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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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

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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

A136630 Triangular array: T(n,k) counts the partitions of the set [n] into k odd sized blocks.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 4, 0, 1, 0, 1, 0, 10, 0, 1, 0, 0, 16, 0, 20, 0, 1, 0, 1, 0, 91, 0, 35, 0, 1, 0, 0, 64, 0, 336, 0, 56, 0, 1, 0, 1, 0, 820, 0, 966, 0, 84, 0, 1, 0, 0, 256, 0, 5440, 0, 2352, 0, 120, 0, 1, 0, 1, 0, 7381, 0, 24970, 0, 5082, 0, 165, 0, 1, 0, 0, 1024, 0, 87296, 0
Offset: 0

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Author

Paul D. Hanna, Jan 14 2008

Keywords

Comments

For partitions into blocks of even size see A156289.
Essentially the unsigned matrix inverse of triangle A121408.
From Peter Bala, Jul 28 2014: (Start)
Define a polynomial sequence x_(n) by setting x_(0) = 1 and for n = 1,2,... setting x_(n) = x*(x + n - 2)*(x + n - 4)*...*(x + n - 2*(n - 1)). Then this table is the triangle of connection constants for expressing the monomial polynomials x^n in terms of the basis x_(k), that is, x^n = sum {k = 0..n} T(n,k)*x_(k) for n = 0,1,2,.... An example is given below.
Let M denote the lower unit triangular array A119467 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/ having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle, omitting the first row and column, equals the infinite matrix product M(0)*M(1)*M(2)*.... (End)
Also the Bell transform of A000035(n+1). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016

Examples

			Triangle begins:
  1;
  0, 1;
  0, 0,   1;
  0, 1,   0,    1;
  0, 0,   4,    0,    1;
  0, 1,   0,   10,    0,     1;
  0, 0,  16,    0,   20,     0,    1;
  0, 1,   0,   91,    0,    35,    0,    1;
  0, 0,  64,    0,  336,     0,   56,    0,   1;
  0, 1,   0,  820,    0,   966,    0,   84,   0,   1;
  0, 0, 256,    0, 5440,     0, 2352,    0, 120,   0, 1;
  0, 1,   0, 7381,    0, 24970,    0, 5082,   0, 165, 0, 1;
T(5,3) = 10. The ten partitions of the set [5] into 3 odd-sized blocks are
(1)(2)(345), (1)(3)(245), (1)(4)(235), (1)(5)(234), (2)(3)(145),
(2)(4)(135), (2)(5)(134), (3)(4)(125), (3)(5)(124), (4)(5)(123).
Connection constants: Row 5 = [0,1,0,10,0,1]. Hence, with the polynomial sequence x_(n) as defined in the Comments section we have x^5 = x_(1) + 10*x_(3) + x_(5) = x + 10*x*(x+1)*(x-1) + x*(x+3)*(x+1)*(x-1)*(x-3).
		

References

  • L. Comtet, Analyse Combinatoire, Presses Univ. de France, 1970, Vol. II, pages 61-62.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 225-226.

Crossrefs

Cf. A121408; A136631 (antidiagonal sums), A003724 (row sums), A136632; A002452 (column 3), A002453 (column 5); A008958 (central factorial triangle), A156289. A185690, A196776.

Programs

  • Maple
    A136630 := proc (n, k) option remember; if k < 0 or n < k then 0 elif k = n then 1 else procname(n-2, k-2) + k^2*procname(n-2, k) end if end proc: seq(seq(A136630(n, k), k = 1 .. n), n = 1 .. 12); # Peter Bala, Jul 27 2014
    # The function BellMatrix is defined in A264428.
    BellMatrix(n -> (n+1) mod 2, 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    t[n_, k_] := Coefficient[ x^k/Product[ 1 - (2*j + k - 2*Quotient[k, 2])^2*x^2, {j, 0, k/2}] + x*O[x]^n, x, n]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 22 2013, after Pari *)
    BellMatrix[f_Function, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len-1}, {k, 0, len-1}]];
    rows = 13;
    M = BellMatrix[Mod[#+1, 2]&, rows];
    Table[M[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 23 2018, after Peter Luschny *)
  • PARI
    {T(n,k)=polcoeff(x^k/prod(j=0,k\2,1-(2*j+k-2*(k\2))^2*x^2 +x*O(x^n)),n)}

