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.

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A185415 Table of coefficients of a polynomial sequence of binomial type related to A080635.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 0, 8, 0, 1, 18, 0, 20, 0, 1, 0, 148, 0, 40, 0, 1, 378, 0, 658, 0, 70, 0, 1, 0, 5040, 0, 2128, 0, 112, 0, 1, 14562, 0, 33992, 0, 5628, 0, 168, 0, 1, 0, 277164, 0, 158480, 0, 12936, 0, 240, 0, 1
Offset: 1

Views

Author

Peter Bala, Jan 27 2011

Keywords

Comments

Define a sequence of polynomials P(n,x) by means of the recurrence relation
(1)... P(n+1,x) = x*{P(n,x-1)-P(n,x)+P(n,x+1)}
with starting value P(0,x) = 1. The first few polynomials are
P(1,x) = x
P(2,x) = x^2
P(3,x) = x*(x^2+2),
P(4,x) = x^2*(x^2+8),
P(5,x) = x*(x^4+20*x^2+18).
This triangle lists the coefficients of these polynomials in ascending powers of x. The triangle has links with A080635, which gives the number of ordered increasing 0-1-2 trees on n nodes (plane unary-binary trees in the notation of [BERGERON et al.]). The number of forests of k such trees on n nodes is given by the formula
... 1/k!*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*P(n,j).
See A185422.
We also have A080635(n) = P(n,1), which can be used to calculate the terms of A080635 - see A185416.
For similarly defined polynomial sequences for other families of trees see A147309 and A185419. See also A185417.
Exponential Riordan array [(3/2)*(1-sqrt(3)*tan((Pi+3*sqrt(3)*x)/6))/(-1+2*sin((Pi-6*sqrt(3)*x)/6)), log((1/2)*(1+sqrt(3)*tan(sqrt(3)*x/2+Pi/6)))]. Production matrix is the exponential Riordan array [2*cosh(x)-1,x] beheaded. A185422*A008277^{-1}.

Examples

			Triangle begins:
  n\k|....1......2......3......4......5......6......7......8
  ==========================================================
  ..1|....1
  ..2|....0......1
  ..3|....2......0......1
  ..4|....0......8......0......1
  ..5|...18......0.....20......0......1
  ..6|....0....148......0.....40......0......1..
  ..7|..378......0....658......0.....70......0......1
  ..8|....0...5040......0...2128......0....112......0......1
		

References

  • F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, in Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1922, pp. 24-48.

Crossrefs

Programs

  • Maple
    #A185415
    P := proc(n,x)
    description 'polynomial sequence P(n,x)'
    if n = 0
    return 1
    else return
    x*(P(n-1,x-1)-P(n-1,x)+P(n-1,x+1))
    end proc:
    with(PolynomialTools):
    for n from 1 to 10
    CoefficientList(P(n,x),x);
    end do;
  • Mathematica
    p[0][x_] = 1; p[n_][x_] := p[n][x] = x*(p[n-1][x-1] - p[n-1][x] + p[n-1][x+1]) // Expand; row[n_] := CoefficientList[ p[n][x], x]; Table[row[n] // Rest, {n, 1, 10}] // Flatten (* Jean-François Alcover, Sep 11 2012 *)

Formula

GENERATING FUNCTION
The e.g.f. is
(1)... F(x, t) = E(t)^x = Sum_{n >= 0} P(n, x) * t^n/n!,
where
E(t) = 1/2+sqrt(3)/2*tan[sqrt(3)/2*t+Pi/6] = 1 + t + t^2/2! + 3*t^3/3! + 9*t^4/4! + ... is the e.g.f. for A080635.
ROW POLYNOMIALS
One easily checks that
... d/dt(F(x,t)) = x*(F(x-1,t)-F(x,t)+F(x+1,t))
and hence the row generating polynomials P(n,x) satisfy the recurrence relation
(2)... P(n+1,x) = x*{P(n,x-1)-P(n,x)+P(n,x+1)}.
RELATIONS WITH OTHER SEQUENCES
A080635(n) = P(n,1).
A185422(n,k) = 1/k!*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*P(n,j).
A185423(n,k) = Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*P(n,j).

A185416 Square array, read by antidiagonals, used to recursively calculate A080635.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 9, 6, 3, 1, 39, 24, 11, 4, 1, 189, 114, 51, 18, 5, 1, 1107, 648, 279, 96, 27, 6, 1, 7281, 4194, 1767, 594, 165, 38, 7, 1, 54351, 30816, 12699, 4176, 1143, 264, 51, 8, 1, 448821, 251586, 101979, 32922, 8865, 2034, 399, 66, 9, 1
Offset: 1

Views

Author

Peter Bala, Jan 28 2011

Keywords

Comments

The table entries T(n,k), n,k>=1, are defined by the recurrence relation
1)... T(n+1,k) = (k-1)*T(n,k-1)-k*T(n,k)+(k+1)*T(n,k+1) with boundary condition T(1,k)=1.
The first column of the table is A080635.
For similar tables to calculate the zigzag numbers, the Springer numbers and the number of minimax trees see A185414, A185418 and A185420, respectively.

Examples

			Triangle begins
n\k|....1......2......3......4......5.......6.......7
=====================================================
..1|....1......1......1......1......1.......1.......1
..2|....1......2......3......4......5.......6.......7
..3|....3......6.....11.....18.....27......38......51
..4|....9.....24.....51.....96....165.....264.....399
..5|...39....114....279....594...1143....2034....3399
..6|..189....648...1767...4176...8865...17304...31563
..7|.1107...4194..12699..32922..76203..161442..318339
..
Examples of the recurrence:
T(4,4) = 96 = 3*T(3,3)-4*T(3,4)+5*T(3,5) = 3*11-4*18+ 5*27;
T(5,1) = 39 = 0*T(4,0)-1*T(4,1)+2*T(4,2) = -1*9+2*24;
		

Crossrefs

Programs

  • Maple
    #A185416
    P := proc(n,x) description 'polynomial sequence P(n,x) A185415'
    if n = 0 return 1
    else return
    x*(P(n-1,x-1)-P(n-1,x)+P(n-1,x+1))
    end proc:
    for n from 1 to 10 do
    seq(P(n,k)/k,k = 1..10);
    end do;
  • PARI
    {T(n, k)=if(n==1, 1, (k-1)*T(n-1, k-1)-k*T(n-1,k)+(k+1)*T(n-1, k+1))}

Formula

(1)... T(n,k) = P(n,k)/k, where P(n,x) are the polynomials defined in A185415.

A357537 a(n) = 2*A080635(n) if n > 0. a(0) = 1.

Original entry on oeis.org

1, 2, 2, 6, 18, 78, 378, 2214, 14562, 108702, 897642, 8171766, 81066258, 871695918, 10091490138, 125189658054, 1656458307522, 23288226400062, 346663764078282, 5447099463010326, 90094171024954098, 1564653992673809358, 28467075416816935098, 541467979789775621094
Offset: 0

Views

Author

Michael Somos, Oct 02 2022

Keywords

Examples

			G.f. = 1 + 2*x + 2*x^2 + 6*x^3 + 18*x^4 + 78*x^5 + 378*x^6 + ...
E.g.f. = 1 + 2*x + x^2 + x^3 + 3/4*x^4 + 13/20*x^5 + 21/40*x^6 + ...
		

Crossrefs

Cf. A080635.

Programs

  • Mathematica
    a[ n_] := If[ n < 0, 0, n! Simplify@SeriesCoefficient[ Sqrt[3] Tan[ Pi/6 + x Sqrt[3]/2], {x, 0, n}]];
    a[ n_] := If[ n < 0, 0, Nest[Expand[(1 + x + x^2) D[#, x]]&, 1 + 2 x, n] /. x->0];
  • PARI
    {a(n) = my(A); if(n<0, 0, A = 1 + 2*x; for( k=1, n, A = A' * (1 + x + x^2)); polcoeff(A, 0))};

Formula

E.g.f.: sqrt(3) tan(Pi/6 + x sqrt(3)/2).
E.g.f. A(x) satisfies 2*A' = 3 + A^2, A'' = A*A'.
Let f(x) = 1 + x + x^2. Then a(n+1) = (f(x)*d/dx)^n f'(x) evaluated at x = 0.

A000111 Euler or up/down numbers: e.g.f. sec(x) + tan(x). Also for n >= 2, half the number of alternating permutations on n letters (A001250).

Original entry on oeis.org

1, 1, 1, 2, 5, 16, 61, 272, 1385, 7936, 50521, 353792, 2702765, 22368256, 199360981, 1903757312, 19391512145, 209865342976, 2404879675441, 29088885112832, 370371188237525, 4951498053124096, 69348874393137901, 1015423886506852352, 15514534163557086905, 246921480190207983616, 4087072509293123892361
Offset: 0

