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 31-40 of 48 results. Next

A261042 Generating function g(0) where g(k) = 1 - x*2*(k+1)*(k+2)/(x*2*(k+1)*(k+2) - 1/g(k+1)).

Original entry on oeis.org

1, 4, 64, 2176, 126976, 11321344, 1431568384, 243680935936, 53725527801856, 14893509177769984, 5070334006399074304, 2079588119566033616896, 1011390382859091900891136, 575501120339508919401447424, 378784713733072451034702413824, 285539131625477547496925147693056
Offset: 0

Views

Author

Peter Luschny, Aug 08 2015

Keywords

Comments

More generally let G(y) defined by the Taylor expansion of the continued fraction
g(y,k) = 1 - (y*x*(k+1)*(k+2)) / ((y*x*(k+1)*(k+2)) - 1/g(y,k+1)). Then
G(1/2) -> A002105, G(1) -> A000182, G(2) -> A261042, G(4) -> A253165 and G(1/8)(n) *2^(n-1+padic(n,2)) -> A002425.

Crossrefs

Cf. A000182, A002105, A002425, A126156 (example section), A253165.

Programs

  • Maple
    eulerCF := proc(f, len) local g, k; g := 1;
    for k from len-2 by -1 to 0 do g := 1 - f(k)/(f(k)-1/g) od;
    PolynomialTools:-CoefficientList(convert(series(g, x, len), polynom), x) end:
    A261042_list := len -> eulerCF(k -> x*2*(k+1)*(k+2), len): A261042_list(16);
    # Alternative:
    ser := series(cos(x/sqrt(2))^(-2), x, 32):
    seq(2^(2*n)*(2*n)!*coeff(ser, x, 2*n), n = 0..15); # Peter Luschny, Sep 03 2022
  • Mathematica
    fracGen[f_, len_] := Module[{g, k}, g[len] = 1; For[k = len-1, k >= 0, k--, g[k] = 1-f[k]/(f[k]-1/g[k+1])]; CoefficientList[g[0] + O[x]^(len+1), x] ]; A261042list[len_] := fracGen[x*2*(#+1)*(#+2)&, len-1]; A261042list[16] (* Jean-François Alcover, Aug 08 2015, after Peter Luschny *)
  • Sage
    def A261042_list(len):
        f = lambda k: x*2*(k+1)*(k+2)
        g = 1
        for k in range(len-2,-1,-1):
            g = (1-f(k)/(f(k)-1/g)).simplify_rational()
        return taylor(g, x, 0, len-1).list()
    A261042_list(16)

Formula

a(n) = 2^(2*n)*(2*n)!*[x^(2*n)] cos(x/sqrt(2))^(-2). - Peter Luschny, Sep 03 2022

A325222 E.g.f.: D(x,k) = dn( i * Integral C(x,k) dx, k) such that C(x,k) = cn( i * Integral C(x,k) dx, k), where D(x,k) = Sum_{n>=0} Sum_{j=0..n} T(n,j) * x^(2*n)*k^(2*j)/(2*n)!, as a triangle of coefficients T(n,j) read by rows.

Original entry on oeis.org

1, 0, 1, 0, 8, 1, 0, 136, 88, 1, 0, 3968, 6240, 816, 1, 0, 176896, 513536, 195216, 7376, 1, 0, 11184128, 51880064, 39572864, 5352544, 66424, 1, 0, 951878656, 6453433344, 8258202240, 2458228480, 139127640, 597864, 1, 0, 104932671488, 978593947648, 1889844670464, 994697838080, 137220256000, 3535586112, 5380832, 1, 0, 14544442556416, 178568645312512, 485265505927168, 398800479698944, 102950036177920, 7233820923904, 88992306208, 48427552, 1
Offset: 0

Views

Author

Paul D. Hanna, Apr 13 2019

Keywords

Comments

Equals a row reversal of triangle A322231.
Compare to dn(x,k) = 1 - k^2 * Integral sn(x,k)*cn(x,k) dx, where sn(x,k), cn(x,k), and dn(x,k) are Jacobi elliptic functions.

