3 log_2 n = 4.754... log_3 n for all n > 1. [Guy, Unsolved Problems in Number Theory, Sect. F26.] - Charles R Greathouse IV, Apr 19 2012 [Comment revised by N. J. A. Sloane, Jul 17 2016]
Zelinsky (2022) improves the upper bound to a(n) <= A*log n where A = 41/log(55296) = 3.754422.... This compares to the constant 2.7307176... of the lower bound. - Charles R Greathouse IV, Dec 11 2022
a(n) >= A007600(n) is a very accurate lower bound. - Mehmet Sarraç, Dec 18 2022

Extensions

More terms from David W. Wilson, May 15 1997

A252739 a(n) = A252738(n) / n.

Original entry on oeis.org

2, 6, 720, 612360000, 1697781042840960000000000, 504261397867001013813789115612253942400000000000000000000000000
Offset: 1

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Author

Antti Karttunen, Dec 21 2014

Keywords

Comments

Note how 6, 720 and 612360000 occur in A244743 as its 0th, 4th and 8th term, from which my bold conjecture that A244743(12) or A244743(16) = 1697781042840960000000000.
According to preliminary results from Janis Iraids, the value of A005245(a(5)) = ||1697781042840960000000000|| = 160, while ||1697781042840960000000000 - 1|| = 169, which lays to rest my naive conjecture above, as 169 - 160 is neither 12 nor 16. Note also how 5, 719 and 612359999 are all primes, while a(5)-1 factorizes as 1697781042840959999999999 = 13 * 89443 * 908669 * 1606890407869. - Antti Karttunen, Dec 20 2015

Crossrefs

Programs

Formula

a(n) = A252738(n) / n.

A319975 Smallest number of complexity n with respect to the operations {1, shift, multiply}.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 10, 11, 14, 19, 22, 23, 38, 43, 58, 59, 89, 107, 134, 167, 179, 263, 347, 383, 537, 713, 719, 1103, 1319, 1439, 2099, 2879, 3833, 4283, 5939, 6299, 9059, 12239, 15118, 19079, 23039, 26459, 44879, 49559, 66239, 78839, 98999, 137339
Offset: 1

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Author

N. J. A. Sloane, Oct 11 2018

Keywords

Comments

The shift operation here is also sometimes called successor, see A263283.
Note this complexity measure counts both operands (the ones) and operators (the shifts and multiplications), whereas most of the complexity measures in the crossrefs count only operands. However, in the presence of successor it would not make sense to count only operands, since any number can be expressed with a single occurrence of 1. - Glen Whitney, Oct 06 2021

Examples

			1 = 1 has complexity 1
2 = S1 has complexity 2
3 = SS1 has complexity 3
4 = SSS1 has complexity 4
5 = SSSS1 has complexity 5
6 = SSSSS1 has complexity 6
7 = SSSSSS1 has complexity 7
10 = S*SS1SS1 = shift(product of (3 and 3)) has complexity 8
(Note that 8 = *S1SSS1 has complexity 7)
11 = SS*SS1SS1 has complexity 9
14 = SS*SS1SSS1 has complexity 10
		

Crossrefs

Smallest number of complexity n (other definitions): A003037, A005520, A244743, A259466, and A265360.
Other definitions of the complexity of n: A005208, A005245, A025280, and A099053.

Programs

  • Python
    def aupton(nn):
        alst, R, allR = [1], {1: {1}}, {1} # R[n] is set reachable using n ops
        for n in range(2, nn+1):
            R[n]  = set(a+1 for a in R[n-1])
            R[n] |= set(a*b for i in range(1, (n+1)//2) for a in R[i] for b in R[n-1-i])
            alst.append(min(R[n] - allR))
            allR |= R[n]
        return alst
    print(aupton(49)) # Michael S. Branicky, Oct 06 2021

A244744 Smallest number with additive excess n.

Original entry on oeis.org

46, 253, 649, 6049, 69989, 166213, 551137, 9064261, 68444596, 347562415, 612220081
Offset: 1

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Author

Juan Arias-de-Reyna, Jul 05 2014

Keywords

Comments

Let m = Product p_j ^ a_j be the standard prime factorization of m. The additive excess of m is defined to be Sum||p_j ^ a_j|| - ||m|| where ||m||=A005245(n) is the complexity of n.
The n-th term of this sequence is the least number m with additive excess n.

Examples

			||253||=17, ||11||=8, ||23||=11.  11 + 8 - 17 = 2. The second term is 253.
		

Crossrefs

Showing 1-4 of 4 results.