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 1-4 of 4 results.

A258925 E.g.f. satisfies: A(x) = Integral 1 + A(x)^5 dx.

Original entry on oeis.org

1, 120, 3024000, 858574080000, 1226178516326400000, 5912338932461445120000000, 75732595735526211882516480000000, 2195068320271703663798288449536000000000, 128322069958974226301129597680106864640000000000
Offset: 0

Views

Author

Paul D. Hanna, Jun 15 2015

Keywords

Examples

			E.g.f.: A(x) = x + 120*x^6/6! + 3024000*x^11/11! + 858574080000*x^16/16! +...
where Series_Reversion(A(x)) = x - x^6/6 + x^11/11 - x^16/16 + x^21/21 +...
Also, A(x) = S(x)/C(x) where
S(x) = x - 24*x^6/6! - 169344*x^11/11! - 25255286784*x^16/16! - 23089632627769344*x^21/21! +...+ A258924(n)*x^(5*n+1)/(5*n+1)! +...
C(x) = 1 - 24*x^5/5! - 169344*x^10/10! - 25255286784*x^15/15! - 23089632627769344*x^20/20! +...+ A258924(n)*x^(5*n)/(5*n)! +...
such that C(x)^5 + S(x)^5 = 1.
		

Crossrefs

Programs

  • PARI
    /* E.g.f. Series_Reversion( Integral 1/(1+x^5) dx ): */
    {a(n) = local(A=x); A = serreverse( intformal( 1/(1 + x^5 + O(x^(5*n+2))) ) ); (5*n+1)!*polcoeff(A, 5*n+1)}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* E.g.f. A(x) = Integral 1 + A(x)^5 dx.: */
    {a(n) = local(A=x); for(i=1, n+1, A = intformal( 1 + A^5 + O(x^(5*n+2)) )); (5*n+1)!*polcoeff(A, 5*n+1)}
    for(n=0, 20, print1(a(n), ", "))

Formula

E.g.f. A(x) satisfies:
(1) A(x) = Series_Reversion( Integral 1/(1+x^5) dx ).
Let C(x) = S'(x) such that S(x) = Series_Reversion( Integral 1/(1-x^5)^(1/5) dx ) is the e.g.f. of A258924, then e.g.f. A(x) of this sequence satisfies:
(2) A(x) = S(x)/C(x),
(3) A(x) = Integral 1/C(x)^5 dx.
From Vaclav Kotesovec, Jun 18 2015: (Start)
a(n) ~ 5^(5*n+5/2) * ((5-sqrt(5))/8)^(5*n/2+5/8) * (5*n)! * n^(1/4) / (sqrt(2) * Gamma(1/4) * Pi^(5*n+5/4)).
a(n) ~ 5^(5*n+5/2) * sin(Pi/5)^(5*n+5/4) * (5*n)! * n^(1/4) / (sqrt(2) * Gamma(1/4) * Pi^(5*n+5/4)).
(End)

A258969 E.g.f.: A'(x) = 1 + A(x)^3, with A(0)=1.

Original entry on oeis.org

1, 2, 6, 42, 390, 4698, 69174, 1203498, 24163110, 549811962, 13982486166, 393026414922, 12099527531910, 404881353252378, 14632253175107574, 567974815524008298, 23567351945550373350, 1040985881615266375482, 48767788927611416600406, 2415210691383917131432842
Offset: 0

Views

Author

Vaclav Kotesovec, Jun 15 2015

Keywords

Comments

Conjecture: A227250(n+1) = a(n).

Examples

			A(x) = 1 + 2*x + 6*x^2/2! + 42*x^3/3! + 390*x^4/4! + 4698*x^5/5! + ...
A'(x) = 2 + 6*x + 21*x^2 + 65*x^3 + 783*x^4/4 + 11529*x^5/20 + ...
1 + A(x)^3 = 2 + 6*x + 21*x^2 + 65*x^3 + 783*x^4/4 + 11529*x^5/20 + ...
		

Crossrefs

Programs

  • Mathematica
    nmax=20; Subscript[a,0]=1; egf=Sum[Subscript[a,k]*x^k, {k,0,nmax+1}]; Table[Subscript[a,k]*k!, {k,0,nmax}] /.Solve[Take[CoefficientList[Expand[1+egf^3-D[egf,x]],x],nmax]==ConstantArray[0,nmax]][[1]]
  • PARI
    {a(n) = local(A=1); A = 1 + serreverse( intformal( 1/((2+x)*(1+x+x^2) +x*O(x^n)) )); n!*polcoeff(A,n)}
    for(n=0,25,print1(a(n),", ")) \\ Paul D. Hanna, Jun 16 2015

Formula

a(n) ~ (3/(Pi/sqrt(3)-log(2)))^(n+1/2) * n^n / exp(n).
E.g.f.: 1 + Series_Reversion( Integral 1/((2+x)*(1+x+x^2)) dx ). - Paul D. Hanna, Jun 16 2015

A258994 E.g.f.: A'(x) = 1 + A(x)^6, with A(0)=1.