Examples

			E.g.f.: D(x,k) = 1 + k^2*x^2/2! + (8*k^2 + 1*k^4)*x^4/4! + (136*k^2 + 88*k^4 + 1*k^6)*x^6/6! + (3968*k^2 + 6240*k^4 + 816*k^6 + 1*k^8)*x^8/8! + (176896*k^2 + 513536*k^4 + 195216*k^6 + 7376*k^8 + 1*k^10)*x^10/10! + (11184128*k^2 + 51880064*k^4 + 39572864*k^6 + 5352544*k^8 + 66424*k^10 + 1*k^12)*x^12/12! + (951878656*k^2 + 6453433344*k^4 + 8258202240*k^6 + 2458228480*k^8 + 139127640*k^10 + 597864*k^12 + 1*k^14)*x^14/14! + ...
such that D(x,k) = dn( i * Integral C(x,k) dx, k) where C(x,k) = cn( i * Integral C(x,k) dx, k).
This triangle of coefficients T(n,j) of x^(2*n)*k^(2*j)/(2*n)! in e.g.f. D(x,k) begins:
1;
0, 1;
0, 8, 1;
0, 136, 88, 1;
0, 3968, 6240, 816, 1;
0, 176896, 513536, 195216, 7376, 1;
0, 11184128, 51880064, 39572864, 5352544, 66424, 1;
0, 951878656, 6453433344, 8258202240, 2458228480, 139127640, 597864, 1;
0, 104932671488, 978593947648, 1889844670464, 994697838080, 137220256000, 3535586112, 5380832, 1;
0, 14544442556416, 178568645312512, 485265505927168, 398800479698944, 102950036177920, 7233820923904, 88992306208, 48427552, 1; ...
RELATED SERIES.
The related series S(x,k), where D(x,k)^2 - k^2*S(x,k)^2 = 1, starts
S(x,k) = x + (2 + 1*k^2)*x^3/3! + (16 + 28*k^2 + 1*k^4)*x^5/5! + (272 + 1032*k^2 + 270*k^4 + 1*k^6)*x^7/7! + (7936 + 52736*k^2 + 36096*k^4 + 2456*k^6 + 1*k^8)*x^9/9! + (353792 + 3646208*k^2 + 4766048*k^4 + 1035088*k^6 + 22138*k^8 + 1*k^10)*x^11/11! + (22368256 + 330545664*k^2 + 704357760*k^4 + 319830400*k^6 + 27426960*k^8 + 199284*k^10 + 1*k^12)*x^13/13! + (1903757312 + 38188155904*k^2 + 120536980224*k^4 + 93989648000*k^6 + 18598875760*k^8 + 702812568*k^10 + 1793606*k^12 + 1*k^14)*x^15/15! + ...
The related series C(x,k), where C(x,k)^2 - S(x,k)^2 = 1, starts
C(x,k) = 1 + x^2/2! + (5 + 4*k^2)*x^4/4! + (61 + 148*k^2 + 16*k^4)*x^6/6! + (1385 + 6744*k^2 + 2832*k^4 + 64*k^6)*x^8/8! + (50521 + 410456*k^2 + 383856*k^4 + 47936*k^6 + 256*k^8)*x^10/10! + (2702765 + 32947964*k^2 + 54480944*k^4 + 17142784*k^6 + 780544*k^8 + 1024*k^10)*x^12/12! + (199360981 + 3402510924*k^2 + 8760740640*k^4 + 5199585280*k^6 + 686711040*k^8 + 12555264*k^10 + 4096*k^12)*x^14/14! + ...
which also satisfies C(x,k) = cn( i * Integral C(x,k) dx, k).
		

Crossrefs

Cf. A325220 (S), A325221(C).
Cf. A322231 (row reversal).

