A001586 -id:A001586 - cp's OEIS frontend

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A104035 Triangle T(n,k), 0 <= k <= n, read by rows, defined by T(0,0) = 1; T(0,k) = 0 if k>0 or if k<0; T(n,k) = kT(n-1,k-1) + (k+1)T(n-1,k+1).

1, 0, 1, 1, 0, 2, 0, 5, 0, 6, 5, 0, 28, 0, 24, 0, 61, 0, 180, 0, 120, 61, 0, 662, 0, 1320, 0, 720, 0, 1385, 0, 7266, 0, 10920, 0, 5040, 1385, 0, 24568, 0, 83664, 0, 100800, 0, 40320, 0, 50521, 0, 408360, 0, 1023120, 0, 1028160, 0, 362880, 50521, 0, 1326122, 0, 6749040

Offset: 0

Philippe Deléham, Apr 06 2005

Or, triangle of coefficients (with exponents in increasing order) in polynomials Q_n(u) defined by d^n sec x / dx^n = Q_n(tan x)*sec x.
Interpolates between factorials and Euler (or secant) numbers. Related to Springer numbers.
Companion triangles are A155100 (derivative polynomials of tangent function) and A185896 (derivative polynomials of squared secant function).
A combinatorial interpretation for the polynomial Q_n(u) as the generating function for a sign change statistic on certain types of signed permutation can be found in [Verges]. A signed permutation is a sequence (x_1,x_2,...,x_n) of integers such that {|x_1|,|x_2|,...,|x_n|} = {1,2,...,n}. They form a group, the hyperoctahedral group of order 2^n*n! = A000165(n), isomorphic to the group of symmetries of the n dimensional cube.
Let x_1,...,x_n be a signed permutation. Adjoin x_0 = 0 to the front of the permutation and x_(n+1) = (-1)^n*(n+1) to the end to form x_0,x_1,...,x_n,x_(n+1). Then x_0,x_1,...,x_n,x_(n+1) is a snake of type S(n;0) when x_0 < x_1 > x_2 < ... x_(n+1). For example, 0 3 -1  2 -4 is a snake of type S(3;0).
Let sc be the number of sign changes through a snake ... sc = #{i, 0 <= i <= n, x_i*x_(i+1) < 0}. For example, the snake 0 3 -1 2 -4 has sc = 3. The polynomial Q_n(u) is the generating function for the sign change statistic on snakes of type S(n;0): ... Q_n(u) = sum {snakes in S(n;0)} u^sc. See the example section below for the cases n = 2 and n = 3.
PRODUCTION MATRIX
Let D = subdiag(1,2,3,...) be the array with the indicated sequence on the first subdiagonal and zeros elsewhere and let C = transpose(D). The production matrix for this triangle is C+D: the first row of (C+D)^n is the n-th row of this triangle. D represents the derivative operator d/dx and C represents the operator p(x) -> x*d/dx(x*p(x)) acting on the basis monomials {x^n}n>=0. See Formula (1) below.

			The polynomials Q_0(u) through Q_6(u) (with exponents in decreasing order) are:
  1
  u
  2*u^2 + 1
  6*u^3 + 5*u
  24*u^4 + 28*u^2 + 5
  120*u^5 + 180*u^3 + 61*u
  720*u^6 + 1320*u^4 + 662*u^2 + 61
Triangle begins:
  1
  0 1
  1 0 2
  0 5 0 6
  5 0 28 0 24
  0 61 0 180 0 120
  61 0 662 0 1320 0 720
  0 1385 0 7266 0 10920 0 5040
  1385 0 24568 0 83664 0 100800 0 40320
  0 50521 0 408360 0 1023120 0 1028160 0 362880
  50521 0 1326122 0 6749040 0 13335840 0 11491200 0 3628800
  0 2702765 0 30974526 0 113760240 0 185280480 0 139708800 0 39916800
  2702765 0 98329108 0 692699304 0 1979524800 0 2739623040 0 1836172800 0 479001600
Examples of sign change statistic sc on snakes of type S(n;0)
= = = = = = = = = = = = = = = = = = = = = =
.....Snakes....# sign changes sc.......u^sc
= = = = = = = = = = = = = = = = = = = = = =
n=2
...0 1 -2 3...........2.................u^2
...0 2  1 3...........0.................1
...0 2 -1 3...........2.................u^2
yields Q_2(u) = 2*u^2 + 1.
n=3
...0 1 -2  3 -4.......3.................u^3
...0 1 -3  2 -4.......3.................u^3
...0 1 -3 -2 -4.......1.................u
...0 2  1  3 -4.......1.................u
...0 2 -1  3 -4.......3.................u^3
...0 2 -3  1 -4.......3.................u^3
...0 2 -3 -2 -4.......1.................u
...0 3  1  2 -4.......1.................u
...0 3 -1  2 -4.......3.................u^3
...0 3 -2  1 -4.......3.................u^3
...0 3 -2 -1 -4.......1.................u
yields Q_3(u) = 6*u^3 + 5*u.

