A206743 G.f.: 1/(1 - x/(1 - x^2/(1 - x^5/(1 - x^12/(1 - x^29/(1 - x^70/(1 -...- x^Pell(n)/(1 -...)))))))), a continued fraction.

1, 1, 1, 2, 3, 5, 8, 13, 22, 36, 60, 99, 164, 272, 450, 746, 1235, 2046, 3389, 5613, 9299, 15402, 25514, 42262, 70005, 115962, 192084, 318182, 527053, 873043, 1446161, 2395504, 3968060, 6572925, 10887788, 18035177, 29874537, 49485965, 81971484, 135782448

Offset: 0

Author

Paul D. Hanna, Feb 12 2012

Keywords

• nonn

Comments

From Clark Kimberling, Jun 12 2016: (Start)

Number of real integers in n-th generation of tree T(2i) defined as follows.

Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*. Let g(n) be the set of nodes in the n-th generation, so that g(0) = {0}, g(1) = {1}, g(2) = {2,x}, g(3) = {3,2x,x+1,x^2}, etc. Let T(r) be the tree obtained by substituting r for x.

For r = 2i, then g(3) = {3,2r,r+1, r^2}, in which the number of real integers is a(3) = 2.

See A274142 for a guide to related sequences. (End)

Examples

			G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 3*x^4 + 5*x^5 + 8*x^6 + 13*x^7 +...
		

Links

• Kenny Lau, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A000621, A206741, A274142.

Programs

Maple

A206743 := proc(r)
local gs,n,gs2,el,a ;
gs := [2,r] ;
for n from 3 do
gs2 := [] ;
for el in gs do
gs2 := [op(gs2),el+1,r*el] ;
end do:
gs := gs2 ;
a := 0 ;
for el in gs do
if type(el,'realcons') then
a := a+1 :
end if;
end do:
print(n,a) ;
end do:
end proc: # R. J. Mathar, Jun 16 2016

Mathematica

z = 18; t = Join[{{0}}, Expand[NestList[DeleteDuplicates[Flatten[Map[{# + 1, x*#} &, #], 1]] &, {1}, z]]]; u = Table[t[[k]] /. x -> 2 I, {k, 1, z}]; Table[Count[Map[IntegerQ, u[[k]]], True], {k, 1, z}] (* Clark Kimberling, Jun 12 2016 *)

PARI

{Pell(n)=polcoeff(x/(1-2*x-x^2+x*O(x^n)),n)}
{a(n)=local(CF=1+x*O(x^n),M=ceil(log(2*n+1)/log(2.4))); for(k=0, M, CF=1/(1-x^Pell(M-k+1)*CF)); polcoeff(CF, n, x)}
for(n=0,55,print1(a(n),", "))

Python

N = 1000
pell = [0,1]
c = 2
while c < N:
    pell.append(c)
    c = pell[-1]*2 + pell[-2]
pell.reverse()
gf = [0]*(N+1)
for p in pell:
    gf = [-x for x in gf]
    gf[0] += 1
    quotient = [0]*(N+1)
    remainder = [0]*(N+1)
    remainder[p] = 1
    for n in range(N+1):
        q = remainder[n]//gf[0]
        for i in range(n,N+1):
            remainder[i] -= q*gf[i-n]
        quotient[n] = q
    gf = quotient
for i in range(N+1):
    print(i,gf[i])
# Kenny Lau, Aug 01 2017

Formula

a(n) ~ c * d^n, where d = 1.6564594309887754808836889708489581749625897572527517021957723319642053908... and c = 0.3844078703275069072126260832303344589497793302955451672191630264983... - Vaclav Kotesovec, Aug 25 2017