A119706 Total length of longest runs of 1's in all bitstrings of length n.
1, 4, 11, 27, 62, 138, 300, 643, 1363, 2866, 5988, 12448, 25770, 53168, 109381, 224481, 459742, 939872, 1918418, 3910398, 7961064, 16190194, 32893738, 66772387, 135437649, 274518868, 556061298, 1125679616, 2277559414, 4605810806, 9309804278, 18809961926
Offset: 1
Keywords
Examples
a(3)=11 because for the 8(2^3) possible runs 0 is longest run of heads once, 1 four times, 2 two times and 3 once and 0*1+1*4+2*2+3*1 = 11.
References
- A. M. Odlyzko, Asymptotic Enumeration Methods, pp. 136-137
- R. Sedgewick and P. Flajolet, Analysis of Algorithms, Addison Wesley, 1996, page 372.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..1000
- Steven Finch, Cantor-solus and Cantor-multus distributions, arXiv:2003.09458 [math.CO], 2020.
Crossrefs
Cf. A334833.
Programs
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Maple
A038374 := proc(n) local nshft, thisr, resul; nshft := n ; resul :=0 ; thisr :=0 ; while nshft > 0 do if nshft mod 2 <> 0 then thisr := thisr+1 ; else resul := max(resul, thisr) ; thisr := 0 ; fi ; nshft := floor(nshft/2) ; od ; resul := max(resul, thisr) ; RETURN(resul) ; end : A119706 := proc(n) local count, c, rlen ; count := array(0..n) ; for c from 0 to n do count[c] := 0 ; od ; for c from 0 to 2^n-1 do rlen := A038374(c) ; count[rlen] := count[rlen]+1 ; od ; RETURN( sum('count[c]*c','c'=0..n) ); end: for n from 1 to 40 do print(n,A119706(n)) ; od : # R. J. Mathar, Jun 15 2006 # second Maple program: b:= proc(n, m) option remember; `if`(n=0, 1, `if`(m=0, add(b(n-j, j), j=1..n), add(b(n-j, min(n-j, m)), j=1..min(n, m)))) end: a:= proc(n) option remember; `if`(n<2, n, 2*a(n-1) +b(n, 0)) end: seq(a(n), n=1..40); # Alois P. Heinz, Dec 19 2014
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Mathematica
nn=10;Drop[Apply[Plus,Table[CoefficientList[Series[1/(1-2x)-(1-x^n)/(1-2x+x^(n+1)),{x,0,nn}],x],{n,1,nn}]],1] (* Geoffrey Critzer, Jan 12 2013 *)
Formula
a(n+1) = 2*a(n) + A007059(n+2)
a(n) > 2*a(n-1). a(n) = Sum_{i=1..(2^n)-1} A038374(i). - R. J. Mathar, Jun 15 2006
From Geoffrey Critzer, Jan 12 2013: (Start)
O.g.f.: Sum_{k>=1} 1/(1-2*x) - (1-x^k)/(1-2*x+x^(k+1)). - Corrected by Steven Finch, May 16 2020
a(n) = Sum_{k=1..n} A048004(n,k) * k.
(End)
Conjecture: a(n) = A102712(n+1)-2^n. - R. J. Mathar, Jun 05 2025
Extensions
More terms from R. J. Mathar, Jun 15 2006
Name edited by Alois P. Heinz, Mar 18 2020
Comments