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.

A224732 G.f.: exp( Sum_{n>=1} binomial(2*n,n)^n * x^n/n ).

Original entry on oeis.org

1, 2, 20, 2704, 6008032, 203263062688, 103724721990326528, 801185400238209125917312, 94088900962948953837864576996352, 168691065596220817138271126002845218561536, 4634314586972355372645450331391809316221983940020224
Offset: 0

Views

Author

Paul D. Hanna, Apr 16 2013

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 20*x^2 + 2704*x^3 + 6008032*x^4 + 203263062688*x^5 +...
where
log(A(x)) = 2*x + 6^2*x^2/2 + 20^3*x^3/3 + 70^4*x^4/4 + 252^5*x^5/5 + 924^6*x^6/6 + 3432^7*x^7/7 + 12870^8*x^8/8 +...+ A000984(n)^n*x^n/n +...
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(exp(sum(k=1,n,binomial(2*k,k)^k*x^k/k)+x*O(x^n)),n)}
    for(n=0,20,print1(a(n),", "))

Formula

Logarithmic derivative yields A224733.
a(n) ~ exp(-1/8) * 2^(2*n^2) / (Pi^(n/2) * n^(1 + n/2)). - Vaclav Kotesovec, Jan 26 2015
a(n) ~ (binomial(2*n,n))^n / n. - Vaclav Kotesovec, Jan 26 2015

A224735 G.f.: exp( Sum_{n>=1} binomial(2*n,n)^3 * x^n/n ).

Original entry on oeis.org

1, 8, 140, 3616, 116542, 4316080, 175593800, 7640774080, 349626142909, 16632958651688, 816163494236860, 41069537125459360, 2110206360805542510, 110346590629125981872, 5857345961837113457864, 314962180518584299711424, 17128125582951726423704502, 940726748732537798295599280
Offset: 0

Views

Author

Paul D. Hanna, Apr 16 2013

Keywords

Examples

			G.f.: A(x) = 1 + 8*x + 140*x^2 + 3616*x^3 + 116542*x^4 + 4316080*x^5 +...
where
log(A(x)) = 2^3*x + 6^3*x^2/2 + 20^3*x^3/3 + 70^3*x^4/4 + 252^3*x^5/5 + 924^3*x^6/6 + 3432^3*x^7/7 + 12870^3*x^8/8 +...+ A000984(n)^3*x^n/n +...
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[Exp[8*x*HypergeometricPFQ[{1, 1, 3/2, 3/2, 3/2}, {2, 2, 2, 2}, 64*x]], {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 27 2025 *)
  • PARI
    {a(n)=polcoeff(exp(sum(k=1,n,binomial(2*k,k)^3*x^k/k)+x*O(x^n)),n)}
    for(n=0,20,print1(a(n),", "))

Formula

Logarithmic derivative yields A002897.

A224736 G.f.: exp( Sum_{n>=1} binomial(2*n,n)^4 * x^n/n ).

Original entry on oeis.org

1, 16, 776, 64384, 7151460, 947788608, 141137282720, 22814994697728, 3918995299504938, 705339416079749024, 131725296229995045840, 25348575698532710671104, 5000341179482293108254824, 1007144334380887781805059200, 206487157000689985136888031296
Offset: 0

Views

Author

Paul D. Hanna, Apr 16 2013

Keywords

Examples

			G.f.: A(x) = 1 + 16*x + 776*x^2 + 64384*x^3 + 7151460*x^4 + 947788608*x^5 +...
where
log(A(x)) = 2^4*x + 6^4*x^2/2 + 20^4*x^3/3 + 70^4*x^4/4 + 252^4*x^5/5 + 924^4*x^6/6 + 3432^4*x^7/7 + 12870^4*x^8/8 +...+ A000984(n)^4*x^n/n +...
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[Exp[16*x*HypergeometricPFQ[{1, 1, 3/2, 3/2, 3/2, 3/2}, {2, 2, 2, 2, 2}, 256*x]], {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 27 2025 *)
  • PARI
    {a(n)=polcoeff(exp(sum(k=1,n,binomial(2*k,k)^4*x^k/k)+x*O(x^n)),n)}
    for(n=0,20,print1(a(n),", "))

Formula

Logarithmic derivative yields A186420.

A158266 G.f.: A(x) = exp( Sum_{n>=1} C(2n-1,n)^2*x^n/n ).

Original entry on oeis.org

1, 1, 5, 38, 352, 3659, 41012, 484739, 5959417, 75523708, 980470867, 12980840984, 174675568464, 2382923659387, 32890264803521, 458576476085929, 6450351908991558, 91437202854436755, 1305115286958337403
Offset: 0

Views

Author

Paul D. Hanna, Apr 09 2009

Keywords

Comments

Compare g.f. to Catalan(x) = exp( Sum_{n>=1} C(2n-1,n)*x^n/n ), where C(2n-1,n) = A001700(n-1) and Catalan(x) is the g.f. of A000108(n) = C(2n,n)/(n+1).

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 38*x^3 + 352*x^4 + 3659*x^5 + 41012*x^6 +...
log(A(x)) = x + 3^2*x^2/2 + 10^2*x^3/3 + 35^2*x^4/4 + 126^2*x^5/5 +...
log(C(x)) = x + 3*x^2/2 + 10*x^3/3 + 35*x^4/4 + 126*x^5/5 +...
C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 +... (g.f. of A000108).
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(exp(sum(m=1,n,binomial(2*m-1,m)^2*x^m/m)+x*O(x^n)),n)}
    
  • PARI
    {a(n)=if(n==0, 1, (1/n)*sum(k=1, n, binomial(2*k-1,k)^2*a(n-k)))}

Formula

a(n) = (1/n)*Sum_{k=1..n} C(2k-1,k)^2 * a(n-k) for n>0, with a(0)=1.
A(x) = exp( Sum_{n >= 1} 1/4*C(2*n,n)^2*x^n/n ). A(x)^4 is the o.g.f. for A224734. - Peter Bala, Jun 04 2015

A362730 a(n) = [x^n] E(x)^n where E(x) = exp( Sum_{k >= 1} binomial(2*k,k)^2*x^k/k ).

Original entry on oeis.org

1, 4, 68, 1336, 27972, 607004, 13478072, 304083224, 6941422916, 159882680452, 3708781743068, 86526900550864, 2028273983776440, 47733938489878528, 1127187050415921304, 26694934151138897336, 633813403549444601156, 15082008687681962081088, 359592614152718921447108
Offset: 0

Views

Author

Peter Bala, May 05 2023

Keywords

Comments

Compare with A359108(n) = [x^n] F(x)^n where F(x) = exp( Sum_{k >= 1} binomial(2*k,k)*x^k/k ).
More generally, we inductively define a family of sequences {a(i,n) : n >= 0}, i >= 0, by setting a(0,n) = binomial(2*n,n)^2 and, for i >= 1, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. In this notation the present sequence is {a(1,n)}.
We conjecture that the sequences {a(i,n) : n >= 0}, i >= 1, satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5, and positive integers n and r.

Crossrefs

Programs

  • Maple
    E(n,x) := series( exp(n*add(binomial(2*k,k)^2*x^k/k, k = 1..20)), x, 21 ):
    seq(coeftayl(E(n,x), x = 0, n), n = 0..20);

Formula

Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 5 and positive integers n and r.
Showing 1-5 of 5 results.