

A080776


Oscillating sequence which rises to 2^(k1) in kth segment (k>=1) then falls back to 0.


1



0, 0, 1, 0, 1, 2, 1, 0, 1, 2, 3, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 31, 30
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OFFSET

0,6


COMMENTS

kth segment has length 2^k (k>=0).


REFERENCES

HsienKuei 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/wpcontent/files/2016/12/aathhrr1.pdf. Also Exact and Asymptotic Solutions of a DivideandConquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585


LINKS

Table of n, a(n) for n=0..97.
R. Stephan, Some divideandconquer sequences ...
R. Stephan, Table of generating functions


FORMULA

G.f.: 1 + 2/(1x) + 1/(1x)^2 * (1 + sum(k>=0, 2t^2(t1), t=x^2^k)). a(n) = A005942(n+2)  3(n+1), n>0.  Ralf Stephan, Sep 13 2003
a(0)=0, a(2n) = a(n) + a(n1) + (n==1), a(2n+1) = 2a(n).  Ralf Stephan, Oct 20 2003


CROSSREFS

Essentially the same as A053646.
Sequence in context: A004074 A245615 A053646 * A065358 A062329 A022958
Adjacent sequences: A080773 A080774 A080775 * A080777 A080778 A080779


KEYWORD

nonn


AUTHOR

N. J. A. Sloane, Mar 11 2003


STATUS

approved



