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.

A004001 Hofstadter-Conway $10000 sequence: a(n) = a(a(n-1)) + a(n-a(n-1)) with a(1) = a(2) = 1.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 4, 4, 5, 6, 7, 7, 8, 8, 8, 8, 9, 10, 11, 12, 12, 13, 14, 14, 15, 15, 15, 16, 16, 16, 16, 16, 17, 18, 19, 20, 21, 21, 22, 23, 24, 24, 25, 26, 26, 27, 27, 27, 28, 29, 29, 30, 30, 30, 31, 31, 31, 31, 32, 32, 32, 32, 32, 32, 33, 34, 35, 36, 37, 38, 38, 39, 40, 41, 42
Offset: 1

Views

Author

Keywords

Comments

On Jul 15 1988 during a colloquium talk at Bell Labs, John Conway stated that he could prove that a(n)/n -> 1/2 as n approached infinity, but that the proof was extremely difficult. He therefore offered $100 to someone who could find an n_0 such that for all n >= n_0, we have |a(n)/n - 1/2| < 0.05, and he offered $10,000 for the least such n_0. I took notes (a scan of my notebook pages appears below), plus the talk - like all Bell Labs Colloquia at that time - was recorded on video. John said afterwards that he meant to say $1000, but in fact he said $10,000. I was in the front row. The prize was claimed by Colin Mallows, who agreed not to cash the check. - N. J. A. Sloane, Oct 21 2015
a(n) - a(n-1) = 0 or 1 (see the D. Newman reference). - Emeric Deutsch, Jun 06 2005
a(A188163(n)) = n and a(m) < n for m < A188163(n). - Reinhard Zumkeller, Jun 03 2011
From Daniel Forgues, Oct 04 2019: (Start)
Conjectures:
a(n) = n/2 iff n = 2^k, k >= 1.
a(n) = 2^(k-1): k times, for n = 2^k - (k-1) to 2^k, k >= 1. (End)

Examples

			If n=4, 2^4=16, a(16-i) = 2^(4-1) = 8 for 0 <= i <= 4-1 = 3, hence a(16)=a(15)=a(14)=a(13)=8.
		

References

  • J. Arkin, D. C. Arney, L. S. Dewald and W. E. Ebel, Jr., Families of recursive sequences, J. Rec. Math., 22 (No. 22, 1990), 85-94.
  • B. W. Conolly, Meta-Fibonacci sequences, in S. Vajda, editor, "Fibonacci and Lucas Numbers and the Golden Section", Halstead Press, NY, 1989, pp. 127-138.
  • R. K. Guy, Unsolved Problems Number Theory, Sect. E31.
  • D. R. Hofstadter, personal communication.
  • C. A. Pickover, Wonders of Numbers, "Cards, Frogs and Fractal sequences", Chapter 96, pp. 217-221, Oxford Univ. Press, NY, 2000.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • S. Vajda, Fibonacci and Lucas Numbers and the Golden Section, Wiley, 1989, see p. 129.
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 129.

Crossrefs

Cf. A005229, A005185, A080677, A088359, A087686, A093879 (first differences), A265332, A266341, A055748 (a chaotic cousin), A188163 (greedy inverse).
Cf. A004074 (A249071), A005350, A005707, A093878. Different from A086841. Run lengths give A051135.
Cf. also permutations A267111-A267112 and arrays A265901, A265903.

Programs

  • Haskell
    a004001 n = a004001_list !! (n-1)
    a004001_list = 1 : 1 : h 3 1  {- memoization -}
      where h n x = x' : h (n + 1) x'
              where x' = a004001 x + a004001 (n - x)
    -- Reinhard Zumkeller, Jun 03 2011
    
  • Magma
    [n le 2 select 1 else Self(Self(n-1))+ Self(n-Self(n-1)):n in [1..75]]; // Marius A. Burtea, Aug 16 2019
    
  • Maple
    A004001 := proc(n) option remember; if n<=2 then 1 else procname(procname(n-1)) +procname(n-procname(n-1)); fi; end;
  • Mathematica
    a[1] = 1; a[2] = 1; a[n_] := a[n] = a[a[n - 1]] + a[n - a[n - 1]]; Table[ a[n], {n, 1, 75}] (* Robert G. Wilson v *)
  • PARI
    a=vector(100);a[1]=a[2]=1;for(n=3,#a,a[n]=a[a[n-1]]+a[n-a[n-1]]);a \\ Charles R Greathouse IV, Jun 10 2011
    
  • PARI
    first(n)=my(v=vector(n)); v[1]=v[2]=1; for(k=3, n, v[k]=v[v[k-1]]+v[k-v[k-1]]); v \\ Charles R Greathouse IV, Feb 26 2017
    
  • Python
    def a004001(n):
        A = {1: 1, 2: 1}
        c = 1 #counter
        while n not in A.keys():
            if c not in A.keys():
                A[c] = A[A[c-1]] + A[c-A[c-1]]
            c += 1
        return A[n]
    # Edward Minnix III, Nov 02 2015
    
  • SageMath
    @CachedFunction
    def a(n): # a = A004001
        if n<3: return 1
        else: return a(a(n-1)) + a(n-a(n-1))
    [a(n) for n in range(1,101)] # G. C. Greubel, Apr 25 2024
  • Scheme
    ;; An implementation of memoization-macro definec can be found for example from: http://oeis.org/wiki/Memoization
    (definec (A004001 n) (if (<= n 2) 1 (+ (A004001 (A004001 (- n 1))) (A004001 (- n (A004001 (- n 1)))))))
    ;; Antti Karttunen, Oct 22 2014
    