Views

Author

Keywords

Comments

Number of linear extensions of the "zig-zag" poset. See ch. 3, prob. 23 of Stanley. - Mitch Harris, Dec 27 2005
Number of increasing 0-1-2 trees on n vertices. - David Callan, Dec 22 2006
Also the number of refinements of partitions. - Heinz-Richard Halder (halder.bichl(AT)t-online.de), Mar 07 2008
The ratio a(n)/n! is also the probability that n numbers x1,x2,...,xn randomly chosen uniformly and independently in [0,1] satisfy x1 > x2 < x3 > x4 < ... xn. - Pietro Majer, Jul 13 2009
For n >= 2, a(n-2) = number of permutations w of an ordered n-set {x_1 < ... x_n} with the following properties: w(1) = x_n, w(n) = x_{n-1}, w(2) > w(n-1), and neither any subword of w, nor its reversal, has the first three properties. The count is unchanged if the third condition is replaced with w(2) < w(n-1). - Jeremy L. Martin, Mar 26 2010
A partition of zigzag permutations of order n+1 by the smallest or the largest, whichever is behind. This partition has the same recurrent relation as increasing 1-2 trees of order n, by induction the bijection follows. - Wenjin Woan, May 06 2011
As can be seen from the asymptotics given in the FORMULA section, one has lim_{n->oo} 2*n*a(n-1)/a(n) = Pi; see A132049/A132050 for the simplified fractions. - M. F. Hasler, Apr 03 2013
a(n+1) is the sum of row n in triangle A008280. - Reinhard Zumkeller, Nov 05 2013
M. Josuat-Verges, J.-C. Novelli and J.-Y. Thibon (2011) give a far-reaching generalization of the bijection between Euler numbers and alternating permutations. - N. J. A. Sloane, Jul 09 2015
Number of treeshelves avoiding pattern T321. Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link, see A278678 for more definitions and examples. - Sergey Kirgizov, Dec 24 2016
Number of sequences (e(1), ..., e(n-1)), 0 <= e(i) < i, such that no three terms are equal. [Theorem 7 of Corteel, Martinez, Savage, and Weselcouch] - Eric M. Schmidt, Jul 17 2017
Number of self-dual edge-labeled trees with n vertices under "mind-body" duality. Also number of self-dual rooted edge-labeled trees with n vertices. See my paper linked below. - Nikos Apostolakis, Aug 01 2018
The ratio a(n)/n! is the volume of the convex polyhedron defined as the set of (x_1,...,x_n) in [0,1]^n such that x_i + x_{i+1} <= 1 for every 1 <= i <= n-1; see the solutions by Macdonald and Nelsen to the Amer. Math. Monthly problem referenced below. - Sanjay Ramassamy, Nov 02 2018
Number of total cyclic orders on {0,1,...,n} such that the triple (i-1,i,i+1) is positively oriented for every 1 <= i <= n-1; see my paper on cyclic orders linked below. - Sanjay Ramassamy, Nov 02 2018
The number of binary, rooted, unlabeled histories with n+1 leaves (following the definition of Rosenberg 2006). Also termed Tajima trees, Tajima genealogies, or binary, rooted, unlabeled ranked trees (Palacios et al. 2015). See Disanto & Wiehe (2013) for a proof. - Noah A Rosenberg, Mar 10 2019
From Gus Wiseman, Dec 31 2019: (Start)
Also the number of non-isomorphic balanced reduced multisystems with n + 1 distinct atoms and maximum depth. A balanced reduced multisystem is either a finite multiset, or a multiset partition with at least two parts, not all of which are singletons, of a balanced reduced multisystem. The labeled version is A006472. For example, non-isomorphic representatives of the a(0) = 1 through a(4) = 5 multisystems are (commas elided):
{1} {12} {{1}{23}} {{{1}}{{2}{34}}} {{{{1}}}{{{2}}{{3}{45}}}}
{{{12}}{{3}{4}}} {{{{1}}}{{{23}}{{4}{5}}}}
{{{{1}{2}}}{{{3}}{{45}}}}
{{{{1}{23}}}{{{4}}{{5}}}}
{{{{12}}}{{{3}}{{4}{5}}}}
Also the number of balanced reduced multisystems with n + 1 equal atoms and maximum depth. This is possibly the meaning of Heinz-Richard Halder's comment (see also A002846, A213427, A265947). The non-maximum-depth version is A318813. For example, the a(0) = 1 through a(4) = 5 multisystems are (commas elided):
{1} {11} {{1}{11}} {{{1}}{{1}{11}}} {{{{1}}}{{{1}}{{1}{11}}}}
{{{11}}{{1}{1}}} {{{{1}}}{{{11}}{{1}{1}}}}
{{{{1}{1}}}{{{1}}{{11}}}}
{{{{1}{11}}}{{{1}}{{1}}}}
{{{{11}}}{{{1}}{{1}{1}}}}
(End)
With s_n denoting the sum of n independent uniformly random numbers chosen from [-1/2,1/2], the probability that the closest integer to s_n is even is exactly 1/2 + a(n)/(2*n!). (See Hambardzumyan et al. 2023, Appendix B.) - Suhail Sherif, Mar 31 2024
The number of permutations of size n+1 that require exactly n passes through a stack (i.e. have reverse-tier n-1) with an algorithm that prioritizes outputting the maximum possible prefix of the identity in a given pass and reverses the remainder of the permutation for prior to the next pass. - Rebecca Smith, Jun 05 2024

Examples

			G.f. = 1 + x + x^2 + 2*x^3 + 5*x^4 + 16*x^5 + 61*x^6 + 272*x^7 + 1385*x^8 + ...
Sequence starts 1,1,2,5,16,... since possibilities are {}, {A}, {AB}, {ACB, BCA}, {ACBD, ADBC, BCAD, BDAC, CDAB}, {ACBED, ADBEC, ADCEB, AEBDC, AECDB, BCAED, BDAEC, BDCEA, BEADC, BECDA, CDAEB, CDBEA, CEADB, CEBDA, DEACB, DEBCA}, etc. - _Henry Bottomley_, Jan 17 2001
		

References

  • M. D. Atkinson: Partial orders and comparison problems, Sixteenth Southeastern Conference on Combinatorics, Graph Theory and Computing, (Boca Raton, Feb 1985), Congressus Numerantium 47, 77-88.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pages 34, 932.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 258-260, section #11.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 110.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 262.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 66.
  • O. Heimo and A. Karttunen, Series help-mates in 8, 9 and 10 moves (Problems 2901, 2974-2976), Suomen Tehtavaniekat (Proceedings of the Finnish Chess Problem Society) vol. 60, no. 2/2006, pp. 75, 77.
  • L. B. W. Jolley, Summation of Series. 2nd ed., Dover, NY, 1961, p. 238.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 444.
  • E. Netto, Lehrbuch der Combinatorik. 2nd ed., Teubner, Leipzig, 1927, p. 110.
  • C. A. Pickover, The Math Book, Sterling, NY, 2009; see p. 184.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 1, 1997 and Vol. 2, 1999; see Problem 5.7.

Crossrefs

Cf. A000364 (secant numbers), A000182 (tangent numbers).
Cf. A181937 for n-alternating permutations.
Cf. A109449 for an extension to an exponential Riordan array.
Column k=2 of A250261.
For 0-1-2 trees with n nodes and k leaves, see A301344.
Matula-Goebel numbers of 0-1-2 trees are A292050.
An overview over generalized Euler numbers gives A349264.

Programs

  • Haskell
    a000111 0 = 1
    a000111 n = sum $ a008280_row (n - 1)
    -- Reinhard Zumkeller, Nov 01 2013
    
  • Maple
    A000111 := n-> n!*coeff(series(sec(x)+tan(x),x,n+1), x, n);
    s := series(sec(x)+tan(x), x, 100): A000111 := n-> n!*coeff(s, x, n);
    A000111:=n->piecewise(n mod 2=1,(-1)^((n-1)/2)*2^(n+1)*(2^(n+1)-1)*bernoulli(n+1)/(n+1),(-1)^(n/2)*euler(n)):seq(A000111(n),n=0..30); A000111:=proc(n) local k: k:=floor((n+1)/2): if n mod 2=1 then RETURN((-1)^(k-1)*2^(2*k)*(2^(2*k)-1)*bernoulli(2*k)/(2*k)) else RETURN((-1)^k*euler(2*k)) fi: end:seq(A000111(n),n=0..30); (C. Ronaldo)
    T := n -> 2^n*abs(euler(n,1/2)+euler(n,1)): # Peter Luschny, Jan 25 2009
    S := proc(n,k) option remember; if k=0 then RETURN(`if`(n=0,1,0)) fi; S(n,k-1)+S(n-1,n-k) end:
    A000364 := n -> S(2*n,2*n);
    A000182 := n -> S(2*n+1,2*n+1);
    A000111 := n -> S(n,n); # Peter Luschny, Jul 29 2009
    a := n -> 2^(n+2)*n!*(sum(1/(4*k+1)^(n+1), k = -infinity..infinity))/Pi^(n+1):
    1, seq(a(n), n = 1..22); # Emeric Deutsch, Aug 17 2009
    # alternative Maple program:
    b:= proc(u, o) option remember;
          `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
        end:
    a:= n-> b(n, 0):
    seq(a(n), n=0..30);  # Alois P. Heinz, Nov 29 2015
  • Mathematica
    n=22; CoefficientList[Series[(1+Sin[x])/Cos[x], {x, 0, n}], x] * Table[k!, {k, 0, n}] (* Jean-François Alcover, May 18 2011, after Michael Somos *)
    a[n_] := If[EvenQ[n], Abs[EulerE[n]], Abs[(2^(n+1)*(2^(n+1)-1)*BernoulliB[n+1])/(n+1)]]; Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Oct 09 2012, after C. Ronaldo *)
    ee = Table[ 2^n*EulerE[n, 1] + EulerE[n] - 1, {n, 0, 26}]; Table[ Differences[ee, n] // First // Abs, {n, 0, 26}] (* Jean-François Alcover, Mar 21 2013, after Paul Curtz *)
    a[ n_] := If[ n < 0, 0, (2 I)^n If[ EvenQ[n], EulerE[n, 1/2], EulerE[n, 0] I]]; (* Michael Somos, Aug 15 2015 *)
    a[ n_] := If[ n < 1, Boole[n == 0], With[{m = n - 1}, m! SeriesCoefficient[ 1 / (1 - Sin[x]), {x, 0, m}]]]; (* Michael Somos, Aug 15 2015 *)
    s[0] = 1; s[] = 0; t[n, 0] := s[n]; t[n_, k_] := t[n, k] = t[n, k-1] + t[n-1, n-k]; a[n_] := t[n, n]; Array[a, 30, 0](* Jean-François Alcover, Feb 12 2016 *)
    a[n_] := If[n == 0, 1, 2*Abs[PolyLog[-n, I]]]; (* Jean-François Alcover, Dec 02 2023, after M. F. Hasler *)
    a[0] := 1; a[1] := 1; a[n_] := a[n] = Sum[Binomial[n - 2, k] a[k] a[n - 1 - k], {k, 0, n - 2}]; Map[a, Range[0, 26]] (* Oliver Seipel, May 24 2024 after Peter Bala *)
    a[0] := 1; a[1] := 1; a[n_] := a[n] = 1/(n (n-1)) Sum[a[n-1-k] a[k] k, {k, 1, n-1}]; Map[#! a[#]&, Range[0, 26]] (* Oliver Seipel, May 27 2024 *)
  • Maxima
    a(n):=sum((if evenp(n+k) then (-1)^((n+k)/2)*sum(j!*stirling2(n,j)*2^(1-j)*(-1)^(n+j-k)*binomial(j-1,k-1),j,k,n) else 0),k,1,n); /* Vladimir Kruchinin, Aug 19 2010 */
    
  • Maxima
    a(n):=if n<2 then 1 else 2*sum(4^m*(sum((i-(n-1)/2)^(n-1)*binomial(n-2*m-1,i-m)*(-1)^(n-i-1),i,m,(n-1)/2)),m,0,(n-2)/2); /* Vladimir Kruchinin, Aug 09 2011 */
    
  • PARI
    {a(n) = if( n<1, n==0, n--; n! * polcoeff( 1 / (1 - sin(x + x * O(x^n))), n))}; \\ Michael Somos, Feb 03 2004
    
  • PARI
    {a(n) = local(v=[1], t); if( n<0, 0, for(k=2, n+2, t=0; v = vector(k, i, if( i>1, t+= v[k+1-i]))); v[2])}; \\ Michael Somos, Feb 03 2004
    
  • PARI
    {a(n) = local(an); if( n<1, n>=0, an = vector(n+1, m, 1); for( m=2, n, an[m+1] = sum( k=0, m-1, binomial(m-1, k) * an[k+1] * an[m-k]) / 2); an[n+1])}; \\ Michael Somos, Feb 03 2004
    
  • PARI
    z='z+O('z^66); egf = (1+sin(z))/cos(z); Vec(serlaplace(egf)) \\ Joerg Arndt, Apr 30 2011
    
