A006165 a(1) = a(2) = 1; thereafter a(2n+1) = a(n+1) + a(n), a(2n) = 2a(n).
1, 1, 2, 2, 3, 4, 4, 4, 5, 6, 7, 8, 8, 8, 8, 8, 9, 10, 11, 12, 13, 14, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43
Offset: 1
Examples
From _Peter Bala_, Aug 01 2022: (Start)
1) The sequence {n - a(a(n)) : n >= 1} begins [0, 1, 2, 3, 3, 4, 5, 6, 6, 6, 7, 8, 9, 10, 11, 12, 12, 12, 12, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 49, ...] has the repeated values 3 (twice), 6 (three times), 12 (five times), 24 (nine times), 48 (seventeen times) ..., conjecturally of the form 3*2^m
2) The sequence {n - a(a(a(n))) : n >= 1} begins [0, 1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 10, 11, 12, 13, 14, 14, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 28, 28, 28, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 56, 56, 56, 56, 56, 56, 56, 56, 57, ...] has the repeated values 7 (twice), 14 (three times), 28 (five times), 56 (nine times) ..., conjecturally of the form 7*2^m. (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), 85-94.
- Hsien-Kuei Hwang, S Janson, TH Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint, 2016; http://140.109.74.92/hk/wp-content/files/2016/12/aat-hhrr-1.pdf. Also Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- T. D. Noe, Table of n, a(n) for n = 1..1024
- J.-P. Allouche and J. Shallit, The Ring of k-regular Sequences, II
- J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
- Dale Gerdemann, Second-Order Josephus Problem (video)
- Jeffrey Shallit, Intertwining of Complementary Thue-Morse Factors, arXiv:2203.02917 [cs.FL], 2022.
- Ralf Stephan, Some divide-and-conquer sequences ...
- Ralf Stephan, Table of generating functions
- Index entries for sequences related to the Josephus Problem
Programs
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Maple
a := proc (n) option remember; if n = 1 then 1 else n - a(n - a(a(n-1))) end if end proc: seq(a(n), n = 1..100); # Peter Bala, Jul 31 2022
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Mathematica
t = {1, 1}; Do[If[OddQ[n], AppendTo[t, t[[Floor[n/2]]] + t[[Ceiling[n/2]]]], AppendTo[t, 2*t[[n/2]]]], {n, 3, 128}] (* T. D. Noe, May 25 2011 *) -
PARI
a(n) = my(i=logint(n,2)-1); if(bittest(n,i), 2<
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PARI
a(n)=if(n<2,1,n-a(n-a(n\2))); \\ Benoit Cloitre, May 12 2024
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Python
from functools import lru_cache @lru_cache(maxsize=None) def A006165(n): return 1 if n <= 2 else A006165(n//2) + A006165((n+1)//2) # Chai Wah Wu, Mar 08 2022
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Python
def A006165(n): return min(n-(m:=1<
Formula
For n >= 2, if a(n) >= A006257(n), i.e., if msb(n) > n - a(n)/2, then a(n+1) = a(n)+1, otherwise a(n+1) = a(n). - Henry Bottomley, Jan 21 2002
a(n+1) = min(msb(n), 1+n-msb(n)/2) for all n (msb = most significant bit, A053644). - Boyko Bantchev (bantchev(AT)math.bas.bg), May 17 2002
a(1)=1, a(n) = n - a(n - a(a(n-1))). - Benoit Cloitre, Nov 08 2002
a(1)=1, a(n) = n - a(n - a(floor(n/2))). - Benoit Cloitre, May 12 2024
For k > 0, 0 <= i <= 2^k-1, a(2^k+i) = 2^(k-1)+i; for 2^k-2^(k-2) <= x <= 2^k a(x) = 2^(k-1); (also a(m*2^k) = a(m)*2^k for m >= 2). - Benoit Cloitre, Dec 16 2002
G.f.: x * (1/(1+x) + (1/(1-x)^2) * Sum_{k>=0} t^2*(1-t)) where t = x^2^k. - Ralf Stephan, Sep 12 2003
From Peter Bala, Jul 31 2022: (Start)
For k a positive integer, define the k-th iterated sequence a^(k) of a by a^(1)(n) = a(n) and setting a^(k)(n) = a^(k-1)(a(n)) for k >= 2. For example, a^(2)(n) = a(a(n)) and a^(3)(n) = a(a(a(n))).
Conjectures: for n >= 2 there holds
(i) a(n) + a(n - a(n - a(n - a(n - a(n))))) = n;
(ii) a(n - a(n - a(n - a(n)))) = a(n - a(n - a(n - a(n - a(n - a(n))))));
(iii) a^2(n) = a(n - a(n - a(n - a(n))));
(iv) n - a(n) = a(n - a^(2)(n));
(v) a(n - a(n)) = a^(2)(n - a^(2)(n - a^(2)(n - a^(2)(n))));
(vi) for k >= 2, a^(k)(n - a^(k)(n)) = a^(k)(n - a^(k)(n - a^(k)(n - a^(k)(n)))).
(vii) for k >= 1, the sequence {n - a^(k)(n) : n >= 1} has first differences either 0 or 1. We conjecture that the repeated values of the sequence are of the form (2^k - 1)*2^m. The number of repeated values appears to always be 2, 3, 5, 9, 17, 35, ..., independent of k, conjecturally A000051. Two examples are given below.
A similar property may hold for the sequences {n - A060973^(k)(n) : n >= 2^(k-1)}, k = 1,2,3,.... (End)
Extensions
More terms from Larry Reeves (larryr(AT)acm.org), Jun 12 2002
Comments