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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A101193 G.f. defined as the limit: A(x) = lim_{n->oo} F(n)^(1/4^(n-1)) where F(n) is the n-th iteration of: F(0) = 1, F(n) = F(n-1)^4 + (4x)^((4^n-1)/3) for n >= 1.

Original entry on oeis.org

1, 4, 0, 0, 0, 256, -3072, 24576, -163840, 983040, -5603328, 32112640, -195035136, 1283457024, -8975810560, 64281903104, -458387095552, 3216662069248, -22225382014976, 152271623028736, -1043452104015872, 7199883459035136, -50175319780360192, 353054558068408320
Offset: 0

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Author

Paul D. Hanna, Dec 07 2004

Keywords

Comments

The Euler transform of the power series A(x) at x=1/4 converges to the constant: c = Sum_{n>=0} (Sum_{k=0..n} C(n,k)*a(k)/4^k)/2^(n+1) = 2.030544704345910171947313128... which is the limit of S(n)^(1/4^(n-1)) where S(0)=1, S(n+1) = S(n)^4 +1.

Examples

			The iteration begins:
F(0) = 1,
F(1) = 1 +  4*x,
F(2) = 1 + 16*x +   96*x^2 +   256*x^3 + 256*x^4 + 1024*x^5,
F(3) = 1 + 64*x + 1920*x^2 + 35840*x^3 + ...     + 4398046511104*x^21.
The 4^(n-1)-th roots of F(n) tend to the limit of A(x):
F(1)^(1/4^0) = 1 + 4*x
F(2)^(1/4^1) = 1 + 4*x + 256*x^5 - 3072*x^6 + 24576*x^7 - 163840*x^8 + ...
F(3)^(1/4^2) = 1 + 4*x + 256*x^5 - 3072*x^6 + 24576*x^7 - 163840*x^8 + ...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(F=1,A,L);if(n==0,A=1,L=ceil(log(n+1)/log(4)); for(k=1,L,F=F^4+(4*x)^((4^k-1)/3)); A=polcoeff((F+x*O(x^n))^(1/4^(L-1)),n));A}

Formula

G.f. begins: A(x) = (1+m*x) + m^m*x^(m+1)/(1+m*x)^(m-1) + ... at m=4.

A101194 G.f. defined as the limit: A(x) = lim_{n->oo} F(n)^(1/5^(n-1)) where F(n) is the n-th iteration of: F(0) = 1, F(n) = F(n-1)^5 + (5x)^((5^n-1)/4) for n >= 1.

Original entry on oeis.org

1, 5, 0, 0, 0, 0, 3125, -62500, 781250, -7812500, 68359375, -546875000, 4082031250, -28417968750, 179443359375, -939941406250, 2685546875000, 23010253906250, -569122314453125, 7669982910156250, -84739685058593750, 836715698242187500, -7611751556396484375
Offset: 0

Views

Author

Paul D. Hanna, Dec 07 2004

Keywords

Comments

The Euler transform of the power series A(x) at x=1/5 converges to the constant: c = Sum_{n>=0} (Sum_{k=0..n} C(n,k)*a(k)/5^k) / 2^(n+1) = 2.012346619142363112612326559... which is the limit of S(n)^(1/5^(n-1)) where S(0)=1, S(n+1) = S(n)^5 + 1.

Examples

			The iteration begins:
F(0) = 1,
F(1) = 1 + 5*x
F(2) = 1 + 25*x + 250*x^2 + 1250*x^3 + 3125*x^4 + 3125*x^5 + 15625*x^6
F(3) = 1 + 125*x + 7500*x^2 + 287500*x^3 + ... + 5^31*x^31.
The 5^(n-1)-th root of F(n) tend to the limit of A(x):
F(1)^(1/5^0) = 1 + 5*x
F(2)^(1/5^1) = 1 + 5*x + 3125*x^6 - 62500*x^7 + 781250*x^8 + ...
F(3)^(1/5^2) = 1 + 5*x + 3125*x^6 - 62500*x^7 + 781250*x^8 + ...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(F=1,A,L);if(n==0,A=1,L=ceil(log(n+1)/log(5)); for(k=1,L,F=F^5+(5*x)^((5^k-1)/4)); A=polcoeff((F+x*O(x^n))^(1/5^(L-1)),n));A}

Formula

G.f. begins: A(x) = (1+m*x) + m^m*x^(m+1)/(1+m*x)^(m-1) + ... at m=5.
Showing 1-2 of 2 results.