Programs

  • PARI
    N=10;
    {S=x; C=1; D=1; for(i=1, 2*N, S = intformal(C^2*D +O(x^(2*N+1))); C = 1 + intformal(S*C*D); D = 1 + k^2*intformal(S*C^2)); }
    {T(n,j) = (2*n)!*polcoeff(polcoeff(D, 2*n, x), 2*j, k)}
    for(n=0, N, for(j=0, n, print1( T(n,j), ", ")) ; print(""))

Formula

E.g.f. D = D(x,k) = Sum_{n>=0} Sum_{j=0..n} T(n,j) * x^(2*n)*k^(2*j)/(2*n)!, along with related series S = S(x,k) and C = C(x,k), satisfies:
(1a) S = Integral C^2*D dx.
(1b) C = 1 + Integral S*C*D dx.
(1c) D = 1 + k^2 * Integral S*C^2 dx.
(2a) C^2 - S^2 = 1.
(2b) D^2 - k^2*S^2 = 1.
(3a) C + S = exp( Integral C*D dx ).
(3b) D + k*S = exp( k * Integral C^2 dx ).
(4a) S = sinh( Integral C*D dx ).
(4b) S = sinh( k * Integral C^2 dx ) / k.
(4c) C = cosh( Integral C*D dx ).
(4d) D = cosh( k * Integral C^2 dx ).
(5a) d/dx S = C^2*D.
(5b) d/dx C = S*C*D.
(5c) d/dx D = k^2 * S*C^2.
Given sn(x,k), cn(x,k), and dn(x,k) are Jacobi elliptic functions, with i^2 = -1, k' = sqrt(1-k^2), then
(6a) S = -i * sn( i * Integral C dx, k),
(6b) C = cn( i * Integral C dx, k),
(6c) D = dn( i * Integral C dx, k).
(7a) S = sc( Integral C dx, k') = sn(Integral C dx, k')/cn(Integral C dx, k'),
(7b) C = nc( Integral C dx, k') = 1/cn(Integral C dx, k'),
(7c) D = dc( Integral C dx, k') = dn(Integral C dx, k')/cn(Integral C dx, k').
Row sums equal ( (2*n)!/(n!*2^n) )^2 = A001818(n), the squares of the odd double factorials.
Column T(n,n+1) = 2^n*A002105(n+1), for n>=0, where A002105 gives the reduced tangent numbers.

A087736 Triangle T(n,k) read by rows given by [0, 1, 3, 6, 10, 15, 21, ...] DELTA [1, 3, 6, 10, 15, 21, 28,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 4, 0, 4, 23, 34, 0, 34, 249, 606, 496, 0, 496, 4354, 14181, 20434, 11056, 0, 11056, 112238, 450097, 894838, 885032, 349504, 0, 349504, 4008024, 18911670, 47136095, 65613780, 48468804, 14873104
Offset: 0

Views

Author

Philippe Deléham, Sep 30 2003, Jul 17 2007

Keywords

Examples

			Triangle begins:
1;
0, 1;
0, 1, 4;
0, 4, 23, 34;
0, 34, 249, 606, 496;
0, 496, 4354, 14181, 20434, 11056;
0, 11056, 112238, 450097, 894838, 885032, 349504;
0, 349504, 4008024, 18911670, 47136095, 65613780, 48468804, 14873104 ;...
		

Crossrefs

Diagonals give A002105: [1, 1, 4, 34, 496, ...] Row sums give A000364 : [1, 1, 5, 61, 1385, ...] Euler numbers.

Formula

Sum_{k, 0<=k<=n}T(n,k)*x^(n-k)=A000012(n), A011782(n), A001147(n), A002105(n+1), A000364(n), A126151(n), A126156(n) for n = -3,-2,-1,0,1,2,3 respectively .

Extensions

Corrected and edited. - Philippe Deléham, Nov 24 2008

A177389 Expansion of o.g.f.: Sum_{n>=0} Product_{k=1..n} tanh(k*arctanh(x)).

