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.

Showing 1-7 of 7 results.

A125054 Central terms of triangle A125053.

Original entry on oeis.org

1, 3, 21, 327, 9129, 396363, 24615741, 2068052367, 225742096209, 31048132997523, 5252064083753061, 1071525520294178007, 259439870666594250489, 73542221109962636293083, 24125551094579137082039181, 9068240688454120376775401247, 3871645204706420218816959159969
Offset: 0

Views

Author

Paul D. Hanna, Nov 21 2006

Keywords

Comments

Triangle A125053 is a variant of triangle A008301 (enumeration of binary trees) such that the leftmost column is the secant numbers (A000364).
Right edge of triangle A210108.
Apparently all terms (except the initial 1) have 3-valuation 1. - F. Chapoton, Aug 02 2021

Crossrefs

Programs

  • Mathematica
    b[n_]:=n!*SeriesCoefficient[Tan[x],{x,0,n}]; Table[Sum[Binomial[n,k]*b[2*k+1],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, May 30 2015 *)

Formula

Binomial transform of A000182 (e.g.f. tan(x)).
a(n) = Sum_{k=0..n} A130847(n,k)*2^k. - Philippe Deléham, Jul 22 2007
G.f.: 1/(1-sqrt(x))/Q(0), where Q(k)= 1 + sqrt(x) - x*(2*k+1)*(2*k+2)/(1 - sqrt(x) - x*(2*k+2)*(2*k+3)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 27 2013
G.f.: Q(0)/(1-3*x), where Q(k) = 1 - 4*x^2*(2*k+1)*(2*k+3)*(k+1)^2/( 4*x^2*(2*k+1)*(2*k+3)*(k+1)^2 - (1 - 8*x*k^2 - 8*x*k -3*x)*(1 - 8*x*k^2 - 24*x*k -19*x)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 23 2013
G.f.: Q(0)/(1-1*x), where Q(k) = 1 - (2*k+1)*(2*k+2)*x/(x*(2*k+1)*(2*k+2) - (1-x)/(1 - (2*k+2)*(2*k+3)*x/(x*(2*k+2)*(2*k+3) - (1-x)/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2013
a(n) ~ 2^(4*n+5) * n^(2*n+3/2) / (exp(2*n) * Pi^(2*n+3/2)). - Vaclav Kotesovec, May 30 2015
From Peter Bala, May 11 2017: (Start)
O.g.f. as an S-fraction: A(x) = 1/(1 - 3*x/(1 - 4*x/(1 - 15*x/(1 - 16*x/(1 - 35*x/(1 - 36*x/(1 - ...))))))), where the unsigned coefficients in the partial numerators [3, 4, 15, 16, 35, 36, ...] come in pairs of the form 4*n^2 - 1, 4*n^2 for n = 1,2,....
A(x) = 1/(1 + 3*x - 6*x/(1 - 2*x/(1 + 3*x - 20*x/(1 - 12*x/(1 + 3*x - 42*x/(1 - 30*x/(1 + 3*x - ...))))))), , where the unsigned coefficients in the partial numerators [6, 2, 20, 12, 42, 30, ...] are obtained from the sequence [2, 6, 12, 20, ..., n*(n + 1), ...] by swapping adjacent terms. (End)

A210111 Left half of triangle A125053.

Original entry on oeis.org

1, 1, 3, 5, 15, 21, 61, 183, 285, 327, 1385, 4155, 6681, 8475, 9129, 50521, 151563, 247065, 325947, 378105, 396363, 2702765, 8108295, 13311741, 17908935, 21517869, 23823015, 24615741, 199360981, 598082943, 985993845, 1341471567, 1643702325, 1874297343
Offset: 0

Views

Author

Reinhard Zumkeller, Mar 17 2012

Keywords

Crossrefs

Cf. A000364 (left edge), A125054 (right edge); A210108.

Programs

  • Haskell
    a210111 n k = a210111_tabl !! n !! k
    a210111_row n = a210111_tabl !! n
    a210111_tabl = zipWith take [1..] a125053_tabf

Formula

T(n,k) = A125053(n,k), 0 <= k <= n.

A125055 Diagonal of symmetric triangle A125053 located immediately below the central terms (A125054).

Original entry on oeis.org

1, 15, 285, 8475, 378105, 23823015, 2018820885, 221605991475, 30596648805105, 5189967817758015, 1061021392126671885, 257296819626005894475, 73023341368629447792105, 23978466652359211809453015
Offset: 0

Views

Author

Paul D. Hanna, Nov 21 2006

Keywords

Comments

Triangle A125053 is a variant of triangle A008301 (enumeration of binary trees) such that the leftmost column gives the secant numbers (A000364).

Crossrefs

Formula

a(n) = A125054(n+1) - 2*A125054(n). - Philippe Deléham, Jul 22 2007

A000364 Euler (or secant or "Zig") numbers: e.g.f. (even powers only) sec(x) = 1/cos(x).