Formula

G.f. for column k: x^k/Product_{j=0..floor(k/2)} (1 - (2*j + k-2*floor(k/2))^2 * x^2).
G.f. for column 2*k: x^(2*k)/Product_{j=0..k} (1 - (2*j)^2*x^2).
G.f. for column 2*k+1: x^(2*k+1)/Product_{j=0..k} (1 - (2*j+1)^2*x^2).
From Peter Bala, Feb 21 2011 (Start)
T(n,k) = 1/(2^k*k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j-k)^n,
Recurrence relation T(n+2,k) = T(n,k-2) + k^2*T(n,k).
E.g.f.: F(x,z) = exp(x*sinh(z)) = Sum_{n>=0} R(n,x)*z^n/n! = 1 + x*z + x^2*z^2/2! + (x+x^3)*z^3/3! + ....
The row polynomials R(n,x) begin
R(1,x) = x
R(2,x) = x^2
R(3,x) = x+x^3.
The e.g.f. F(x,z) satisfies the partial differential equation d^2/dz^2(F) = x^2*F + x*F' + x^2*F'' where ' denotes differentiation w.r.t. x.
Hence the row polynomials satisfy the recurrence relation R(n+2,x) = x^2*R(n,x) + x*R'(n,x) + x^2*R''(n,x) with R(0,x) = 1.
The recurrence relation for T(n,k) given above follows from this.
(End)
For the corresponding triangle of ordered partitions into odd-sized blocks see A196776. Let P denote Pascal's triangle A070318 and put M = 1/2*(P-P^-1). M is A162590 (see also A131047). Then the first column of exp(t*M) lists the row polynomials for the present triangle. - Peter Bala, Oct 06 2011
Row generating polynomials equal D^n(exp(x*t)) evaluated at x = 0, where D is the operator sqrt(1+x^2)*d/dx. Cf. A196776. - Peter Bala, Dec 06 2011
From Peter Bala, Jul 28 2014: (Start)
E.g.f.: exp(t*sinh(x)) = 1 + t*x + t^2*x^2/2! + (t + t^3)*x^3/3! + ....
Hockey-stick recurrence: T(n+1,k+1) = Sum_{i = 0..floor((n-k)/2)} binomial(n,2*i)*T(n-2*i,k).
Recurrence equation for the row polynomials R(n,t):
R(n+1,t) = t*Sum_{k = 0..floor(n/2)} binomial(n,2*k)*R(n-2*k,t) with R(0,t) = 1. (End)

A006154 Number of labeled ordered partitions of an n-set into odd parts.

Original entry on oeis.org

1, 1, 2, 7, 32, 181, 1232, 9787, 88832, 907081, 10291712, 128445967, 1748805632, 25794366781, 409725396992, 6973071372547, 126585529106432, 2441591202059281, 49863806091395072, 1074927056650469527, 24392086908129247232, 581176736647853024581
Offset: 0

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Author

Keywords

Comments

Conjecture: taking the sequence modulo an integer k gives an eventually periodic sequence. For example, the sequence taken modulo 10 is [1, 1, 2, 7, 2, 1, 2, 7, 2, 1, 2, 7, 2, ...], with an apparent period [1, 2, 7, 2] beginning at a(1), of length 4. Cf. A000670. - Peter Bala, Apr 12 2023

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    readlib(coeftayl):
    with(combinat, bell);
    A:=series(1/(1-sinh(x)),x,20);
    G(x):=A : f[0]:=G(x): for n from 0 to 21 do f[n]:=coeftayl(G(x), x=0, n);;
    p[n]:=f[n]*((n)!) od: x:=0:seq(p[n], n=0..20); # Sergei N. Gladkovskii, Jun 01 2012
    # second Maple program:
    a:= proc(n) option remember; `if`(n=0, 1, add((i->
          a(n-i)*binomial(n, i))(2*j+1), j=0..(n-1)/2))
        end:
    seq(a(n), n=0..23);  # Alois P. Heinz, Feb 01 2022
  • Mathematica
    a[n_] := Sum[ (-1)^i*(k - 2*i)^n*Binomial[k, i]/2^k, {k, 1, n}, {i, 0, k}]; a[0] = 1; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Dec 07 2011, after Vladimir Kruchinin *)
    With[{nn=20},CoefficientList[Series[1/(1-Sinh[x]),{x,0,nn}],x]Range[0,nn]!] (* Harvey P. Dale, Nov 16 2012 *)
  • Maxima
    a(n):=sum(sum((-1)^i*(k-2*i)^n*binomial(k,i),i,0,k)/2^k,k,1,n); /* Vladimir Kruchinin, May 28 2011 */
  • PARI
    a(n)=if(n<2,n>=0,sum(k=1,ceil(n/2),binomial(n,2*k-1)*a(n-2*k+1))) \\ Ralf Stephan
    

