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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A185418 Square array, read by antidiagonals, used to recursively calculate the Springer numbers A001586.

Original entry on oeis.org

1, 1, 1, 3, 3, 1, 11, 11, 5, 1, 57, 57, 27, 7, 1, 361, 361, 175, 51, 9, 1, 2763, 2763, 1353, 413, 83, 11, 1, 24611, 24611, 12125, 3801, 819, 123, 13, 1, 250737, 250737, 123987, 39487, 8857, 1441, 171, 15, 1, 2873041, 2873041, 1424215, 458331, 105489, 18057, 2327, 227, 17, 1
Offset: 0

Views

Author

Peter Bala, Jan 30 2011

Keywords

Comments

The table entries T(n,k), n,k>=0, are defined by the recurrence relation:
1)... T(n+1,k) = k*T(n,k-1)+(k+1)*T(n,k+1) with boundary condition T(0,k) = 1.
The first column of the table produces the sequence of Springer numbers A001586.
For similarly defined tables see A185414, A185416 and A185420.

Examples

			Square array begins
n\k|.....0......1.......2.......3........4........5........6
============================================================
..0|.....1......1.......1.......1........1........1........1
..1|.....1......3.......5.......7........9.......11.......13
..2|.....3.....11......27......51.......83......123......171
..3|....11.....57.....175.....413......819.....1441.....2327
..4|....57....361....1353....3801.....8857....18057....33321
..5|...361...2763...12125...39487...105489...244211...507013
..6|..2763..24611..123987..458331..1379003..3569523..8229891
..
Examples of recurrence relation:
T(4,3) = 3801 = 3*T(3,2) + 4*T(3,4) = 3*175 + 4*819;
T(5,1) = 2763 = 1*T(4,0)+ 2*T(4,2) = 1*57 + 2*1353.
		

Crossrefs

Programs

  • Maple
    # A185418
    S := proc(n, x) option remember; description `polynomials S(n, x)`;
    if n = 0 then 1 else x*S(n-1,x-1)+(x+1)*S(n-1,x+1) end if end proc:
    for n from 0 to 10 do seq(S(n, k), k = 0..10) end do;
  • Mathematica
    T[n_, k_] := T[n, k] = If[n<0 || k<0, 0, If[n == 0, 1, k T[n-1, k-1] + (k+1)*T[n-1, k+1]]];
    Table[T[n-k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 22 2021 *)
  • PARI
    {T(n,k)=if(n<0||k<0,0,if(n==0,1,k*T(n-1,k-1)+(k+1)*T(n-1,k+1)))}

Formula

(1)... T(n,k) = S(n,k) with S(n,x) the polynomials described in A185417.
(2)... First column: T(n,0) = A001586(n).
(3)... Second column: T(n,1) = A001586(n+1).
(4)... Second row: T(1,k) = A005408(k).
(5)... Third row: T(2,k) = A164897(k).

A214554 Triangle read by rows, coefficients of polynomials related to the Springer numbers A001586.

Original entry on oeis.org

1, -1, 2, -3, -4, 4, 11, -18, -12, 8, 57, 88, -72, -32, 16, -361, 570, 440, -240, -80, 32, -2763, -4332, 3420, 1760, -720, -192, 64, 24611, -38682, -30324, 15960, 6160, -2016, -448, 128, 250737, 393776, -309456, -161728, 63840, 19712, -5376, -1024, 256
Offset: 0

Views

Author

Peter Luschny, Jul 30 2012

Keywords

Comments

The polynomials might be called Springer polynomials because both p{n}(0) and p{n}(1) are signed versions of the Springer numbers. p{n}(0) is the first column of the triangle (A212435 with e.g.f. exp(-x)/cosh(2x)) and p{n}(1) are the row sums (A188458 with e.g.f. exp(x)/cosh(2x)).

Examples

			[0]     1,
[1]    -1,      2,
[2]    -3,     -4,      4,
[3]    11,    -18,    -12,     8,
[4]    57,     88,    -72,   -32,   16,
[5]  -361,    570,    440,  -240,  -80,    32,
[6] -2763,  -4332,   3420,  1760, -720,  -192,   64,
[7] 24611, -38682, -30324, 15960, 6160, -2016, -448, 128.
		

Programs

  • Sage
    @CachedFunction
    def SpringerPoly(n,x) :
        if n == 0 : return 1
        return add(2^(n-k)*SpringerPoly(k,1/2)*binomial(n,k)*((x-1/2)^(n-k)+n%2-1) for k in range(n)[::2])
    R = PolynomialRing(ZZ, 'x')
    def A214554_row(n) : return R(SpringerPoly(n,x)).coeffs()
    for n in (0..7) : A214554_row(n)

Formula

p{0}(x) = 1 and for n>0
p{n}(x) = Sum_{0<=k

A256679 A signed triangle of V. I. Arnold for the Springer numbers (A001586).

Original entry on oeis.org

1, 1, 0, -2, -3, -3, -8, -6, -3, 0, 40, 48, 54, 57, 57, 256, 216, 168, 114, 57, 0, -1952, -2208, -2424, -2592, -2706, -2763, -2763, -17408, -15456, -13248, -10824, -8232, -5526, -2763, 0, 177280, 194688, 210144, 223392, 234216, 242448, 247974, 250737, 250737
Offset: 0

Author

Vladimir Kruchinin, Apr 07 2015

Keywords

Comments

This triangle is denoted R(b) on page 6 of the Arnold reference.
Unsigned version of triangle is triangle of A202818.
First column (m=0) is A000828. - Robert Israel, Apr 08 2015

Examples

			Triangle begins:
    1;
    1,   0;
   -2,  -3,  -3;
   -8,  -6,  -3,   0;
   40,  48,  54,  57, 57;
  256, 216, 168, 114, 57, 0;
		

Programs

  • Maple
    T:= (n,m) -> add(add(4^i*euler(2*i)*binomial(2*k,2*i)*binomial(n-m,2*k-m),i=0..k),k=floor(m/2)..floor(n/2)):
    seq(seq(T(n,m),m=0..n),n=0..10); # Robert Israel, Apr 08 2015
    # Second program, about 100 times faster than the first for the first 100 rows.
    Triangle := proc(len) local s, A, n, k;
    A := Array(0..len-1,[1]); lprint(A[0]);
    for n from 1 to len-1 do
       if n mod 2 = 1 then s := 0 else
       s := 2^(3*n+1)*(Zeta(0,-n,1/8)-Zeta(0,-n,5/8)) fi;
       A[n] := s;
       for k from n-1 by -1 to 0 do
           s := s + A[k]; A[k] := s od;
       lprint(seq(A[k], k=0..n));
    od end:
    Triangle(100); # Peter Luschny, Apr 08 2015
  • Mathematica
    T[n_, m_] := Sum[4^i EulerE[2i] Binomial[2k, 2i] Binomial[n-m, 2k-m], {k, Floor[m/2], n/2}, {i, 0, k}];
    Table[T[n, m], {n, 0, 8}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 12 2019 *)
  • Maxima
    T(n,m):=(sum((sum(4^i*euler(2*i)*binomial(2*k,2*i),i,0,k))*binomial(n-m,2*k-m),k,floor(m/2),n/2));
    
  • Sage
    def triangle(len):
        L = [1]; print(L)
        for n in range(1,len):
            if is_even(n):
                s = 2^(3*n+1)*(hurwitz_zeta(-n,1/8)-hurwitz_zeta(-n,5/8))
            else: s = 0
            L.append(s)
            for k in range(n-1,-1,-1):
                s = s + L[k]; L[k] = s
            print(L)
    triangle(7) # Peter Luschny, Apr 08 2015

Formula

E.g.f.: cosh(x+y)/cosh(2*(x+y))*exp(x).
T(n,m) = Sum_{k=floor(m/2)..floor(n/2)} Sum_{i=0..k} 4^i*E(2*i)*C(2*k,2*i)*C(n-m,2*k-m), where E(n) are the Euler secant numbers A122045.

