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Partial sums of A005811.
11

%I #30 Aug 27 2021 11:03:10

%S 0,1,3,4,6,9,11,12,14,17,21,24,26,29,31,32,34,37,41,44,48,53,57,60,62,

%T 65,69,72,74,77,79,80,82,85,89,92,96,101,105,108,112,117,123,128,132,

%U 137,141,144,146,149,153,156,160,165,169,172,174,177,181,184,186,189

%N Partial sums of A005811.

%C Partial sums of number of runs in binary expansion of n (n>0). Partial sums of number of 1's in Gray code for n. The subsequence of squares in this partial sum begins 0, 1, 4, 9, 144, 169, 256, 289, 324 (since we also have 32 and 128 I wonder about why so many powers). The subsequence of primes in this partial sum begins: 3, 11, 17, 29, 31, 37, 41, 53, 79, 89, 101, 137, 149, 181, 191, 197, 229, 271.

%C Note: A227744 now gives the squares, which occur at positions given by A227743. - _Antti Karttunen_, Jul 27 2013

%H Antti Karttunen, <a href="/A173318/b173318.txt">Table of n, a(n) for n = 0..8191</a>

%H Richard Blecksmith and Purushottam W. Laud, <a href="http://www.jstor.org/stable/2975267">Some Exact Number Theory Computations via Probability Mechanisms</a>, American Mathematical Monthly, volume 102, number 10, December 1995, pages 893-903, with a(n) = Sum_{j=0..n} b_j as calculated in section 2.

%H Hsien-Kuei Hwang, Svante Janson and Tsung-Hsi Tsai, <a href="https://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, volume 13, number 4, December 2017, article number 47, pages 1-43. Also <a href="http://algo.stat.sinica.edu.tw/hk/?p=1043">first authors' copy</a>, 2016. See example 5.5.

%H Kevin Ryde, <a href="http://user42.tuxfamily.org/dragon/index.html">Iterations of the Dragon Curve</a>, see index "DirCumul".

%F a(n) = sum(i=0..n) A005811(i) = sum(i=0..n) (A037834(i)+1) = sum(i=0..n) (A069010(i) + A033264(i)).

%F a(A000225(n)) = A001787(n) = A000788(A000225(n)). - _Antti Karttunen_, Jul 27 2013 & Aug 09 2013

%F a(2n) = 2*a(n) + n - 2*(ceiling(A005811(n)/2) - (n mod 2)), a(2n+1) = 2*a(n) + n + 1. - _Ralf Stephan_, Aug 11 2013

%e 1 has 1 run in its binary representation "1".

%e 2 has 2 runs in its binary representation "10".

%e 3 has 1 run in its binary representation "11".

%e 4 has 2 runs in its binary representation "100".

%e 5 has 3 runs in its binary representation "101".

%e Thus a(1) = 1, a(2) = 1+2 = 3, a(3) = 1+2+1 = 4, a(4) = 1+2+1+2 = 6, a(5) = 1+2+1+2+3 = 9.

%t Accumulate[Join[{0},Table[Length[Split[IntegerDigits[n,2]]],{n,110}]]] (* _Harvey P. Dale_, Jul 29 2013 *)

%o (PARI) a(n) = my(v=binary(n+1),d=0,e=4); for(i=1,#v, if(v[i], v[i]=#v-i+d;d+=e;e=0, e=4)); fromdigits(v,2)>>1; \\ _Kevin Ryde_, Aug 27 2021

%Y Cf. A005811, A000788, A056539, A014707, A014577, A082410, A000975, A034947.

%Y Cf. also A227737, A227741, A227742.

%Y Cf. A227744 (squares occurring), A227743 (indices of squares).

%K easy,nonn

%O 0,3

%A _Jonathan Vos Post_, Feb 16 2010