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.

Showing 1-5 of 5 results.

A101190 G.f.: A(x) = Sum_{n>=0} a(n)/2^A005187(n) * x^n = lim_{n->oo} F(n)^(1/2^n) where F(n) is defined by F(n) = F(n-1)^2 + x^(2^n-1) for n >= 1 with F(0) = 1.

Original entry on oeis.org

1, 1, -1, 5, -53, 127, -677, 2221, -61133, 205563, -1394207, 4852339, -68586849, 243751723, -1741612525, 6265913725, -363239625661, 1323861506899, -9699189175227, 35700526467479, -527987675255931, 1960112858076289, -14606721595781139, 54604708004873403
Offset: 0

Views

Author

Paul D. Hanna, Dec 03 2004

Keywords

Examples

			G.f.: A(x) = 1 + 1/2*x - 1/8*x^2 + 5/16*x^3 - 53/128*x^4 + 127/256*x^5 - 677/1024*x^6 + 2221/2048*x^7 + ... + a(n)/2^A005187(n)*x^n + ...
where 2^A005187(n) is also the denominator of [x^n] 1/sqrt(1-x).
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 + 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-th roots of F(n) tend to the limit of the g.f.:
F(1)^(1/2^1) = 1 + 1/2*x - 1/8*x^2 + 1/16*x^3 - 5/128*x^4 + 7/256*x^5 - 21/1024*x^6 + 33/2048*x^7 - 429/32768*x^8 + ...
F(2)^(1/2^2) = 1 + 1/2*x - 1/8*x^2 + 5/16*x^3 - 53/128*x^4 + 127/256*x^5 - 677/1024*x^6 + 1965/2048*x^7 - 46797/32768*x^8 + ...
F(3)^(1/2^3) = 1 + 1/2*x - 1/8*x^2 + 5/16*x^3 - 53/128*x^4 + 127/256*x^5 - 677/1024*x^6 + 2221/2048*x^7 - 61133/32768*x^8 + ...
...
The limit of this process equals the g.f. A(x) of this sequence.
Note: the sum of the coefficients in F(n) equals A003095(n):
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).
POWERS OF A(x).
The coefficients of x^k in the 2^n powers of the g.f. A(x) begin:
A^(2^0) = [1, 1/2, -1/8, 5/16, -53/128, 127/256, -677/1024, 2221/2048, ...],
A^(2^1) = [1, 1, 0, 1/2, -1/2, 1/2, -5/8, 9/8, -2, 53/16, -89/16, 155/16, ...],
A^(2^2) = [1, 2, 1, 1, 0, 0, 0, 1/2, -1, 3/2, -5/2, 9/2, -8, 14, -197/8, 44, ...],
A^(2^3) = [1, 4, 6, 6, 5, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1/2, -2, 5, ...],
A^(2^4) = [1, 8, 28, 60, 94, 116, 114, 94, 69, 44, 26, 14, 5, 2, 1, 1, 0, 0, ...].

Crossrefs

Cf. A101189, A101191, A005187, A003095, A000108.

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 + x^(2^k-1) +x*O(x^n)); A = polcoeff(F^(1/2^L),n)); numerator(A)}
for(n=0,32, print1(a(n),", "))

Formula

G.f. A(x) = ( Sum_{n>=0} A101191(n)/2^A004134(n) * x^n )^2.
G.f. A(x) satisfies A(2*x)^2 = Sum_{n>=0} A101189(n)*(2*x)^n.

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

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

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.

A101192 G.f. defined as the limit: A(x) = lim_{n->oo} F(n)^(1/3^(n-1)) where F(n) is the n-th iteration of: F(0) = 1, F(n) = F(n-1)^3 + (3x)^((3^n-1)/2) for n >= 1.

Original entry on oeis.org

1, 3, 0, 0, 27, -162, 729, -2916, 10206, -28431, 39366, 216513, -2506302, 16395939, -87687765, 419838390, -1879883964, 8098629399, -33997343652, 136405492911, -478000355922, 987247848321, 4754553381171, -85842565710012, 782970953914944, -5641921802462517, 34830591205459716
Offset: 0

Views

Author

Paul D. Hanna, Dec 07 2004

Keywords

Comments

The Euler transform of the power series A(x) at x=1/3 converges to the constant: c = Sum_{n>=0} (Sum_{k=0..n} C(n,k)*a(k)/3^k)/2^(n+1) = 2.080400667750319352117745232... which is the limit of S(n)^(1/3^(n-1)) where S(0)=1, S(n+1) = S(n)^3 + 1.

Examples

			The iteration begins:
F(0) = 1,
F(1) = 1 +  3*x,
F(2) = 1 +  9*x +  27*x^2 +   27*x^3 +    81*x^4,
F(3) = 1 + 27*x + 324*x^2 + 2268*x^3 + 10449*x^4 + ... + 1594323*x^13.
The 3^(n-1)-th roots of F(n) tend to the limit of A(x):
F(1)^(1/3^0) = 1 + 3*x
F(2)^(1/3^1) = 1 + 3*x + 27*x^4 - 162*x^5 + 729*x^6 - 2916*x^7 + ...
F(3)^(1/3^2) = 1 + 3*x + 27*x^4 - 162*x^5 + 729*x^6 - 2916*x^7 + ...

Crossrefs

Cf. A101189, A101193, A101194.

Programs

  • PARI
    {a(n)=local(F=1,A,L);if(n==0,A=1,L=ceil(log(n+1)/log(3)); for(k=1,L,F=F^3+(3*x)^((3^k-1)/2)); A=polcoeff((F+x*O(x^n))^(1/3^(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=3.

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

Views

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

Cf. A101189, A101192, A101194.

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

Cf. A101189, A101192, A101193.

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 1-5 of 5 results.

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