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

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

A003881 Decimal expansion of Pi/4.

Original entry on oeis.org

7, 8, 5, 3, 9, 8, 1, 6, 3, 3, 9, 7, 4, 4, 8, 3, 0, 9, 6, 1, 5, 6, 6, 0, 8, 4, 5, 8, 1, 9, 8, 7, 5, 7, 2, 1, 0, 4, 9, 2, 9, 2, 3, 4, 9, 8, 4, 3, 7, 7, 6, 4, 5, 5, 2, 4, 3, 7, 3, 6, 1, 4, 8, 0, 7, 6, 9, 5, 4, 1, 0, 1, 5, 7, 1, 5, 5, 2, 2, 4, 9, 6, 5, 7, 0, 0, 8, 7, 0, 6, 3, 3, 5, 5, 2, 9, 2, 6, 6, 9, 9, 5, 5, 3, 7
Offset: 0

Keywords

Comments

Also the ratio of the area of a circle to the circumscribed square. More generally, the ratio of the area of an ellipse to the circumscribed rectangle. Also the ratio of the volume of a cylinder to the circumscribed cube. - Omar E. Pol, Sep 25 2013
Also the surface area of a quarter-sphere of diameter 1. - Omar E. Pol, Oct 03 2013
Least positive solution to sin(x) = cos(x). - Franklin T. Adams-Watters, Jun 17 2014
Dirichlet L-series of the non-principal character modulo 4 (A101455) at 1. See e.g. Table 22 of arXiv:1008.2547. - R. J. Mathar, May 27 2016
This constant is also equal to the infinite sum of the arctangent functions with nested radicals consisting of square roots of two. Specifically, one of the Viete-like formulas for Pi is given by Pi/4 = Sum_{k = 2..oo} arctan(sqrt(2 - a_{k - 1})/a_k), where the nested radicals are defined by recurrence relations a_k = sqrt(2 + a_{k - 1}) and a_1 = sqrt(2) (see the article [Abrarov and Quine]). - Sanjar Abrarov, Jan 09 2017
Pi/4 is the area enclosed between circumcircle and incircle of a regular polygon of unit side. - Mohammed Yaseen, Nov 29 2023

Examples

			0.785398163397448309615660845819875721049292349843776455243736148...
N = 2, m = 6: Pi/4 = 4!*3^4 Sum_{k >= 0} (-1)^k/((2*k - 11)*(2*k - 5)*(2*k + 1)*(2*k + 7)*(2*k + 13)). - _Peter Bala_, Nov 15 2016
		

References

  • Jörg Arndt and Christoph Haenel, Pi: Algorithmen, Computer, Arithmetik, Springer 2000, p. 150.
  • Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 437.
  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Sections 6.3 and 8.4, pp. 429 and 492.
  • Douglas R. Hofstadter, Gödel, Escher, Bach: An Eternal Golden Braid, Basic Books, p. 408.
  • J. Rivaud, Analyse, Séries, équations différentielles, Mathématiques supérieures et spéciales, Premier cycle universitaire, Vuibert, 1981, Exercice 3, p. 136.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 119.

Crossrefs

Cf. A006752 (beta(2)=Catalan), A153071 (beta(3)), A175572 (beta(4)), A175571 (beta(5)), A175570 (beta(6)), A258814 (beta(7)), A258815 (beta(8)), A258816 (beta(9)).
Cf. A001622.

Programs

  • Haskell
    -- see link: Literate Programs
    import Data.Char (digitToInt)
    a003881_list len = map digitToInt $ show $ machin `div` (10 ^ 10) where
       machin = 4 * arccot 5 unity - arccot 239 unity
       unity = 10 ^ (len + 10)
       arccot x unity = arccot' x unity 0 (unity `div` x) 1 1 where
         arccot' x unity summa xpow n sign
        | term == 0 = summa
        | otherwise = arccot'
          x unity (summa + sign * term) (xpow `div` x ^ 2) (n + 2) (- sign)
        where term = xpow `div` n
    -- Reinhard Zumkeller, Nov 20 2012
    
  • Magma
    R:= RealField(100); Pi(R)/4; // G. C. Greubel, Mar 08 2018
  • Maple
    evalf(Pi/4) ;
  • Mathematica
    RealDigits[N[Pi/4,6! ]]  (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)
    (* PROGRAM STARTS *)
    (* Define the nested radicals a_k by recurrence *)
    a[k_] := Nest[Sqrt[2 + #1] & , 0, k]
    (* Example of Pi/4 approximation at K = 100 *)
    Print["The actual value of Pi/4 is"]
    N[Pi/4, 40]
    Print["At K = 100 the approximated value of Pi/4 is"]
    K := 100;  (* the truncating integer *)
    N[Sum[ArcTan[Sqrt[2 - a[k - 1]]/a[k]], {k, 2, K}], 40] (* equation (8) *)
    (* Error terms for Pi/4 approximations *)
    Print["Error terms for Pi/4"]
    k := 1; (* initial value of the index k *)
    K := 10; (* initial value of the truncating integer K *)
    sqn := {}; (* initiate the sequence *)
    AppendTo[sqn, {"Truncating integer K ", " Error term in Pi/4"}];
    While[K <= 30,
    AppendTo[sqn, {K,
       N[Pi/4 - Sum[ArcTan[Sqrt[2 - a[k - 1]]/a[k]], {k, 2, K}], 1000] //
        N}]; K++]
    Print[MatrixForm[sqn]]
    (* Sanjar Abrarov, Jan 09 2017 *)
  • PARI
    Pi/4 \\ Charles R Greathouse IV, Jul 07 2014
    
  • SageMath
    # Leibniz/Cohen/Villegas/Zagier/Arndt/Haenel
    def FastLeibniz(n):
        b = 2^(2*n-1); c = b; s = 0
        for k in range(n-1,-1,-1):
            t = 2*k+1
            s = s + c/t if is_even(k) else s - c/t
            b *= (t*(k+1))/(2*(n-k)*(n+k))
            c += b
        return s/c
    A003881 = RealField(3333)(FastLeibniz(1330))
    print(A003881)  # Peter Luschny, Nov 20 2012
    