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998, p. 287.
S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see pp. 445 and 469.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
K. Boyadzhiev, Derivative Polynomials for tanh, tan, sech and sec in Explicit Form, arXiv:0903.0117 [math.CA], 2009-2010.
M.-P. Grosset and A. P. Veselov, Bernoulli numbers and solitons, arXiv:math/0503175 [math.GM], 2005.
Gordon Haigh, A "natural" approach to Pick's theorem, Math. Gaz. 64 (1980), no. 429, 173-180.
Michael E. Hoffman, Derivative polynomials for tangent and secant, Amer. Math. Monthly, 102 (1995), 23-30.
Michael E. Hoffman, Derivative Polynomials, Euler Polynomials, and Associated Integer Sequences, The Electronic Journal of Combinatorics, Volume 6.1 (1999): Research paper R21, 13 p.
M. Josuat-Vergès, Enumeration of snakes and cycle-alternating permutations, arXiv:1011.0929 [math.CO], 2010.
Donald E. Knuth and Thomas J. Buckholtz, Computation of tangent, Euler and Bernoulli numbers, Math. Comp. 21 1967 663-688.

See A008294 for another version of this triangle.
Setting u=0,1,2,3,4 gives A000364, A001586, A156129, A156131, A156132.
Setting u=sqrt(2) gives A156134 and A156138; u=sqrt(3) gives A002437 and A002439.
Cf. A060187, A155100, A185896, A186492.

a104035 n k = a104035_tabl !! n !! k
a104035_row n = a104035_tabl !! n
a104035_tabl = iterate f [1] where
   f xs = zipWith (+)
     (zipWith (*) [1..] (tail xs) ++ [0,0]) ([0] ++ zipWith (*) [1..] xs)
-- Reinhard Zumkeller, Apr 27 2014

nmax = 10; t[n_, k_] := t[n, k] = k*t[n-1, k-1] + (k+1)*t[n-1, k+1]; t[0, 0] = 1; t[0, ] = 0; Flatten[ Table[t[n, k], {n, 0, nmax}, {k, 0, n}]] (* _Jean-François Alcover, Nov 14 2011 *)

T(n, n) = n!; T(n, 0) = 0 if n = 2m + 1; T(n, 0) = A000364(m) if n = 2m.
Sum_{k>=0} T(m, k)*T(n, k) = T(m+n, 0).
Sum_{k>=0} T(n, k) = A001586(n): Springer numbers.
G.f.: Sum_{n >= 0} Q_n(u)*t^n/n! = 1/(cos t - u sin t).
From Peter Bala: (Start)
RECURRENCE RELATION
For n>=0,
(1)... Q_(n+1)(u) = d/du Q_n(u) + u*d/du(u*Q_n(u))
... = (1+u^2)*d/du Q_n(u) + u*Q_n(u),
with starting condition Q_0(u) = 1. Compare with Formula (4) of A186492.
RELATION WITH TYPE B EULERIAN NUMBERS
(2)... Q_n(u) = ((u+i)/2)^n*B(n,(u-i)/(u+i)), where i = sqrt(-1) and
[B(n,u)]n>=0 = [1,1+u,1+6*u+u^2,1+23*u+23*u^2+u^3,...] is the sequence of type B Eulerian polynomials (with a factor of u removed) - see A060187.
(End)
T(n,0) = abs(A122045(n)). - Reinhard Zumkeller, Apr 27 2014

Entry revised by N. J. A. Sloane, Nov 06 2009

A000281 Expansion of cos(x)/cos(2x).

Original entry on oeis.org

1, 3, 57, 2763, 250737, 36581523, 7828053417, 2309644635483, 898621108880097, 445777636063460643, 274613643571568682777, 205676334188681975553003, 184053312545818735778213457, 193944394596325636374396208563

Offset: 0

N. J. A. Sloane

a(n) is (2n)! times the coefficient of x^(2n) in the Taylor series for cos(x)/cos(2x).

			cos x / cos 2*x = 1 + 3*x^2/2 + 19*x^4/8 + 307*x^6/80 + ...

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).

