

A004001


HofstadterConway $10000 sequence: a(n) = a(a(n1)) + a(na(n1)) with a(1) = a(2) = 1.
(Formerly M0276)


210



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,3


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(n1) = 0 or 1 (see the D. Newman reference).  Emeric Deutsch, Jun 06 2005
Conjectures:
a(n) = n/2 iff n = 2^k, k >= 1.
a(n) = 2^(k1): k times, for n = 2^k  (k1) to 2^k, k >= 1. (End)


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), 8594.
B. W. Conolly, MetaFibonacci sequences, in S. Vajda, editor, "Fibonacci and Lucas Numbers and the Golden Section", Halstead Press, NY, 1989, pp. 127138.
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. 217221, 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.


LINKS

D. R. Hofstadter, Analogies and Sequences: Intertwined Patterns of Integers and Patterns of Thought Processes, Lecture in DIMACS Conference on Challenges of Identifying Integer Sequences, Rutgers University, Oct 10 2014; Part 1, Part 2.
A. Isgur, R. Lech, S. Moore, S. Tanny, Y. Verberne, and Y. Zhang, Constructing New Families of Nested Recursions with Slow Solutions, SIAM J. Discrete Math., 30(2), 2016, 11281147. (20 pages); DOI:10.1137/15M1040505.


FORMULA

Limit_{n>infinity} a(n)/n = 1/2 and as special cases, if n > 0, a(2^ni) = 2^(n1) for 0 <= i < = n1; a(2^n+1) = 2^(n1) + 1.  Benoit Cloitre, Aug 04 2002 [Corrected by Altug Alkan, Apr 03 2017]


EXAMPLE

If n=4, 2^4=16, a(16i) = 2^(41) = 8 for 0 <= i <= 41 = 3, hence a(16)=a(15)=a(14)=a(13)=8.


MAPLE

A004001 := proc(n) option remember; if n<=2 then 1 else procname(procname(n1)) +procname(nprocname(n1)); 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 *)


PROG

(Haskell)
a004001 n = a004001_list !! (n1)
a004001_list = 1 : 1 : h 3 1 { memoization }
where h n x = x' : h (n + 1) x'
where x' = a004001 x + a004001 (n  x)
(PARI) a=vector(100); a[1]=a[2]=1; for(n=3, #a, a[n]=a[a[n1]]+a[na[n1]]); 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[k1]]+v[kv[k1]]); v \\ Charles R Greathouse IV, Feb 26 2017
(Scheme)
;; An implementation of memoizationmacro definec can be found for example from: http://oeis.org/wiki/Memoization
(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[c1]] + A[cA[c1]]
c += 1
return A[n]
(Magma) I:=[1, 1]; [n le 2 select I[n] else Self(Self(n1))+ Self(nSelf(n1)):n in [1..75]]; // Marius A. Burtea, Aug 16 2019


CROSSREFS



KEYWORD

nonn,easy,nice


AUTHOR



STATUS

approved