Formula

Equals Integral_{x=0..oo} sin(2x)/(2x) dx.
Equals lim_{n->oo} n*A001586(n-1)/A001586(n) (conjecture). - Mats Granvik, Feb 23 2011
Equals Integral_{x=0..1} 1/(1+x^2) dx. - Gary W. Adamson, Jun 22 2003
Equals Integral_{x=0..Pi/2} sin(x)^2 dx, or Integral_{x=0..Pi/2} cos(x)^2 dx. - Jean-François Alcover, Mar 26 2013
Equals (Sum_{x=0..oo} sin(x)*cos(x)/x) - 1/2. - Bruno Berselli, May 13 2013
Equals (-digamma(1/4) + digamma(3/4))/4. - Jean-François Alcover, May 31 2013
Equals Sum_{n>=0} (-1)^n/(2*n+1). - Geoffrey Critzer, Nov 03 2013
Equals Integral_{x=0..1} Product_{k>=1} (1-x^(8*k))^3 dx [cf. A258414]. - Vaclav Kotesovec, May 30 2015
Equals Product_{k in A071904} (if k mod 4 = 1 then (k-1)/(k+1)) else (if k mod 4 = 3 then (k+1)/(k-1)). - Dimitris Valianatos, Oct 05 2016
From Peter Bala, Nov 15 2016: (Start)
For N even: 2*(Pi/4 - Sum_{k = 1..N/2} (-1)^(k-1)/(2*k - 1)) ~ (-1)^(N/2)*(1/N - 1/N^3 + 5/N^5 - 61/N^7 + 1385/N^9 - ...), where the sequence of unsigned coefficients [1, 1, 5, 61, 1385, ...] is A000364. See Borwein et al., Theorem 1 (a).
For N odd: 2*(Pi/4 - Sum_{k = 1..(N-1)/2} (-1)^(k-1)/(2*k - 1)) ~ (-1)^((N-1)/2)*(1/N - 1/N^2 + 2/N^4 - 16/N^6 + 272/N^8 - ...), where the sequence of unsigned coefficients [1, 1, 2, 16, 272, ...] is A000182 with an extra initial term of 1.
For N = 0,1,2,... and m = 1,3,5,... there holds Pi/4 = (2*N)! * m^(2*N) * Sum_{k >= 0} ( (-1)^(N+k) * 1/Product_{j = -N..N} (2*k + 1 + 2*m*j) ); when N = 0 we get the Madhava-Gregory-Leibniz series for Pi/4.
For examples of asymptotic expansions for the tails of these series representations for Pi/4 see A024235 (case N = 1, m = 1), A278080 (case N = 2, m = 1) and A278195 (case N = 3, m = 1).
For N = 0,1,2,..., Pi/4 = 4^(N-1)*N!/(2*N)! * Sum_{k >= 0} 2^(k+1)*(k + N)!* (k + 2*N)!/(2*k + 2*N + 1)!, follows by applying Euler's series transformation to the above series representation for Pi/4 in the case m = 1. (End)
From Peter Bala, Nov 05 2019: (Start)
For k = 0,1,2,..., Pi/4 = k!*Sum_{n = -oo..oo} 1/((4*n+1)*(4*n+3)* ...*(4*n+2*k+1)), where Sum_{n = -oo..oo} f(n) is understood as lim_{j -> oo} Sum_{n = -j..j} f(n).
Equals Integral_{x = 0..oo} sin(x)^4/x^2 dx = Sum_{n >= 1} sin(n)^4/n^2, by the Abel-Plana formula.
Equals Integral_{x = 0..oo} sin(x)^3/x dx = Sum_{n >= 1} sin(n)^3/n, by the Abel-Plana formula. (End)
From Amiram Eldar, Aug 19 2020: (Start)
Equals arcsin(1/sqrt(2)).
Equals Product_{k>=1} (1 - 1/(2*k+1)^2).
Equals Integral_{x=0..oo} x/(x^4 + 1) dx.
Equals Integral_{x=0..oo} 1/(x^2 + 4) dx. (End)
With offset 1, equals 5 * Pi / 2. - Sean A. Irvine, Aug 19 2021
Equals (1/2)!^2 = Gamma(3/2)^2. - Gary W. Adamson, Aug 23 2021
Equals Integral_{x = 0..oo} exp(-x)*sin(x)/x dx (see Rivaud reference). - Bernard Schott, Jan 28 2022
From Amiram Eldar, Nov 06 2023: (Start)
Equals beta(1), where beta is the Dirichlet beta function.
Equals Product_{p prime >= 3} (1 - (-1)^((p-1)/2)/p)^(-1). (End)
Equals arctan( F(1)/F(4) ) + arctan( F(2)/F(3) ), where F(1), F(2), F(3), and F(4) are any four consecutive Fibonacci numbers. - Gary W. Adamson, Mar 03 2024
Pi/4 = Sum_{n >= 1} i/(n*P(n, i)*P(n-1, i)) = (1/2)*Sum_{n >= 1} (-1)^(n+1)*4^n/(n*A006139(n)*A006139(n-1)), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The n-th summand of the series is O( 1/(3 + 2*sqrt(3))^n ). - Peter Bala, Mar 16 2024
Equals arctan( phi^(-3) ) + arctan(phi^(-1) ). - Gary W. Adamson, Mar 27 2024
Equals Sum_{n>=1} eta(n)/2^n, where eta(n) is the Dirichlet eta function. - Antonio Graciá Llorente, Oct 04 2024
Equals Product_{k>=2} ((k + 1)^(k*(2*k + 1))*(k - 1)^(k*(2*k - 1)))/k^(4*k^2). - Antonio Graciá Llorente, Apr 12 2025
Equals Integral_{x=sqrt(2)..oo} dx/(x*sqrt(x^2 - 1)). - Kritsada Moomuang, May 29 2025

Extensions

a(98) and a(99) corrected by Reinhard Zumkeller, Nov 20 2012

A153641 Nonzero coefficients of the Swiss-Knife polynomials for the computation of Euler, tangent, and Bernoulli numbers (triangle read by rows).

Original entry on oeis.org

1, 1, 1, -1, 1, -3, 1, -6, 5, 1, -10, 25, 1, -15, 75, -61, 1, -21, 175, -427, 1, -28, 350, -1708, 1385, 1, -36, 630, -5124, 12465, 1, -45, 1050, -12810, 62325, -50521, 1, -55, 1650, -28182, 228525, -555731, 1, -66, 2475, -56364, 685575, -3334386, 2702765, 1
Offset: 0

Author

Peter Luschny, Dec 29 2008

Keywords

Comments

In the following the expression [n odd] is 1 if n is odd, 0 otherwise.
(+) W_n(0) = E_n are the Euler (or secant) numbers A122045.
(+) W_n(1) = T_n are the signed tangent numbers, see A009006.
(+) W_{n-1}(1) n / (4^n - 2^n) = B_n gives for n > 1 the Bernoulli number A027641/A027642.
(+) W_n(-1) 2^{-n}(n+1) = G_n the Genocchi number A036968.
(+) W_n(1/2) 2^{n} are the signed generalized Euler (Springer) number, see A001586.
(+) | W_n([n odd]) | the number of alternating permutations A000111.
(+) | W_n([n odd]) / n! | for 0<=n the Euler zeta number A099612/A099617 (see Wikipedia on Bernoulli number). - Peter Luschny, Dec 29 2008
The diagonals in the full triangle (with zero coefficients) of the polynomials have the general form E(k)*binomial(n+k,k) (k>=0 fixed, n=0,1,...) where E(n) are the Euler numbers in the enumeration A122045. For k=2 we find the triangular numbers A000217 and for k=4 A154286. - Peter Luschny, Jan 06 2009
From Peter Bala, Jun 10 2009: (Start)
The Swiss-Knife polynomials W_n(x) may be expressed in terms of the Bernoulli polynomials B(n,x) as
... W_n(x) = 4^(n+1)/(2*n+2)*[B(n+1,(x+3)/4) - B(n+1,(x+1)/4)].
The Swiss-Knife polynomials are, apart from a multiplying factor, examples of generalized Bernoulli polynomials.
Let X be the Dirichlet character modulus 4 defined by X(4*n+1) = 1, X(4*n+3) = -1 and X(2*n) = 0. The generalized Bernoulli polynomials B(X;n,x), n = 1,2,..., associated with the character X are defined by means of the generating function
... t*exp(x*t)*(exp(t)-exp(3*t))/(exp(4*t)-1) = sum {n = 1..inf} B(X;n,x)*t^n/n!.
The first few values are B(X;1,x) = -1/2, B(X;2,x) = -x, B(X,3,x) = -3/2*(x^2-1) and B(X;4,x) = -2*(x^3-3*x).
In general, W_n(x) = -2/(n+1)*B(X;n+1,x).
For the theory of generalized Bernoulli polynomials associated to a periodic arithmetical function see [Cohen, Section 9.4].
The generalized Bernoulli polynomials may be used to evaluate twisted sums of k-th powers. For the present case the result is
sum{n = 0..4*N-1} X(n)*n^k = 1^k - 3^k + 5^k - 7^k + ... - (4*N-1)^k
= [B(X;k+1,4*N) - B(X;k+1,0)]/(k+1) = [W_k(0) - W_k(4*N)]/2.
For the proof apply [Cohen, Corollary 9.4.17 with m = 4 and x = 0].
The generalized Bernoulli polynomials and the Swiss-Knife polynomials are also related to infinite sums of powers through their Fourier series - see the formula section below. For a table of the coefficients of generalized Bernoulli polynomials attached to a Dirichlet character modulus 8 see A151751.
(End)
The Swiss-Knife polynomials provide a general formula for alternating sums of powers similar to the formula which are provided by the Bernoulli polynomials for non-alternating sums of powers (see the Luschny link). Sequences covered by this formula include A001057, A062393, A062392, A011934, A144129, A077221, A137501, A046092. - Peter Luschny, Jul 12 2009
The greatest common divisor of the nonzero coefficients of the decapitated Swiss-Knife polynomials is exp(Lambda(n)), where Lambda(n) is the von Mangoldt function for odd primes, symbolically:
gcd(coeffs(SKP_{n}(x) - x^n)) = A155457(n) (n>1). - Peter Luschny, Dec 16 2009
Another version is at A119879. - Philippe Deléham, Oct 26 2013