Original entry on oeis.org

1, 1, 5, 61, 1385, 50521, 2702765, 199360981, 19391512145, 2404879675441, 370371188237525, 69348874393137901, 15514534163557086905, 4087072509293123892361, 1252259641403629865468285, 441543893249023104553682821, 177519391579539289436664789665
Offset: 0

Views

Author

Keywords

Comments

Inverse Gudermannian gd^(-1)(x) = log(sec(x) + tan(x)) = log(tan(Pi/4 + x/2)) = arctanh(sin(x)) = 2 * arctanh(tan(x/2)) = 2 * arctanh(csc(x) - cot(x)). - Michael Somos, Mar 19 2011
a(n) is the number of downup permutations of [2n]. Example: a(2)=5 counts 4231, 4132, 3241, 3142, 2143. - David Callan, Nov 21 2011
a(n) is the number of increasing full binary trees on vertices {0,1,2,...,2n} for which the leftmost leaf is labeled 2n. - David Callan, Nov 21 2011
a(n) is the number of unordered increasing trees of size 2n+1 with only even degrees allowed and degree-weight generating function given by cosh(t). - Markus Kuba, Sep 13 2014
a(n) is the number of standard Young tableaux of skew shape (n+1,n,n-1,...,3,2)/(n-1,n-2,...2,1). - Ran Pan, Apr 10 2015
Since cos(z) has a root at z = Pi/2 and no other root in C with a smaller |z|, the radius of convergence of the e.g.f. (intended complex-valued) is Pi/2 = A019669 (see also A028296). - Stanislav Sykora, Oct 07 2016
All terms are odd. - Alois P. Heinz, Jul 22 2018
The sequence starting with a(1) is periodic modulo any odd prime p. The minimal period is (p-1)/2 if p == 1 mod 4 and p-1 if p == 3 mod 4 [Knuth & Buckholtz, 1967, Theorem 2]. - Allen Stenger, Aug 03 2020
Conjecture: 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 [1, 5, 19, 20, 16, 2, 1, 5, 19, 20, 16, 2, 1, 5, 19, 20, 16, 2, 1, 5, 19, ...] with an apparent period of 6 = phi(21)/2. - Peter Bala, May 08 2023

Examples

			G.f. = 1 + x + 5*x^2 + 61*x^3 + 1385*x^4 + 50521*x^5 + 2702765*x^6 + 199360981*x^7 + ...
sec(x) = 1 + 1/2*x^2 + 5/24*x^4 + 61/720*x^6 + ...
From _Gary W. Adamson_, Jul 18 2011: (Start)
The first few rows of matrix M are:
   1,  1,  0,  0,  0, ...
   4,  4,  4,  0,  0, ...
   9,  9,  9,  9,  0, ...
  16, 16, 16, 16, 16, ... (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 810; gives a version with signs: E_{2n} = (-1)^n*a(n) (this is A028296).
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
  • J. M. Borwein and D. M. Bailey, Mathematics by Experiment, Peters, Boston, 2004; p. 49
  • J. M. Borwein, D. H. Bailey, and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 141.
  • Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 420.
  • G. Chrystal, Algebra, Vol. II, p. 342.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 49.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 110.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 69.
  • L. Euler, Inst. Calc. Diff., Section 224.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 444.
  • L. Seidel, Über eine einfache Entstehungsweise der Bernoulli'schen Zahlen und einiger verwandten Reihen, Sitzungsberichte der mathematisch-physikalischen Classe der königlich bayerischen Akademie der Wissenschaften zu München, volume 7 (1877), 157-187.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapters 5 and 33, pages 41, 314.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 269.

Crossrefs

Essentially same as A028296 and A122045.
First column of triangle A060074.
Two main diagonals of triangle A060058 (as iterated sums of squares).
Absolute values of row sums of A160485. - Johannes W. Meijer, Jul 06 2009
Left edge of triangle A210108, see also A125053, A076552. Cf. A255881.
Bisection (even part) of A317139.
The sequences [(-k^2)^n*Euler(2*n, 1/k), n = 0, 1, ...] are: A000007 (k=1), A000364 (k=2), |A210657| (k=3), A000281 (k=4), A272158 (k=5), A002438 (k=6), A273031 (k=7).

