A048625 -id:A048625 - cp's OEIS frontend

A048626 Pisot sequence P(6,9).

6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873, 2745, 4023, 5896, 8641, 12664, 18560, 27201, 39865, 58425, 85626, 125491, 183916, 269542, 395033, 578949, 848491, 1243524, 1822473, 2670964, 3914488, 5736961, 8407925, 12322413, 18059374

Offset: 0

David W. Wilson

nonn

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,1).

Subsequence of A048625, A000930. See A008776 for definitions of Pisot sequences.

Cf. A020721.

LinearRecurrence[{1, 0, 1}, {6, 9, 13}, 42] (* Robert G. Wilson v, Jul 12 2014 *)

pisotP(nmax, a1, a2) = {
  a=vector(nmax); a[1]=a1; a[2]=a2;
  for(n=3, nmax, a[n] = ceil(a[n-1]^2/a[n-2]-1/2));
  a
}
pisotP(50, 6, 9) \\ Colin Barker, Aug 08 2016

a(n) = a(n-1) + a(n-3) (Checked up to n = 1000)

a(n) = A048625(n+1). - R. J. Mathar, Oct 15 2008

A118645 Number of binary strings of length n such that there exist three consecutive digits where at least two of them are 1's.

Original entry on oeis.org

0, 0, 1, 4, 10, 23, 51, 109, 228, 471, 964, 1960, 3967, 8003, 16107, 32362, 64941, 130200, 260866, 522415, 1045831, 2093129, 4188408, 8379967, 16764552, 33535872, 67081663, 134177863, 268377031, 536785286, 1073616333, 2147299732

Offset: 0

Tanya Khovanova, May 10 2006

We set a(2) = 1 by convention; there is one string of length 2 which has two consecutive 1's, namely 11. This also makes various formulas simpler.
For n>=3, a(n) = 2^n - the sum of all terms in the (n-3)rd power of the 4 X 4 matrix [[1 1 0 0] [0 0 1 0] [0 0 0 1] [1 1 0 0]] because this matrix represents the transitions from the state where the last three bits are 000, 001, 010, 100 to the state after the next bit, always avoiding two 1's out of the last three bits. - Joshua Zucker, Aug 04 2006
Complementary to A048625 which starts 4,6,9,13,19,28,41,60,88,129,189. For n >= 3, a(n) + A048625(n-3) = 2^n. A048625 is a subsequence of A000930, A068921 and A078012. All of them satisfy the recurrence a(n) = a(n-1) + a(n-3). - Tanya Khovanova, Aug 22 2006

			a(4) = 10 because we have: 0011, 0101, 0110, 0111, 1010, 1011, 1100, 1101, 1110, 1111. - _Geoffrey Critzer_, Jan 19 2014

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2,1,-2).

nn=31;r=Solve[{s==1+x s+x b,a==x s,b==x a,c==x a+x b+2x c},{s,a,b,c}]; CoefficientList[Series[c/.r,{x,0,nn}],x] (* Geoffrey Critzer, Jan 19 2014 *)
LinearRecurrence[{3,-2,1,-2},{0,0,1,4},40] (* Harvey P. Dale, Dec 15 2014 *)

x='x+O('x^50); concat([0,0], Vec((x^3 + x^2)/(2*x^4 - x^3 + 2*x^2 - 3*x + 1))) \\ G. C. Greubel, May 02 2017

a(n) = 3*2^(n-3) + a(n-1) + a(n-3) for n >= 3. - Tanya Khovanova, Aug 22 2006
From R. J. Mathar, Oct 03 2011: (Start)
G.f.: (x^3 + x^2)/(2*x^4 - x^3 + 2*x^2 - 3*x + 1).
G.f.: x^2 * (x+1)/((2*x-1)*(x^3+x-1)).
a(n) = 2^n - A000930(n+2). (End)

More terms from Joshua Zucker, Aug 04 2006

Edited by Franklin T. Adams-Watters, Sep 30 2011

cp's OEIS Frontend

A048626 Pisot sequence P(6,9).

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