cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A101189 G.f. A(x) is defined as the limit A(x) = lim_{n->oo} F(n)^(1/2^(n-1)) where F(n) is defined by F(n) = F(n-1)^2 + (2*x)^(2^n-1) for n >= 1 with F(0) = 1.

Original entry on oeis.org

1, 2, 0, 4, -8, 16, -40, 144, -512, 1696, -5696, 19840, -70048, 247744, -880128, 3152768, -11386624, 41389568, -151273728, 555794944, -2052141056, 7610274816, -28331018240, 105833345024, -396594444800, 1490425179136, -5615651143680, 21209004267520, -80276663808000
Offset: 0

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Author

Paul D. Hanna, Dec 03 2004

Keywords

Comments

Sequences A101190 and A101191 are related to doubly exponential numbers A003095 and to Catalan numbers (A000108).

Examples

			G.f.: A(x) = 1 + 2*x + 4*x^3 - 8*x^4 + 16*x^5 - 40*x^6 + 144*x^7 - 512*x^8 + 1696*x^9 - 5696*x^10 + 19840*x^11 - 70048*x^12 + ...
GENERATING METHOD.
We can illustrate the generating method for g.f. A(x) as follows.
Given F(n) = F(n-1)^2 + (2*x)^(2^n-1) for n >= 1 with F(0) = 1,
the first few polynomials generated by F(n) begin
F(0) = 1,
F(1) = F(0)^2 + (2*x)^(2^1-1) = 1 + 2*x,
F(2) = F(1)^2 + (2*x)^(2^2-1) = 1 + 4*x + 4*x^2 + 8*x^3,
F(3) = F(2)^2 + (2*x)^(2^3-1) = 1 + 8*x + 24*x^2 + 48*x^3 + 80*x^4 + 64*x^5 + 64*x^6 + 128*x^7,
F(4) = F(3)^2 + (2*x)^(2^4-1) = = 1 + 16*x + 112*x^2 + 480*x^3 + 1504*x^4 + 3712*x^5 + 7296*x^6 + 12032*x^7 + 17664*x^8 + 22528*x^9 + 26624*x^10 + 28672*x^11 + 20480*x^12 + 16384*x^13 + 16384*x^14 + 32768*x^15,
...
and the 2^(n-1)-th root of F(n) yields the series shown by
F(1)^(1/2^0) = 1 + 2*x,
F(2)^(1/2^1) = 1 + 2*x + 4*x^3 - 8*x^4 + 16*x^5 - 40*x^6 + 112*x^7 - 320*x^8 + 928*x^9 - 2752*x^10 + 8320*x^11 - 25504*x^12 + ...,
F(3)^(1/2^2) = 1 + 2*x + 4*x^3 - 8*x^4 + 16*x^5 - 40*x^6 + 144*x^7 - 512*x^8 + 1696*x^9 - 5696*x^10 + 19840*x^11 - 70048*x^12 + ...,
F(4)^(1/2^3) = 1 + 2*x + 4*x^3 - 8*x^4 + 16*x^5 - 40*x^6 + 144*x^7 - 512*x^8 + 1696*x^9 - 5696*x^10 + 19840*x^11 - 70048*x^12 + ...,
...
The limit of this process tends to the g.f. A(x).
		

Crossrefs

Programs

  • PARI
    {a(n) = my(F=1,A,L); if(n==0,A=1, L = ceil(log(n+1)/log(2)); for(k=1,L, F = F^2 + (2*x)^(2^k-1)  +x*O(x^n)); A = polcoeff(F^(1/(2^(L-1))),n)); A}
    for(n=0,32, print1(a(n),", "))

Formula

G.f. A(x) = ( Sum_{n>=0} A101190(n)/2^A005187(n) * (2*x)^n )^2.
G.f. A(x) = ( Sum_{n>=0} A101191(n)/2^A004134(n) * (2*x)^n )^4.

Extensions

Entry revised by Paul D. Hanna, Mar 05 2024

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.

Original entry on oeis.org

1, 1, -3, 23, -525, 2695, -29687, 191991, -10488701, 70977675, -968279181, 6752850945, -191225421641, 1363019302883, -19538003443615, 140961586090743, -16379289413266717, 119621607825995891, -1755802638936696081, 12944528671963135869, -383361262914445548739
Offset: 0

Views

Author

Paul D. Hanna, Dec 03 2004

Keywords

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...

Examples

			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).
		

Crossrefs

Programs

  • 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)}

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.
Showing 1-2 of 2 results.