Examples

			1
x
x^2  -1
x^3  -3x
x^4  -6x^2   +5
x^5 -10x^3  +25x
x^6 -15x^4  +75x^2  -61
x^7 -21x^5 +175x^3 -427x
		

References

  • H. Cohen, Number Theory - Volume II: Analytic and Modern Tools, Graduate Texts in Mathematics. Springer-Verlag. [From Peter Bala, Jun 10 2009]

Crossrefs

W_n(k), k=0,1,...
W_0: 1, 1, 1, 1, 1, 1, ........ A000012
W_1: 0, 1, 2, 3, 4, 5, ........ A001477
W_2: -1, 0, 3, 8, 15, 24, ........ A067998
W_3: 0, -2, 2, 18, 52, 110, ........ A121670
W_4: 5, 0, -3, 32, 165, 480, ........
W_n(k), n=0,1,...
k=0: 1, 0, -1, 0, 5, 0, -61, ... A122045
k=1: 1, 1, 0, -2, 0, 16, 0, ... A155585
k=2: 1, 2, 3, 2, -3, 2, 63, ... A119880
k=3: 1, 3, 8, 18, 32, 48, 128, ... A119881
k=4: 1, 4, 15, 52, 165, 484, ........ [Peter Luschny, Jul 07 2009]

Programs

  • Maple
    w := proc(n,x) local v,k,pow,chen; pow := (a,b) -> if a = 0 and b = 0 then 1 else a^b fi; chen := proc(m) if irem(m+1,4) = 0 then RETURN(0) fi; 1/((-1)^iquo(m+1,4) *2^iquo(m,2)) end; add(add((-1)^v*binomial(k,v)*pow(v+x+1,n)*chen(k),v=0..k), k=0..n) end:
    # Coefficients with zeros:
    seq(print(seq(coeff(i!*coeff(series(exp(x*t)*sech(t),t,16),t,i),x,i-n),n=0..i)), i=0..8);
    # Recursion
    W := proc(n,z) option remember; local k,p;
    if n = 0 then 1 else p := irem(n+1,2);
    z^n - p + add(`if`(irem(k,2)=1,0,
    W(k,0)*binomial(n,k)*(power(z,n-k)-p)),k=2..n-1) fi end:
    # Peter Luschny, edited and additions Jul 07 2009, May 13 2010, Oct 24 2011
  • Mathematica
    max = 9; rows = (Reverse[ CoefficientList[ #, x]] & ) /@ CoefficientList[ Series[ Exp[x*t]*Sech[t], {t, 0, max}], t]*Range[0, max]!; par[coefs_] := (p = Partition[ coefs, 2][[All, 1]]; If[ EvenQ[ Length[ coefs]], p, Append[ p, Last[ coefs]]]); Flatten[ par /@ rows] (* Jean-François Alcover, Oct 03 2011, after g.f. *)
    sk[n_, x_] := Sum[Binomial[n, k]*EulerE[k]*x^(n-k), {k, 0, n}]; Table[CoefficientList[sk[n, x], x] // Reverse // Select[#, # =!= 0 &] &, {n, 0, 13}] // Flatten (* Jean-François Alcover, May 21 2013 *)
    Flatten@Table[Binomial[n, 2k] EulerE[2k], {n, 0, 12}, {k, 0, n/2}](* Oliver Seipel, Jan 14 2025 *)
  • Sage
    def A046978(k):
        if k % 4 == 0:
            return 0
        return (-1)**(k // 4)
    def A153641_poly(n, x):
        return expand(add(2**(-(k // 2))*A046978(k+1)*add((-1)**v*binomial(k,v)*(v+x+1)**n for v in (0..k)) for k in (0..n)))
    for n in (0..7): print(A153641_poly(n, x))  # Peter Luschny, Oct 24 2011

Formula

W_n(x) = Sum_{k=0..n}{v=0..k} (-1)^v binomial(k,v)*c_k*(x+v+1)^n where c_k = frac((-1)^(floor(k/4))/2^(floor(k/2))) [4 not div k] (Iverson notation).
From Peter Bala, Jun 10 2009: (Start)
E.g.f.: 2*exp(x*t)*(exp(t)-exp(3*t))/(1-exp(4*t))= 1 + x*t + (x^2-1)*t^2/2! + (x^3-3*x)*t^3/3! + ....
W_n(x) = 1/(2*n+2)*Sum_{k=0..n+1} 1/(k+1)*Sum_{i=0..k} (-1)^i*binomial(k,i)*((x+4*i+3)^(n+1) - (x+4*i+1)^(n+1)).
Fourier series expansion for the generalized Bernoulli polynomials:
B(X;2*n,x) = (-1)^n*(2/Pi)^(2*n)*(2*n)! * {sin(Pi*x/2)/1^(2*n) - sin(3*Pi*x/2)/3^(2*n) + sin(5*Pi*x/2)/5^(2*n) - ...}, valid for 0 <= x <= 1 when n >= 1.
B(X;2*n+1,x) = (-1)^(n+1)*(2/Pi)^(2*n+1)*(2*n+1)! * {cos(Pi*x/2)/1^(2*n+1) - cos(3*Pi*x/2)/3^(2*n+1) + cos(5*Pi*x/2)/5^(2*n+1) - ...}, valid for 0 <= x <= 1 when n >= 1 and for 0 <= x < 1 when n = 0.
(End)
E.g.f.: exp(x*t) * sech(t). - Peter Luschny, Jul 07 2009
O.g.f. as a J-fraction: z/(1-x*z+z^2/(1-x*z+4*z^2/(1-x*z+9*z^2/(1-x*z+...)))) = z + x*z^2 + (x^2-1)*z^3 + (x^3-3*x)*z^4 + .... - Peter Bala, Mar 11 2012
Conjectural o.g.f.: Sum_{n >= 0} (1/2^((n-1)/2))*cos((n+1)*Pi/4)*( Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 - (k + x)*t) ) = 1 + x*t + (x^2 - 1)*t^2 + (x^3 - 3*x)*t^3 + ... (checked up to O(t^13)), which leads to W_n(x) = Sum_{k = 0..n} 1/2^((k - 1)/2)*cos((k + 1)*Pi/4)*( Sum_{j = 0..k} (-1)^j*binomial(k, j)*(j + x)^n ). - Peter Bala, Oct 03 2016