G. C. Greubel, Table of n, a(n) for n = 0..215 (terms 0..50 from T. D. Noe)
P. Bala, A triangle for calculating A000281
P. Bala, Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)
Kwang-Wu Chen, An Interesting Lemma for Regular C-fractions, J. Integer Seqs., Vol. 6, 2003.
D. Dumont, Further triangles of Seidel-Arnold type and continued fractions related to Euler and Springer numbers, Adv. Appl. Math., 16 (1995), 275-296.
Matthieu Josuat-Vergès and Jang Soo Kim, Touchard-Riordan formulas, T-fractions, and Jacobi's triple product identity, arXiv:1101.5608 [math.CO] (2011).
D. Shanks, Generalized Euler and class numbers. Math. Comp. 21 (1967) 689-694, sequence c(2,n).
D. Shanks, Corrigenda to: "Generalized Euler and class numbers", Math. Comp. 22 (1968), 699.
D. Shanks, Generalized Euler and class numbers, Math. Comp. 21 (1967), 689-694; 22 (1968), 699. [Annotated scanned copy]

Cf. A000364 A086646, A002437.
Cf. A064069. Bisection of A000825, A001586.
Cf. A098432. Cf. A093954, A188458, A212435, A235605, A255883.

a := n -> (-1)^n*2^(6*n+1)*(Zeta(0,-2*n,1/8)-Zeta(0,-2*n,5/8)):
seq(a(n), n=0..13); # Peter Luschny, Mar 11 2015

With[{nn=30},Take[CoefficientList[Series[Cos[x]/Cos[2x],{x,0,nn}],x] Range[0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Oct 06 2011 *)

{a(n) = if( n<0, 0, n*=2; n! * polcoeff( cos(x + x * O(x^n)) / cos(2*x + x * O(x^n)), n))}; /* Michael Somos, Feb 09 2006 */

a(n) = Sum_{k=0..n} (-1)^k*binomial(2n, 2k)*A000364(n-k)*4^(n-k). - Philippe Deléham, Jan 26 2004
E.g.f.: Sum_{k>=0} a(k)x^(2k)/(2k)! = cos(x)/cos(2x).
a(n-1) is approximately 2^(4*n-3)*(2*n-1)!*sqrt(2)/((Pi^(2*n-1))*(2*n-1)). The approximation is quite good a(250) is of the order of 10^1181 and this formula is accurate to 238 digits. - Simon Plouffe, Jan 31 2007
G.f.: 1 / (1 - 1*3*x / (1 - 4*4*x / (1 - 5*7*x / (1 - 8*8*x / (1 - 9*11*x / ... ))))). - Michael Somos, May 12 2012
G.f.: 1/E(0) where E(k) = 1 - 3*x - 16*x*k*(2*k+1) - 16*x^2*(k+1)^2*(4*k+1)*(4*k+3)/E(k+1) (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 17 2012
G.f.: T(0)/(1-3*x), where T(k) = 1 - 16*x^2*(4*k+1)*(4*k+3)*(k+1)^2/( 16*x^2*(4*k+1)*(4*k+3)*(k+1)^2 - (32*x*k^2+16*x*k+3*x-1 )*(32*x*k^2+80*x*k+51*x -1)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 11 2013
From Peter Bala, Mar 09 2015: (Start)
a(n) = (-1)^n*4^(2*n)*E(2*n,1/4), where E(n,x) denotes the n-th Euler polynomial.
O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 + x*(4*k + 1)^2) = 1 + 3*x + 57*x^2 + 2763*x^3 + ....
We appear to have the asymptotic expansion Pi/(2*sqrt(2)) - Sum {k = 0..n - 1} (-1)^floor(k/2)/(2*k + 1) ~ 1/(2*n) - 3/(2*n)^3 + 57/(2*n)^5 - 2763/(2*n)^7 + .... See A093954.
Bisection of A001586. See also A188458 and A212435. Second row of A235605 (read as a square array).
The expansion of exp( Sum_{n >= 1} a(n)*x^n/n ) appears to have integer coefficients. See A255883. (End)
From Peter Luschny, Mar 11 2015: (Start)
a(n) = ((-64)^n/((n+1/2)))*(B(2*n+1,7/8)-B(2*n+1,3/8)), B(n,x) Bernoulli polynomials.
a(n) = 2*(-16)^n*LerchPhi(-1, -2*n, 1/4).
a(n) = (-1)^n*Sum_{0..2*n} 2^k*C(2*n,k)*E(k), E(n) the Euler secant numbers A122045.
a(n) = (-4)^n*SKP(2*n,1/2) where SKP are the Swiss-Knife polynomials A153641.
a(n) = (-1)^n*2^(6*n+1)*(Zeta(-2*n,1/8) - Zeta(-2*n,5/8)), where Zeta(a,z) is the generalized Riemann zeta function. (End)
From Peter Bala, May 13 2017: (Start)
G.f.: 1/(1 + x - 4*x/(1 - 12*x/(1 + x - 40*x/(1 - 56*x/(1 + x - ... - 4*n(4*n - 3)*x/(1 - 4*n(4*n - 1)*x/(1 + x - ...
G.f.: 1/(1 + 9*x - 12*x/(1 - 4*x/(1 + 9*x - 56*x/(1 - 40*x/(1 + 9*x - ... - 4*n(4*n - 1)*x/(1 - 4*n(4*n - 3)*x/(1 + 9*x - .... (End)
From Peter Bala, Nov 08 2019: (Start)
a(n) = sqrt(2)*4^n*Integral_{x = 0..inf} x^(2*n)*cosh(Pi*x/2)/cosh(Pi*x) dx. Cf. A002437.
The L-series 1 + 1/3^(2*n+1) - 1/5^(2*n+1) - 1/7^(2*n+1) + + - - ... = sqrt(2)*(Pi/4)^(2*n+1)*a(n)/(2*n)! (see Shanks), which gives a(n) ~ (1/sqrt(2))*(2*n)!*(4/Pi)^(2*n+1). (End)

A188458 Expansion of e.g.f. exp(x)/cosh(2*x).

