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.

Previous Showing 11-15 of 15 results.

A207397 G.f.: Sum_{n>=0} Product_{k=1..n} (q^k - 1) where q = (1+x)/(1+x^2).

Original entry on oeis.org

1, 1, 1, 2, 11, 74, 557, 4799, 47004, 516717, 6302993, 84502346, 1235198136, 19552296646, 333212892221, 6083009119262, 118433569748072, 2449663066933397, 53643715882853914, 1239875630317731463, 30163779836127304106, 770476745704778418686
Offset: 0

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Author

Paul D. Hanna, Feb 17 2012

Keywords

Comments

Motivated by Peter Bala's identity described in A158690:
Sum_{n>=0} Product_{k=1..n} (q^k - 1) =
Sum_{n>=0} q^(-n^2) * Product_{k=1..n} (q^(2*k-1) - 1),
here q = (1+x)/(1+x^2). See cross-references for other examples.
At present Bala's identity is conjectural and needs formal proof.

Examples

			G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 11*x^4 + 74*x^5 + 557*x^6 + 4799*x^7 +...
Let q = (1+x)/(1+x^2), then
A(x) = 1 + (q-1) + (q-1)*(q^2-1) + (q-1)*(q^2-1)*(q^3-1) + (q-1)*(q^2-1)*(q^3-1)*(q^4-1) + (q-1)*(q^2-1)*(q^3-1)*(q^4-1)*(q^5-1) +...
which also is proposed to equal:
A(x) = 1 + (q-1)/q + (q-1)*(q^3-1)/q^4 + (q-1)*(q^3-1)*(q^5-1)/q^9 + (q-1)*(q^3-1)*(q^5-1)*(q^7-1)/q^16 + (q-1)*(q^3-1)*(q^5-1)*(q^7-1)*(q^9-1)/q^25 +...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(A=1+x,q=(1+x)/(1+x^2 +x*O(x^n))); A=sum(m=0,n,prod(k=1,m,(q^k-1)));polcoeff(A,n)}
    
  • PARI
    {a(n)=local(A=1+x,q=(1+x)/(1+x^2 +x*O(x^n))); A=sum(m=0,n,q^(-m^2)*prod(k=1,m,(q^(2*k-1)-1)));polcoeff(A,n)}
    for(n=0,25,print1(a(n),", "))

Formula

G.f.: Sum_{n>=0} q^(-n^2) * Product_{k=1..n} (q^(2*k-1) - 1) where q = (1+x)/(1+x^2). [Based on Peter Bala's conjecture in A158690]
a(n) ~ c * 12^n * n! / Pi^(2*n), where c = 6*sqrt(2) / (Pi^2 * exp(Pi^2/8)) = 0.250367043877216848533826021231826... . - Vaclav Kotesovec, May 06 2014, updated Aug 22 2017

A207433 G.f.: Sum_{n>=0} Product_{k=1..n} (q^k - 1) where q = (1-x^3)/(1-x).

Original entry on oeis.org

1, 1, 3, 11, 56, 350, 2609, 22582, 222625, 2462969, 30219676, 407276420, 5981197376, 95073427910, 1626294895274, 29788176027819, 581704672430937, 12064521684969823, 264843222932272690, 6135057298705027024, 149559103545555671423, 3827360866024134614644
Offset: 0

Views

Author

Paul D. Hanna, Feb 17 2012

Keywords

Comments

Motivated by Peter Bala's identity described in A158690:
Sum_{n>=0} Product_{k=1..n} (q^k - 1) =
Sum_{n>=0} q^(-n^2) * Product_{k=1..n} (q^(2*k-1) - 1),
here q = (1+x)/(1+x^2). See cross-references for other examples.
At present Bala's identity is conjectural and needs formal proof.
a(n) = number of upper triangular matrices with entries from {0,1,2} with no zero rows such that the sum of the entries is n, that is, row Fishburn matrices of size n with entries from {0,1,2}. Cf. A179525. - Peter Bala, Nov 05 2017