Original entry on oeis.org

1, 1, 2, 6, 22, 98, 514, 3110, 21334, 163650, 1388162, 12902086, 130391830, 1423632546, 16699055490, 209432697830, 2796597560150, 39613075175554, 593253347702530, 9366042608039814, 155466234198142998
Offset: 0

Views

Author

Paul D. Hanna, May 15 2010

Keywords

Examples

			O.g.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 22*x^4 + 98*x^5 + 514*x^6 + ...
Let G(x) = Sum_{n>=0} A002105(n+1)*x^n/n!, so that
G(x) = 1 + x + 4*x^2/2! + 34*x^3/3! + 496*x^4/4! + 11056*x^5/5! + ...
then A(x) = G(arctanh(x)).
G.f.: 1 + x + x*(2x/(1+x^2)) + x*(2x/(1+x^2))*((3x+x^3)/(1+3x^2)) + x*(2x/(1+x^2))*((3x+x^3)/(1+3x^2))*((4x+4x^3)/(1+6x^2+x^4)) + ... - _Paul D. Hanna_, May 22 2010
		

Crossrefs

Cf. A002105.

Programs

  • PARI
    {a(n)=local(X=x+x*O(x^n),Egf);Egf=sum(m=0,n,prod(k=1,m,tanh(k*atanh(X))));polcoeff(Egf,n)}
    
  • PARI
    {a(n)=polcoeff(sum(m=0,n,prod(k=1,m,((1+x)^k-(1-x)^k)/((1+x)^k+(1-x)^k+x*O(x^n)))),n)} \\ Paul D. Hanna, May 22 2010

Formula

O.g.f.: A(x) = Sum_{n>=0} A002105(n+1)*arctanh(x)^n/n!, where A002105 is the reduced tangent numbers.
G.f.: Sum_{n>=0} Product_{k=1..n} ((1+x)^k - (1-x)^k)/((1+x)^k + (1-x)^k). - Paul D. Hanna, May 22 2010
a(n) ~ 2^(3*n+9/2) * n^(n+1) / (exp(n) * Pi^(2*n+2)). - Vaclav Kotesovec, Nov 06 2014

A218260 E.g.f.: Sum_{n>=0} Product_{k=1..n} tanh((2*k-1)*x).

Original entry on oeis.org

1, 1, 6, 88, 2280, 92416, 5393376, 428428288, 44450655360, 5836916064256, 946245183223296, 185613384522661888, 43330332249288714240, 11871318610487327850496, 3772031142226151742038016, 1375871976238663365598117888
Offset: 0

Views

Author

Paul D. Hanna, Oct 24 2012

Keywords

Examples

			E.g.f.: A(x) = 1 + x + 6*x^2/2! + 88*x^3/3! + 2280*x^4/4! + 92416*x^5/5! +...
where
A(x) = 1 + tanh(x) + tanh(x)*tanh(3*x) + tanh(x)*tanh(3*x)*tanh(5*x) + tanh(x)*tanh(3*x)*tanh(5*x)*tanh(7*x) +...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(X=x+x*O(x^n),Egf);Egf=sum(m=0,n,prod(k=1,m,tanh((2*k-1)*X)));n!*polcoeff(Egf,n)}
    for(n=0,30,print1(a(n),", "))

Formula

E.g.f.: Sum_{n>=0} Product_{k=1..n} (1 - exp(-2*(2*k-1)*x)) / (1 + exp(-2*(2*k-1)*x)).
a(n) ~ 2^(4*n+7/2) * n^(2*n+1) / (exp(2*n) * Pi^(2*n+1)). - Vaclav Kotesovec, Nov 02 2014

A305533 Expansion of 1/(1 - x/(1 - 1*x/(1 - 3*x/(1 - 6*x/(1 - 10*x/(1 - ... - (k*(k + 1)/2)*x/(1 - ...))))))), a continued fraction.

Original entry on oeis.org

1, 1, 2, 7, 47, 592, 12287, 374857, 15639302, 851542747, 58536120467, 4953497262712, 505784457870707, 61300510121162077, 8698776162350603222, 1428545280744850604767, 268795232754158224790687, 57445320930331531152213232, 13837791987711934467999437927
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 04 2018

Keywords

Comments

Invert transform of reduced tangent numbers (A002105).

