A081254 Numbers k such that A081252(m)/m^2 has a local maximum for m = k.

1, 3, 6, 13, 26, 53, 106, 213, 426, 853, 1706, 3413, 6826, 13653, 27306, 54613, 109226, 218453, 436906, 873813, 1747626, 3495253, 6990506, 13981013, 27962026, 55924053, 111848106, 223696213, 447392426, 894784853, 1789569706, 3579139413

Offset: 1

Author

Klaus Brockhaus, Mar 17 2003

Keywords

• nonn
• easy

Comments

The limit of the local maxima, lim_{m->inf} A081252(m)/m^2 = 1/10. For local minima cf. A081253.

Row sums of the triangle A181971. - Reinhard Zumkeller, Jul 09 2012

Examples

			13 is a term since A081252(12)/12^2 = 15/144 = 0.104..., A081252(13)/13^2 = 18/169 = 0.106..., A081252(14)/14^2 = 20/196 = 0.102....
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Thomas Baruchel, Properties of the cumulated deficient binary digit sum, arXiv:1908.02250 [math.NT], 2019.
• Klaus Brockhaus, Illustration for A053646, A081252, A081253 and A081254
• Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Crossrefs

Cf. A000975, A020714, A020989, A048573, A052549, A053646, A072197, A081252, A081253, A266219 (binary).

Programs

Magma

[Floor(2^(n-1)*5/3): n in [1..40]]; // Vincenzo Librandi, Apr 04 2012

Maple

seq(floor(2^(n-1)*5/3),n=1..35); # Muniru A Asiru, Sep 20 2018

Mathematica

Rest@CoefficientList[Series[-(x^2 - x - 1)*x/((x - 1)*(x + 1)*(2*x - 1)), {x, 0, 32}], x] (* Vincenzo Librandi, Apr 04 2012 *)
a[n_]:=Floor[2^(n-1)*5/3]; Array[a,33,1] (* Stefano Spezia, Sep 01 2018 *)

PARI

a(n) = 2^(n-1)*5\3; \\ Altug Alkan, Sep 21 2018

Formula

a(n) = floor(2^(n-1)*5/3). [corrected by Michel Marcus, Sep 21 2018]

a(n) = a(n-2) + 5*2^(n-3) for n > 2;

a(n+2) - a(n) = A020714(n-1);

a(n) + a(n-1) = A052549(n-1) for n > 1;

a(2*n+1) = A020989(n); a(2n) = A072197(n-1);

a(n+1) - a(n) = A048573(n-1).

G.f.: -(x^2 - x - 1)*x/((x - 1)*(x + 1)*(2*x - 1)).

a(n) = 5*2^(n-1)/3 + (-1)^n/6-1/2. a(n) = 2*a(n-1) + (1+(-1)^n)/2, a(1)=1. - Paul Barry, Mar 24 2003

a(2n) = 2*a(2*n-1) + 1, a(2*n+1) = 2*a(2*n), a(1)=1. a(n) = A000975(n-1) + 2^(n-1). - Philippe Deléham, Oct 15 2006

a(n) = A005578(n) + A000225(n-1). - Yuchun Ji, Sep 21 2018

a(n) - a(n-2) = 2 * (a(n-1) - a(n-3)), with a(0..2)=[1,3,6]. - Yuchun Ji, Mar 18 2020