

A122248


a(n)  a(n1) = A113474(n).


11



0, 1, 3, 5, 9, 13, 18, 23, 31, 39, 48, 57, 68, 79, 91, 103, 119, 135, 152, 169, 188, 207, 227, 247, 270, 293, 317, 341, 367, 393, 420, 447, 479, 511, 544, 577, 612, 647, 683, 719, 758, 797, 837, 877, 919, 961, 1004, 1047, 1094, 1141, 1189
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OFFSET

0,3


COMMENTS

First differences are A113474.
The following sequences all appear to have the same parity: A003071, A029886, A061297, A092524, A093431, A102393, A104258, A122248, A128975.  Jeremy Gardiner, Dec 28 2008


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..50.
Tanya Khovanova, There are no coincidences, arXiv preprint 1410.2193, 2014


FORMULA

G.f.: (1/(1x))*Sum{k=0..infinity, x^(2^k)/((1x)*(1x^(2^k)))}x^2/(1x)^3; a(n)=sum{k=1..n, sum{j=0..n, floor(k/2^j)}}binomial(n,2); a(n)=A122247(n)binomial(n,2);


CROSSREFS

Sequence in context: A036713 A260733 A265429 * A024403 A129230 A203567
Adjacent sequences: A122245 A122246 A122247 * A122249 A122250 A122251


KEYWORD

easy,nonn


AUTHOR

Paul Barry, Aug 27 2006


STATUS

approved