A046802 T(n, k) = Sum_{j=k..n} binomial(n, j)*E1(j, j-k), where E1 are the Eulerian numbers A173018. Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 15, 33, 15, 1, 1, 31, 131, 131, 31, 1, 1, 63, 473, 883, 473, 63, 1, 1, 127, 1611, 5111, 5111, 1611, 127, 1, 1, 255, 5281, 26799, 44929, 26799, 5281, 255, 1, 1, 511, 16867, 131275, 344551, 344551, 131275, 16867, 511, 1, 1, 1023, 52905
Offset: 0

Keywords

Comments

T(n,k) is the number of positroid cells of the totally nonnegative Grassmannian G+(k,n) (cf. Postnikov/Williams). It is the triangle of the h-vectors of the stellahedra. - Tom Copeland, Oct 10 2014
See A248727 for a simple transformation of the row polynomials of this entry that produces the umbral compositional inverses of the polynomials of A074909, related to the face polynomials of the simplices. - Tom Copeland, Jan 21 2015
From Tom Copeland, Jan 24 2015: (Start)
The reciprocal of this entry's e.g.f. is [t e^(-xt) - e^(-x)] / (t-1) = 1 - (1+t) x + (1+t+t^2) x^2/2! - (1+t+t^2+t^3) x^3/3! + ... = e^(q.(0;t)x), giving the base sequence (q.(0;t))^n = q_n(0;t) = (-1)^n [1-t^(n+1)] / (1-t) for the umbral compositional inverses (q.(0;t)+z)^n = q_n(z;t) of the Appell polynomials associated with this entry, p_n(z;t) below, i.e., q_n(p.(z;t)) = z^n = p_n(q.(z;t)), in umbral notation. The relations in A133314 then apply between the two sets of base polynomials. (Inserted missing index in a formula - Mar 03 2016.)
The associated o.g.f. for the umbral inverses is Og(x) = x / (1-x q.(0:t)) = x / [(1+x)(1+tx)] = x / [1+(1+t)x+tx^2]. Applying A134264 to h(x) = x / Og(x) = 1 + (1+t) x + t x^2 leads to an o.g.f. for the Narayana polynomials A001263 as the comp. inverse Oginv(x) = [1-(1+t)x-sqrt[1-2(1+t)x+((t-1)x)^2]] / (2xt). Note that Og(x) gives the signed h-polynomials of the simplices and that Oginv(x) gives the h-polynomials of the simplicial duals of the Stasheff polynomials, or type A associahedra. Contrast this with A248727 = A046802 * A007318, which has o.g.f.s related to the corresponding f-polynomials. (End)
The Appell polynomials p_n(x;t) in the formulas below specialize to the Swiss-knife polynomials of A119879 for t = -1, so the Springer numbers A001586 are given by 2^n p_n(1/2;-1). - Tom Copeland, Oct 14 2015
The row polynomials are the h-polynomials associated to the stellahedra, whose f-polynomials are the row polynomials of A248727. Cf. page 60 of the Buchstaber and Panov link. - Tom Copeland, Nov 08 2016
The row polynomials are the h-polynomials of the stellohedra, which enumerate partial permutations according to descents. Cf. Section 10.4 of the Postnikov-Reiner-Williams reference. - Lauren Williams, Jul 05 2022
From p. 60 of the Buchstaber and Panov link, S = P * C / T where S, P, C, and T are the bivariate e.g.f.s of the h vectors of the stellahedra, permutahedra, hypercubes, and (n-1)-simplices, respectively. - Tom Copeland, Jan 09 2017
The number of Le-diagrams of type (k, n) this means the diagram uses the bounding box size k x (n-k), equivalently the number of Grassmann necklaces of type (k, n) and also the number of decorated permutations with k anti-exceedances. - Thomas Scheuerle, Dec 29 2024

Examples

			The triangle T(n, k) begins:
n\k 0   1     2      3      4      5      6     7
0:  1
1:  1   1
2:  1   3     1
3:  1   7     7      1
4:  1  15    33     15      1
5:  1  31   131    131     31      1
6:  1  63   473    883    473     63      1
7:  1 127  1611   5111   5111   1611    127     1
... Reformatted. - _Wolfdieter Lang_, Feb 14 2015
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, Holland, 1974, page 245 [From Roger L. Bagula, Nov 21 2009]
  • D. Singh, The numbers L(m,n) and their relations with prepared Bernoulli and Eulerian numbers, Math. Student, 20 (1952), 66-70.

Programs

  • Maple
    T := (n, k) -> add(binomial(n, r)*combinat:-eulerian1(r, r-k), r = k .. n):
    for n from 0 to 8 do seq(T(n, k), k=0..n) od; # Peter Luschny, Jun 27 2018
  • Mathematica
    t[, 1] = 1; t[n, n_] = 1; t[n_, 2] = 2^(n-1)-1;
    t[n_, k_] = Sum[((i-k+1)^i*(k-i)^(n-i-1) - (i-k+2)^i*(k-i-1)^(n-i-1))*Binomial[n-1, i], {i, 0, k-1}];
    T[n_, k_] := t[n+1, k+1]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten
    (* Jean-François Alcover, Jan 22 2015, after Tom Copeland *)
    T[ n_, k_] := Coefficient[n! SeriesCoefficient[(1-x) Exp[t] / (1 - x Exp[(1-x) t]), {t, 0, n}] // Simplify, x, k];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* Michael Somos, Jan 22 2015 *)

Formula

E.g.f.: (y-1)*exp(x*y)/(y-exp((y-1)*x)). - Vladeta Jovovic, Sep 20 2003
p(t,x) = (1 - x)*exp(t)/(1 - x*exp(t*(1 - x))). - Roger L. Bagula, Nov 21 2009
With offset=0, T(n,0)=1 otherwise T(n,k) = sum_{i=0..k-1} C(n,i)((i-k)^i*(k-i+1)^(n-i) - (i-k+1)^i*(k-i)^(n-i)) (cf. Williams). - Tom Copeland, Oct 10 2014
With offset 0, T = A007318 * A123125. Second column is A000225; 3rd, appears to be A066810. - Tom Copeland, Jan 23 2015
A raising operator (with D = d/dx) associated with this entry's row polynomials is R = x + t + (1-t) / [1-t e^{(1-t)D}] = x + t + 1 + t D + (t+t^2) D^2/2! + (t+4t^2+t^3) D^3/3! + ... , containing the e.g.f. for the Eulerian polynomials of A123125. Then R^n 1 = (p.(0;t)+x)^n = p_n(x;t) are the Appell polynomials with this entry's row polynomials p_n(0;t) as the base sequence. Examples of this formalism are given in A028246 and A248727. - Tom Copeland, Jan 24 2015
With offset 0, T = A007318*(padded A090582)*(inverse of A097805) = A007318*(padded A090582)*(padded A130595) = A007318*A123125 = A007318*(padded A090582)*A007318*A097808*A130595, where padded matrices are of the form of padded A007318, which is A097805. Inverses of padded matrices are just the padded versions of inverses of the unpadded matrices. This relates the face vectors, or f-vectors, and h-vectors of the permutahedra / permutohedra to those of the stellahedra / stellohedra. - Tom Copeland, Nov 13 2016
Umbrally, the row polynomials (offset 0) are r_n(x) = (1 + q.(x))^n, where (q.(x))^k = q_k(x) are the row polynomials of A123125. - Tom Copeland, Nov 16 2016
From the previous umbral statement, OP(x,d/dy) y^n = (y + q.(x))^n, where OP(x,y) = exp[y * q.(x)] = (1-x)/(1-x*exp((1-x)y)), the e.g.f. of A123125, so OP(x,d/dy) y^n evaluated at y = 1 is r_n(x), the n-th row polynomial of this entry, with offset 0. - Tom Copeland, Jun 25 2018
Consolidating some formulas in this entry and A248727, in umbral notation for concision, with all offsets 0: Let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of this entry (A046802, the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of A248727 (the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020
From Peter Luschny, Apr 30 2021: (Start)
Sum_{k=0..n} (-1)^k*T(n, k) = A122045(n).
Sum_{k=0..n} 2^(n-k)*T(n,k) = A007047(n).
Sum_{k=0..n} T(n, n-k) = A000522(n).
Sum_{k=0..n} T(n-k, k) = Sum_{k=0..n} (n - k)^k = A026898(n-1) for n >= 1.
Sum_{k=0..n} k*T(n, k) = A036919(n) = floor(n*n!*e/2).
(End)

