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.
%I A067699 #32 Aug 08 2022 20:56:59 %S A067699 0,4,8,14,18,24,30,38,42,48,54,62,68,76,84,94,98,104,110,118,124,132, %T A067699 140,150,156,164,172,182,190,200,210,222,226,232,238,246,252,260,268, %U A067699 278,284,292,300,310,318,328,338,350,356,364,372,382 %N A067699 Number of comparisons made in a version of the sorting algorithm QuickSort for an array of size n with n identical elements. %D A067699 Thomas H. Cormen, Charles E. Leiserson and Ronald L. Rivest. Introduction to Algorithms. McGraw-Hill Book Company, 2000. (Gives description of this version of QuickSort.) %H A067699 Michael S. Branicky, <a href="/A067699/b067699.txt">Table of n, a(n) for n = 1..10000</a> %H A067699 Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, <a href="http://140.109.74.92/hk/wp-content/files/2016/12/aat-hhrr-1.pdf">Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications</a>, preprint, 2016. %H A067699 Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, <a href="http://doi.org/10.1145/3127585">Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications</a>, ACM Transactions on Algorithms 13:4 (2017), #47. %H A067699 Ralf Stephan, <a href="/somedcgf.html">Some divide-and-conquer sequences with (relatively) simple generating functions</a>, 2004. %H A067699 Ralf Stephan, <a href="/A079944/a079944.ps">Table of generating functions (ps file)</a>. %H A067699 Ralf Stephan, <a href="/A067699/a067699.pdf">Table of generating functions (pdf file)</a>. %F A067699 a(n) = 2*ceiling((n+1)/2) + a(ceiling(n/2)) + a(floor(n/2)) with a(1) = 0, a(2) = 4 and a(3) = 8. %F A067699 From _Ralf Stephan_, Oct 24 2003: (Start) %F A067699 a(n) = A076178(n-1) + 4*(n-1) for n >= 1. %F A067699 a(n) = b(n-1) for n >= 1, where b(0) = 0, b(2*n) = b(n) + b(n-1) + 2*n + 2, b(2*n+1) = 2*b(n) + 2*n + 4. %F A067699 (End) %o A067699 (Python) %o A067699 from functools import cache %o A067699 @cache %o A067699 def b(n): return 0 if n == 0 else b(n//2) + b((n-1)//2) + n + 2 + (n&1) %o A067699 def a(n): return b(n-1) %o A067699 print([a(n) for n in range(1, 53)]) # _Michael S. Branicky_, Aug 08 2022 %Y A067699 Cf. A063090, A076178, A093418, A096620, A115107, A288964. %K A067699 nonn %O A067699 1,2 %A A067699 Karla J. Oty (oty(AT)uscolo.edu), Feb 05 2002