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 1-3 of 3 results.

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

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Author

Keywords

Comments

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

			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"
		

References

  • 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).

Crossrefs

Programs

  • MATLAB
    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
    
  • Maple
    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
  • Mathematica
    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 *)
  • Python
    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

Extensions

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

A189123 Positive integers n such that A005245(n-i) + A005245(i) < A005245(n-1) + 1 for some i, 1 < i < n.

Original entry on oeis.org

21080618, 63241604, 67139098, 116385658, 117448688, 126483083, 152523860, 189724562, 212400458, 229762259, 318689258, 348330652, 353942783, 366873514, 373603732, 379448999, 385159320, 404764540, 409108300, 460759642, 465722100, 477258719, 498197068, 511069678, 516743639, 519835084, 538858312, 545438698, 545790940, 546853138, 574842670, 575550972, 581106238, 590785918
Offset: 1

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Author

Max Alekseyev, Apr 17 2011

Keywords

Comments

Counterexamples to the first conjecture of David Wilson on A005245.
a(1) was found by Igor aka CD_Eater from Moscow in January, 2008
Prime terms are listed in A189125.

Crossrefs

A189124 Positive integers n such that A005245(n) < min{ A005245(n-1) + 1, A005245(x) + A005245(y) } where x,y range over positive integers with x*y = n.

Original entry on oeis.org

353942783, 516743639, 1163385647, 1542243239, 1932319583, 2336924879, 3113713259, 3444631199, 3878989487, 4103787551, 4166809919, 4937621453, 5123340683, 5170931639, 5184740299, 5200683263, 5390865059, 5455982879, 5467766947, 5570566315
Offset: 1

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Author

Max Alekseyev, Apr 17 2011

Keywords

Comments

Counterexamples to the second conjecture of David Wilson on A005245.
a(1), a(2) were found by Martin Fuller; a(3) was found by Janis Iraids.
Prime terms are given by A189125.

Crossrefs

Extensions

More terms from Janis Iraids, Apr 20 2011
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