Extensions

More terms from Vladeta Jovovic, Sep 20 2003
First formula corrected by Wolfdieter Lang, Feb 14 2015
Offset set to 0 and edited by Peter Luschny, Apr 30 2021

A349264 Generalized Euler numbers, a(n) = n!*[x^n](sec(4*x)*(sin(4*x) + 1)).

Original entry on oeis.org

1, 4, 16, 128, 1280, 16384, 249856, 4456448, 90767360, 2080374784, 52975108096, 1483911200768, 45344872202240, 1501108249821184, 53515555843342336, 2044143848640217088, 83285910482761809920, 3605459138582973251584, 165262072909347030040576, 7995891855149741436305408
Offset: 0

Author

Peter Luschny, Nov 20 2021

Keywords

Examples

			Exponential generating functions of generalized Euler numbers in context:
egf1 = sec(1*x)*(sin(x) + 1).
   [A000111, A000364, A000182]
egf2 = sec(2*x)*(sin(x) + cos(x)).
   [A001586, A000281, A000464]
egf3 = sec(3*x)*(sin(2*x) + cos(x)).
   [A007289, A000436, A000191]
egf4 = sec(4*x)*(sin(4*x) + 1).
   [A349264, A000490, A000318]
egf5 = sec(5*x)*(sin(x) + sin(3*x) + cos(2*x) + cos(4*x)).
   [A349265, A000187, A000320]
egf6 = sec(6*x)*(sin(x) + sin(5*x) + cos(x) + cos(5*x)).
   [A001587, A000192, A000411]
egf7 = sec(7*x)*(-sin(2*x) + sin(4*x) + sin(6*x) + cos(x) + cos(3*x) - cos(5*x)).
   [A349266, A064068, A064072]
egf8 = sec(8*x)*2*(sin(4*x) + cos(4*x)).
   [A349267, A064069, A064073]
egf9 = sec(9*x)*(4*sin(3*x) + 2)*cos(3*x)^2.
   [A349268, A064070, A064074]
		

Programs

  • Maple
    sec(4*x)*(sin(4*x) + 1): series(%, x, 20): seq(n!*coeff(%, x, n), n = 0..19);
  • Mathematica
    m = 19; CoefficientList[Series[Sec[4*x] * (Sin[4*x] + 1), {x, 0, m}], x] * Range[0, m]! (* Amiram Eldar, Nov 20 2021 *)
  • PARI
    seq(n)={my(x='x + O('x^(n+1))); Vec(serlaplace((sin(4*x) + 1)/cos(4*x)))} \\ Andrew Howroyd, Nov 20 2021

A000464 Expansion of e.g.f. sin(x)/cos(2*x).

Original entry on oeis.org

1, 11, 361, 24611, 2873041, 512343611, 129570724921, 44110959165011, 19450718635716001, 10784052561125704811, 7342627959965776406281, 6023130568334172003579011, 5858598896811701995459355761, 6667317162352419006959182803611, 8776621742176931117228228227924441
Offset: 0

Keywords

Comments

From Peter Bala, Dec 22 2021: (Start)
Conjectures:
1) Taking the sequence (a(n))n>=1 modulo an integer k gives a purely periodic sequence with period dividing phi(k). For example, the sequence taken modulo 21 begins [11, 4, 20, 10, 17, 1, 11, 4, 20, 10, 17, 1, 11, 4, 20, 10, 17, 1, ...] with an apparent period of length 6, which divides phi(21) = 12.
2) For i >= 0, define a_i(n) = a(n+i). Then for each 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 the expansion of exp(Sum_{n >= 1} a_i(n)*x^n/n) has integer coefficients.
3) a(m*n) == a(m)^n (mod 2^k) for k = 2*v_2(m) + 4, where v_p(i) denotes the p-adic valuation of i.
4)(i) a(2*m*n) == a(n)^(2*m) (mod 2^k) for k = v_2(m) + 4
(ii) a((2*m+1)*n) == a(n)^(2*m+1) (mod 2^k) for k = v_2(m) + 4. (End)

References

  • H. Cohen, Number Theory - Volume II: Analytic and Modern Tools, Graduate Texts in Mathematics. Springer-Verlag.
  • J. W. L. Glaisher, "On the coefficients in the expansions of cos x/ cos 2x and sin x/ cos 2x", Quart. J. Pure and Applied Math., 45 (1914), 187-222.
  • I. J. Schwatt, Intro. to Operations with Series, Chelsea, p. 278.
  • 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).

Crossrefs

Row 2 of A235606.
Cf. A064073. Bisection of A000822, A001586.

Programs

  • Maple
    a := n -> (-1)^n*2^(6*n+4)*(Zeta(0, -2*n-1, 5/8)-Zeta(0, -2*n-1, 7/8)):
    seq(a(n), n=0..12); # Peter Luschny, Oct 15 2015
  • Mathematica
    With[{nn=30},Take[CoefficientList[Series[Sin[x]/Cos[2x],{x,0,nn}],x] Range[0,nn-1]!,{2,-1,2}]] (* Harvey P. Dale, Mar 23 2012 *)
    nmax = 15; km0 = 10; d[n_, km_] := Round[(2^(4n-1/2) (2n-1)! Sum[ JacobiSymbol[2, 2k+1]/(2k+1)^(2n), {k, 0, km}])/Pi^(2n)]; dd[km_] := dd[km] = Table[d[n, km], {n, 1, nmax}]; dd[km0]; dd[km = 2*km0]; While[dd[km] != dd[km/2, km = 2*km]]; A000464 = dd[km] (* Jean-François Alcover, Feb 08 2016 *)
  • PARI
    a(n)=if(n<0, 0, n+=n+1; n!*polcoeff(sin(x+x*O(x^n))/cos(2*x+x*O(x^n)),n)) /* Michael Somos, Feb 09 2006 */