Formula

E.g.f.: 1/(1 - sinh(x)).
With alternating signs, e.g.f.: 1/(1+sinh(x)). - Ralf Stephan, Apr 29 2004
a(0) = a(1) = 1, a(n) = Sum_{k=1..ceiling(n/2)} C(n,2*k-1)*a(n-2*k+1). - Ralf Stephan, Apr 29 2004
a(n) ~ (sqrt(2)/2)*n!/log(1+sqrt(2))^(n+1). - Conjectured by Simon Plouffe, Feb 17 2007.
From Andrew Hone, Feb 22 2007: (Start)
This formula can be proved using the techniques in the article by Philippe Flajolet (see links) [see Theorem 5 and Table 2, noting that 1/(1-sinh(x)) just has a simple pole at x=log(1+sqrt(2))]. (End)
a(n) = Sum_{k=1..n} Sum_{i=0..k} (-1)^i*(k-2*i)^n*binomial(k,i)/2^k, n > 0, a(0)=1. - Vladimir Kruchinin, May 28 2011
Row sums (apart from a(0)) of A196776. - Peter Bala, Oct 06 2011
Row sums of A193474. - Peter Luschny, Oct 07 2011
a(n) = D^n(1/(1-x)) evaluated at x = 0, where D is the operator sqrt(1+x^2)*d/dx. Cf. A003724 and A000111. - Peter Bala, Dec 06 2011
From Sergei N. Gladkovskii, Jun 01 2012: (Start)
Let E(x) be the e.g.f., then
E(x) = -1/x + 1/(x*(1-x))+ x^3/((1-x)*((1-x)*G(0) - x^2)); G(k) = (2*k+2)*(2*k+3)+x^2-(2*k+2)*(2*k+3)*x^2/G(k+1); (continued fraction).
E(x) = -1/x + 1/(x*(1-x))+ x^3/((1-x)*((1-x)*G(0) - x^2)); G(k) = 8*k+6+x^2/(1 + (2*k+2)*(2*k+3)/G(k+1)); (continued fraction).
E(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))); (continued fraction).
(End).
E.g.f. 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))); (continued fraction). - Sergei N. Gladkovskii, Oct 01 2012
O.g.f A(x) satisfies A(x) = 1 + ( A(x/(1-x))/(1-x) - A(x/(1+x))/(1+x) )/2. - Paul D. Hanna, Aug 19 2024

Extensions

More terms from Christian G. Bower, Oct 15 1999

A162590 Polynomials with e.g.f. exp(x*t)/csch(t), triangle of coefficients read by rows.

Original entry on oeis.org

0, 1, 0, 0, 2, 0, 1, 0, 3, 0, 0, 4, 0, 4, 0, 1, 0, 10, 0, 5, 0, 0, 6, 0, 20, 0, 6, 0, 1, 0, 21, 0, 35, 0, 7, 0, 0, 8, 0, 56, 0, 56, 0, 8, 0, 1, 0, 36, 0, 126, 0, 84, 0, 9, 0, 0, 10, 0, 120, 0, 252, 0, 120, 0, 10, 0, 1, 0, 55, 0, 330, 0, 462, 0, 165, 0, 11, 0, 0, 12, 0, 220, 0, 792, 0, 792, 0
Offset: 0

