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A101191 G.f.: A(x) = Sum_{n>=0}a(n)/2^A004134(n)*x^n = limit_{n->oo} F(n)^(1/2^(n+1)) where F(n) is the n-th iteration of: F(0) = 1, F(n) = F(n-1)^2 + x^(2^n-1) for n>=1. 2
1, 1, -3, 23, -525, 2695, -29687, 191991, -10488701, 70977675, -968279181, 6752850945, -191225421641, 1363019302883, -19538003443615, 140961586090743, -16379289413266717, 119621607825995891, -1755802638936696081, 12944528671963135869, -383361262914445548739 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Although the power series for the g.f. A(x) diverges at x=1, the Euler transform of the power series A(x) at x=1 converges to the constant A076949: Sum_{n>=0}[Sum_{k=0..n}C(n,k)*a(k))/2^A004134(n) ]/2^(n+1) = 1.2259024435...

LINKS

Table of n, a(n) for n=0..20.

FORMULA

G.f. A(x) satisfies: A(x)^2 = Sum_{n>=0} A101190(n)/2^A005187(n)*x^n. G.f. A(x) satisfies: A(2*x)^4 = Sum_{n>=0} A101189(n)*(2x)^n.

EXAMPLE

The iteration begins:

F(0) = 1,

F(1) = F(0)^2 + x^(2^1-1) = 1 +x,

F(2) = F(1)^2 + x^(2^2-1) = 1 +2*x +x^2 +x^3,

F(3) = F(2)^2 + x^(2^3-1) = 1 +4*x +6*x^2 +6*x^3 +5*x^4 +2*x^5 +x^6 +x^7.

The 2^(n+1)-th roots of F(n) tend to the limit of the g.f.:

F(1)^(1/2^2) = 1 +1/4*x -3/32*x^2 +7/128*x^3 -77/2048*x^4 +231/8192*x^5 +...

F(2)^(1/2^3) = 1 +1/4*x -3/32*x^2 +23/128*x^3 -525/2048*x^4 +2695/8192*x^5 +...

F(3)^(1/2^4) = 1 +1/4*x -3/32*x^2 +23/128*x^3 -525/2048*x^4 +2695/8192*x^5 +...

The limit of this process is the g.f. A(x) of this sequence.

The coefficients of x^k in the 2^n powers of the g.f. A(x) begin:

A^(2^0)=[1,1/4,-3/32,23/128,-525/2048,2695/8192,-29687/65536,...],

A^(2^1)=[1,1/2,-1/8,5/16,-53/128,127/256,-677/1024,2221/2048,...],

A^(2^2)=[1,1,0,1/2,-1/2,1/2,-5/8,9/8,-2,53/16,-89/16,155/16,...],

A^(2^3)=[1,2,1,1,0,0,0,1/2,-1,3/2,-5/2,9/2,-8,14,-197/8,44,...],

A^(2^4)=[1,4,6,6,5,2,1,1,0,0,0,0,0,0,0,1/2,-2,5,...],

A^(2^5)=[1,8,28,60,94,116,114,94,69,44,26,14,5,2,1,1,0,0,...].

Note: the sum of the coefficients of x^k in F(n) equals A003095(n+1):

1, 2=1+1, 5=1+2+1+1, 26=1+4+6+6+5+2+1+1, ...

The last n coefficients in F(n) read backwards are Catalan numbers (A000108).

PROG

(PARI) {a(n)=local(F=1, A, L); if(n==0, A=1, L=ceil(log(n+1)/log(2)); for(k=1, L, F=F^2+x^(2^k-1)); A=polcoeff(F^(1/2^(L+1))+x*O(x^n), n)); numerator(A)}

CROSSREFS

Cf. A101189, A101190, A005187, A003095, A076949.

Sequence in context: A171777 A154896 A134050 * A127900 A229266 A143985

Adjacent sequences:  A101188 A101189 A101190 * A101192 A101193 A101194

KEYWORD

sign

AUTHOR

Paul D. Hanna, Dec 03 2004

STATUS

approved

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Last modified September 20 08:10 EDT 2021. Contains 347577 sequences. (Running on oeis4.)