Formula

E.g.f.: Sum_{k>=0} a(k)x^(2k+1)/(2k+1)! = sin(x)/cos(2x).
a(n) = (-1)^n*L(X,-2n+1) where L(X,z) is the Dirichlet L-function L(X,z) = Sum_{k>=0} X(k)/k^z and where X(k) is the Dirichlet character Legendre(k,2) which begins 1,0,-1,0,-1,0,1,0,1,0,-1,0,-1,0,1,0,1,0,-1,0.... - Benoit Cloitre, Mar 22 2009 [This Dirichlet character is A091337. - Jianing Song, Oct 22 2023]
From Peter Bala, Mar 24 2009: (Start)
Basic hypergeometric generating function:
2*exp(-t)*Sum_{n = 0..inf} (Product_{k = 1..n} (1-exp(-16*k*t))/Product_{k = 1..n+1} (1+exp(-(16*k-8)*t))) = 1 + 11*t + 361*t^2/2! + 24611*t^3/3! + .... For other sequences with generating functions of a similar type see A000364, A002105, A002439, A079144 and A158690.
a(n) = (-1)^(n+1)*L(-2*n-1), where L(s) is the Dirichlet L-function L(s) = 1 - 1/3^s - 1/5^s + 1/7^s + - - + ... [Andrews et al., Theorem 5]. (End)
From Peter Bala, Jun 18 2009: (Start)
a(n) = (-1)^n*B_(2*n+2)(X)/(2*n+2), where B_n(X) denotes the X-Bernoulli number with X a Dirichlet character modulus 8 given by X(8*n+1) = X(8*n+7) = 1, X(8*n+3) = X(8*n+5) = -1 and X(2*n) = 0. See A161722 for the values of B_n(X).
For the theory and properties of the generalized Bernoulli numbers B_n(X) and the associated generalized Bernoulli polynomials B_n(X,x) see [Cohen, Section 9.4].
The present sequence also occurs in the evaluation of the finite sum of powers Sum_{i = 0..m-1} {(8*i+1)^n - (8*i+3)^n - (8*i+5)^n + (8*i+7)^n}, n = 1,2,... - see A151751 for details. (End)
G.f. 1/G(0) where G(k) = 1 + x - x*(4*k+3)*(4*k+4)/(1 - (4*k+4)*(4*k+5)*x/G(k+1)); (continued fraction, 2-step). - Sergei N. Gladkovskii, Aug 11 2012
G.f.: 1/E(0) where E(k) = 1 - 11*x - 32*x*k*(k+1) - 16*x^2*(k+1)^2*(4*k+3)*(4*k+5)/E(k+1) (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 17 2012
a(n) ~ (2*n+1)! * 2^(4*n+7/2) / Pi^(2*n+2). - Vaclav Kotesovec, May 03 2014
a(n) = (-1)^n*2^(6*n+4)*(Zeta(-2*n-1,5/8)-Zeta(-2*n-1,7/8)). - Peter Luschny, Oct 15 2015
From Peter Bala, May 11 2017: (Start)
G.f. A(x) = 1 + 11*x + 361*x^2 + ... = 1/(1 + x - 12*x/(1 - 20*x/(1 + x - 56*x/(1 - 72*x/(1 + x - ... - 4*n*(4*n - 1)*x/(1 - 4*n*(4*n + 1)*x/((1 + x) - ...))))))).
A(x) = 1/(1 + 9*x - 20*x/(1 - 12*x/(1 + 9*x - 72*x/(1 - 56*x/(1 + 9*x - ... - 4*n*(4*n + 1)*x/(1 - 4*n*(4*n - 1)*x/(1 + 9*x - ...))))))).
It follows that the first binomial transform of A(x) and the ninth binomial transform of A(x) have continued fractions of Stieltjes-type (S-fractions). (End)
a(n) = (-1)^(n+1)*4^(2*n+1)*E(2*n+1,1/4), where E(n,x) is the n-th Euler polynomial. Cf. A002439. - Peter Bala, Aug 13 2017
From Peter Bala, Dec 04 2021: (Start)
F(x) = exp(x)*(exp(2*x) - 1)/(exp(4*x) + 1) = x - 11*x^3/3! + 361x^5/5! - 24611*x^7/7! + ... is the e.g.f. for the sequence [1, 0, -11, 0, 361, 0, -24611, 0, ...], a signed and aerated version of this sequence.
The binomial transform exp(x)*F(x) = x + 2*x^2/2! - 8*x^3/3! - 40*x^4/4! + + - - is an e.g.f. for a signed version of A000828 (omitting the initial term). (End)
From Peter Bala, Dec 22 2021: (Start)
a(1) = 1, a(n) = (-1)^(n-1) - Sum_{k = 1..n} (-4)^k*C(2*n-1,2*k)*a(n-k).
a(n) == 1 (mod 10); a(5*n+1) == 0 mod(11);
a(n) == - 23^(n+1) (mod 108); a(n) == (7^2)*59^n (mod 144);
a(n) == 11^n (mod 240); a(n) == (11^2)*131^n (mod 360). (End)

Extensions

Better description, new reference, Aug 15 1995

A080635 Number of permutations on n letters without double falls and without initial falls.

Original entry on oeis.org

1, 1, 1, 3, 9, 39, 189, 1107, 7281, 54351, 448821, 4085883, 40533129, 435847959, 5045745069, 62594829027, 828229153761, 11644113200031, 173331882039141, 2723549731505163, 45047085512477049, 782326996336904679, 14233537708408467549, 270733989894887810547
Offset: 0

Author

Emanuele Munarini, Feb 28 2003

Keywords

Comments

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).
exp(x*(1-y+y^2)*D_y)*f(y)|_{y=0} = f(1-E(-x)) for any function f with a Taylor series. D_y means differentiation with respect to y and E(x) is the e.g.f. given below. For a proof of exp(x*g(y)*D_y)*f(y) = f(F^{-1}(x+F(y))) with the compositional inverse F^{-1} of F(y)=int(1/g(y),y) with F(0)=0 see, e.g., the Datolli et al. reference.
Number of increasing ordered trees on vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <= 2. - David Callan, Mar 30 2007
Number of increasing colored 1-2 trees of order n with choice of two colors for the right branches of the vertices of outdegree 2. - Wenjin Woan, May 21 2011

Examples

			E.g.f. = 1 + x + (1/2)*x^2 + (1/2)*x^3 + (3/8)*x^4 + (13/40)*x^5 + (21/80)*x^6 + ...
G.f. = 1 + x + x^2 + 3*x^3 + 9*x^4 + 39*x^5 + 189*x^6 + 1107*x^7 + ...
For n = 3: 123, 132, 231. For n = 4: 1234, 1243, 1324, 1342, 1423, 2314, 2341, 2413, 3412.
a(4)=9. The 9 plane (ordered) increasing unary-binary trees are
...................................................................
..4................................................................
..|................................................................
..3..........4...4...............4...4...............3...3.........
..|........./.....\............./.....\............./.....\........
..2....2...3.......3...2...3...2.......2...3...4...2.......2...4...
..|.....\./.........\./.....\./.........\./.....\./.........\./....
..1......1...........1.......1...........1.......1...........1.....
...................................................................
..3...4...4...3....................................................
...\./.....\./.....................................................
....2.......2......................................................
....|.......|......................................................
....1.......1......................................................
...................................................................
		