Original entry on oeis.org

1, 1, -3, -11, 57, 361, -2763, -24611, 250737, 2873041, -36581523, -512343611, 7828053417, 129570724921, -2309644635483, -44110959165011, 898621108880097, 19450718635716001, -445777636063460643, -10784052561125704811, 274613643571568682777

Offset: 0

Paul D. Hanna, Apr 01 2011

sign

A signed version of A001586 (Springer numbers).

Equals the logarithmic derivative of A188514 (ignoring the initial term of this sequence); note that the unsigned version (A001586) does not form a logarithmic derivative of an integer sequence.

			E.g.f.: exp(x)/cosh(2*x) = 1 + x - 3*x^2/2! - 11*x^3/3! + 57*x^4/4! + 361*x^5/5! +...
Illustration of other generating functions.
E.g.f.: 1 = exp(-x) + exp(x)*x - 3*exp(-x)*x^2/2! - 11*exp(x)*x^3/3! +...
L.g.f.: log(1+x) = x/(1-x) - 3*(x^2/2)/(1+x)^2 - 11*(x^3/3)/(1-x)^3 +...
G.f.: 1 = 1/(1+x) + 1*x/(1-x)^2 - 3*x^2/(1+x)^3 - 11*x^3/(1-x)^4 +...
G.f.: 1 = 1/(1+x)^2 + 1*2*x/(1-x)^3 - 3*3*x^2/(1+x)^4 - 11*4*x^3/(1-x)^5 +...
G.f.: 1 = 1/(1+x)^3 + 1*3*x/(1-x)^4 - 3*6*x^2/(1+x)^5 - 11*10*x^3/(1-x)^6 +...

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A001586, A188514.

seq(4^n*euler(n,3/4), n=0..20); # Peter Luschny, Apr 19 2014

CoefficientList[Series[E^x/Cosh[2*x], {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Oct 07 2013 *)

{a(n)=local(X=x+x*O(x^n));n!*polcoeff(exp(X)/cosh(2*X),n)}

{a(n)=n!*polcoeff(1-sum(k=0, n-1, a(k)*exp(-(-1)^k*x+x*O(x^n))*x^k/k!), n)}

{a(n)=polcoeff(1-sum(k=0, n-1, a(k)*x^k/(1+(-1)^k*x+x*O(x^n))^(k+1)), n)}

/* Holds for m>=1: */
{a(n)=local(m=1); polcoeff(1-sum(k=0, n-1, a(k)*binomial(m+k-1, k)*x^k/(1+(-1)^k*x+x*O(x^n))^(k+m)), n)/binomial(m+n-1, n)}

/* Recurrence: */
{a(n)=if(n<0,0,if(n==0,1, sum(k=1, n, -(-1)^(n*k)*binomial(n, k)*a(n-k))))}

{EULER(n)=n!*polcoeff(1/cosh(x+x*O(x^n)),n)}
{a(n)=sum(k=0,n,2^k*binomial(n,k)*EULER(k))}

{a(n)=(-1)^(n\2)*((1+I)/2)^n*sum(k=0, n, ((1-I)/(1+I))^k*sum(j=0, k, (-1)^(k-j)*binomial(n+1, k-j)*(2*j+1)^n))}