Programs

  • Maple
    series(sec(x),x,40): SERIESTOSERIESMULT(%): subs(x=sqrt(y),%): seriestolist(%);
    # end of program
    A000364_list := proc(n) local S,k,j; S[0] := 1;
    for k from 1 to n do S[k] := k*S[k-1] od;
    for k from  1 to n do
        for j from k to n do
            S[j] := (j-k)*S[j-1]+(j-k+1)*S[j] od od;
    seq(S[j], j=1..n)  end:
    A000364_list(16);  # Peter Luschny, Apr 02 2012
    A000364 := proc(n)
        abs(euler(2*n)) ;
    end proc: # R. J. Mathar, Mar 14 2013
  • Mathematica
    Take[ Range[0, 32]! * CoefficientList[ Series[ Sec[x], {x, 0, 32}], x], {1, 32, 2}] (* Robert G. Wilson v, Apr 23 2006 *)
    Table[Abs[EulerE[2n]], {n, 0, 30}] (* Ray Chandler, Mar 20 2007 *)
    a[ n_] := If[ n < 0, 0, With[{m = 2 n}, m! SeriesCoefficient[ Sec[ x], {x, 0, m}]]]; (* Michael Somos, Nov 22 2013 *)
    a[ n_] := If[ n < 0, 0, With[{m = 2 n + 1}, m! SeriesCoefficient[ InverseGudermannian[ x], {x, 0, m}]]]; (* Michael Somos, Nov 22 2013 *)
    a[n_] := Sum[Sum[Binomial[k, m] (-1)^(n+k)/(2^(m-1)) Sum[Binomial[m, j]* (2j-m)^(2n), {j, 0, m/2}] (-1)^(k-m), {m, 0, k}], {k, 1, 2n}]; Table[ a[n], {n, 0, 16}] (* Jean-François Alcover, Jun 26 2019, after Vladimir Kruchinin *)
    a[0] := 1; a[n_] := a[n] = -Sum[a[n - k]/(2 k)!, {k, 1, n}]; Map[(-1)^# (2 #)! a[#] &, Range[0, 16]] (* Oliver Seipel, May 18 2024 *)
  • Maxima
    a(n):=sum(sum(binomial(k,m)*(-1)^(n+k)/(2^(m-1))*sum(binomial(m,j)*(2*j-m)^(2*n),j,0,m/2)*(-1)^(k-m),m,0,k),k,1,2*n); /* Vladimir Kruchinin, Aug 05 2010 */
    
  • Maxima
    a[n]:=if n=0 then 1 else sum(sum((i-k)^(2*n)*binomial(2*k, i)*(-1)^(i+k+n), i, 0, k-1)/ (2^(k-1)), k, 1, 2*n); makelist(a[n], n, 0, 16); /* Vladimir Kruchinin, Oct 05 2012 */
    
  • PARI
    {a(n)=local(CF=1+x*O(x^n));if(n<0,return(0), for(k=1,n,CF=1/(1-(n-k+1)^2*x*CF));return(Vec(CF)[n+1]))} \\ Paul D. Hanna Oct 07 2005
    
  • PARI
    {a(n) = if( n<0, 0, (2*n)! * polcoeff( 1 / cos(x + O(x^(2*n + 1))), 2*n))}; /* Michael Somos, Jun 18 2002 */
    
  • PARI
    {a(n) = my(A); if( n<0, 0, n = 2*n+1 ; A = x * O(x^n); n! * polcoeff( log(1 / cos(x + A) + tan(x + A)), n))}; /* Michael Somos, Aug 15 2007 */
    
  • PARI
    {a(n)=polcoeff(sum(m=0, n, (2*m)!/2^m * x^m/prod(k=1, m, 1+k^2*x+x*O(x^n))), n)} \\ Paul D. Hanna, Sep 20 2012
    
  • PARI
    list(n)=my(v=Vec(1/cos(x+O(x^(2*n+1)))));vector(n,i,v[2*i-1]*(2*i-2)!) \\ Charles R Greathouse IV, Oct 16 2012
    
  • PARI
    a(n)=subst(bernpol(2*n+1),'x,1/4)*4^(2*n+1)*(-1)^(n+1)/(2*n+1) \\ Charles R Greathouse IV, Dec 10 2014
    
  • PARI
    a(n)=abs(eulerfrac(2*n)) \\ Charles R Greathouse IV, Mar 23 2022
    
  • PARI
    \\ Based on an algorithm of Peter Bala, cf. link in A110501.
    upto(n) = my(v1, v2, v3); v1 = vector(n+1, i, 0); v1[1] = 1; v2 = vector(n, i, i^2); v3 = v1; for(i=2, n+1, for(j=2, i-1, v1[j] += v2[i-j+1]*v1[j-1]); v1[i] = v1[i-1]; v3[i] = v1[i]); v3 \\ Mikhail Kurkov, Aug 30 2025
    
  • Python
    from functools import lru_cache
    from math import comb
    @lru_cache(maxsize=None)
    def A000364(n): return 1 if n == 0 else (1 if n % 2 else -1)*sum((-1 if i % 2 else 1)*A000364(i)*comb(2*n,2*i) for i in range(n)) # Chai Wah Wu, Jan 14 2022
    
  • Python
    # after Mikhail Kurkov, based on an algorithm of Peter Bala, cf. link in A110501.
    def euler_list(len: int) -> list[int]:
        if len == 0: return []
        v1 = [1] + [0] * (len - 1)
        v2 = [i**2 for i in range(1, len + 1)]
        result = [0] * len
        result[0] = 1
        for i in range(1, len):
            for j in range(1, i):
                v1[j] += v2[i - j] * v1[j - 1]
            v1[i] = v1[i - 1]
            result[i] = v1[i]
        return result
    print(euler_list(1000))  # Peter Luschny, Aug 30 2025
  • Sage
    # Algorithm of L. Seidel (1877)
    # n -> [a(0), a(1), ..., a(n-1)] for n > 0.
    def A000364_list(len) :
        R = []; A = {-1:0, 0:1}; k = 0; e = 1
        for i in (0..2*len-1) :
            Am = 0; A[k + e] = 0; e = -e
            for j in (0..i) : Am += A[k]; A[k] = Am; k += e
            if e < 0 : R.append(A[-i//2])
        return R
    A000364_list(17) # Peter Luschny, Mar 31 2012
    