Views

Author

Peter Luschny, Jul 07 2009

Keywords

Comments

Comment from Peter Bala (Dec 06 2011): "Let P denote Pascal's triangle A070318 and put M = 1/2*(P-P^-1). M is A162590 (see also A131047). Then the first column of (I-t*M)^-1 (apart from the initial 1) lists the row polynomials for" A196776(n,k), which gives the number of ordered partitions of an n set into k odd-sized blocks. - Peter Luschny, Dec 06 2011
The n-th row of the triangle is formed by multiplying by 2^(n-1) the elements of the first row of the limit as k approaches infinity of the stochastic matrix P^(2k-1) where P is the stochastic matrix associated with the Ehrenfest model with n balls. The elements of a stochastic matrix P give the probability of arriving in a state j given the previous state i. In particular the sum of every row of the matrix must be 1, and so the sum of the terms in the n-th row of this triangle is 2^(n-1). Furthermore, by the properties of Markov chains, we can interpret P^(2k) as the (2k)-step transition matrix of the Ehrenfest model and its limit exists and it is again a stochastic matrix. The rows of the triangle divided by 2^(n-1) are the even rows (second, fourth, ...) and the odd rows (first, third, ...) of the limit matrix P^(2k). - Luca Onnis, Oct 29 2023

Examples

			Triangle begins:
  0
  1,  0
  0,  2,  0
  1,  0,  3,  0
  0,  4,  0,  4,  0
  1,  0, 10,  0,  5,  0
  0,  6,  0, 20,  0,  6,  0
  1,  0, 21,  0, 35,  0,  7,  0
  ...
  p[0](x) = 0;
  p[1](x) = 1
  p[2](x) = 2*x
  p[3](x) = 3*x^2 +  1
  p[4](x) = 4*x^3 +  4*x
  p[5](x) = 5*x^4 + 10*x^2 +  1
  p[6](x) = 6*x^5 + 20*x^3 +  6*x
  p[7](x) = 7*x^6 + 35*x^4 + 21*x^2 + 1
  p[8](x) = 8*x^7 + 56*x^5 + 56*x^3 + 8*x
.
Cf. the triangle of odd-numbered terms in rows of Pascal's triangle (A034867).
p[n] (k), n=0,1,...
k=0:  0, 1,  0,   1,    0,     1, ... A000035, (A059841)
k=1:  0, 1,  2,   4,    8,    16, ... A131577, (A000079)
k=2:  0, 1,  4,  13,   40,   121, ... A003462
k=3:  0, 1,  6,  28,  120,   496, ... A006516
k=4:  0, 1,  8,  49,  272,  1441, ... A005059
k=5:  0, 1, 10,  76,  520,  3376, ... A081199, (A016149)
k=6:  0, 1, 12, 109,  888,  6841, ... A081200, (A016161)
k=7:  0, 1, 14, 148, 1400, 12496, ... A081201, (A016170)
k=8:  0, 1, 16, 193, 2080, 21121, ... A081202, (A016178)
k=9:  0, 1, 18, 244, 2952, 33616, ... A081203, (A016186)
k=10: 0, 1, 20, 301, 4040, 51001, ... ......., (A016190)
.
p[n] (k), k=0,1,...
p[0]: 0,  0,   0,    0,    0,     0, ... A000004
p[1]: 1,  1,   1,    1,    1,     1, ... A000012
p[2]: 0,  2,   4,    6,    8,    10, ... A005843
p[3]: 1,  4,  13,   28,   49,    76, ... A056107
p[4]: 0,  8,  40,  120,  272,   520, ... A105374
p[5]: 1, 16, 121,  496, 1441,  3376, ...
p[6]: 0, 32, 364, 2016, 7448, 21280, ...
		

Crossrefs

Cf. A119467.

Programs

  • Maple
    # Polynomials: p_n(x)
    p := proc(n,x) local k;
    pow := (n,k) -> `if`(n=0 and k=0,1,n^k);
    add((k mod 2)*binomial(n,k)*pow(x,n-k),k=0..n) end;
    # Coefficients: a(n)
    seq(print(seq(coeff(i!*coeff(series(exp(x*t)/csch(t), t,16),t,i),x,n), n=0..i)), i=0..8);
  • Mathematica
    p[n_, x_] := Sum[Binomial[n, 2*k-1]*x^(n-2*k+1), {k, 0, n+2}]; row[n_] := CoefficientList[p[n, x], x] // Append[#, 0]&; Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Jun 28 2013 *)
    n = 15; "n-th row"
    mat = Table[Table[0, {j, 1, n + 1}], {i, 1, n + 1}];
    mat[[1, 2]] = 1;
    mat[[n + 1, n]] = 1;
    For[i = 2, i <= n, i++, mat[[i, i - 1]] = (i - 1)/n ];
    For[i = 2, i <= n, i++, mat[[i, i + 1]] = (n - i + 1)/n];
    mat // MatrixForm;
    P2 = Dot[mat, mat];
    R1 = Simplify[
      Eigenvectors[Transpose[P2]][[1]]/
       Total[Eigenvectors[Transpose[P2]][[1]]]]
    R2 = Table[Dot[R1, Transpose[mat][[k]]], {k, 1, n + 1}]
    even = R1*2^(n - 1) (* Luca Onnis, Oct 29 2023 *)