a(n) = Sum_{k=1..n} -(-1)^(n*k)*C(n, k)*a(n-k) for n>0 with a(0)=1.
L.g.f.: log(1+x) = Sum_{n>=1} a(n)*(x^n/n)/(1 + (-1)^n*x)^n.
E.g.f.: 1 = Sum_{n>=0} a(n)*exp(-(-1)^n*x)*x^n/n!.
G.f.: 1 = Sum_{n>=0} a(n)*x^n/(1 + (-1)^n*x)^(n+1).
G.f.: 1 = Sum_{n>=0} a(n)*C(n+m-1,n)*x^n/(1 + (-1)^n*x)^(n+m) for m>=1.
a(n) = Sum_{k=0..n} 2^k C(n,k) Euler(k). - Peter Luschny
a(n) = (-1)^[n/2]*((1+I)/2)^n * Sum_{k=0..n} ((1-I)/(1+I))^k * Sum_{j=0..k} (-1)^(k-j)*C(n+1, k-j)*(2*j+1)^n. - Peter Bala
O.g.f.: 1/(1-x/(1+4*x/(1-x- 4*x/(1+4*x/(1+x- 6*x/(1+6*x/(1+x- 8*x/(1+8*x/(1+x- 10*x/(1+10*x/(1+x- 12*x/(1+12*x/(1+x- ...))))))))))))) (continued fraction).
E.g.f.: E(x) = exp(x)/cosh(2*x) = 2/G(0) where G(k)= 1 -((-1)^k)*3^k/(1 - x/(x + (k+1)*((-1)^k)*3^k/G(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Jun 07 2012
a(n) ~ n! * (cos(n*Pi/2) + sin(n*Pi/2)) * 2^(2*n+3/2) / Pi^(n+1). - Vaclav Kotesovec, Oct 07 2013
G.f.: conjecture T(0)/(1-x), where T(k) = 1 - 4*x^2*(k+1)^2/(4*x^2*(k+1)^2 + (1-x)^2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 12 2013
From Peter Luschny, Apr 19 2014: (Start)
a(n) = 2^n*skp(n, 1/2), where skp(n,x) are the Swiss-Knife polynomials A153641.
a(n) = 4^n*E(n, 3/4), where E(n,x) are Euler polynomials.
a(n) = (8^n/((n+1)/2))*(B(n+1, 7/8) - B(n+1, 3/8)), where B(n,x) are the Bernoulli polynomials. (End)
a(n) = 2^(3*n+1)*(Zeta(-n,3/8)-Zeta(-n,7/8)). - Peter Luschny, Oct 15 2015

A007788 Number of augmented Andre 3-signed permutations: E.g.f. (1-sin(3*x))^(-1/3).

Original entry on oeis.org

1, 1, 4, 19, 136, 1201, 13024, 165619, 2425216, 40132801, 740882944, 15091932019, 336257744896, 8134269015601, 212309523595264, 5946914908771219, 177934946000306176, 5663754614516217601, 191097349696090537984, 6812679868133940475219, 255885704427935576621056

Offset: 0

R. Ehrenborg (ehrenbor(AT)lacim.uqam.ca) and M. A. Readdy (readdy(AT)lacim.uqam.ca)

nonn

It appears that all members are of the form 3k+1. - Ralf Stephan, Nov 12 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..200
R. Ehrenborg and M. A. Readdy, Sheffer posets and r-signed permutations, Preprint submitted to Ann. Sci. Math. Quebec, 1994. (Annotated scanned copy)
R. Ehrenborg and M. A. Readdy, Sheffer posets and r-signed permutations, Ann. Sci. Math. Québec, 19 (1995), no. 2, 173-196.
R. Ehrenborg and M. A. Readdy, The r-cubical lattice and a generalization of the cd-index, European J. Combin. 17 (1996), no. 8, 709-725.

Cf. A001586, A144015, A230134, A227544, A235128, A230114.
Cf. A007863, A235135, A235132.
Cf. A007559, A136630.

R:=PowerSeriesRing(Rationals(), 20); Coefficients(R!(Laplace( (1-Sin(3*x))^(-1/3) ))); // G. C. Greubel, Mar 05 2020

m:=20; S:=series( (1-sin(3*x))^(-1/3), x, m+1): seq(j!*coeff(S, x, j), j=0..m); # G. C. Greubel, Mar 05 2020

With[{nn=20},CoefficientList[Series[(1-Sin[3x])^(-1/3),{x,0,nn}], x] Range[0,nn]!] (* Harvey P. Dale, Nov 23 2011 *)

Vec(serlaplace( (1-sin(3*x))^(-1/3) +O('x^20) )) \\ G. C. Greubel, Mar 05 2020

a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
a007559(n) = prod(k=0, n-1, 3*k+1);
a(n) = sum(k=0, n, a007559(k)*(3*I)^(n-k)*a136630(n, k)); \\ Seiichi Manyama, Jun 24 2025

m=20;
def A007788_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( (1-sin(3*x))^(-1/3) ).list()
a=A007788_list(m+1); [factorial(n)*a[n] for n in (0..m)] # G. C. Greubel, Mar 05 2020

E.g.f.: (1-sin(3*x))^(-1/3).
a(n) ~ n! * 2*6^n/(Pi^(n+2/3)*n^(1/3)*Gamma(2/3)). - Vaclav Kotesovec, Jun 25 2013
a(n) = Sum_{k=0..n} A007559(k) * (3*i)^(n-k) * A136630(n,k), where i is the imaginary unit. - Seiichi Manyama, Jun 24 2025

A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+1).