Formula

E.g.f.: Sum_{n >= 0} a(n) * x^(2*n) / (2*n)! = sec(x). - Michael Somos, Aug 15 2007
E.g.f.: Sum_{n >= 0} a(n) * x^(2*n+1) / (2*n+1)! = gd^(-1)(x). - Michael Somos, Aug 15 2007
E.g.f.: Sum_{n >= 0} a(n)*x^(2*n+1)/(2*n+1)! = 2*arctanh(cosec(x)-cotan(x)). - Ralf Stephan, Dec 16 2004
Pi/4 - [Sum_{k=0..n-1} (-1)^k/(2*k+1)] ~ (1/2)*[Sum_{k>=0} (-1)^k*E(k)/(2*n)^(2k+1)] for positive even n. [Borwein, Borwein, and Dilcher]
Also, for positive odd n, log(2) - Sum_{k = 1..(n-1)/2} (-1)^(k-1)/k ~ (-1)^((n-1)/2) * Sum_{k >= 0} (-1)^k*E(k)/n^(2*k+1), where E(k) is the k-th Euler number, by Borwein, Borwein, and Dilcher, Lemma 2 with f(x) := 1/(x + 1/2), h := 1/2 and then replace x with (n-1)/2. - Peter Bala, Oct 29 2016
Let M_n be the n X n matrix M_n(i, j) = binomial(2*i, 2*(j-1)) = A086645(i, j-1); then for n>0, a(n) = det(M_n); example: det([1, 1, 0, 0; 1, 6, 1, 0; 1, 15, 15, 1; 1, 28, 70, 28 ]) = 1385. - Philippe Deléham, Sep 04 2005
This sequence is also (-1)^n*EulerE(2*n) or abs(EulerE(2*n)). - Paul Abbott (paul(AT)physics.uwa.edu.au), Apr 14 2006
a(n) = 2^n * E_n(1/2), where E_n(x) is an Euler polynomial.
a(k) = a(j) (mod 2^n) if and only if k == j (mod 2^n) (k and j are even). [Stern; see also Wagstaff and Sun]
E_k(3^(k+1)+1)/4 = (3^k/2)*Sum_{j=0..2^n-1} (-1)^(j-1)*(2j+1)^k*[(3j+1)/2^n] (mod 2^n) where k is even and [x] is the greatest integer function. [Sun]
a(n) ~ 2^(2*n+2)*(2*n)!/Pi^(2*n+1) as n -> infinity. [corrected by Vaclav Kotesovec, Jul 10 2021]
a(n) = Sum_{k=0..n} A094665(n, k)*2^(n-k). - Philippe Deléham, Jun 10 2004
Recurrence: a(n) = -(-1)^n*Sum_{i=0..n-1} (-1)^i*a(i)*binomial(2*n, 2*i). - Ralf Stephan, Feb 24 2005
O.g.f.: 1/(1-x/(1-4*x/(1-9*x/(1-16*x/(...-n^2*x/(1-...)))))) (continued fraction due to T. J. Stieltjes). - Paul D. Hanna, Oct 07 2005
a(n) = (Integral_{t=0..Pi} log(tan(t/2)^2)^(2n)dt)/Pi^(2n+1). - Logan Kleinwaks (kleinwaks(AT)alumni.princeton.edu), Mar 15 2007
From Peter Bala, Mar 24 2009: (Start)
Basic hypergeometric generating function: 2*exp(-t)*Sum {n >= 0} Product_{k = 1..n} (1-exp(-(4*k-2)*t))*exp(-2*n*t)/Product_{k = 1..n+1} (1+exp(-(4*k-2)*t)) = 1 + t + 5*t^2/2! + 61*t^3/3! + .... For other sequences with generating functions of a similar type see A000464, A002105, A002439, A079144 and A158690.
a(n) = 2*(-1)^n*L(-2*n), where L(s) is the Dirichlet L-function L(s) = 1 - 1/3^s + 1/5^s - + .... (End)
Sum_{n>=0} a(n)*z^(2*n)/(4*n)!! = Beta(1/2-z/(2*Pi),1/2+z/(2*Pi))/Beta(1/2,1/2) with Beta(z,w) the Beta function. - Johannes W. Meijer, Jul 06 2009
a(n) = Sum_(Sum_(binomial(k,m)*(-1)^(n+k)/(2^(m-1))*Sum_(binomial(m,j)*(2*j-m)^(2*n),j,0,m/2)*(-1)^(k-m),m,0,k),k,1,2*n), n>0. - Vladimir Kruchinin, Aug 05 2010
If n is prime, then a(n)==1 (mod 2*n). - Vladimir Shevelev, Sep 04 2010
From Peter Bala, Jan 21 2011: (Start)
(1)... a(n) = (-1/4)^n*B(2*n,-1),
where {B(n,x)}n>=1 = [1, 1+x, 1+6*x+x^2, 1+23*x+23*x^2+x^3, ...] is the sequence of Eulerian polynomials of type B - see A060187. Equivalently,