Formula

Limit_{n->infinity} a(n)/n = 1/2 and as special cases, if n > 0, a(2^n-i) = 2^(n-1) for 0 <= i < = n-1; a(2^n+1) = 2^(n-1) + 1. - Benoit Cloitre, Aug 04 2002 [Corrected by Altug Alkan, Apr 03 2017]

A004074 a(n) = 2*A004001(n) - n, where A004001 is the Hofstadter-Conway $10000 sequence.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 1, 0, 1, 2, 3, 2, 3, 2, 1, 0, 1, 2, 3, 4, 3, 4, 5, 4, 5, 4, 3, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 4, 5, 6, 7, 6, 7, 8, 7, 8, 7, 6, 7, 8, 7, 8, 7, 6, 7, 6, 5, 4, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 5, 6, 7, 8, 9, 8, 9, 10, 11, 10, 11, 12, 11, 12, 11, 10, 11, 12, 13, 12, 13, 14, 13, 14, 13, 12
Offset: 1

Views

Author

Keywords

Comments

The sequence is 0 at 2^n for n = 1, 2, 3, ... The maximum value between 2^n and 2^(n+1) appears to be A072100(n). - T. D. Noe, Jun 04 2012
Hofstadter shows the plot of sequence A004001(n)-(n/2) at point 10:52 of the part two of DIMACS lecture. This sequence is obtained by doubling those values, thus producing only integers. Cf. also A249071. - Antti Karttunen, Oct 22 2014

Crossrefs

Cf. also A249071 (gives the even bisection halved), A233270 (also has a similar Blancmange curve appearance).

Programs

  • Mathematica
    Clear[a]; a[1] = 1; a[2] = 1; a[n_] := a[n] = a[a[n - 1]] + a[n - a[n - 1]]; Table[2*a[n] - n, {n, 100}] (* T. D. Noe, Jun 04 2012 *)
  • Scheme
    (define (A004074 n) (- (* 2 (A004001 n)) n)) ;; Other code as in A004001. - Antti Karttunen, Oct 22 2014

Formula

a(2^n)=0; for n >= 1, Sum_{i=2^(n-1)..2^n} a(i) = A082590(n-2). - Benoit Cloitre, Jun 04 2004

Extensions

More terms from Benoit Cloitre, Jun 04 2004

A283655 a(n) = b(b(n+1)) - b(n-b(n)+1) where b(n) = A004001(n).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 4, 4, 4, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 6, 6, 6, 6, 7, 7, 7, 7, 7
Offset: 1

Views

Author

Altug Alkan, Mar 13 2017

Keywords

Examples

			a(1) = 0 since a(1) = A004001(A004001(2)) - A004001(1-A004001(1)+1) = 1 - 1 = 0.
		

Crossrefs

Programs

  • Mathematica
    b[1] = 1; b[2] = 1; b[n_] := b[n] = b[b[n - 1]] + b[n - b[n - 1]]; a[n_] := b[b[n + 1]] - b[n - b[n] + 1]; Array[a, 100] (* Robert G. Wilson v, Mar 13 2017 *)
  • PARI
    a=vector(1001); a[1]=a[2]=1; for(n=3, #a, a[n]=a[a[n-1]]+a[n-a[n-1]]); va = vector(1000, n, a[a[n+1]]-a[n+1-a[n]])

A367076 Irregular triangle read by rows: T(n,k) (0 <= n, 0 <= k < 2^n). T(n,k) = -Sum_{i=0..k} A365968(n,i).

Original entry on oeis.org

0, 1, 0, 3, 4, 3, 0, 6, 10, 12, 12, 12, 10, 6, 0, 10, 18, 24, 28, 32, 34, 34, 32, 34, 34, 32, 28, 24, 18, 10, 0, 15, 28, 39, 48, 57, 64, 69, 72, 79, 84, 87, 88, 89, 88, 85, 80, 85, 88, 89, 88, 87, 84, 79, 72, 69, 64, 57, 48, 39, 28, 15, 0, 21, 40, 57, 72, 87
Offset: 0

Views

Author

John Tyler Rascoe, Nov 05 2023

Keywords

Examples

			Triangle begins:
    k=0   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15
n=0:  0;
n=1:  1,  0;
n=2:  3,  4,  3,  0;
n=3:  6, 10, 12, 12, 12, 10,  6,  0;
n=4; 10, 18, 24, 28, 32, 34, 34, 32, 34, 34, 32, 28, 24, 18, 10,  0;
		

Crossrefs

Cf. A000217 (column k=0), A028552 (column k=1), A192021 (row sums).

Programs

  • Mathematica
    nmax=10; row[n_]:=Join[CoefficientList[Series[1/(1-x)*Sum[ i/(1+x^2^(i-1))*Product[1+x^2^j,{j,0,i-2}],{i,n}],{x,0,2^n-1}],x],{0}]; Array[row,6,0] (* Stefano Spezia, Dec 23 2023 *)
  • Python
    def row_gen(n):
        x = 0
        for k in range(2**n):
            b = bin(k)[2:].zfill(n)
            x += sum((-1)**(int(b[n-i])+1)*i for i in range(1,n+1))
            yield(-x)
    def A367076_row_n(n): return(list(row_gen(n)))

Formula

T(n,k) = Sum_{i=0..n} abs(k + 1 - (2^i) * round((k+1)/2^i)) * i.
G.f. for n-th row: 1/(1-x) * Sum_{i=1..n} (i/(1+x^2^(i-1)) * Product_{j=0..i-2} 1 + x^2^j).
Showing 1-4 of 4 results.