Crossrefs

Programs

  • Maple
    a:= proc(n) if n < 2 then 1 else n! * sum((sqrt(3)/(2*Pi*(k+1/3)))^(n+1), k=-infinity..infinity) fi end: seq(a(n), n=0..30); # Richard Ehrenborg, Dec 09 2013
    a := proc(n) option remember; local k; if n < 3 then 1 else
    add(binomial(n-1, k)*a(k)*a(n-k-1), k = 0..n-2) fi end:
    seq(a(n), n = 0..23); # Peter Luschny, May 24 2024
  • Mathematica
    Table[n!, {n, 0, 40}]*CoefficientList[Series[ (1 + 1/Sqrt[3] Tan[Sqrt[3]/2 x])/(1 - 1/Sqrt[3] Tan[Sqrt[3]/2 x]), {x, 0, 40}], x]
    a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ 1/2 + Sqrt[3]/2 Tan[ Pi/6 + Sqrt[3] x/2], {x, 0, n}]]; (* Michael Somos, May 22 2011 *)
    Join[{1}, FullSimplify[Table[3^((n+1)/2) * n! * (Zeta[n+1, 1/3] - (-1)^n*Zeta[n+1, 2/3]) / (2*Pi)^(n+1), {n, 1, 20}]]] (* Vaclav Kotesovec, Aug 06 2021 *)
  • Maxima
    a(n):=if n=0 then 1 else sum((-3)^((n-k)/2)*((-1)^(n-k)+1)*sum(binomial(j+k-1,j)*(j+k)!*2^(-j-k)*(-1)^(j)*stirling2(n,j+k),j,0,n-k),k,1,n); /* Vladimir Kruchinin, Feb 13 2019 */
  • PARI
    {a(n) = my(A); if( n<1, n==0, A = O(x); for( k=1, n, A = intformal( 1 + A + A^2)); n! * polcoeff( A, n))}; /* Michael Somos, Oct 04 2003 */
    
  • PARI
    {a(n) = n! * polcoeff( exp( serreverse( intformal( 1/(2*cosh(x +x*O(x^n)) - 1) ) )), n)}
    for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Feb 22 2016
    
  • Sage
    @CachedFunction
    def c(n,k) :
        if n==k: return 1
        if k<1 or k>n: return 0
        return ((n-k)//2+1)*c(n-1,k-1)+2*k*c(n-1,k+1)
    def A080635(n):
        return add(c(n,k) for k in (0..n))
    [A080635(n) for n in (0..23)] # Peter Luschny, Jun 10 2014
    

Formula

E.g.f.: (1 + 1/sqrt(3) * tan(sqrt(3)/2 * x)) / (1 - 1/sqrt(3) * tan( sqrt(3)/2 * x)).
Recurrence: a(n+1) = (Sum_{k=0..n} binomial(n, k) * a(k) * a(n-k)) - a(n) + 0^n.
E.g.f.: A(x) satisfies A' = 1 - A + A^2. - Michael Somos, Oct 04 2003
E.g.f.: E(x) = (3*cos((1/2)*3^(1/2)*x) + (3^(1/2))*sin((1/2)*3^(1/2)*x))/(3*cos((1/2)*3^(1/2)*x) - (3^(1/2))* sin((1/2)*3^(1/2)*x)). See the Michael Somos comment. - Wolfdieter Lang, Sep 12 2005
O.g.f.: A(x) = 1+x/(1-x-2*x^2/(1-2*x-2*3*x^2/(1-3*x-3*4*x^2/(1-... -n*x-n*(n+1)*x^2/(1- ...))))) (continued fraction). - Paul D. Hanna, Jan 17 2006
From Peter Bala: (Start)
An alternative form of the e.g.f. for this sequence taken from [Bergeron et al.] is
(1)... (sqrt(3)/2)*tan((sqrt(3)/2)*x+Pi/6) [with constant term 1/2].
By comparing the egf for this sequence with the egf for the Eulerian numbers A008292 we can show that
(2)... a(n) = A(n,w)/(1+w)^(n-1) for n >= 1,
where w = exp(2*Pi*i/3) 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,
(3)... a(n) = (-i*sqrt(3))^(n-1)*Sum_{k=1..n} k!*Stirling2(n,k)*(-1/2 + sqrt(3)*i/6)^(k-1) for n >= 1, and
(4)... a(n) = (-i*sqrt(3))^(n-1)*Sum_{k=1..n} (-1/2 + sqrt(3)*i/6)^(k-1)* Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*j^n for n >= 1.
This explicit formula for a(n) may be used to obtain various congruence results. For example,
(5a)... a(p) == 1 (mod p) for prime p = 6*n+1,
(5b)... a(p) == -1 (mod p) for prime p = 6*n+5.
For the corresponding results for the case of non-plane unary-binary trees see A000111. For type B results see A001586. For a related sequence of polynomials see A185415. See also A185416 for a recursive method to compute this sequence. For forests of plane increasing unary binary trees see A185422 and A185423. (End)
O.g.f.: A(x) = x - (1/2)*x^2 + (1/2)*x^3 - (3/8)*x^4 + (13/40)*x^5 - (21/80)*x^6 + (123/560)*x^7 - (809/4480)*x^8 + (671/4480)*x^9 - (5541/44800)*x^10 + .... - Vladimir Kruchinin, Jan 18 2011
Let f(x) = 1+x+x^2. Then a(n+1) = (f(x)*d/dx)^n f(x) evaluated at x = 0. - Peter Bala, Oct 06 2011
From Sergei N. Gladkovskii, May 06 2013 - Dec 24 2013: (Start)
Continued fractions:
G.f.: 1 + 1/Q(0), where Q(k) = 1/(x*(k+1)) - 1 - 1/Q(k+1).
E.g.f.: 1 + 2*x/(W(0)-x), where W(k) = 4*k + 2 - 3*x^2/W(k+1).
G.f.: 1 + x/Q(0), m=1, where Q(k) = 1 - m*x*(2*k+1) - m*x^2*(2*k+1)*(2*k+2)/( 1 - m*x*(2*k+2) - m*x^2*(2*k+2)*(2*k+3)/Q(k+1) ).
G.f.: 1 + x/Q(0), where Q(k) = 1 - x*(k+1) - x^2*(k+1)*(k+2)/Q(k+1).
G.f.: 1 + T(0)*x/(1-x), where T(k) = 1 - x^2*(k+1)*(k+2)/( x^2*(k+1)*(k+2) - (1-x*(k+1))*(1-x*(k+2))/T(k+1) ).
G.f.: 1 + x/(G(0)-x), where G(k) = 1 + x*(k+1) - x*(k+1)/(1 - x*(k+2)/G(k+1) ). (End)
a(n) ~ 3^(3*(n+1)/2) * n^(n+1/2) / (exp(n)*(2*Pi)^(n+1/2)). - Vaclav Kotesovec, Oct 05 2013
a(n) = n! * Sum_{k=-oo..oo} (sqrt(3)/(2*Pi*(k+1/3)))^(n+1) for n >= 1. - Richard Ehrenborg, Dec 09 2013
From Peter Bala, Sep 11 2015: (Start)
The e.g.f. A(x) = (sqrt(3)/2)*tan((sqrt(3)/2)*x + Pi/6) satisfies the differential equation A"(x) = 2*A(x)*A'(x) with A(0) = 1/2 and A'(0) = 1, leading to the recurrence a(0) = 1/2, a(1) = 1, else a(n) = 2*Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the sequence [1/2, 1, 1, 3, 9, 39, 189, 1107, ...].
Note, the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence n! and with a(0) = 0 and a(1) = 1 produces A000182. Cf. A002105, A234797. (End)
E.g.f.: exp( Series_Reversion( Integral 1/(2*cosh(x) - 1) dx ) ). - Paul D. Hanna, Feb 22 2016
a(n) = Sum_{k=1..n} (-3)^((n-k)/2)*((-1)^(n-k)+1)*Sum_{j=0..n-k} C(j+k-1,j)*(j+k)!*2^(-j-k)*(-1)^j*Stirling2(n,j+k),n>0, a(0)=1. - Vladimir Kruchinin, Feb 13 2019
For n > 0, a(n) = 3^((n+1)/2) * n! * (zeta(n+1, 1/3) - (-1)^n*zeta(n+1, 2/3)) / (2*Pi)^(n+1). - Vaclav Kotesovec, Aug 06 2021

Extensions

Several typos corrected by Olivier Gérard, Mar 26 2011
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