(2)... a(n) = Sum_{k = 0..2*n} Sum_{j = 0..k} (-1)^(n-j) *binomial(2*n+1,k-j)*(j+1/2)^(2*n).
We also have
(3)... a(n) = 2*A(2*n,i)/(1+i)^(2*n+1),
where i = sqrt(-1) and where {A(n,x)}n>=1 = [x, x + x^2, x + 4*x^2 + x^3, ...] denotes the sequence of Eulerian polynomials - see A008292. Equivalently,
(4)... a(n) = i*Sum_{k = 1..2*n} (-1)^(n+k)*k!*Stirling2(2*n,k) *((1+i)/2)^(k-1)
= i*Sum_{k = 1..2*n} (-1)^(n+k)*((1+i)/2)^(k-1) Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*j^(2*n).
Either this explicit formula for a(n) or (2) above may be used to obtain congruence results for a(n). For example, for prime p
(5a)... a(p) = 1 (mod p)
(5b)... a(2*p) = 5 (mod p)
and for odd prime p
(6a)... a((p+1)/2) = (-1)^((p-1)/2) (mod p)
(6b)... a((p-1)/2) = -1 + (-1)^((p-1)/2) (mod p).
(End)
a(n) = (-1)^n*2^(4*n+1)*(zeta(-2*n,1/4) - zeta(-2*n,3/4)). - Gerry Martens, May 27 2011
a(n) may be expressed as a sum of multinomials taken over all compositions of 2*n into even parts (Vella 2008): a(n) = Sum_{compositions 2*i_1 + ... + 2*i_k = 2*n} (-1)^(n+k)* multinomial(2*n, 2*i_1, ..., 2*i_k). For example, there are 4 compositions of the number 6 into even parts, namely 6, 4+2, 2+4 and 2+2+2, and hence a(3) = 6!/6! - 6!/(4!*2!) - 6!/(2!*4!) + 6!/(2!*2!*2!) = 61. A companion formula expressing a(n) as a sum of multinomials taken over the compositions of 2*n-1 into odd parts has been given by Malenfant 2011. - Peter Bala, Jul 07 2011
a(n) = the upper left term in M^n, where M is an infinite square production matrix; M[i,j] = A000290(i) = i^2, i >= 1 and 1 <= j <= i+1, and M[i,j] = 0, i >= 1 and j >= i+2 (see examples). - Gary W. Adamson, Jul 18 2011
E.g.f. A'(x) satisfies the differential equation A'(x)=cos(A(x)). - Vladimir Kruchinin, Nov 03 2011
From Peter Bala, Nov 28 2011: (Start)
a(n) = D^(2*n)(cosh(x)) evaluated at x = 0, where D is the operator cosh(x)*d/dx. a(n) = D^(2*n-1)(f(x)) evaluated at x = 0, where f(x) = 1+x+x^2/2! and D is the operator f(x)*d/dx.
Other generating functions: cosh(Integral_{t = 0..x} 1/cos(t)) dt = 1 + x^2/2! + 5*x^4/4! + 61*x^6/6! + 1385*x^8/8! + .... Cf. A012131.
A(x) := arcsinh(tan(x)) = log( sec(x) + tan(x) ) = x + x^3/3! + 5*x^5/5! + 61*x^7/7! + 1385*x^9/9! + .... A(x) satisfies A'(x) = cosh(A(x)).
B(x) := Series reversion( log(sec(x) + tan(x)) ) = x - x^3/3! + 5*x^5/5! - 61*x^7/7! + 1385*x^9/9! - ... = arctan(sinh(x)). B(x) satisfies B'(x) = cos(B(x)). (End)
HANKEL transform is A097476. PSUM transform is A173226. - Michael Somos, May 12 2012
a(n+1) - a(n) = A006212(2*n). - Michael Somos, May 12 2012
a(0) = 1 and, for n > 0, a(n) = (-1)^n*((4*n+1)/(2*n+1) - Sum_{k = 1..n} (4^(2*k)/2*k)*binomial(2*n,2*k-1)*A000367(k)/A002445(k)); see the Bucur et al. link. - L. Edson Jeffery, Sep 17 2012
O.g.f.: Sum_{n>=0} (2*n)!/2^n * x^n / Product_{k=1..n} (1 + k^2*x). - Paul D. Hanna, Sep 20 2012
From Sergei N. Gladkovskii, Oct 31 2011 to Oct 11 2013: (Start)