Original entry on oeis.org

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

Offset: 0

Jonathan D. B. Hodgson, Oct 01 2010, Oct 06 2010

			0.86697298733991103757399516388287071365217536734524490433....
At N = 100000 the truncated series 2*Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 1) = 1.7339(3)5974(5)7982(5)075(25)79(846)27(404)7... to 32 digits. The bracketed numbers show where this decimal expansion differs from that of 2*A181048. The numbers 1, 1, -3, -11, 57, 361 must be added to the bracketed numbers to give the correct decimal expansion to 32 digits: 2*( (log(1 + sqrt(2)) + Pi/2)/(2*sqrt(2)) ) = 1.7339(4)5974(6)7982(2)075(14)79(903)27(765)7.... - _Peter Bala_, Sep 23 2016

Jolley, Summation of Series, Dover (1961) eq 82 page 16.
Murray R. Spiegel, Seymour Lipschutz, John Liu. Mathematical Handbook of Formulas and Tables, 3rd Ed. Schaum's Outline Series. New York: McGraw-Hill (2009): p. 135, equation 21.17

Ivan Panchenko, Table of n, a(n) for n = 0..1000
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Eric W. Weisstein, Euler's Series Transformation.
Herbert S. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics & Theoretical Computer Science, Vol 3, No 4 (1999).

Cf. A001586, A003881, A024396, A091648, A093954, A113476, A181049, A181122, A188458, A193534, A262246.

RealDigits[(Pi Sqrt[2])/8 + (Sqrt[2] Log[1 + Sqrt[2]])/4, 10, 100][[1]] (* Alonso del Arte, Aug 11 2011 *)

(log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) \\ G. C. Greubel, Jul 05 2017

(asinh(1)+Pi/2)/sqrt(8) \\ Charles R Greathouse IV, Jul 06 2017

Equals (A093954 + A091648/sqrt(2))/2.
Integral_{x = 0..1} 1/(1+x^4) = Sum_{k >= 0} (-1)^k/(4*k+1) = (log(1+sqrt(2)) + Pi/2)/(2*sqrt(2)).
1 - 1/5 + 1/9 - 1/13 + 1/17 - ... = (Pi*sqrt(2))/8 + (sqrt(2)*log(1 + sqrt(2)))/4 = (Pi + 2*log(1 + sqrt(2)))/(4 sqrt(2)). The first two are the formulas as given in Spiegel et al., the third is how Mathematica rewrites the infinite sum. - Alonso del Arte, Aug 11 2011
Let N be a positive integer divisible by 4. We have the asymptotic expansion 2*( (log(1 + sqrt(2)) + Pi/2)/(2*sqrt(2)) - Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 1) ) ~ 1/N + 1/N^2 - 3/N^3 - 11/N^4 + 57/N^5 + 361/N^6 - ..., where the sequence of coefficients [1, 1, -3, -11, 57, 361, ...] is A188458. This follows from Borwein et al., Lemma 2 with f(x) = 1/x and then set x = N/4 and h = 1/4. An example is given below. Cf. A181049. - Peter Bala, Sep 23 2016
Equals Sum_{n >= 0} 2^(n-1)*n!/(Product_{k = 0..n} 4*k + 1) = Sum_{n >= 0} 2^(n-1)*n!/A007696(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(4*k + 1)). - Peter Bala, Dec 01 2021
From Peter Bala, Oct 23 2023: (Start)
The slowly converging series representation Sum_{n >= 0} (-1)^n/(4*n + 1) for the constant can be accelerated to give the following faster converging series:
1/2 + 2*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5));
7/10 + 8*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9));
71/90 + 48*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9)*(4*n + 13));
971/1170 + 384*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9)*(4*n + 13)*(4*n + 17)).
These results may be easily verified by taking the partial fraction expansions of the summands. The general result appears to be that for r >= 0, the constant equals
C(r) + (2^r)*r!*Sum_{n >= 0} (-1)^n/((4*n + 1)*(4*n + 5)*...*(4*n + 4*r + 1)), where C(r) is the rational number Sum_{k = 0..r-1} 2^(k-1)*k!/(1*5*9*...*(4*k + 1)). [added 19 Feb 2024: the general result can be proved by the WZ method as described in Wilf.]
In the limit as r -> oo we find that the constant equals Sum_{k >= 0} 2^(k-1)*k!/(Product_{i = 0..k} 4*i + 1) as noted above. (End)
From Peter Bala, Mar 03 2024: (Start)
Continued fraction: 1/(1 + 1^2/(4 + 5^2/(4 + 9^2/(4 + 13^2/(4 + ... ))))) due to Euler.
Equals hypergeom([1/4, 1], [5/4], -1).
Gauss's continued fraction: 1/(1 + 1^2/(5 + 4^2/(9 + 5^2/(13 + 8^2/(17 + 9^2/(21 + 12^2/(25 + 13^2/(29 + 16^2/(33 + 17^2/(37 + ... )))))))))). (End)

A212435 Expansion of e.g.f.: exp(-x) / cosh(2*x).