  • PARI
    A000111(n)={my(k);sum(m=0,n\2,(-1)^m*sum(j=0,k=n+1-2*m,binomial(k,j)*(-1)^j*(k-2*j)^(n+1))/k>>k)}  \\ M. F. Hasler, May 19 2012
    
  • PARI
    A000111(n)=if(n,2*abs(polylog(-n,I)),1)  \\ M. F. Hasler, May 20 2012
    
  • Python
    # requires python 3.2 or higher
    from itertools import accumulate
    A000111_list, blist = [1,1], [1]
    for n in range(10**2):
        blist = list(reversed(list(accumulate(reversed(blist))))) + [0] if n % 2 else [0]+list(accumulate(blist))
        A000111_list.append(sum(blist)) # Chai Wah Wu, Jan 29 2015
    
  • Python
    from mpmath import *
    mp.dps = 150
    l = chop(taylor(lambda x: sec(x) + tan(x), 0, 26))
    [int(fac(i) * li) for i, li in enumerate(l)]  # Indranil Ghosh, Jul 06 2017
    
  • Python
    from sympy import bernoulli, euler
    def A000111(n): return abs(((1<Chai Wah Wu, Nov 13 2024
  • Sage
    # Algorithm of L. Seidel (1877)
    def A000111_list(n) :
        R = []; A = {-1:0, 0:1}; k = 0; e = 1
        for i in (0..n) :
            Am = 0; A[k + e] = 0; e = -e
            for j in (0..i) : Am += A[k]; A[k] = Am; k += e
            R.append(Am)
        return R
    A000111_list(22) # Peter Luschny, Mar 31 2012 (revised Apr 24 2016)
    

Formula

E.g.f.: (1+sin(x))/cos(x) = tan(x) + sec(x).
E.g.f. for a(n+1) is 1/(cos(x/2) - sin(x/2))^2 = 1/(1-sin(x)) = d/dx(sec(x) + tan(x)).
E.g.f. A(x) = -log(1-sin(x)), for a(n+1). - Vladimir Kruchinin, Aug 09 2010
O.g.f.: A(x) = 1+x/(1-x-x^2/(1-2*x-3*x^2/(1-3*x-6*x^2/(1-4*x-10*x^2/(1-... -n*x-(n*(n+1)/2)*x^2/(1- ...)))))) (continued fraction). - Paul D. Hanna, Jan 17 2006
E.g.f. A(x) = y satisfies 2y' = 1 + y^2. - Michael Somos, Feb 03 2004
a(n) = P_n(0) + Q_n(0) (see A155100 and A104035), defining Q_{-1} = 0. Cf. A156142.
2*a(n+1) = Sum_{k=0..n} binomial(n, k)*a(k)*a(n-k).
Asymptotics: a(n) ~ 2^(n+2)*n!/Pi^(n+1). For a proof, see for example Flajolet and Sedgewick.
a(n) = (n-1)*a(n-1) - Sum_{i=2..n-2} (i-1)*E(n-2, n-1-i), where E are the Entringer numbers A008281. - Jon Perry, Jun 09 2003
a(2*k) = (-1)^k euler(2k) and a(2k-1) = (-1)^(k-1)2^(2k)(2^(2k)-1) Bernoulli(2k)/(2k). - C. Ronaldo (aga_new_ac(AT)hotmail.com), Jan 17 2005
|a(n+1) - 2*a(n)| = A000708(n). - Philippe Deléham, Jan 13 2007
a(n) = 2^n|E(n,1/2) + E(n,1)| where E(n,x) are the Euler polynomials. - Peter Luschny, Jan 25 2009
a(n) = 2^(n+2)*n!*S(n+1)/(Pi)^(n+1), where S(n) = Sum_{k = -inf..inf} 1/(4k+1)^n (see the Elkies reference). - Emeric Deutsch, Aug 17 2009
a(n) = i^(n+1) Sum_{k=1..n+1} Sum_{j=0..k} binomial(k,j)(-1)^j (k-2j)^(n+1) (2i)^(-k) k^{-1}. - Ross Tang (ph.tchaa(AT)gmail.com), Jul 28 2010
a(n) = sum((if evenp(n+k) then (-1)^((n+k)/2)*sum(j!*Stirling2(n,j)*2^(1-j)*(-1)^(n+j-k)*binomial(j-1,k-1),j,k,n) else 0),k,1,n), n>0. - Vladimir Kruchinin, Aug 19 2010
If n==1(mod 4) is prime, then a(n)==1(mod n); if n==3(mod 4) is prime, then a(n)==-1(mod n). - Vladimir Shevelev, Aug 31 2010
For m>=0, a(2^m)==1(mod 2^m); If p is prime, then a(2*p)==1(mod 2*p). - Vladimir Shevelev, Sep 03 2010
From Peter Bala, Jan 26 2011: (Start)
a(n) = A(n,i)/(1+i)^(n-1), where i = sqrt(-1) and {A(n,x)}n>=1 = [1,1+x,1+4*x+x^2,1+11*x+11*x^2+x^3,...] denotes the sequence of Eulerian polynomials.
Equivalently, a(n) = i^(n+1)*Sum_{k=1..n} (-1)^k*k!*Stirling2(n,k) * ((1+i)/2)^(k-1) = i^(n+1)*Sum_{k = 1..n} (-1)^k*((1+i)/2)^(k-1)* Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*j^n.
This explicit formula for a(n) can be used to obtain congruence results. For example, for odd prime p, a(p) = (-1)^((p-1)/2) (mod p), as noted by Vladimir Shevelev above.
For the corresponding type B results see A001586. For the corresponding results for plane increasing 0-1-2 trees see A080635.
For generalized Eulerian, Stirling and Bernoulli numbers associated with the zigzag numbers see A145876, A147315 and A185424, respectively. For a recursive triangle to calculate a(n) see A185414.
(End)
a(n) = I^(n+1)*2*Li_{-n}(-I) for n > 0. Li_{s}(z) is the polylogarithm. - Peter Luschny, Jul 29 2011
a(n) = 2*Sum_{m=0..(n-2)/2} 4^m*(Sum_{i=m..(n-1)/2} (i-(n-1)/2)^(n-1)*binomial(n-2*m-1,i-m)*(-1)^(n-i-1)), n > 1, a(0)=1, a(1)=1. - Vladimir Kruchinin, Aug 09 2011
a(n) = D^(n-1)(1/(1-x)) evaluated at x = 0, where D is the operator sqrt(1-x^2)*d/dx. Cf. A006154. a(n) equals the alternating sum of the nonzero elements of row n-1 of A196776. This leads to a combinatorial interpretation for a(n); for example, a(4*n+2) gives the number of ordered set partitions of 4*n+1 into k odd-sized blocks, k = 1 (mod 4), minus the number of ordered set partitions of 4*n+1 into k odd-sized blocks, k = 3 (mod 4). Cf A002017. - Peter Bala, Dec 06 2011
From Sergei N. Gladkovskii, Nov 14 2011 - Dec 23 2013: (Start)
Continued fractions:
E.g.f.: tan(x) + sec(x) = 1 + x/U(0); U(k) = 4k+1-x/(2-x/(4k+3+x/(2+x/U(k+1)))).
E.g.f.: for a(n+1) is E(x) = 1/(1-sin(x)) = 1 + x/(1 - x + x^2/G(0)); G(k) = (2*k+2)*(2*k+3)-x^2+(2*k+2)*(2*k+3)*x^2/G(k+1).
E.g.f.: for a(n+1) is E(x) = 1/(1-sin(x)) = 1/(1 - x/(1 + x^2/G(0))) ; G(k) = 8*k+6-x^2/(1 + (2*k+2)*(2*k+3)/G(k+1)).
E.g.f.: for a(n+1) is E(x) = 1/(1 - sin(x)) = 1/(1 - x*G(0)); G(k) = 1 - x^2/(2*(2*k+1)*(4*k+3) - 2*x^2*(2*k+1)*(4*k+3)/(x^2 - 4*(k+1)*(4*k+5)/G(k+1))).
E.g.f.: for a(n+1) is E(x) = 1/(1 - sin(x)) = 1/(1 - x*G(0)) where G(k)= 1 - x^2/( (2*k+1)*(2*k+3) - (2*k+1)*(2*k+3)^2/(2*k+3 - (2*k+2)/G(k+1))).
E.g.f.: tan(x) + sec(x) = 1 + 2*x/(U(0)-x) where U(k) = 4k+2 - x^2/U(k+1).
E.g.f.: tan(x) + sec(x) = 1 + 2*x/(2*U(0)-x) where U(k) = 4*k+1 - x^2/(16*k+12 - x^2/U(k+1)).
E.g.f.: tan(x) + sec(x) = 4/(2-x*G(0))-1 where G(k) = 1 - x^2/(x^2 - 4*(2*k+1)*(2*k+3)/G(k+1)).
G.f.: 1 + x/Q(0), m=+4, u=x/2, where Q(k) = 1 - 2*u*(2*k+1) - m*u^2*(k+1)*(2*k+1)/(1 - 2*u*(2*k+2) - m*u^2*(k+1)*(2*k+3)/Q(k+1)).
G.f.: conjecture: 1 + T(0)*x/(1-x), where T(k) = 1 - x^2*(k+1)*(k+2)/(x^2*(k+1)*(k+2) - 2*(1-x*(k+1))*(1-x*(k+2))/T(k+1)).
E.g.f.: 1+ 4*x/(T(0) - 2*x), where T(k) = 4*(2*k+1) - 4*x^2/T(k+1):
E.g.f.: T(0)-1, where T(k) = 2 + x/(4*k+1 - x/(2 - x/( 4*k+3 + x/T(k+1)))). (End)
E.g.f.: tan(x/2 + Pi/4). - Vaclav Kotesovec, Nov 08 2013
Asymptotic expansion: 4*(2*n/(Pi*e))^(n+1/2)*exp(1/2+1/(12*n) -1/(360*n^3) + 1/(1260*n^5) - ...). (See the Luschny link.) - Peter Luschny, Jul 14 2015
From Peter Bala, Sep 10 2015: (Start)
The e.g.f. A(x) = tan(x) + sec(x) satisfies A''(x) = A(x)*A'(x), hence the recurrence a(0) = 1, a(1) = 1, else a(n) = Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i).
Note, the same recurrence, but with the initial conditions a(0) = 0 and a(1) = 1, produces the sequence [0,1,0,1,0,4,0,34,0,496,...], an aerated version of A002105. (End)
a(n) = A186365(n)/n for n >= 1. - Anton Zakharov, Aug 23 2016
From Peter Luschny, Oct 27 2017: (Start)
a(n) = abs(2*4^n*(H(((-1)^n - 3)/8, -n) - H(((-1)^n - 7)/8, -n))) where H(z, r) are the generalized harmonic numbers.
a(n) = (-1)^binomial(n + 1, 2)*2^(2*n + 1)*(zeta(-n, 1 + (1/8)*(-7 + (-1)^n)) - zeta(-n, 1 + (1/8)*(-3 + (-1)^n))). (End)
a(n) = i*(i^n*Li_{-n}(-i) - (-i)^n*Li_{-n}(i)), where i is the imaginary unit and Li_{s}(z) is the polylogarithm. - Peter Luschny, Aug 28 2020
Sum_{n>=0} 1/a(n) = A340315. - Amiram Eldar, May 29 2021
a(n) = n!*Re([x^n](1 + I^(n^2 - n)*(2 - 2*I)/(exp(x) + I))). - Peter Luschny, Aug 09 2021

Extensions

Edited by M. F. Hasler, Apr 04 2013
Title corrected by Geoffrey Critzer, May 18 2013

A000182 Tangent (or "Zag") numbers: e.g.f. tan(x), also (up to signs) e.g.f. tanh(x).