Formula

p_n(x) = Sum_{k=0..n} (k mod 2)*binomial(n,k)*x^(n-k).
E.g.f.: exp(x*t)/csch(t) = 0*(t^0/0!) + 1*(t^1/1!) + (2*x)*(t^2/2!) + (3*x^2+1)*(t^3/3!) + ...
The 'co'-polynomials with generating function exp(x*t)*sech(t) are the Swiss-Knife polynomials (A153641).

A381181 Expansion of e.g.f. (1/x) * Series_Reversion( x / (1 + sin(x)) ).

Original entry on oeis.org

1, 1, 2, 5, 8, -79, -1584, -20539, -223616, -1855295, -1736960, 435730789, 14511117312, 338965239601, 6202042886144, 71638247035109, -714560796196864, -84697775518956799, -3650903032332091392, -115829159202293866939, -2739961030150105333760, -29414406825401517785039
Offset: 0

Views

Author

Seiichi Manyama, Feb 16 2025

Keywords

Crossrefs

Programs

  • PARI
    a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
    a(n) = sum(k=0, n, k!*binomial(n+1, k)*I^(n-k)*a136630(n, k))/(n+1);

Formula

E.g.f. A(x) satisfies A(x) = 1 + sin(x * A(x)).
a(n) = (1/(n+1)) * Sum_{k=0..n} k! * binomial(n+1,k) * i^(n-k) * A136630(n,k), where i is the imaginary unit.

A193474 Table read by rows: The coefficients of the polynomials P(n, x) = Sum{k=0..n} Sum{j=0..k} (-1)^j * 2^(-k) * binomial(k, j) * (k-2*j)^n * x^(n-k).

Original entry on oeis.org

1, 1, 0, 2, 0, 0, 6, 0, 1, 0, 24, 0, 8, 0, 0, 120, 0, 60, 0, 1, 0, 720, 0, 480, 0, 32, 0, 0, 5040, 0, 4200, 0, 546, 0, 1, 0, 40320, 0, 40320, 0, 8064, 0, 128, 0, 0, 362880, 0, 423360, 0, 115920, 0, 4920, 0, 1, 0, 3628800, 0, 4838400, 0, 1693440, 0, 130560, 0, 512, 0, 0
Offset: 1

Views

Author

Peter Luschny, Aug 01 2011

Keywords

Comments

See A196776 for a row reversed form of this triangle. - Peter Bala, Oct 06 2011

Examples

			The sequence of polynomials P(n, x) begins:
[0]    1;
[1]    1;
[2]    2;
[3]    6 +      x^2;
[4]   24 +    8*x^2;
[5]  120 +   60*x^2 +     x^4;
[6]  720 +  480*x^2 +  32*x^4;
[7] 5040 + 4200*x^2 + 546*x^4 + x^6.
		

Crossrefs

Programs

  • Maple
    A193474_polynom := proc(n,x) local k, j;
    add(add((-1)^j*2^(-k)*binomial(k,j)*(k-2*j)^n*x^(n-k),j=0..k),k=0..n) end: seq(seq(coeff(A193474_polynom(n,x),x,i),i=0..n),n=0..10);
  • Mathematica
    p[n_, x_] := Sum[(-1)^j*2^(-k)*Binomial[k, j]*(k-2*j)^n*x^(n-k), {k, 0, n}, {j, 0, k}]; t[n_, k_] := Coefficient[p[n, x], x, k]; t[0, 0] = 1; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 27 2014 *)

Formula

P(n, 0) = A000142(n).
P(n, 1) = A006154(n).
P(n, 2) = A191277(n).
P(n, i) = A000111(n+1), where i is the imaginary unit.
P(n, i)*2^n = A000828(n+1).
P(n, 1/2)*2^n = A000557(n).
P(n, 1/3)*3^n = A107403(n).
P(n, i/2)*2^n = A007289(n).
G(m, x) = 1/(1 - m*sinh(x)) is the generating function of m^n*P(n, 1/m).
GI(m, x) = 1/(1 - m*sin(x)) is the generating function of m^n*P(n, i/m).
[x^2] P(n+1, x) = A005990(n).