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 11*x^3 + 56*x^4 + 350*x^5 + 2609*x^6 +...
Let q = (1-x^3)/(1-x) = 1 + x + x^2, then
A(x) = 1 + (q-1) + (q-1)*(q^2-1) + (q-1)*(q^2-1)*(q^3-1) + (q-1)*(q^2-1)*(q^3-1)*(q^4-1) + (q-1)*(q^2-1)*(q^3-1)*(q^4-1)*(q^5-1) +...
Also, we have the identity:
A(x) = 1 + (q-1)/q + (q-1)*(q^3-1)/q^4 + (q-1)*(q^3-1)*(q^5-1)/q^9 + (q-1)*(q^3-1)*(q^5-1)*(q^7-1)/q^16 + (q-1)*(q^3-1)*(q^5-1)*(q^7-1)*(q^9-1)/q^25 +...
From _Peter Bala_, Nov 05 2017: (Start)
a(3) = 11: The eleven row Fishburn matrices of size 3 with entries in {0,1,2} are
/1 0\ /2 0\ /0 1\ /0 2\ /1 1\
\0 2/ \0 1/ \0 2/ \0 1/ \0 1/
and
/1 0 0\ /0 1 0\ /0 0 1\ /1 0 0\ /0 1 0\ /0 0 1\
|0 1 0| |0 1 0| |0 0 1| |0 0 1| |0 0 1| |0 0 1|.
\0 0 1/ \0 0 1/ \0 0 1/ \0 0 1/ \0 0 1/ \0 0 1/
(End)
		

Crossrefs

Programs

  • PARI
    {a(n)=local(A=1+x,q=(1+x+x^2 +x*O(x^n))); A=sum(m=0,n,prod(k=1,m,(q^k-1)));polcoeff(A,n)}
    
  • PARI
    {a(n)=local(A=1+x,q=(1+x+x^2 +x*O(x^n))); A=sum(m=0,n,q^(-m^2)*prod(k=1,m,(q^(2*k-1)-1)));polcoeff(A,n)}
    for(n=0,25,print1(a(n),", "))

Formula

G.f.: Sum_{n>=0} q^(-n^2) * Product_{k=1..n} (q^(2*k-1) - 1) where q = (1-x^3)/(1-x). [Based on Peter Bala's conjecture in A158690]
a(n) ~ exp(Pi^2/24) * 2^(2*n+3/2) * 3^(n+1) * n! / Pi^(2*n+2). - Vaclav Kotesovec, Aug 22 2017

A186737 G.f. satisfies: A(x) = Sum_{n>=0} Product_{k=1..n} ((1 + x*A(x))^k - 1).

Original entry on oeis.org

1, 1, 3, 14, 82, 563, 4390, 38273, 370090, 3951524, 46436236, 598606795, 8431113021, 129134793211, 2139840753676, 38161937079832, 728817179712481, 14837170081079164, 320656534559643735, 7329984468062242588, 176662137501934534082, 4476371962949911027001, 118946746110478235868525
Offset: 0

Views

Author

Paul D. Hanna, Apr 23 2013

Keywords

Comments

This is an example of Peter Bala's identity (cf. A158690):
Sum_{n>=0} Product_{k=1..n} (q^k - 1) = Sum_{n>=0} q^(-n^2) * Product_{k=1..n} (q^(2*k-1) - 1) at q = 1+x*A(x).

Examples

			G.f.: A(X) = 1 + x + 3*x^2 + 14*x^3 + 82*x^4 + 563*x^5 + 4390*x^6 +...
where
A(x) = 1 + ((1+x*A(x))-1) + ((1+x*A(x))-1)*((1+x*A(x))^2-1) + ((1+x*A(x))-1)*((1+x*A(x))^2-1)*((1+x*A(x))^3-1) + ((1+x*A(x))-1)*((1+x*A(x))^2-1)*((1+x*A(x))^3-1)*((1+x*A(x))^4-1) +...
Let q = 1+x*A(x), then:
A(x) = 1 + (q-1)/q + (q-1)*(q^3-1)/q^4 + (q-1)*(q^3-1)*(q^5-1)/q^9 + (q-1)*(q^3-1)*(q^5-1)*(q^7-1)/q^16 + (q-1)*(q^3-1)*(q^5-1)*(q^7-1)*(q^9-1)/q^25 +...
		

Crossrefs

Cf. A179525.

Programs

  • Mathematica
    nmax = 22; sol = {a[0] -> 1};
    Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x] - Sum[Product[ ((1 + x A[x])^k - 1), {k, 1, j}], {j, 0, n}] + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
    sol /. Rule -> Set;
    a /@ Range[0, nmax] (* Jean-François Alcover, Nov 02 2019 *)
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=sum(m=0,n,prod(k=1,m,(1+x*A+x*O(x^n))^k - 1)));polcoeff(A,n)}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    {a(n)=local(A=1+x, q); for(i=1,n,q=(1+x*A +x*O(x^n)); A=sum(m=0, n, q^(-m^2)*prod(k=1, m, (q^(2*k-1)-1)))); polcoeff(A, n)}
    for(n=0, 20, print1(a(n), ", "))

Formula

G.f. satisfies:
(1) A(x) = Sum_{n>=0} (1+x*A(x))^(-n^2) * Product_{k=1..n} ((1+x*A(x))^(2*k-1) - 1).
(2) A(x) = 1/(1 - (q-1)/(q - (q^2-1)/(q^2 - (q^3-1)/(q^3 - (q^4-1)/(q^4 - (q^5-1)/(q^5 -...)))))), a continued fraction, where q = 1+x*A(x).
a(n) ~ 6*sqrt(2) * exp(Pi^2/24) * 12^n * n! / Pi^(2*n+2). - Vaclav Kotesovec, May 05 2014

A207214 E.g.f.: Sum_{n>=0} exp(n*x) * Product_{k=1..n} (exp(k*x) - 1).