Original entry on oeis.org

1, 2, 16, 272, 7936, 353792, 22368256, 1903757312, 209865342976, 29088885112832, 4951498053124096, 1015423886506852352, 246921480190207983616, 70251601603943959887872, 23119184187809597841473536, 8713962757125169296170811392, 3729407703720529571097509625856
Offset: 1

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Author

Keywords

Comments

Number of Joyce trees with 2n-1 nodes. Number of tremolo permutations of {0,1,...,2n}. - Ralf Stephan, Mar 28 2003
The Hankel transform of this sequence is A000178(n) for n odd = 1, 12, 34560, ...; example: det([1, 2, 16; 2, 16, 272, 16, 272, 7936]) = 34560. - Philippe Deléham, Mar 07 2004
a(n) is the number of increasing labeled full binary trees with 2n-1 vertices. Full binary means every non-leaf vertex has two children, distinguished as left and right; labeled means the vertices are labeled 1,2,...,2n-1; increasing means every child has a label greater than its parent. - David Callan, Nov 29 2007
From Micha Hofri (hofri(AT)wpi.edu), May 27 2009: (Start)
a(n) was found to be the number of permutations of [2n] which when inserted in order, to form a binary search tree, yield the maximally full possible tree (with only one single-child node).
The e.g.f. is sec^2(x)=1+tan^2(x), and the same coefficients can be manufactured from the tan(x) itself, which is the e.g.f. for the number of trees as above for odd number of nodes. (End)
a(n) is the number of increasing strict binary trees with 2n-1 nodes. For more information about increasing strict binary trees with an associated permutation, see A245894. - Manda Riehl, Aug 07 2014
For relations to alternating permutations, Euler and Bernoulli polynomials, zigzag numbers, trigonometric functions, Fourier transform of a square wave, quantum algebras, and integrals over and in n-dimensional hypercubes and over Green functions, see Hodges and Sukumar. For further discussion on the quantum algebra, see the later Hodges and Sukumar reference and the paper by Hetyei presenting connections to the general combinatorial theory of Viennot on orthogonal polynomials, inverse polynomials, tridiagonal matrices, and lattice paths (thereby related to continued fractions and cumulants). - Tom Copeland, Nov 30 2014
The Zigzag Hankel transform is A000178. That is, A000178(2*n - k) = det( [a(i+j - k)]{i,j = 1..n} ) for n>0 and k=0,1. - _Michael Somos, Mar 12 2015
a(n) is the number of standard Young tableaux of skew shape (n,n,n-1,n-2,...,3,2)/(n-1,n-2,n-3,...,2,1). - Ran Pan, Apr 10 2015
For relations to the Sheffer Appell operator calculus and a Riccati differential equation for generating the Meixner-Pollaczek and Krawtchouk orthogonal polynomials, see page 45 of the Feinsilver link and Rzadkowski. - Tom Copeland, Sep 28 2015
For relations to an elliptic curve, a Weierstrass elliptic function, the Lorentz formal group law, a Lie infinitesimal generator, and the Eulerian numbers A008292, see A155585. - Tom Copeland, Sep 30 2015
Absolute values of the alternating sums of the odd-numbered rows (where the single 1 at the apex of the triangle is counted as row #1) of the Eulerian triangle, A008292. The actual alternating sums alternate in sign, e.g., 1, -2, 16, -272, etc. (Even-numbered rows have alternating sums always 0.) - Gregory Gerard Wojnar, Sep 28 2018
The sequence is periodic modulo any odd prime p. The minimal period is (p-1)/2 if p == 1 mod 4 and p-1 if p == 3 mod 4 [Knuth & Buckholtz, 1967, Theorem 1]. - Allen Stenger, Aug 03 2020
From Peter Bala, Dec 24 2021: (Start)
Conjectures:
1) The sequence taken modulo any integer k eventually becomes periodic with period dividing phi(k).
2) The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k, except when p = 2, n = 1 and k = 1 or 2.
3) For i >= 1 define a_i(n) = a(n+i). The Gauss congruences a_i(n*p^k) == a_i(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k. If true, then for each i >= 1 the expansion of exp(Sum_{n >= 1} a_i(n)*x^n/n) has integer coefficients. For an example, see A262145.(End)

Examples

			tan(x) = x + 2*x^3/3! + 16*x^5/5! + 272*x^7/7! + ... = x + 1/3*x^3 + 2/15*x^5 + 17/315*x^7 + 62/2835*x^9 + O(x^11).
tanh(x) = x - 1/3*x^3 + 2/15*x^5 - 17/315*x^7 + 62/2835*x^9 - 1382/155925*x^11 + ...
(sec x)^2 = 1 + x^2 + 2/3*x^4 + 17/45*x^6 + ...
a(3)=16 because we have: {1, 3, 2, 5, 4}, {1, 4, 2, 5, 3}, {1, 4, 3, 5, 2},
  {1, 5, 2, 4, 3}, {1, 5, 3, 4, 2}, {2, 3, 1, 5, 4}, {2, 4, 1, 5, 3},
  {2, 4, 3, 5, 1}, {2, 5, 1, 4, 3}, {2, 5, 3, 4, 1}, {3, 4, 1, 5, 2},
  {3, 4, 2, 5, 1}, {3, 5, 1, 4, 2}, {3, 5, 2, 4, 1}, {4, 5, 1, 3, 2},
  {4, 5, 2, 3, 1}. - _Geoffrey Critzer_, May 19 2013
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 111.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 69.
  • L. M. Milne-Thompson, Calculus of Finite Differences, 1951, p. 148 (the numbers |C^{2n-1}|).
  • J. W. Milnor and J. D. Stasheff, Characteristic Classes, Princeton, 1974, p. 282.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 444.
  • H. Rademacher, Topics in Analytic Number Theory, Springer, 1973, Chap. 1, p. 20.
  • L. Seidel, Über eine einfache Entstehungsweise der Bernoullischen Zahlen und einiger verwandten Reihen, Sitzungsberichte der mathematisch-physikalischen Classe der königlich bayerischen Akademie der Wissenschaften zu München, volume 7 (1877), 157-187.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • E. van Fossen Conrad, Some continued fraction expansions of elliptic functions, PhD thesis, The Ohio State University, 2002, p. 28.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, pp. 267-268.

Crossrefs

A350972 is essentially the same sequence.
a(n)=2^(n-1)*A002105(n). Apart from signs, 2^(2n-2)*A001469(n) = n*a(n).
Cf. A001469, A002430, A036279, A000364 (secant numbers), A000111 (secant-tangent numbers), A024283, A009764. First diagonal of A059419 and of A064190.
Equals A002425(n) * 2^A101921(n).
Equals leftmost column of A162005. - Johannes W. Meijer, Jun 27 2009

Programs

  • Maple
    series(tan(x),x,40);
    with(numtheory): a := n-> abs(2^(2*n)*(2^(2*n)-1)*bernoulli(2*n)/(2*n));
    A000182_list := proc(n) local T,k,j; T[1] := 1;
    for k from 2 to n do T[k] := (k-1)*T[k-1] od;
       for k from 2 to n do
           for j from k to n do
               T[j] := (j-k)*T[j-1]+(j-k+2)*T[j] od od;
    seq(T[j], j=1..n)  end:
    A000182_list(15);  # Peter Luschny, Apr 02 2012
  • Mathematica
    Table[ Sum[2^(2*n + 1 - k)*(-1)^(n + k + 1)*k!*StirlingS2[2*n + 1, k], {k, 1, 2*n + 1}], {n, 0, 7}] (* Victor Adamchik, Oct 05 2005 *)
    v[1] = 2; v[n_] /; n >= 2 := v[n] = Sum[ Binomial[2 n - 3, 2 k - 2] v[k] v[n - k], {k, n - 1}]; Table[ v[n]/2, {n, 15}] (* Zerinvary Lajos, Jul 08 2009 *)
    Rest@ Union[ Range[0, 29]! CoefficientList[ Series[ Tan[x], {x, 0, 30}], x]] (* Harvey P. Dale, Oct 19 2011; modified by Robert G. Wilson v, Apr 02 2012 *)
    t[1, 1] = 1; t[1, 0] = 0; t[n_ /; n > 1, m_] := t[n, m] = m*(m+1)*Sum[t[n-1, k], {k, m-1, n-1}]; a[n_] := t[n, 1]; Table[a[n], {n, 1, 15}]  (* Jean-François Alcover, Jan 02 2013, after A064190 *)
    a[ n_] := If[ n < 1, 0, With[{m = 2 n - 1}, m! SeriesCoefficient[ Tan[x], {x, 0, m}]]]; (* Michael Somos, Mar 12 2015 *)
    a[ n_] := If[ n < 1, 0, ((-16)^n - (-4)^n) Zeta[1 - 2 n]]; (* Michael Somos, Mar 12 2015 *)
    Table[2 PolyGamma[2n - 1, 1/2]/Pi^(2n), {n, 1, 10}] (* Vladimir Reshetnikov, Oct 18 2015 *)
    a[ n_] := a[n] = If[ n < 2, Boole[n == 1], Sum[Binomial[2 n - 2, 2 k - 1] a[k] a[n - k], {k, n - 1}]]; (* Michael Somos, Aug 02 2018 *)
    a[n_] := (2^(2*n)*(2^(2*n) - 1)*Abs[BernoulliB[2*n]])/(2*n); a /@  Range[20]  (* Stan Wagon, Nov 21 2022 *)
  • Maxima
    a(n):=sum(sum(binomial(k,r)*sum(sum(binomial(l,j)/2^(j-1)*sum((-1)^(n)*binomial(j,i)*(j-2*i)^(2*n),i,0,floor((j-1)/2))*(-1)^(l-j),j,1,l)*(-1)^l*binomial(r+l-1,r-1),l,1,2*n)*(-1)^(1-r),r,1,k)/k,k,1,2*n); /* Vladimir Kruchinin, Aug 23 2010 */
    