Crossrefs

Programs

  • Maple
    T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1)
    else (((k - n - 1)*(k - n)) / 2) * T(n, k - 1) + T(n - 1, k) fi fi end:
    a := n -> T(n, n): seq(a(n), n = 0..18);  # Peter Luschny, Oct 01 2023
  • Mathematica
    nmax = 18; CoefficientList[Series[1/(1 - x/(1 + ContinuedFractionK[-k (k + 1)/2 x, 1, {k, 1, nmax}])), {x, 0, nmax}], x]
    nmax = 18; CoefficientList[Series[1/(1 - Sum[PolyGamma[2 k - 1, 1/2]/(2^(k - 2) Pi^(2 k)) x^k, {k, 1, nmax}]), {x, 0, nmax}], x]
    a[0] = 1; a[n_] := a[n] = Sum[2^k (2^(2 k) - 1) Abs[BernoulliB[2 k]]/k a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 18}]

Formula

a(n) ~ 2^(3*n + 2) * n^(2*n - 1/2) / (exp(2*n) * Pi^(2*n - 1/2)). - Vaclav Kotesovec, Jun 08 2019

A309522 Generalized Blasius numbers, square array read by ascending antidiagonals, A(n, k) for n, k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 4, 6, 14, 1, 1, 11, 34, 24, 42, 1, 1, 36, 375, 496, 120, 132, 1, 1, 127, 6306, 27897, 11056, 720, 429, 1, 1, 463, 129256, 3156336, 3817137, 349504, 5040, 1430, 1, 1, 1717, 2877883, 514334274, 3501788976, 865874115, 14873104, 40320, 4862
Offset: 0

Views

Author

Keywords

Comments

The generalized Blasius o.d.e. of order n whose infinite series solution involves row n of this square array appears in Salié (1955). Rows n = 2 and n = 3 of this array appear in Kuba and Panholzer (2014, 2016), who give combinatorial interpretations to the numbers in those two rows.
Eq. (22) in Kuba and Panholzer (2014, p. 23) and Eq. (5) in Kuba and Panholzer (2016, p. 233) are general o.d.e.s for generating infinite sequences of numbers with some combinatorial properties. Even though they bear some similarity to Salié's general o.d.e., it is not clear whether either one can be used to give combinatorial interpretation to the numbers in rows n >= 4 of the current square array.

Examples

			Table A(n,k) (with rows n >= 0 and columns k >= 0) begins as follows:
[0] 1, 1,  2,    5,      14,         42,           132, ...  A000108
[1] 1, 1,  2,    6,      24,        120,           720, ...  A000142
[2] 1, 1,  4,   34,     496,      11056,        349504, ...  A002105
[3] 1, 1, 11,  375,   27897,    3817137,     865874115, ...  A018893
[4] 1, 1, 36, 6306, 3156336, 3501788976, 7425169747776, ...
     A260878|
		

Crossrefs

Rows include A000108, A000142, A002105 (shifted), A018893.
Columns include A260878.
Cf. A256522 (Blasius constant), A260876.

Programs

  • Maple
    A := proc(n, k) option remember; if k = 0 then 1 else
    add(binomial(n*k-1, n*v)*A(n, v)*A(n, k-1-v), v=0..k-1) fi end:
    seq(seq(A(n-k, k), k=0..n), n=0..9);
  • Mathematica
    A[n_, k_] := A[n, k] = If[k == 0, 1, Sum[Binomial[n*k - 1, n*v]*A[n, v]* A[n, k - 1 - v], {v, 0, k - 1}]];
    Table[A[n - k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 26 2019, from Maple *)

Formula

A(n, k) = Sum_{v=0..k-1} binomial(n*k-1, n*v)*A(n, v)*A(n, k-1-v) for k > 0 and A(n, 0) = 1.
A(n, 2) = A260876(n, 2) = binomial(2*n - 1, n) + 1 for n >= 0.
A(n, 3) = A260876(n, 2) + A260876(n, 3) - 1 = (binomial(3*n - 1, 2*n) + 1) * (binomial(2*n - 1, n) + 1) + binomial(3*n - 1, n) for n >= 1.