Continued fractions:
E.g.f.: (sec(x)) = 1+x^2/T(0), T(k) = 2(k+1)(2k+1) - x^2 + x^2*(2k+1)(2k+2)/T(k+1).
E.g.f.: 2/Q(0) where Q(k) = 1 + 1/(1 - x^2/(x^2 - 2*(k+1)*(2*k+1)/Q(k+1))).
G.f.: 1/Q(0) where Q(k) = 1 + x*k*(3*k-1) - x*(k+1)*(2*k+1)*(x*k^2+1)/Q(k+1).
E.g.f.: (2 + x^2 + 2*U(0))/(2 + (2 - x^2)*U(0)) where U(k)= 4*k + 4 + 1/( 1 + x^2/(2 - x^2 + (2*k+3)*(2*k+4)/U(k+1))).
E.g.f.: 1/cos(x) = 8*(x^2+1)/(4*x^2 + 8 - x^4*U(0)) where U(k) = 1 + 4*(k+1)*(k+2)/(2*k+3 - x^2*(2*k+3)/(x^2 - 8*(k+1)*(k+2)*(k+3)/U(k+1))).
G.f.: 1/U(0) where U(k) = 1 + x - x*(2*k+1)*(2*k+2)/(1 - x*(2*k+1)*(2*k+2)/U(k+1)).
G.f.: 1 + x/G(0) where G(k) = 1 + x - x*(2*k+2)*(2*k+3)/(1 - x*(2*k+2)*(2*k+3)/G(k+1)).
Let F(x) = sec(x^(1/2)) = Sum_{n>=0} a(n)*x^n/(2*n)!, then F(x)=2/(Q(0) + 1) where Q(k)= 1 - x/(2*k+1)/(2*k+2)/(1 - 1/(1 + 1/Q(k+1))).
G.f.: Q(0), where Q(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 1/Q(k+1)).
E.g.f.: 1/cos(x) = 1 + x^2/(2-x^2)*Q(0), where Q(k) = 1 - 2*x^2*(k+1)*(2*k+1)/( 2*x^2*(k+1)*(2*k+1)+ (12-x^2 + 14*k + 4*k^2)*(2-x^2 + 6*k + 4*k^2)/Q(k+1)). (End)
a(n) = Sum_{k=1..2*n} (Sum_{i=0..k-1} (i-k)^(2*n)*binomial(2*k,i)*(-1)^(i+k+n)) / 2^(k-1) for n>0, a(0)=1. - Vladimir Kruchinin, Oct 05 2012
It appears that a(n) = 3*A076552(n -1) + 2*(-1)^n for n >= 1. Conjectural congruences: a(2*n) == 5 (mod 60) for n >= 1 and a(2*n+1) == 1 (mod 60) for n >= 0. - Peter Bala, Jul 26 2013
From Peter Bala, Mar 09 2015: (Start)
O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 - sqrt(-x)*(2*k + 1)) = Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 + x*(2*k + 1)^2).
O.g.f. is 1 + x*d/dx(log(F(x))), where F(x) = 1 + x + 3*x^2 + 23*x^3 + 371*x^4 + ... is the o.g.f. for A255881. (End)
Sum_(n >= 1, A034947(n)/n^(2d+1)) = a(d)*Pi^(2d+1)/(2^(2d+2)-2)(2d)! for d >= 0; see Allouche and Sondow, 2015. - Jonathan Sondow, Mar 21 2015
Asymptotic expansion: 4*(4*n/(Pi*e))^(2*n+1/2)*exp(1/2+1/(24*n)-1/(2880*n^3) +1/(40320*n^5)-...). (See the Luschny link.) - Peter Luschny, Jul 14 2015
a(n) = 2*(-1)^n*Im(Li_{-2n}(i)), where Li_n(x) is polylogarithm, i=sqrt(-1). - Vladimir Reshetnikov, Oct 22 2015
Limit_{n->infinity} ((2*n)!/a(n))^(1/(2*n)) = Pi/2. - Stanislav Sykora, Oct 07 2016
O.g.f.: 1/(1 + x - 2*x/(1 - 2*x/(1 + x - 12*x/(1 - 12*x/(1 + x - 30*x/(1 - 30*x/(1 + x - ... - (2*n - 1)*(2*n)*x/(1 - (2*n - 1)*(2*n)*x/(1 + x - ... ))))))))). - Peter Bala, Nov 09 2017
For n>0, a(n) = (-PolyGamma(2*n, 1/4) / 2^(2*n - 1) - (2^(2*n + 2) - 2) * Gamma(2*n + 1) * zeta(2*n + 1)) / Pi^(2*n + 1). - Vaclav Kotesovec, May 04 2020
a(n) ~ 2^(4*n + 3) * n^(2*n + 1/2) / (Pi^(2*n + 1/2) * exp(2*n)) * exp(Sum_{k>=1} bernoulli(k+1) / (k*(k+1)*2^k*n^k)). - Vaclav Kotesovec, Mar 05 2021
Peter Bala's conjectured congruences, a(2n) == 5 (mod 60) for n >= 1 and a(2n + 1) == 1 (mod 60), hold due to the results of Stern (mod 4) and Knuth & Buckholtz (mod 3 and 5). - Charles R Greathouse IV, Mar 23 2022