  • Maxima
    a[n]:=if n=1 then 1 else 2*sum(sum(binomial(2*j,j+k)*(-4*k^2)^(n-1)*(-1)^k/(4^j),k,1,j),j,1,n-1);
    makelist(a[n],n,1,30); /* Tani Akinari, Sep 20 2023 */
    
  • PARI
    {a(n) = if( n<1, 0, ((-4)^n - (-16)^n) * bernfrac(2*n) / (2*n))};
    
  • PARI
    {a(n) = my(an); if( n<2, n==1, an = vector(n, m, 1); for( m=2, n, an[m] = sum( k=1, m-1, binomial(2*m - 2, 2*k - 1) * an[k] * an[m-k])); an[n])}; /* Michael Somos */
    
  • PARI
    {a(n) = if( n<1, 0, (2*n - 1)! * polcoeff( tan(x + O(x^(2*n + 2))), 2*n - 1))}; /* Michael Somos */
    
  • PARI
    {a(n) = my(X=x+x*O(x^n),Egf); Egf = x*sum(m=0,n, prod(k=1,m, tanh(2*k*X))); (n-1)!*polcoeff(Egf,n)} /* Paul D. Hanna, May 11 2010 */
    
  • PARI
    /* Continued Fraction for the e.g.f. tan(x), from Paul D. Hanna: */
    {a(n)=local(CF=1+O(x)); for(i=1, n, CF=1/(2*(n-i+1)-1-x^2*CF)); (2*n-1)!*polcoeff(x*CF, 2*n-1)}
    
  • PARI
    /* O.g.f. Sum_{n>=1} a(n)*x^n, from Paul D. Hanna Feb 05 2013: */
    {a(n)=polcoeff( x+2*x*sum(m=1, n, x^m*prod(k=1, m, (2*k-1)^2/(1+(2*k-1)^2*x +x*O(x^n))) ), n)}
    
  • Python
    # The objective of this implementation is efficiency.
    # n -> [0, a(1), a(2), ..., a(n)] for n > 0.
    def A000182_list(n):
        T = [0 for i in range(1, n+2)]
        T[1] = 1
        for k in range(2, n+1):
            T[k] = (k-1)*T[k-1]
        for k in range(2, n+1):
            for j in range(k, n+1):
                T[j] = (j-k)*T[j-1]+(j-k+2)*T[j]
        return T
    print(A000182_list(100)) # Peter Luschny, Aug 07 2011
    
  • Python
    from sympy import bernoulli
    def A000182(n): return abs(((2-(2<<(m:=n<<1)))*bernoulli(m)<Chai Wah Wu, Apr 14 2023
    
  • Sage
    # Algorithm of L. Seidel (1877)
    # n -> [a(1), ..., a(n)] for n >= 1.
    def A000182_list(len) :
        R = []; A = {-1:0, 0:1}; k = 0; e = 1
        for i in (0..2*len-1) :
            Am = 0; A[k + e] = 0; e = -e
            for j in (0..i) : Am += A[k]; A[k] = Am; k += e
            if e > 0 : R.append(A[i//2])
        return R
    A000182_list(15) # Peter Luschny, Mar 31 2012

Formula

E.g.f.: log(sec x) = Sum_{n > 0} a(n)*x^(2*n)/(2*n)!.
E.g.f.: tan x = Sum_{n >= 0} a(n+1)*x^(2*n+1)/(2*n+1)!.
E.g.f.: (sec x)^2 = Sum_{n >= 0} a(n+1)*x^(2*n)/(2*n)!.
2/(exp(2x)+1) = 1 + Sum_{n>=1} (-1)^(n+1) a(n) x^(2n-1)/(2n-1)! = 1 - x + x^3/3 - 2*x^5/15 + 17*x^7/315 - 62*x^9/2835 + ...
a(n) = 2^(2*n) (2^(2*n) - 1) |B_(2*n)| / (2*n) where B_n are the Bernoulli numbers (A000367/A002445 or A027641/A027642).
Asymptotics: a(n) ~ 2^(2*n+1)*(2*n-1)!/Pi^(2*n).
Sum[2^(2*n + 1 - k)*(-1)^(n + k + 1)*k!*StirlingS2[2*n + 1, k], {k, 1, 2*n + 1}]. - Victor Adamchik, Oct 05 2005
a(n) = abs[c(2*n-1)] where c(n)= 2^(n+1) * (1-2^(n+1)) * Ber(n+1)/(n+1) = 2^(n+1) * (1-2^(n+1)) * (-1)^n * Zeta(-n) = [ -(1+EN(.))]^n = 2^n * GN(n+1)/(n+1) = 2^n * EP(n,0) = (-1)^n * E(n,-1) = (-2)^n * n! * Lag[n,-P(.,-1)/2] umbrally = (-2)^n * n! * C{T[.,P(.,-1)/2] + n, n} umbrally for the signed Euler numbers EN(n), the Bernoulli numbers Ber(n), the Genocchi numbers GN(n), the Euler polynomials EP(n,t), the Eulerian polynomials E(n,t), the Touchard / Bell polynomials T(n,t), the binomial function C(x,y) = x!/[(x-y)!*y! ] and the polynomials P(j,t) of A131758. - Tom Copeland, Oct 05 2007
a(1) = A094665(0,0)*A156919(0,0) and a(n) = Sum_{k=1..n-1} 2^(n-k-1)*A094665(n-1, k)*A156919(k,0) for n = 2, 3, .., see A162005. - Johannes W. Meijer, Jun 27 2009
G.f.: 1/(1-1*2*x/(1-2*3*x/(1-3*4*x/(1-4*5*x/(1-5*6*x/(1-... (continued fraction). - Paul Barry, Feb 24 2010
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-2x-12x^2/(1-18x-240x^2/(1-50x-1260x^2/(1-98x-4032x^2/(1-162x-9900x^2/(1-... (continued fraction);
coefficient sequences given by 4*(n+1)^2*(2n+1)*(2n+3) and 2(2n+1)^2 (see Van Fossen Conrad reference). (End)
E.g.f.: x*Sum_{n>=0} Product_{k=1..n} tanh(2*k*x) = Sum_{n>=1} a(n)*x^n/(n-1)!. - Paul D. Hanna, May 11 2010 [corrected by Paul D. Hanna, Sep 28 2023]
a(n) = (-1)^(n+1)*Sum_{j=1..2*n+1} j!*Stirling2(2*n+1,j)*2^(2*n+1-j)*(-1)^j for n >= 0. Vladimir Kruchinin, Aug 23 2010: (Start)
If n is odd such that 2*n-1 is prime, then a(n) == 1 (mod (2*n-1)); if n is even such that 2*n-1 is prime, then a(n) == -1 (mod (2*n-1)). - Vladimir Shevelev, Sep 01 2010
Recursion: a(n) = (-1)^(n-1) + Sum_{i=1..n-1} (-1)^(n-i+1)*C(2*n-1,2*i-1)* a(i). - Vladimir Shevelev, Aug 08 2011
E.g.f.: tan(x) = Sum_{n>=1} a(n)*x^(2*n-1)/(2*n-1)! = x/(1 - x^2/(3 - x^2/(5 - x^2/(7 - x^2/(9 - x^2/(11 - x^2/(13 -...))))))) (continued fraction from J. H. Lambert - 1761). - Paul D. Hanna, Sep 21 2011
From Sergei N. Gladkovskii, Oct 31 2011 to Oct 09 2013: (Start)
Continued fractions:
E.g.f.: (sec(x))^2 = 1+x^2/(x^2+U(0)) where U(k) = (k+1)*(2k+1) - 2x^2 + 2x^2*(k+1)*(2k+1)/U(k+1).
E.g.f.: tan(x) = x*T(0) where T(k) = 1-x^2/(x^2-(2k+1)*(2k+3)/T(k+1)).
E.g.f.: tan(x) = x/(G(0)+x) where G(k) = 2*k+1 - 2*x + x/(1 + x/G(k+1)).
E.g.f.: tanh(x) = x/(G(0)-x) where G(k) = k+1 + 2*x - 2*x*(k+1)/G(k+1).
E.g.f.: tan(x) = 2*x - x/W(0) where W(k) = 1 + x^2*(4*k+5)/((4*k+1)*(4*k+3)*(4*k+5) - 4*x^2*(4*k+3) + x^2*(4*k+1)/W(k+1)).
E.g.f.: tan(x) = x/T(0) where T(k) = 1 - 4*k^2 + x^2*(1 - 4*k^2)/T(k+1).
E.g.f.: tan(x) = -3*x/(T(0)+3*x^2) where T(k)= 64*k^3 + 48*k^2 - 4*k*(2*x^2 + 1) - 2*x^2 - 3 - x^4*(4*k -1)*(4*k+7)/T(k+1).
G.f.: 1/G(0) where G(k) = 1 - 2*x*(2*k+1)^2 - x^2*(2*k+1)*(2*k+2)^2*(2*k+3)/G(k+1).
G.f.: 2*Q(0) - 1 where Q(k) = 1 + x^2*(4*k + 1)^2/(x + x^2*(4*k + 1)^2 - x^2*(4*k + 3)^2*(x + x^2*(4*k + 1)^2)/(x^2*(4*k + 3)^2 + (x + x^2*(4*k + 3)^2)/Q(k+1) )).
G.f.: (1 - 1/G(0))*sqrt(-x), where G(k) = 1 + sqrt(-x) - x*(k+1)^2/G(k+1).
G.f.: Q(0), where Q(k) = 1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/Q(k+1)). (End)
O.g.f.: x + 2*x*Sum_{n>=1} x^n * Product_{k=1..n} (2*k-1)^2 / (1 + (2*k-1)^2*x). - Paul D. Hanna, Feb 05 2013
a(n) = (-4)^n*Li_{1-2*n}(-1). - Peter Luschny, Jun 28 2012
a(n) = (-4)^n*(4^n-1)*Zeta(1-2*n). - Jean-François Alcover, Dec 05 2013
Asymptotic expansion: 4*((2*(2*n-1))/(Pi*e))^(2*n-1/2)*exp(1/2+1/(12*(2*n-1))-1/(360*(2*n-1)^3)+1/(1260*(2*n-1)^5)-...). (See Luschny link.) - Peter Luschny, Jul 14 2015
From Peter Bala, Sep 11 2015: (Start)
The e.g.f. A(x) = tan(x) satisfies the differential equation A''(x) = 2*A(x)*A'(x) with A(0) = 0 and A'(0) = 1, leading to the recurrence a(0) = 0, a(1) = 1, else a(n) = 2*Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the aerated sequence [0, 1, 0, 2, 0, 16, 0, 272, ...].
Note, the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence n! and with a(0) = 1/2 and a(1) = 1 produces A080635. Cf. A002105, A234797. (End)
a(n) = 2*polygamma(2*n-1, 1/2)/Pi^(2*n). - Vladimir Reshetnikov, Oct 18 2015
a(n) = 2^(2n-2)*|p(2n-1,-1/2)|, where p_n(x) are the shifted row polynomials of A019538. E.g., a(2) = 2 = 2^2 * |1 + 6(-1/2) + 6(-1/2)^2|. - Tom Copeland, Oct 19 2016
From Peter Bala, May 05 2017: (Start)
With offset 0, the o.g.f. A(x) = 1 + 2*x + 16*x^2 + 272*x^3 + ... has the property that its 4th binomial transform 1/(1 - 4*x) A(x/(1 - 4*x)) has the S-fraction representation 1/(1 - 6*x/(1 - 2*x/(1 - 20*x/(1 - 12*x/(1 - 42*x/(1 - 30*x/(1 - ...))))))), where the coefficients [6, 2, 20, 12, 42, 30, ...] in the partial numerators of the continued fraction are obtained from the sequence [2, 6, 12, 20, ..., n*(n + 1), ...] by swapping adjacent terms. Compare with the S-fraction associated with A(x) given above by Paul Barry.
A(x) = 1/(1 + x - 3*x/(1 - 4*x/(1 + x - 15*x/(1 - 16*x/(1 + x - 35*x/(1 - 36*x/(1 + x - ...))))))), where the unsigned coefficients in the partial numerators [3, 4, 15, 16, 35, 36,...] come in pairs of the form 4*n^2 - 1, 4*n^2 for n = 1,2,.... (End)
a(n) = Sum_{k = 1..n-1} binomial(2*n-2, 2*k-1) * a(k) * a(n-k), with a(1) = 1. - Michael Somos, Aug 02 2018
a(n) = 2^(2*n-1) * |Euler(2*n-1, 0)|, where Euler(n,x) are the Euler polynomials. - Daniel Suteu, Nov 21 2018 (restatement of one of Copeland's 2007 formulas.)
x - Sum_{n >= 1} (-1)^n*a(n)*x^(2*n)/(2*n)! = x - log(cosh(x)). The series reversion of x - log(cosh(x)) is (1/2)*x - (1/2)*log(2 - exp(x)) = Sum_{n >= 0} A000670(n)*x^(n+1)/(n+1)!. - Peter Bala, Jul 11 2022
For n > 1, a(n) = 2*Sum_{j=1..n-1} Sum_{k=1..j} binomial(2*j,j+k)*(-4*k^2)^(n-1)*(-1)^k/(4^j). - Tani Akinari, Sep 20 2023
a(n) = A110501(n) * 4^(n-1) / n (Han and Liu, 2018). - Amiram Eldar, May 17 2024