A113897 Triangle read by rows: number of simsun n-permutations with k descents.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 11, 4, 1, 26, 34, 1, 57, 180, 34, 1, 120, 768, 496, 1, 247, 2904, 4288, 496, 1, 502, 10194, 28768, 11056, 1, 1013, 34096, 166042, 141584, 11056, 1, 2036, 110392, 868744, 1372088, 349504, 1, 4083, 349500, 4247720, 11204160, 6213288, 349504
Offset: 1

Views

Author

Chak-On Chow (cchow(AT)alum.mit.edu), Jan 28 2006

Keywords

Comments

Is this A094503 after removal of the top row? - R. J. Mathar, Aug 13 2008
Yes. See formula of Peter Bala, Jun 26 2012 in A094503. - Stefano Spezia, Aug 09 2023

Examples

			Triangle begins
   1;
   1,   1;
   1,   4;
   1,  11,   4;
   1,  26,  34;
   1,  57, 180,  34;
   ...
		

Crossrefs

Programs

  • Mathematica
    Table[SeriesCoefficient[(2t-1)*(Sec[x*Sqrt[2t-1]/2]/(Sqrt[2t-1]- Tan[x*Sqrt[2t-1]/2]))^2,{x,0,n},{t,0,k}]n!,{n,11},{k,0,Floor[n/2]}]//Flatten (* Stefano Spezia, Aug 09 2023 *)

Formula

T(n, k) = (k+1)*T(n-1, k) + (n-2k+1)*T(n-1, k-1);
Row g.f.: T(n, t) = Sum_{k=0..floor(n/2)} T(n, k)*t^k,
T(n, t) = ((n-1)*t + 1)*T(n-1, t) + t*(1-2t)*T(n-1, t)'.
E.g.f.: Sum_{n>=1} T(n, t)*x^n/n! = (2t-1)*(sec(x*sqrt(2t-1)/2)/(sqrt(2t-1) - tan(x*sqrt(2t-1)/2)))^2.

Extensions

Corrected and extended by Vladeta Jovovic, Jan 30 2006

A187802 E.g.f.: Sum_{n>=0} Product_{k=1..n} tanh(n*k*x).

Original entry on oeis.org

1, 1, 16, 970, 146176, 44183536, 23478931456, 20054284098640, 25800626187206656, 47592874959658936576, 121099500781576410628096, 411996060596290629454466560, 1826628916277875316651443879936, 10329535274999799577516027932553216, 73156530986984637348101331408897703936
Offset: 0

Views

Author

Paul D. Hanna, Jan 06 2013

Keywords

Comments

Compare to the e.g.f. of A002105, the reduced tangent numbers:
Sum_{n>=0} Product_{k=1..n} tanh(k*x).

Examples

			E.g.f.: A(x) = 1 + x + 16*x^2/2! + 970*x^3/3! + 146176*x^4/4! +...
where
A(x) = 1 + tanh(x) + tanh(2*1*x)*tanh(2*2*x) + tanh(3*1*x)*tanh(3*2*x)*tanh(3*3*x) + tanh(4*1*x)*tanh(4*2*x)*tanh(4*3*x)*tanh(4*4*x) +...
		

Crossrefs

Cf. A002105.

Programs

  • Mathematica
    With[{nn=20},CoefficientList[Series[Sum[Product[Tanh[n*k*x],{k,n}],{n,0,nn}],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Jan 26 2024 *)
  • PARI
    {a(n)=local(X=x+x*O(x^n), Egf); Egf=sum(m=0, n, prod(k=1, m, tanh(m*k*X))); n!*polcoeff(Egf, n)}
    for(n=0, 20, print1(a(n), ", "))

Formula

a(n) ~ c * d^n * (n!)^3 / sqrt(n), where d = 2.67441747301630303932685879..., c = 0.4405132627693901422580367... . - Vaclav Kotesovec, Nov 02 2014

A203827 Table of coefficients of up-down polynomials P_n(m) = Sum_{i=0..floor(log_2(2n))} binomial(m,i).