A381180 E.g.f. A(x) satisfies A(x) = 1 + sin(x*A(x)) / A(x).

Original entry on oeis.org

1, 1, 0, -1, -8, -19, 64, 1091, 7680, -1415, -650752, -8575865, -35559424, 857890021, 21380186112, 203548592651, -1615715926016, -95486152906639, -1599622990659584, -1397194164399601, 657963431581974528, 18168041375501245021, 157453907927886725120, -6059840564222790027821
Offset: 0

Views

Author

Seiichi Manyama, Feb 16 2025

Keywords

Crossrefs

Programs

  • PARI
    a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
    a(n) = sum(k=0, n, k!*binomial(n-k+1, k)/(n-k+1)*I^(n-k)*a136630(n, k));

Formula

a(n) = Sum_{k=0..n} k! * binomial(n-k+1,k)/(n-k+1) * i^(n-k) * A136630(n,k), where i is the imaginary unit.

A381182 E.g.f. A(x) satisfies A(x) = 1/( 1 - A(x) * sin(x * A(x)) ).

Original entry on oeis.org

1, 1, 6, 71, 1288, 31661, 984640, 37085075, 1641305472, 83497838425, 4801347029504, 307975150996831, 21802395720298496, 1688562016007776261, 142023935786330431488, 12892154760586821775019, 1256251152910271399624704, 130793914073764385411654321, 14490427167940362294881615872
Offset: 0

Views

Author

Seiichi Manyama, Feb 16 2025

Keywords

Crossrefs

Programs

  • PARI
    a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
    a(n) = sum(k=0, n, k!*binomial(n+2*k+1, k)/(n+2*k+1)*I^(n-k)*a136630(n, k));

Formula

a(n) = Sum_{k=0..n} k! * binomial(n+2*k+1,k)/(n+2*k+1) * i^(n-k) * A136630(n,k), where i is the imaginary unit.

A381177 E.g.f. A(x) satisfies A(x) = 1/( 1 - A(x) * sinh(x * A(x)) ).

Original entry on oeis.org

1, 1, 6, 73, 1352, 33861, 1072000, 41083477, 1849680768, 95708731945, 5597075177984, 365091888890433, 26281788308598784, 2069729710424907181, 177006820644852031488, 16337090667286093559821, 1618592591411194127089664, 171337824188415839421148881, 19299478529228162963028508672
Offset: 0

Views

Author

Seiichi Manyama, Feb 16 2025

Keywords

Crossrefs

Programs

  • PARI
    a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
    a(n) = sum(k=0, n, k!*binomial(n+2*k+1, k)/(n+2*k+1)*a136630(n, k));

Formula

a(n) = Sum_{k=0..n} k! * binomial(n+2*k+1,k)/(n+2*k+1) * A136630(n,k).

A381179 E.g.f. A(x) satisfies A(x) = 1 + sinh(x*A(x)) / A(x).

Original entry on oeis.org

1, 1, 0, 1, 8, 21, 64, 1093, 8448, 47785, 654848, 9402537, 94222336, 1264390141, 23392960512, 363389219053, 5722054885376, 117602664867921, 2434091053613056, 47867013812467921, 1080303165427679232, 26716998341391367141, 645003218568158904320, 16403742152044108508181
Offset: 0

Views

Author

Seiichi Manyama, Feb 16 2025

Keywords

Crossrefs

Programs

  • PARI
    a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
    a(n) = sum(k=0, n, k!*binomial(n-k+1, k)/(n-k+1)*a136630(n, k));

Formula

a(n) = Sum_{k=0..n} k! * binomial(n-k+1,k)/(n-k+1) * A136630(n,k).
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