Extensions

Typo in name corrected by Anders Claesson, Dec 01 2015

A126155 Symmetric triangle, read by rows of 2*n+1 terms, similar to triangle A008301. Second term 5 times first term.

Original entry on oeis.org

1, 1, 5, 1, 7, 35, 55, 35, 7, 139, 695, 1195, 1415, 1195, 695, 139, 5473, 27365, 48145, 63365, 69025, 63365, 48145, 27365, 5473, 357721, 1788605, 3175705, 4343885, 5126905, 5403005, 5126905, 4343885, 3175705, 1788605, 357721
Offset: 0

Views

Author

Paul D. Hanna, Dec 20 2006

Keywords

Examples

			The triangle begins:
  1;
  1, 5, 1;
  7, 35, 55, 35, 7;
  139, 695, 1195, 1415, 1195, 695, 139;
  5473, 27365, 48145, 63365, 69025, 63365, 48145, 27365, 5473;
  357721, 1788605, 3175705, 4343885, 5126905, 5403005, 5126905, 4343885, 3175705, 1788605, 357721; ...
If we write the triangle like this:
  .......................... ....1;
  ................... ....1, ....5, ....1;
  ............ ....7, ...35, ...55, ...35, ....7;
  ..... ..139, ..695, .1195, .1415, .1195, ..695, ..139;
  .5473, 27365, 48145, 63365, 69025, 63365, 48145, 27365, .5473;
then the first term in each row is the sum of the previous row:
  5473 = 139 + 695 + 1195 + 1415 + 1195 + 695 + 139
the next term is 5 times the first:
  27365 = 5*5473,
and the remaining terms in each row are obtained by the rule illustrated by:
  48145 = 2*27365 - 5473 - 8*139;
  63365 = 2*48145 - 27365 - 8*695;
  69025 = 2*63365 - 48145 - 8*1195;
  63365 = 2*69025 - 63365 - 8*1415;
  48145 = 2*63365 - 69025 - 8*1195;
  27365 = 2*48145 - 63365 - 8*695;
  5473 = 2*27365 - 48145 - 8*139.
An alternate recurrence is illustrated by:
  27365 = 5473 + 4*(139 + 695 + 1195 + 1415 + 1195 + 695 + 139);
  48145 = 27365 + 4*(695 + 1195 + 1415 + 1195 + 695);
  63365 = 48145 + 4*(1195 + 1415 + 1195);
  69025 = 63365 + 4*(1415);
and then for k>n, T(n,k) = T(n,2n-k).
		

Crossrefs

Cf. A126156 (column 0); diagonals: A126157, A126158; A126159; variants: A008301 (p=1), A125053 (p=2), A126150 (p=3).

Programs

  • Maple
    T := proc(n,k) option remember; local j;
      if n = 1 then 1
    elif k = 1 then add(T(n-1, j), j=1..2*n-3)
    elif k = 2 then 5*T(n, 1)
    elif k > n then T(n, 2*n-k)
    else 2*T(n, k-1)-T(n, k-2)-8*T(n-1, k-2)
      fi end:
    seq(print(seq(T(n,k), k=1..2*n-1)), n=1..6); # Peter Luschny, May 12 2014
  • Mathematica
    T[n_, k_] := T[n, k] = Which[n==1, 1, k==1, Sum[T[n-1, j], {j, 1, 2n-3}], k==2, 5 T[n, 1], k>n, T[n, 2n-k], True, 2 T[n, k-1] - T[n, k-2] - 8 T[n-1, k-2]];
    Table[T[n, k], {n, 1, 6}, {k, 1, 2n-1}] (* Jean-François Alcover, Jun 15 2019, from Maple *)
  • PARI
    {T(n,k) = local(p=4);if(2*n
    				
  • PARI
    /* Alternate Recurrence: */
    {T(n,k) = local(p=4);if(2*n
    				
  • SageMath
    from functools import cache
    @cache
    def R(n, k):
          return (1 if n == 1 else sum(R(n-1, j) for j in range(1, 2*n-2))
                    if k == 1 else 5*R(n, 1) if k == 2 else R(n, 2*n-k)
                    if k > n else 2*R(n, k-1) - R(n, k-2) - 8*R(n-1, k-2))
    def A126155(n, k): return R(n+1, k+1)
    for n in range(5): print([A126155(n, k) for k in range(2*n+1)])
    # Peter Luschny, Dec 14 2023

Formula

Sum_{k=0..2n} (-1)^k*C(2n,k)*T(n,k) = (-8)^n.

A126150 Symmetric triangle, read by rows of 2*n+1 terms, similar to triangle A008301. Second term 4 times first term.