Original entry on oeis.org

1, 2, 12, 192, 4272, 124992, 4531392, 195869952, 9832326912, 562125837312, 36056880110592, 2564230500421632, 200237330428342272, 17032391106795159552, 1567547894591436275712, 155196096043697480466432, 16447362605632117421309952, 1857733260790463501532659712
Offset: 0

Views

Author

Vaclav Kotesovec, Jun 16 2015

Keywords

Comments

In general, for k>1, if e.g.f. satisfies A'(x) = 1 + A(x)^k, with A(0)=1, then a(n) ~ n! * d^(n + 1/(k-1)) / ((k-1)^(1/(k-1)) * Gamma(1/(k-1)) * n^(1-1/(k-1))), where d = 1 / Sum_{j>=1} (-1)^(j+1)/(k*j-1).

Examples

			A(x) = 1 + 2*x + 12*x^2/2! + 192*x^3/3! + 4272*x^4/4! + 124992*x^5/5! + ...
A'(x) = 2 + 12*x + 96*x^2 + 712*x^3 + 5208*x^4 + 188808*x^5/5 + ...
1 + A(x)^6 = 2 + 12*x + 96*x^2 + 712*x^3 + 5208*x^4 + 188808*x^5/5 + ...
		

Crossrefs

Cf. A000831 (k=2), A258969 (k=3), A258970 (k=4), A258971 (k=5), A258927.

Programs

  • Mathematica
    nmax=20; Subscript[a,0]=1; egf=Sum[Subscript[a,k]*x^k,{k,0,nmax+1}]; Table[Subscript[a,k]*k!,{k,0,nmax}] /.Solve[Take[CoefficientList[Expand[1+egf^6-D[egf,x]],x],nmax]==ConstantArray[0,nmax]][[1]]
  • PARI
    {a(n) = local(A=1); A = 1 + serreverse( intformal( 1/(1 + (1+x)^6 +x*O(x^n)) )); n!*polcoeff(A, n)}
    for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Jun 16 2015

Formula

a(n) ~ n! * d^(n+1/5) / (5^(1/5) * Gamma(1/5) * n^(4/5)), where d = 1 / Sum_{j>=1} (-1)^(j+1)/(6*j-1) = 6/(Pi - sqrt(3)*log(2+sqrt(3))) = 6.97224737278326506475991855023425659249063565...
E.g.f.: 1 + Series_Reversion( Integral 1/(1 + (1+x)^6) dx ). - Paul D. Hanna, Jun 16 2015

A258970 E.g.f.: A'(x) = 1 + A(x)^4, with A(0)=1.

Original entry on oeis.org

1, 2, 8, 80, 1088, 19328, 422912, 10987520, 330555392, 11300913152, 432717037568, 18344259092480, 852932666851328, 43157160112160768, 2360748463307620352, 138821061188696145920, 8732741520836633034752, 585172975239737913638912, 41612642758392039581155328
Offset: 0

Views

Author

Vaclav Kotesovec, Jun 15 2015

Keywords

Examples

			A(x) = 1 + 2*x + 8*x^2/2! + 80*x^3/3! + 1088*x^4/4! + 19328*x^5/5! + ...
A'(x) = 2 + 8*x + 40*x^2 + 544*x^3/3 + 2416*x^4/3 + 52864*x^5/15 + ...
1 + A(x)^4 = 2 + 8*x + 40*x^2 + 544*x^3/3 + 2416*x^4/3 + 52864*x^5/15 + ...
		

Crossrefs

Programs

  • Mathematica
    nmax=20; Subscript[a,0]=1; egf=Sum[Subscript[a,k]*x^k, {k,0,nmax+1}]; Table[Subscript[a,k]*k!, {k,0,nmax}] /.Solve[Take[CoefficientList[Expand[1+egf^4-D[egf,x]],x],nmax]==ConstantArray[0,nmax]][[1]]
  • PARI
    {a(n) = local(A=1); A = 1 + serreverse( intformal( 1/(1 + (1+x)^4 +x*O(x^n)) )); n!*polcoeff(A, n)}
    for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Jun 16 2015

Formula

a(n) ~ n! * d^(n+1/3) / (3^(1/3) * GAMMA(1/3) * n^(2/3)), where d = 1 / Sum_{j>=1} (-1)^(j+1)/(4*j-1) = 4*sqrt(2)/(Pi + log(3-2*sqrt(2))) = 4.10260201986929...
E.g.f.: 1 + Series_Reversion( Integral 1/(1 + (1+x)^4) dx ). - Paul D. Hanna, Jun 16 2015
Showing 1-4 of 4 results.