Original entry on oeis.org

1, -1, 1, -1, 0, 1, 1, -1, 1, -1, 0, 0, 1, 1, -1, 0, 2, 1, 0, -1, 2, -1, 1, -1, 1, -1, 0, 0, 0, 1, 1, -1, 0, 0, 3, 1, 0, -1, 0, 5, -1, 1, -1, 0, 3, 1, 0, 0, -1, 3, -1, 1, 0, -2, 5, -1, 0, 1, -2, 3, 1, -1, 1, -1, 1, -1, 0, 0, 0, 0, 1, 1, -1, 0, 0, 0, 4, 1, 0, -1, 0, 0, 9
Offset: 0

Views

Author

Vladimir Shevelev, Jan 06 2012

Keywords

Comments

For a permutation s = (s_1,...,s_m), the number n = Sum_{j=1..m-1} b_j*2^(m-i-1), where b_j=1, if s_(j+1) > s_j, and b_j=0, if s_(j+1) < s_j, is called index of s. Up-down polynomial P_n(m) gives the number of permutations with index n.
If n = 2^(k_1-1) + 2^(k_2-1) + ... + 2^(k_r-1), k_1 > ... > k_r >= 1, then k_1,k_2,...,k_r are only positive roots of the polynomial P_n(m).
If F(m,x) = Sum_{n>=0} P_n(m)*x^n and t(x) = Sum_{n>=0} t_n*x^n (|x|<1), where t_n = (-1)^A010060(n), then F(m,x)/t(x) is a rational function.
The sequence {n_k} for which P_(n_k)(m) has a root m=-1 begins 2, 5, 8, 11, 23, ...
If n is in A089633, then P_n(m) has only real roots.
Remark from the author. Ivan Niven posed the problem of enumeration of permutations of n elements with a given up-down structure. He introduced (n-1)-dimensional parameter (Niven's signature) and did the enumeration in a determinant form, but did not find simple relations. Therefore, the title of his paper includes the word "problem". Instead of his many-dimensional parameter, the author introduced one-dimensional parameter (index). It allowed us to find many simple relations and properties for the enumeration numbers which is called up-down coefficients since they have many close properties with the binomial coefficients. In particular, they give a decomposition of Eulerian numbers. Many other relations will appear in the paper by the author and U. Spilker (to appear), where, in particular, we prove that the enumeration numbers are maximal when the index corresponds to the alternating (or Andre) permutations.

Examples

			Table begins
   1
  -1  1
  -1  0  1
   1 -1  1
  -1  0  0  1
   1 -1  0  2
   1  0 -1  2
  -1  1 -1  1
  -1  0  0  0  1
   1 -1  0  0  3
   1  0 -1  0  5
  -1  1 -1  0  3
   1  0  0 -1  3
  -1  1  0 -2  5
  -1  0  1 -2  3
   1 -1  1 -1  1
  -1  0  0  0  0  1
   1 -1  0  0  0  4
   1  0 -1  0  0  9
		

References

  • I. Niven, A combinatorial problem of finite sequences, Nieuw Arch. Wisk. 16(1968), 116-123.

Crossrefs

Formula

Sum_{n=0..2^(m-1)} P_n(m) = m!;
Sum_{n=0..2^m-1} (-1)^n*P_n(m) = 0.
P_{2^m-1}(2*m) = binomial(2*m-1, m-1);
P_{2^m-1}(2*m+1) = binomial(2*m, m).
If m is an odd prime, then
(1) P_n(m) == t_n (mod m), where t_n = (-1)^A010060(n);
(2) P_((2^(m+1)-4)/6)(m) == (-1)^((m-1)/2) (mod m);
(3) P_((2^(2*m+1)-2)/6)(2*m) == 1 (mod 2*m).
For m >= 1, P_((2^(2^m+1)-2)/6)(2^m) == 1 (mod 2^m).
P_((4^m-1)/3)(2*m) = |E_(2*m)| (cf. A000364);
P_((2^(2*m-1)-1)/3) = |B_(2*m)|*4^m(4^m-1)/(2*m) (cf. A002105).
If n = 2^(k_1-1) + 2^(k_2-1) + ... + 2^(k_r-1), k_1 > k_2 > ... > k_r >= 1, then
(Recursion 1) P_n(m) = (-1)^r + Sum_{i=1..r} binomial(m,k_i)*P_(n-2^(k_i-1))(k_i) and
(Recursion 2) for h > k_1, P_(n+2^(h-1))(m) = binomial(m,h)*P_n(h) - P_n(m).
Previous Showing 31-40 of 48 results. Next