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.

A207652 G.f.: Sum_{n>=0} Product_{k=1..n} ((1+x)^k - 1)/(1 - x^k).

Original entry on oeis.org

1, 1, 3, 10, 45, 249, 1709, 13912, 131168, 1402706, 16757321, 221018769, 3188425939, 49925523804, 843121969923, 15272776193787, 295372123082865, 6073931908657770, 132329525329523223, 3044691799670213778, 73771773281455834427, 1877511491197391256001
Offset: 0

Views

Author

Paul D. Hanna, Feb 19 2012

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 10*x^3 + 45*x^4 + 249*x^5 + 1709*x^6 +...
such that, by definition,
A(x) = 1 + ((1+x)-1)/(1-x) + ((1+x)-1)*((1+x)^2-1)/((1-x)*(1-x^2)) + ((1+x)-1)*((1+x)^2-1)*((1+x)^3-1)/((1-x)*(1-x^2)*(1-x^3)) +...
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(sum(m=0,n,prod(k=1,m,((1+x)^k-1)/(1-x^k +x*O(x^n)) )),n)}
    for(n=0,40,print1(a(n),", "))

Formula

From Vaclav Kotesovec, Oct 31 2014: (Start)
a(n) ~ 6*sqrt(2) * 12^n * n! / (exp(Pi^2/24) * Pi^(2*n+2)).
a(n) ~ 12^(n+1) * n^(n+1/2) / (exp(n + Pi^2/24) * Pi^(2*n+3/2)).
(End)

A207653 G.f.: Sum_{n>=0} Product_{k=1..n} (1 - (1-x)^(2*k-1))/(1 - x^(2*k-1)).

Original entry on oeis.org

1, 1, 4, 16, 77, 460, 3287, 27561, 265307, 2880875, 34821316, 463543454, 6737545832, 106158368798, 1802204594518, 32793160634292, 636683459975767, 13137118248246982, 287070448575006268, 6622644707103106925, 160846900060253917905, 4102379491083664461080
Offset: 0

Views

Author

Paul D. Hanna, Feb 19 2012

Keywords

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 16*x^3 + 77*x^4 + 460*x^5 + 3287*x^6 +...
such that, by definition,
A(x) = 1 + (1-(1-x))/(1-x) + (1-(1-x))*(1-(1-x)^3)/((1-x)*(1-x^3)) + (1-(1-x))*(1-(1-x)^3)*(1-(1-x)^5)/((1-x)*(1-x^3)*(1-x^5)) +...
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(sum(m=0,n,prod(k=1,m,(1-(1-x)^(2*k-1))/(1-x^(2*k-1) +x*O(x^n)) )),n)}
    for(n=0,40,print1(a(n),", "))

Formula

From Vaclav Kotesovec, Oct 31 2014: (Start)
a(n) ~ 6*sqrt(2) * exp(Pi^2/24) * 12^n * n! / Pi^(2*n+2).
a(n) ~ exp(Pi^2/24) * 12^(n+1) * n^(n+1/2) / (exp(n) * Pi^(2*n+3/2)).
(End)

A207654 G.f.: Sum_{n>=0} Product_{k=1..n} ((1+x)^(2*k-1) - 1)/(1 - x^(2*k-1)).

Original entry on oeis.org

1, 1, 4, 22, 173, 1816, 23659, 367573, 6622465, 135637477, 3111148862, 78984029782, 2198423489832, 66562555228478, 2177861372888738, 76571625673934064, 2878937040339348981, 115260759545001030638, 4895471242828376133806, 219853190410155476470763
Offset: 0

Views

Author

Paul D. Hanna, Feb 19 2012

Keywords

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 22*x^3 + 173*x^4 + 1816*x^5 + 23659*x^6 +...
such that, by definition,
A(x) = 1 + ((1+x)-1)/(1-x) + ((1+x)-1)*((1+x)^3-1)/((1-x)*(1-x^3)) + ((1+x)-1)*((1+x)^3-1)*((1+x)^5-1)/((1-x)*(1-x^3)*(1-x^5)) +...
		

Crossrefs

Programs

  • Mathematica
    With[{nn=20},CoefficientList[Series[Sum[Product[((1+x)^(2k-1)-1)/(1- x^(2k-1)),{k,n}],{n,0,nn}],{x,0,nn}],x]] (* Harvey P. Dale, Sep 06 2015 *)
  • PARI
    {a(n)=polcoeff(sum(m=0,n,prod(k=1,m,((1+x)^(2*k-1)-1)/(1-x^(2*k-1) +x*O(x^n)) )),n)}
    for(n=0,25,print1(a(n),", "))

Formula

From Vaclav Kotesovec, Oct 31 2014: (Start)
a(n) ~ sqrt(6) * 24^n * n! / (exp(Pi^2/48) * sqrt(n) * Pi^(2*n+3/2)).
a(n) ~ 2^n * 12^(n+1/2) * n^n / (exp(n + Pi^2/48) * Pi^(2*n+1)).
(End)

A193548 Decimal expansion of exp(Pi^2/12).

Original entry on oeis.org

2, 2, 7, 6, 1, 0, 8, 1, 5, 1, 6, 2, 5, 7, 3, 4, 0, 9, 4, 7, 9, 1, 0, 6, 1, 4, 1, 2, 0, 3, 1, 4, 9, 7, 4, 4, 6, 6, 9, 7, 9, 7, 4, 2, 6, 0, 3, 0, 0, 2, 3, 7, 7, 5, 6, 1, 5, 5, 1, 6, 1, 7, 0, 9, 8, 2, 7, 5, 0, 6, 3, 7, 2, 8, 6, 3, 0, 1, 4, 3, 1, 8, 6, 6, 8, 4, 6, 5, 7
Offset: 1

Views

Author

John M. Campbell, Jul 30 2011

Keywords

Crossrefs

Programs

  • Mathematica
    Product[Product[k^((1/(n+1))*(-1)^(k)*Binomial[n,k-1]*HarmonicNumber[n]),{k,1,n+1}],{n,1,Infinity}]
    RealDigits[E^(Pi^2/12), 10, 100]
  • PARI
    exp(Pi^2/12) \\ Charles R Greathouse IV, Jul 30 2011

Formula

exp(Pi^2/12) = Product_{n>=1} Product_{k=1..n+1} k^(1/(n+1)) * H(n) * (-1)^k * binomial(n, k-1) where H(n) is the n-th harmonic number.
exp(Pi^2/12) = lim_{n -> infinity} Product_{k=1..n} (1 + k/n)^(1/k). - Peter Bala, Feb 14 2015
Showing 1-4 of 4 results.