A049774 Number of permutations of n elements not containing the consecutive pattern 123.

Original entry on oeis.org

1, 1, 2, 5, 17, 70, 349, 2017, 13358, 99377, 822041, 7477162, 74207209, 797771521, 9236662346, 114579019469, 1516103040833, 21314681315998, 317288088082405, 4985505271920097, 82459612672301846, 1432064398910663705, 26054771465540507273, 495583804405888997218
Offset: 0

Views

Author

Tuwani A. Tshifhumulo (tat(AT)caddy.univen.ac.za)

Keywords

Comments

Permutations on n letters without double falls. A permutation w has a double fall at k if w(k) > w(k+1) > w(k+2) and has an initial fall if w(1) > w(2).
Hankel transform is A055209. - Paul Barry, Jan 12 2009
Increasing colored 1-2 trees of order n with choice of two colors for the right branches of the vertices of outdegree 2 except those vertices on the path from the root to the leftmost leaf. - Wenjin Woan, May 21 2011

Examples

			Permutations without double increase and without pattern 123:
a(3) = 5: 132, 213, 231, 312, 321.
a(4) = 17: 1324, 1423, 1432, 2143, 2314, 2413, 2431, 3142, 3214, 3241, 3412, 3421, 4132, 4213, 4231, 4312, 4321.
		

References

  • F. N. David and D. E. Barton, Combinatorial Chance, Hafner, New York, 1962, pp. 156-157.
  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983, (5.2.17).

Crossrefs

Column k=0 of A162975.
Column k=3 of A242784.
Equals 1 + A000303. - Greg Dresden, Feb 22 2020

Programs

  • Maple
    b:= proc(u, o, t) option remember;
         `if`(u+o=0, 1, add(b(u-j, o+j-1, 0), j=1..u)+
         `if`(t=1, 0,   add(b(u+j-1, o-j, 1), j=1..o)))
        end:
    a:= n-> b(n, 0$2):
    seq(a(n), n=0..23);  # Alois P. Heinz, Nov 04 2021
  • Mathematica
    Table[Simplify[ n! SeriesCoefficient[ Series[ Sqrt[3] Exp[x/2]/(Sqrt[3] Cos[Sqrt[3]/2 x] - Sin[Sqrt[3]/2 x]), {x, 0, n}], n] ], {n, 0, 40}]
    (* Second program: *)
    b[u_, o_, t_, k_] := b[u, o, t, k] = If[t == k, (u + o)!, If[Max[t, u] + o < k, 0, Sum[b[u + j - 1, o - j, t + 1, k], {j, 1, o}] + Sum[b[u - j, o + j - 1, 1, k], {j, 1, u}]]];
    a[n_] := b[0, n, 0, 2] - b[0, n, 0, 3] + 1;
    a /@ Range[0, 40] (* Jean-François Alcover, Nov 09 2020, after Alois P. Heinz in A000303 *)

Formula

E.g.f.: 1/Sum_{i>=0} (x^(3*i)/(3*i)! - x^(3*i+1)/(3*i+1)!). [Corrected g.f. --> e.g.f. by Vaclav Kotesovec, Feb 15 2015]
Equivalently, e.g.f.: exp(x/2) * r / sin(r*x + (2/3)*Pi) where r = sqrt(3)/2. This has simple poles at (3*m+1)*x0 where x0 = Pi/sqrt(6.75) = 1.2092 approximately and m is an arbitrary integer. This yields the asymptotic expansion a(n)/n! ~ x0^(-n-1) * Sum((-1)^m * E^(3*m+1) / (3*m+1)^(n+1)) where E = exp(x0/2) = 1.8305+ and m ranges over all integers. - Noam D. Elkies, Nov 15 2001
E.g.f.: sqrt(3)*exp(x/2)/(sqrt(3)*cos(x*sqrt(3)/2) - sin(x*sqrt(3)/2) ); a(n+1) = Sum_{k=0..n} binomial(n, k)*a(k)*b(n-k) where b(n) = number of n-permutations without double falls and without initial falls. - Emanuele Munarini, Feb 28 2003
O.g.f.: A(x) = 1/(1 - x - x^2/(1 - 2*x - 4*x^2/(1 - 3*x - 9*x^2/(1 - ... - n*x - n^2*x^2/(1 - ...))))) (continued fraction). - Paul D. Hanna, Jan 17 2006
a(n) = leftmost column term of M^n*V, where M = an infinite tridiagonal matrix with (1,2,3,...) in the super, sub, and main diagonals and the rest zeros. V = the vector [1,0,0,0,...]. - Gary W. Adamson, Jun 16 2011
E.g.f.: A(x)=1/Q(0); Q(k)=1-x/((3*k+1)-(x^2)*(3*k+1)/((x^2)-3*(3*k+2)*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 25 2011
a(n) ~ n! * exp(Pi/(3*sqrt(3))) * (3*sqrt(3)/(2*Pi))^(n+1). - Vaclav Kotesovec, Jul 28 2013
E.g.f.: T(0)/(1-x), where T(k) = 1 - x^2*(k+1)^2/( x^2*(k+1)^2 - (1-x-x*k)*(1-2*x-x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 17 2013

Extensions

Corrected and extended by Vladeta Jovovic, Apr 14 2001

A177523 Number of permutations of 1..n avoiding adjacent step pattern up, up, up, up.

Original entry on oeis.org

1, 1, 2, 6, 24, 119, 709, 4928, 39144, 349776, 3472811, 37928331, 451891992, 5832672456, 81074690424, 1207441809209, 19181203110129, 323753459184738, 5785975294622694, 109149016813544376, 2167402030585724571, 45190632809497874161, 987099099863360190632
Offset: 0

Views

Author

R. H. Hardin, May 10 2010

Keywords

Comments

a(n) is the number of permutations of length n that avoid the consecutive pattern 12345 (or equivalently 54321).

Examples

			E.g.f.: A(x) = 1 + x + 2*x^2/2! + 6*x^3/3! + 24*x^4/4! + 119*x^5/5! + 709*x^6/6! +...
where A(x) = 1/(1 - x + x^5/5! - x^6/6! + x^10/10! - x^11/11! + x^15/15! - x^16/16! + x^20/20! +...).
		

Crossrefs

Column k=15 of A242784.

Programs

  • Mathematica
    Table[n!*SeriesCoefficient[1/(Sum[x^(5*k)/(5*k)!-x^(5*k+1)/(5*k+1)!,{k,0,n}]),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Dec 11 2013 *)
    FullSimplify[CoefficientList[Series[10*E^((1+Sqrt[5])*x/4) / ((5+Sqrt[5]) * Cos[Sqrt[(5-Sqrt[5])/2]*x/2] + (5-Sqrt[5]) * E^(Sqrt[5]*x/2) * Cos[Sqrt[(5+Sqrt[5])/2]*x/2] - Sqrt[2*(5-Sqrt[5])] * Sin[Sqrt[(5-Sqrt[5])/2]*x/2] - Sqrt[2*(5+Sqrt[5])] * E^(Sqrt[5]*x/2) * Sin[Sqrt[(5+Sqrt[5])/2]*x/2]),{x,0,20}],x]*Range[0,20]!] (* Vaclav Kotesovec, Aug 29 2014 *)
  • PARI
    {a(n)=n!*polcoeff(1/sum(m=0, n\5+1, x^(5*m)/(5*m)!-x^(5*m+1)/(5*m+1)!+x^2*O(x^n)), n)}