Original entry on oeis.org

1, -1, -3, 11, 57, -361, -2763, 24611, 250737, -2873041, -36581523, 512343611, 7828053417, -129570724921, -2309644635483, 44110959165011, 898621108880097, -19450718635716001, -445777636063460643, 10784052561125704811, 274613643571568682777

Offset: 0

Michael Somos, Jun 21 2012

sign

			G.f. = 1 - x - 3*x^2 + 11*x^3 + 57*x^4 - 361*x^5 - 2763*x^6 + 24611*x^7 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A001586, A046717, A188458, A188514, A255883.

m:=30; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(Exp(-x)/Cosh(2*x))); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Aug 10 2018

CoefficientList[Series[2*E^x/(E^(4*x)+1), {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Feb 25 2014 *)
a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ Exp[ -x] / Cosh[ 2 x], {x, 0, n}]]; (* Michael Somos, Aug 26 2015 *)

{a(n) = my(A); if( n<0, 0, A = x * O(x^n); n! * polcoeff( exp(-x + A) / cosh( 2*x + A), n))};

@CachedFunction
def p(n,x) :
    if n == 0 : return 1
    w = -1 if n%2 == 0 else  0
    v =  1 if n%2 == 0 else -1
    return v*add(p(k,0)*binomial(n,k)*(x^(n-k)+w) for k in range(n)[::2])
def A212435(n) : return 2^n*p(n, 1/2)
[A212435(n) for n in (0..20)]  # Peter Luschny, Jul 19 2012

E.g.f.: 2 * exp(x) / (exp(4*x) + 1).
E.g.f. is the reciprocal of the e.g.f. of A046717.
a(n) = (-1)^n * A188458(n) = (-1)^floor((n + 1) / 2) * A001586(n).
E.g.f.: 2/E(0), where E(k) = 1 + (-1)^k/(3^k - 3*9^k*x/(3*3^k*x + (-1)^k*(k+1)/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Oct 17 2013
G.f.: conjecture T(0)/(1+x), where T(k) = 1 - 4*x^2*(k+1)^2/(4*x^2*(k+1)^2 + (1+ x)^2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 12 2013
a(n) ~ n! * (cos(Pi*n/2)-sin(Pi*n/2)) * 2^(2*n+3/2) / Pi^(n+1). - Vaclav Kotesovec, Feb 25 2014
From Peter Bala, Mar 10 2015: (Start)
a(n) = 4^n*E(n,1/4).
O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 - x*(4*k + 1)).
The series expansion exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 - x - x^2 + 5*x^3 + 11*x^4 - 91*x^5 - 391*x^6 + ... appears to have integer coefficients. Cf. A188514, A255883. (End)

A181049 Decimal expansion of (Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+3).

Original entry on oeis.org

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

Offset: 0

Jonathan D. B. Hodgson, Oct 01 2010, Oct 05 2010

Let N be a positive integer divisible by 4. We have the asymptotic expansion 2*((Pi/2 - log(1 + sqrt(2)))/(2*sqrt(2)) - Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 3)) ~ 1/N - 1/N^2 - 3/N^3 + 11/N^4 + 57/N^5 - ..., where the sequence of coefficients [1, -1, -3, 11, 57, ...] is A212435. This follows from Borwein et al., Lemma 2 with f(x) = 1/x and then set x = N/4 and h = 3/4. An example is given below. Cf. A181048. - Peter Bala, Sep 23 2016

			0.2437477471996805241799750836323027110...
From _Peter Bala_, Sep 23 2016: (Start)
At N = 100000 the truncated series 2*Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 3) = 0.4874(8)5494(4)9936(4)048(24)99(444)67(625)6... to 32 digits. The bracketed numbers show where this decimal expansion differs from that of 2*A181049. The numbers 1, -1, -3, 11, 57, -361 must be added to the bracketed numbers to give the correct decimal expansion to 32 digits: 2*( Pi/2 - log(1+sqrt(2)))/(2*sqrt(2) ) = 0.4874(9)5494(3)9936(1)048(35)99(501)67(264)6.... (End)

Gheorghe Coserea, Table of n, a(n) for n = 0..2015
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Eric W. Weisstein, Euler's Series Transformation.

Cf. A113476, A001586, A093954, A181048, A212435, A247719.

C := ComplexField(); [(Pi(C)/2 - Log(1+Sqrt(2)))/(2*Sqrt(2))]; // G. C. Greubel, Nov 28 2017

Mathematica

First@ RealDigits[N[(Pi/2 - Log[1 + Sqrt@ 2])/(2 Sqrt@ 2), 105]] (* Michael De Vlieger, Oct 07 2015 *)

PARI

default(realprecision, 106); eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(4*n+3)))), "3..-2")) \\ Gheorghe Coserea, Oct 06 2015

PARI

(Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) \\ G. C. Greubel, Nov 28 2017

Formula

Equals Integral_{x=0..1} (x^2 dx)/(1+x^4).