Original entry on oeis.org

1, 1, 7, 85, 1759, 55621, 2501407, 151984645, 12004046719, 1196068161541, 146792747463007, 21762540250822405, 3834791755438306879, 792270319634586707461, 189687840256042278859807, 52103089179906338874671365, 16275196750916467736633834239
Offset: 0

Views

Author

Paul D. Hanna, Feb 16 2012

Keywords

Comments

Compare the e.g.f. to the identity:
exp(-x) = Sum_{n>=0} exp(n*x) * Product_{k=1..n} (1 - exp(k*x)).

Examples

			E.g.f.: A(x) = 1 + x + 7*x^2/2! + 85*x^3/3! + 1759*x^4/4! + 55621*x^5/5! +...
such that, by definition,
A(x) = 1 + exp(x) * (exp(x)-1) + exp(2*x) * (exp(x)-1)*(exp(2*x)-1)
+ exp(3*x) * (exp(x)-1)*(exp(2*x)-1)*(exp(3*x)-1)
+ exp(4*x) * (exp(x)-1)*(exp(2*x)-1)*(exp(3*x)-1)*(exp(4*x)-1) +...
The related e.g.f. of A158690 equals the series:
G(x) = 1 + (exp(x)-1) + (exp(x)-1)*(exp(2*x)-1)
+ (exp(x)-1)*(exp(2*x)-1)*(exp(3*x)-1)
+ (exp(x)-1)*(exp(2*x)-1)*(exp(3*x)-1)*(exp(4*x)-1) +...
or, more explicitly,
G(x) = 1 + x + 5*x^2/2! + 55*x^3/3! + 1073*x^4/4! + 32671*x^5/5! +...
such that G(x) satisfies:
G(x) = (1 + exp(x)*A(x))/2.
		

Crossrefs

Cf. A158690.

Programs

  • PARI
    {a(n)=n!*polcoeff(sum(m=0,n+1,exp(m*x+x*O(x^n))*prod(k=1,m,exp(k*x+x*O(x^n))-1)),n)}
    for(n=0,20,print1(a(n),", "))

Formula

E.g.f. A(x) satisfies: A(x) = exp(-x)*(2*G(x) - 1),
where G(x) = Sum_{n>=0} Product_{k=1..n} (exp(k*x) - 1) = e.g.f. of A158690.
a(n) ~ sqrt(2) * 12^(n+1) * (n!)^2 / Pi^(2*n+2). - Vaclav Kotesovec, May 05 2014

A209832 Expansion of the q-series Sum_{n>=0} (-1)^n*q^(n+1)*Product_{k = 1..n} (1 - q^(2*k-1)), q = exp(t), as a formal Taylor series in t.

Original entry on oeis.org

1, 2, 12, 200, 6576, 353312, 28032192, 3077502080, 446470392576, 82695752049152, 19038594625539072, 5332477132779407360, 1785375992372231909376, 704147423230177089953792, 323094378183013059349757952, 170643791820813252598723543040
Offset: 0

Views

Author

Peter Bala, Mar 14 2012

Keywords

Comments

Compare with A158690.

Crossrefs

Cf. A158690.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Sum[(-1)^n*Exp[x*(n + 1)] * Product[ (1 - Exp[(2*k - 1)*x]), {k, 1, n}], {n, 0, nmax}], {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Oct 09 2023 *)

Formula

E.g.f.: Sum_{n>=0} exp((n+1)*t) * Product_{k = 1..n} (exp((2*k-1)*t) - 1) = exp(t) + exp(2*t)*(exp(t) - 1) + exp(3*t)*(exp(t) - 1)*(exp(3*t) - 1) + ... = 1 + 2*t + 12*t^2/2! + 200*t^3/3! + ...
Conjectural S-fraction expansion for the o.g.f.:
1/(1-2*x/(1-4*x/(1-16*x/(1-20*x/(1-...-2*n(3*n-2)*x/(1-2*n(3*n-1)*x/(1-...
a(n) ~ 2^(3*n + 2) * 3^(n + 1/2) * n^(2*n + 1/2) / (exp(2*n) * Pi^(2*n + 1/2)). - Vaclav Kotesovec, Oct 09 2023
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