Formula

E.g.f.: 1/( Sum_{n>=0} x^(5*n)/(5*n)! - x^(5*n+1)/(5*n+1)! ).
a(n)/n! ~ c * (1/r)^n, where r = 1.007187547786015395418998654... is the root of the equation Sum_{n>=0} (r^(5*n)/(5*n)! - r^(5*n+1)/(5*n+1)!) = 0, c = 1.02806793756750152.... - Vaclav Kotesovec, Dec 11 2013
Equivalently, r = 1.00718754778601539541899865400272701484... is the root of the equation (5+sqrt(5)) * cos(sqrt((5-sqrt(5))/2)*r/2) + (5-sqrt(5)) * exp(sqrt(5)*r/2) * cos(sqrt((5+sqrt(5))/2)*r/2) - sqrt(2*(5-sqrt(5))) * sin(sqrt((5-sqrt(5))/2)*r/2) - sqrt(2*(5+sqrt(5))) * exp(sqrt(5)*r/2) * sin(sqrt((5+sqrt(5))/2)*r/2) = 0. - Vaclav Kotesovec, Aug 29 2014
E.g.f.: 10*exp((1+sqrt(5))*x/4) / ((5+sqrt(5)) * cos(sqrt((5-sqrt(5))/2)*x/2) + (5-sqrt(5)) * exp(sqrt(5)*x/2) * cos(sqrt((5+sqrt(5))/2)*x/2) - sqrt(2*(5-sqrt(5))) * sin(sqrt((5-sqrt(5))/2)*x/2) - sqrt(2*(5+sqrt(5))) * exp(sqrt(5)*x/2) * sin(sqrt((5+sqrt(5))/2)*x/2)). - Vaclav Kotesovec, Aug 29 2014
In closed form, c = 5*exp((1+sqrt(5))*r/4) / (r*((5 + sqrt(5)) * cos(sqrt((5 - sqrt(5))/2)*r/2) + (5 - sqrt(5)) * exp(sqrt(5)*r/2) * cos(sqrt((5 + sqrt(5))/2)*r/2))) = 1.0280679375675015201596831656779442465978511664638... . Vaclav Kotesovec, Feb 01 2015

Extensions

More terms from Ray Chandler, Dec 06 2011
a(0)=1 prepended by Alois P. Heinz, Jan 13 2015

A139605 Weights for expansion of iterated derivatives, powers of a Lie derivative, or infinitesimal generator in vector form, (f(x)D_x)^n; coefficients of A-polynomials of Comtet; Scherk partition polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 4, 1, 4, 7, 6, 1, 1, 7, 4, 11, 1, 5, 30, 15, 10, 25, 10, 1, 1, 11, 15, 32, 34, 26, 1, 6, 57, 34, 146, 31, 15, 120, 90, 20, 65, 15, 1, 1, 16, 26, 15, 76, 192, 34, 122, 180, 57, 1, 7, 98, 140, 406, 462, 588, 63, 21, 252, 154, 896, 301
Offset: 1

Views

Author

Tom Copeland, Jun 12 2008

Keywords

Comments

This entry and the references differ slightly among themselves in the order of coefficients for higher order terms. Table on p. 167 of Comtet has at least one index error.
Let F[FI(x)] = FI[F(x)] = x (i.e., F and FI are a compositional inverse pair) about x=0 with F(0)=FI(0)=0. Define f(x) = 1/[dFI(x)/dx], then for w(x) analytic about the origin, exp[t f(x)d/dx] w(x) = w{F[t+FI(x)]} = q(t,x) with q{t,F[s+FI(x)]} = q(t+s,x). See A145271 for w(x)=x and note that A145271 is embedded in A139605. E.g.f. of the binomial Sheffer sequence associated to F(x) is exp[x f(z)d/dz] exp(t*z)= exp{t*F[x+FI(z)]} evaluated at z=0. - Tom Copeland, Oct 19 2011
dq(t,x)/dt - f(x)dq(t,x)/dx = 0, so (1,-f(x)) gives the components of a vector orthogonal to the gradient of q and therefore tangent to the contour of q at (t,x). - Tom Copeland, Oct 26 2011
The formula exp[t f(x)d/dx] w(x)= w{F[t+FI(x)]} above is implicit in the chain rule formulas on pages 10 and 12 of Mathemagical Forests. Another derivation is alluded to in the Dattoli reference in A080635 (repeated below). - Tom Copeland, Nov 28 2011
Let f(x) and g(x) be two infinitely differential functions. Denote f_0 = f(x), f_1 = df_0/dx, f_2 = df_1/dx, and so on. Same with g_0 = g(x). Define the linear operator L(u(x)) = g(x) * du(x)/dx. Denote F_1 = L(f(x)), F_2 = L(F_1), and so on. When n>0, F_n is a linear combination of f_1, ..., f_n where each f_k is multiplied by a homogeneous polynomial (P(n,k)) of degree n in g_0, ..., g_{n-1}. The triangle of the sum of the coefficients of P(n,k) is A130534. - Michael Somos, Mar 23 2014
Triangle with row n length A000070(n+1) and row n consists of the coefficients: P(n,1), ..., P(n,n). The order of coefficients in P(n,k) is Abramowitz and Stegun order for partitions of n-k with parts g_1, ..., g_{n-k}. - Michael Somos, Mar 23 2014
A130534(n,k) gives the number of rooted trees with (k+1) trunks that are associated with D^(k+1) in the forest of "naturally grown" rooted trees with (n+2) nodes, or vertices, that are associated with R^(n+1) in the example below. Cf. MF link. - Tom Copeland, Mar 23 2014
These partition polynomials appeared in 1823 in a dissertation by Heinrich Scherk. See p. 76 of Blasiak and Flajolet. - Tom Copeland, Jul 14 2021
Schröder made use of iterated derivatives, or iterated infinitesimal generators (IGs), ((1/f') D)^n in his investigations of functional iteration, or iterated functional composition, related to extensions of Newton's method of finding zeros of an equation. He constructs the series, in terms of the IGs, for finv[t+f(z)] evaluated at t = -f(z), giving z_1 = finv(0) although he doesn't present his analysis this way. - Tom Copeland, Jul 19 2021

Examples

			Let R = f(x)d/dx = f(x)D and (j,k) = [(d/dx)^j f(x)]^k ; then
R^0  = 1.
R^1  = (0,1)D.
R^2  = (0,1)(1,1)D + (0,2)D^2.
R^3  = [(0,1)(1,2) + (0,2)(2,1)]D + 3 (0,2)(1,1)D^2 + (0,3)D^3.
R^4  = [(0,1)(1,3) + 4 (0,2)(1,1)(2,1) + (0,3)(3,1)]D +
       [7 (0,2)(1,2) + 4 (0,3)(2,1)]D^2 + 6 (0,3)(1,1)D^3 + (0,4)D^4. - _Tom Copeland_, Jun 12 2008
R^5  = [(0,1)(1,4) + 11 (0,2)(1,2)(2,2) + 4 (0,3)(2,2) + (0,4)(4,1) + 7 (0,3)(1,1)(3,1)]D + [15 (0,2)(1,3) + 30 (0,3)(1,1)(2,1) + 5 (0,4)(1,3)]D^2 + [25 (0,3)(1,2) + 10 (0,4)(2,1) + 25 (0,3)(1,2)]D^3  + 10 (0,4)(1,1)D^4 + (0,5)D^5. - _Tom Copeland_, Jul 17 2016
R^6  = [(0,1)(1,5) + 26 (0,2)(1,3)(2,1) + 34 (0,3)(1,1)(2,2) + 32 (0,3)(1,2)(3,1) + 11 (0,4)(1,1)(4,1) + 15 (0,4)(2,1)(3,1) + (0,5)(1,5)]D + [31 (0,2)(1,4) + 146 (0,3)(1,2)(2,1) + 57 (0,4)(1,1)(3,1) + 34 (0,4)(2,2) + 6 (0,5)(4,1)]D^2 + [90 (0,3)(1,3) + 120 (0,4)(1,1)(2,1) + 15 (0,5)(3,1)]D^3 + [65 (0,4)(1,2) + 20 (0,5)(2,1)]D^4 + 15 (0,5)(1,1)D^5 + (0,6)D^6. - _Tom Copeland_, Jul 17 2016
------------
F_1 = (1*g_0) * f_1, F_2 = (1*g_0*g_1) * f_1 + (1*g_0^2) * f_2, F_3 = (1*g_0*g_1^2 + 1*g_0^2*g_2) * f_1 + (3*g_0^2*g_1) * f_2 + (1*g_0^3) * f_3. - _Michael Somos_, Mar 23 2014
P(4,2) = 4*g0^3*g2 + 7*g0^2*g1^2. P(5,2) = 5*g0^4*g3 + 30*g0^3*g1*g2 + 15*g0^2*g1^3. - _Michael Somos_, Mar 23 2014
1
1 , 1
1 1 , 3 , 1
1 4 1 , 4 7 , 6 , 1
1 7 4 11 1, 5 30 15 , 10 25 , 10 , 1
1 11 15 32 34 26 1 , 6 57 34 146 31 , 15 120 90 , 20 65 , 15 , 1
		

References

  • F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-like Structures, (1997), Cambridge University Press, p. 386.
  • H. Davis, The Theory of Linear Operators, (1936), The Principia Press, p. 13.
  • T. Mansour and M. Schork, Commutation Relations, Normal Ordering, and Stirling Numbers, Chapman and Hall/CRC, 2015.

Crossrefs

Cf. A000070 (number of distinct terms for each order).
Cf. A130534 (sum of numerical coefficients of the derivatives).

Programs

  • Mathematica
    row[n_] := With[{pn = CoefficientRules[Nest[g[x] D[#, x] &, f[x], n], Derivative[#][f][x] & /@ Range[n]][[;; , 2]] /. {Derivative[k_][g][x] :> h[k], g[x] -> 1}}, Table[Coefficient[pn[[k]], Product[h[x], {x, p}]], {k, n - 1}, {p, Sort[Sort /@ IntegerPartitions[n - k]]}]~Join~{{1}}];
    Table[row[n], {n, 7}] // Flatten (* Andrey Zabolotskiy, Mar 08 2024 *)

Formula

Equivalent matrix computation: Multiply the n-th diagonal (with n=0 the main diagonal) of the lower triangular Pascal matrix by f_n = (d/dx)^n f(x) to obtain the matrix VP with VP(n,k) = binomial(n,k) f_(n-k). Then R^n = (1, 0, 0, 0,..) [VP * S]^n (1, D, D^2, ..)^T, where S is the shift matrix A129185, representing differentiation in the basis x^n/n!. Cf. A145271. - Tom Copeland, Jul 17 2016
A formula for the coefficients of this matrix is presented in Ihara, obtained from Comtet. - Tom Copeland, Mar 25 2020
Elaborating on my 2011 comments: Let exp[x F(t)] = exp[t p.(x)] be the e.g.f. for the binomial Sheffer sequence of polynomials (p.(x))^n = p_n(x). Then, evaluated at x = t = 0, the coefficient p_(n,k) = (D_x^k/k!) p_n(x) = D_t^n [F(t)]^k/k! = (f(x)D_x)^n x^k/k! = R^n x^k/k!, and so p_(n,k) is the coefficient of D^k of the operator R^n below evaluated at x=0. - Tom Copeland, May 14 2020
Per earlier comments, the action of the differentials of this entry on an exponential is exp(x g(u)D_u) e^(ut) = e^(t H^{(-1)}(H(u)+x)) with g(x) = 1/D(H(x)) and H^{(-1)}(x) the compositional inverse of H(x). With H^{(-1)}(x) = -log(1-x), the inverse about x=0 is H(x) = 1-e^(-x), giving g(x) = e^x and the resulting action e^(-t log(1-x)) = (1-x)^(-t) for u=0, an e.g.f. for the unsigned Stirling numbers of the first kind (see A008275, A048994, and A130534). Consequently, summing the coefficients of this entry over each associated derivative gives these Stirling numbers. E.g., the fifth row in the examples reduces to (1+4+1) D + (7+4) D^2 + 6 D^3 + D^4 = 6 D + 11 D^2 + 6 D^3 + D^4. The Connes-Moscovici weights A139002 are a refinement of this entry. - Tom Copeland, Jul 14 2021

Extensions

Title expanded by Tom Copeland, Mar 17 2014
Sequence terms rearranged in Abramowitz and Stegun order by Michael Somos, Mar 23 2014
Title expanded by Tom Copeland, Jul 14 2021

A177553 Number of permutations of 1..n avoiding adjacent step pattern up, up, up, up, up, up.