Original entry on oeis.org

1, 1, 4, 1, 6, 24, 36, 24, 6, 96, 384, 636, 744, 636, 384, 96, 2976, 11904, 20256, 26304, 28536, 26304, 20256, 11904, 2976, 151416, 605664, 1042056, 1407024, 1650456, 1736064, 1650456, 1407024, 1042056, 605664, 151416, 11449296, 45797184
Offset: 0

Views

Author

Paul D. Hanna, Dec 19 2006

Keywords

Examples

			Triangle begins:
  1;
  1, 4, 1;
  6, 24, 36, 24, 6;
  96, 384, 636, 744, 636, 384, 96;
  2976, 11904, 20256, 26304, 28536, 26304, 20256, 11904, 2976;
  151416, 605664, 1042056, 1407024, 1650456, 1736064, 1650456, 1407024, 1042056, 605664, 151416; ...
If we write the triangle like this:
  .......................... ....1;
  ................... ....1, ....4, ....1;
  ............ ....6, ...24, ...36, ...24, ....6;
  ..... ...96, ..384, ..636, ..744, ..636, ..384, ...96;
  .2976, 11904, 20256, 26304, 28536, 26304, 20256, 11904, .2976;
then the first term in each row is the sum of the previous row:
  2976 = 96 + 384 + 636 + 744 + 636 + 384 + 96
the next term is 4 times the first:
  11904 = 4*2976,
and the remaining terms in each row are obtained by the rule illustrated by:
  20256 = 2*11904 - 2976 - 6*96;
  26304 = 2*20256 - 11904 - 6*384;
  28536 = 2*26304 - 20256 - 6*636;
  26304 = 2*28536 - 26304 - 6*744;
  20256 = 2*26304 - 28536 - 6*636;
  11904 = 2*20256 - 26304 - 6*384;
  2976 = 2*11904 - 20256 - 6*96.
An alternate recurrence is illustrated by:
  11904 = 2976 + 3*(96 + 384 + 636 + 744 + 636 + 384 + 96);
  20256 = 11904 + 3*(384 + 636 + 744 + 636 + 384);
  26304 = 20256 + 3*(636 + 744 + 636);
  28536 = 26304 + 3*(744);
and then for k>n, T(n,k) = T(n,2n-k).
		

Crossrefs

Cf. A126151 (column 0); diagonals: A126152, A126153; A126154; variants: A008301, A125053, A126155.

Programs

  • PARI
    T(n,k)=local(p=3);if(2*n
    				
  • PARI
    /* Alternate Recurrence: */ T(n,k)=local(p=3);if(2*n
    				

Formula

Sum_{k=0..2n} (-1)^k*C(2n,k)*T(n,k) = (-6)^n.

A236934 Triangle of Poupard numbers g_n(k) read by rows, n>=1, 1<=k<=2n-1.

Original entry on oeis.org

1, 0, 2, 0, 0, 4, 8, 4, 0, 0, 32, 64, 80, 64, 32, 0, 0, 544, 1088, 1504, 1664, 1504, 1088, 544, 0, 0, 15872, 31744, 45440, 54784, 58112, 54784, 45440, 31744, 15872, 0, 0, 707584, 1415168, 2059264, 2576384, 2911744, 3027968, 2911744, 2576384, 2059264, 1415168, 707584, 0
Offset: 1

Views

Author

N. J. A. Sloane, Feb 17 2014

Keywords

Examples

			Triangle begins:
1,
0, 2, 0,
0, 4, 8, 4, 0,
0, 32, 64, 80, 64, 32, 0,
0, 544, 1088, 1504, 1664, 1504, 1088, 544, 0,
...
		

Crossrefs

Cf. A000182 (row sums), A008282, A125053.

Programs

  • Maple
    T := proc(n,k) option remember; local j;
      if n = 1 then 1
    elif k = 1 then 0
    elif k = 2 then 2*add(T(n-1, j), j=1..2*n-3)
    elif k > n then T(n, 2*n-k)
    else 2*T(n, k-1)-T(n, k-2)-4*T(n-1, k-2)
      fi end:
    seq(print(seq(T(n,k), k=1..2*n-1)), n=1..6); # Peter Luschny, May 11 2014
  • Mathematica
    T[n_, k_] /; 1 <= k <= 2n-1 := T[n, k] = Which[n == 1, 1, k == 1, 0, k == 2, 2 Sum[T[n-1, j], {j, 1, 2n-3}], k > n, T[n, 2n-k], True, 2 T[n, k-1] - T[n, k-2] - 4 T[n-1, k-2]]; T[, ] = 0;
    Table[T[n, k], {n, 1, 7}, {k, 1, 2n-1}] // Flatten (* Jean-François Alcover, Jul 08 2019, from Maple *)

Formula

4^(-n)*sum(k=1..2*n+1, binomial(2*n,k-1)*T(n+1,k)) = A000364(n), n>=0. - Peter Luschny, May 11 2014
(-1)^n*sum(k=1..2*n+1, (-1)^(k-1)*binomial(2*n,k-1)*T(n+1,k)) = A000302(n), n>=0. - Peter Luschny, May 11 2014

Extensions

More terms from Peter Luschny, May 11 2014
Showing 1-7 of 7 results.