Original entry on oeis.org

1, 1, 2, 6, 24, 120, 720, 5039, 40305, 362682, 3626190, 39881160, 478490760, 6219298800, 87055051511, 1305598835941, 20885951018102, 354999461960226, 6388879812001704, 121367620532150280, 2426930566055020080, 50956684690331669759, 1120852238721212726609
Offset: 0

Views

Author

R. H. Hardin, May 10 2010

Keywords

Crossrefs

Column k=63 of A242784.

Programs

  • Maple
    b:= proc(u, o, t) option remember; `if`(u+o=0, 1,
          `if`(t<5, add(b(u+j-1, o-j, t+1), j=1..o), 0)+
          add(b(u-j, o+j-1, 0), j=1..u))
        end:
    a:= n-> b(n, 0, 0):
    seq(a(n), n=0..30);  # Alois P. Heinz, Oct 07 2013
  • Mathematica
    nn=20;r=6;a=Apply[Plus,Table[Normal[Series[y x^(r+1)/(1-Sum[y x^i,{i,1,r}]),{x,0,nn}]][[n]]/(n+r)!,{n,1,nn-r}]]/.y->-1;Range[0,nn]! CoefficientList[Series[1/(1-x-a),{x,0,nn}],x] (* Geoffrey Critzer, Feb 25 2014 *)
    Table[n!*SeriesCoefficient[1/(Sum[x^(7*k)/(7*k)!-x^(7*k+1)/(7*k+1)!,{k,0,n}]),{x,0,n}],{n,1,20}] (* Vaclav Kotesovec, Aug 29 2014 *)

Formula

a(n)/n! ~ c * (1/r)^n, where r = 1.0001738181531504504518260962714687775785823593018886... is the root of the equation Sum_{n>=0} (r^(7*n)/(7*n)! - r^(7*n+1)/(7*n+1)!) = 0, c = 1.0010191104259450282450770594076722424772755532278.... - Vaclav Kotesovec, Aug 29 2014
E.g.f.: -(7/(2*((-cos(x*cos(3*Pi/14)))*cosh(x*sin(3*Pi/14)) + cos(x*cos(3*Pi/14))*cosh(x*sin(3*Pi/14))* sin(3*Pi/14) - cosh(x*sin(Pi/14))* (cos(x*cos(Pi/14))*(1 + sin(Pi/14)) - cos(Pi/14)*sin(x*cos(Pi/14))) + cos(3*Pi/14)*cosh(x*sin(3*Pi/14))* sin(x*cos(3*Pi/14)) - cosh(x*cos(Pi/7))* ((1 + cos(Pi/7))*cos(x*sin(Pi/7)) - sin(Pi/7)*sin(x*sin(Pi/7))) + cos(x*sin(Pi/7))* sinh(x*cos(Pi/7)) + cos(Pi/7)*cos(x*sin(Pi/7))* sinh(x*cos(Pi/7)) - sin(Pi/7)*sin(x*sin(Pi/7))* sinh(x*cos(Pi/7)) + cos(x*cos(Pi/14))* sinh(x*sin(Pi/14)) + cos(x*cos(Pi/14))*sin(Pi/14)* sinh(x*sin(Pi/14)) - cos(Pi/14)*sin(x*cos(Pi/14))* sinh(x*sin(Pi/14)) - cos(x*cos(3*Pi/14))* sinh(x*sin(3*Pi/14)) + cos(x*cos(3*Pi/14))* sin(3*Pi/14)*sinh(x*sin(3*Pi/14)) + cos(3*Pi/14)*sin(x*cos(3*Pi/14))* sinh(x*sin(3*Pi/14))))). - Vaclav Kotesovec, Jan 31 2015
In closed form, c = 7 / (r * (2*cos(r*sin(Pi/7))*cosh(r*cos(Pi/7)) + cos(Pi/7 - r*sin(Pi/7)) * cosh(r*cos(Pi/7)) + cos(Pi/7 - r*sin(Pi/7)) * cosh(r*cos(Pi/7)) + 2*cos(r*cos(Pi/14)) * cosh(r*sin(Pi/14)) + 2*cos(r*cos(3*Pi/14)) * cosh(r*sin(3*Pi/14)) + 2*cosh(r*sin(Pi/14)) * sin(Pi/14 + r*cos(Pi/14)) - 2*cosh(r*sin(3*Pi/14)) * sin(3*Pi/14 - r*cos(3*Pi/14)) - 2*cos(r*sin(Pi/7)) * sinh(r*cos(Pi/7)) - cos(Pi/7 - r*sin(Pi/7)) * sinh(r*cos(Pi/7)) - cos(Pi/7 - r*sin(Pi/7)) * sinh(r*cos(Pi/7)) - 2*cos(r*cos(Pi/14)) * sinh(r*sin(Pi/14)) - 2*sin(Pi/14 + r*cos(Pi/14))*sinh(r*sin(Pi/14)) + 2*cos(r*cos(3*Pi/14)) * sinh(r*sin(3*Pi/14)) - 2*sin((3*Pi)/14 - r*cos(3*Pi/14)) * sinh(r*sin(3*Pi/14)))). - Vaclav Kotesovec, Feb 01 2015

Extensions

a(18)-a(22) from Alois P. Heinz, Oct 07 2013
a(0)=1 prepended by Alois P. Heinz, Aug 08 2018

A080795 Number of minimax trees on n nodes.

Original entry on oeis.org

1, 1, 4, 20, 128, 1024, 9856, 110720, 1421312, 20525056, 329334784, 5812797440, 111923560448, 2334639652864, 52444850814976, 1262260748288000, 32405895451246592, 883950436237705216, 25530268718794276864
Offset: 0

Views

Author

Emanuele Munarini, Mar 14 2003

Keywords

Comments

A minimax tree is (i) rooted, (ii) binary (i.e., each node has at most two sons), (iii) topological (i.e., the left son is different from the right son), (iv) labeled (i.e., there is a bijection between the nodes and a finite totally ordered set). Moreover it has the following property: (v) the label of each node x is the minimum or the maximum of all the labels of the nodes of the subtree whose root is x.

Crossrefs

Programs

  • Maple
    w := (sqrt(2) - 1)/2:
    seq(simplify((2*sqrt(2))^(n-1)*add(k!*Stirling2(n, k)*w^(k-1), k = 1..n)), n = 1..20); # Peter Bala, Oct 31 2024
  • Mathematica
    Range[0, 18]! CoefficientList[ Series[ Tanh[ ArcTanh[ Sqrt[2]] - Sqrt[2] x]/Sqrt[2], {x, 0, 18}], x] (* Robert G. Wilson v *)
  • PARI
    {Stirling2(n,k)=(1/k!)*sum(j=0,k,(-1)^j*binomial(k,j)*(k-j)^n)}
    /* Finite sum given by Peter Bala: */
    {a(n)=local(w=(sqrt(2)-1)/2);if(n==0,1,round((2*sqrt(2))^(n-1)*sum(k=1,n,k!*Stirling2(n,k)*w^(k-1))))}

Formula

E.g.f.: ( tanh(arctanh(sqrt(2)) - sqrt(2)*x) )/sqrt(2) = sqrt(2)/2* (1 + (3-2*sqrt(2))* exp(2*sqrt(2)*x) )/( 1 - (3-2*sqrt(2))* exp(2*sqrt(2)*x) ).
Recurrence: a(n+1) = 2*(Sum_{k=0..n} binomial(n,k)*a(k)*a(n-k)) - 0^n.
a(2*n) = 2^n * A006154(2*n), n>0 (conjectured). - Ralf Stephan, Apr 29 2004
For n>0, a(n) = sqrt(2)^(3*n+1)*Sum_{k>=0} k^n/(1+sqrt(2))^(2*k). - Benoit Cloitre, Jan 12 2005
From Peter Bala, Jan 30 2011: (Start)
A finite sum equivalent to the previous formula of Benoit Cloitre is
a(n) = (2*sqrt(2))^(n-1)*Sum_{k = 1..n} k!*Stirling2(n,k)*w^(k-1), for n >= 1, with w = (sqrt(2) - 1)/2.
This formula can be used to prove congruences for a(n). For example, a(p) == (-1)^((p^2-1)/8) (mod p) for odd prime p.
For similar formulas for labeled plane and non-plane unary-binary trees see A080635 and A000111 respectively.
For a sequence of related polynomials see A185419. For a recursive table to calculate a(n) see A185420.
The e.g.f. A(x) satisfies the autonomous differential equation d/dx (A(x)) = 2*A(x)^2 - 1. (End)
From Peter Bala, Aug 26 2011: (Start)
The inverse function A(x)^(-1) of the generating function A(x) satisfies A(x)^(-1) = Integral_{t = 1..x} 1/(2*t^2 - 1) dt.
Let f(x) = 2*x^2 - 1. Define the nested derivative D^n[f](x) by means of the recursion D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0 (see A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x)). Then by [Dominici, Theorem 4.1] we have a(n+1) = D^n[f](1).
For n >= 1 we have a(n) = (2 + sqrt(2))^(n-1)*A(n, 3 - 2*sqrt(2)), where {A(n, x)}n>=1 = [1, 1 + x, 1 + 4*x + x^2, 1 + 11*x + 11*x^2 + x^3, ...] denotes the sequence of Eulerian polynomials (see A008292).
a(n+1) = (-1)^n*(sqrt(-2))^n * R(n, sqrt(-2)) where R(n, x) are the polynomials defined in A185896 (derivative polynomials associated with the function sec^2(x)). (End)
G.f.: 1 + x/G(0) where G(k) = 1 - 4*x*(k+1) - 2*x^2*(k+1)*(k+2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 11 2013
G.f.: 1 + x/(G(0) -x), where G(k) = 1 - x*(k+1) - 2*x*(k+1)/(1 - x*(k+2)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
E.g.f.: sqrt(2)*( -1/2 + (3+2*sqrt(2))/(4 + 2*sqrt(2)- E(0) )), where E(k) = 2 + 2*sqrt(2)*x/( 2*k+1 - 2*sqrt(2)*x/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 27 2013
a(n) ~ n! * 2^((3*n+1)/2) / (log(3+2*sqrt(2)))^(n+1). - Vaclav Kotesovec, Feb 25 2014
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