Equals (1/2) * Integral_{x = 0..Pi/4} sqrt(tan(x)) dx. Cf. A247719. - Peter Bala, Sep 23 2016

Equals Sum_{n >= 0} 2^(n-1)*n!/(Product_{k = 0..n} 4*k + 3) = Sum_{n >= 0} 2^(n-1)*n!/A008545(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(4*k+3)). - Peter Bala, Dec 01 2021

From Peter Bala, Mar 03 2024: (Start)

Continued fraction: 1/(3 + 3^2/(4 + 7^2/(4 + 11^2/(4 + 15^2/(4 + ... ))))) due to Euler.

Equals (1/3)*hypergeom([3/4, 1], [7/4], -1).

Gauss's continued fraction: 1/(3 + 3^2/(7 + 4^2/(11 + 7^2/(15 + 8^2/(19 + 11^2/(23 + 12^2/(27 + 15^2/(31 + 16^2/(35 + 19^2/(39 + ... )))))))))). (End)

Equals Integral_{x=1..oo, y=1..oo} 1/(x^4 + y^4) dx. - Vaclav Kotesovec, Jun 13 2024

A349271 Array A(n, k) that generalizes Euler numbers, class numbers, and tangent numbers, read by ascending antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 3, 2, 2, 4, 8, 11, 5, 2, 4, 16, 46, 57, 16, 1, 6, 30, 128, 352, 361, 61, 2, 8, 46, 272, 1280, 3362, 2763, 272, 2, 8, 64, 522, 3522, 16384, 38528, 24611, 1385, 2, 12, 96, 904, 7970, 55744, 249856, 515086, 250737, 7936

Offset: 1

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Author

Peter Luschny, Nov 23 2021

Keywords

nonn
tabl

Examples

Seen as an array: [1] 1, 1, 1, 2, 5, 16, 61, 272, ... [A000111] [2] 1, 1, 3, 11, 57, 361, 2763, 24611, ... [A001586] [3] 1, 2, 8, 46, 352, 3362, 38528, 515086, ... [A007289] [4] 1, 4, 16, 128, 1280, 16384, 249856, 4456448, ... [A349264] [5] 2, 4, 30, 272, 3522, 55744, 1066590, 23750912, ... [A349265] [6] 2, 6, 46, 522, 7970, 152166, 3487246, 93241002, ... [A001587] [7] 1, 8, 64, 904, 15872, 355688, 9493504, 296327464, ... [A349266] [8] 2, 8, 96, 1408, 29184, 739328, 22634496, 806453248, ... [A349267] [9] 2, 12, 126, 2160, 49410, 1415232, 48649086, 1951153920, ... [A349268] . Seen as a triangle: [1] 1; [2] 1, 1; [3] 1, 1, 1; [4] 1, 2, 3, 2; [5] 2, 4, 8, 11, 5; [6] 2, 4, 16, 46, 57, 16; [7] 1, 6, 30, 128, 352, 361, 61; [8] 2, 8, 46, 272, 1280, 3362, 2763, 272; [9] 2, 8, 64, 522, 3522, 16384, 38528, 24611, 1385;

Links

William Y. C. Chen, Neil J. Y. Fan, and Jeffrey Y. T. Jia, The generating function for the Dirichlet series Lm(s), Mathematics of Computation, Vol. 81, No. 278, pp. 1005-1023, April 2012.

Ruth Lawrence and Don Zagier, Modular forms and quantum invariants of 3-manifolds, Asian J. Math. 3 (1999), no. 1, 93-107.

D. Shanks, Generalized Euler and class numbers, Math. Comp. 21 (1967) 689-694.

D. Shanks, Corrigendum: Generalized Euler and class numbers, Math. Comp. 22, (1968) 699.

D. Shanks, Generalized Euler and class numbers, Math. Comp. 21 (1967), 689-694; 22 (1968), 699. [Annotated scanned copy]

Crossrefs

A235605 (array generalized Euler secant numbers).

A235606 (array generalized Euler tangent numbers).

A349264 (overview generating functions).

Rows: A000111, A001586, A007289, A349265, A001587, A349266, A349267, A349268.

Columns: A000003 (class numbers), A000061, A000233, A000176, A000362, A000488, A000508, A000518.

Cf. A349263 (main diagonal).

A144015 Expansion of e.g.f. 1/(1 - sin(4*x))^(1/4).

Original entry on oeis.org

1, 1, 5, 29, 265, 3001, 42125, 696149, 13296145, 287706481, 6959431445, 186061833869, 5448382252825, 173418192216361, 5961442393047965, 220112963745653189, 8687730877758518305, 365023930617143804641, 16266420334783460443685, 766297734521812843642109

Offset: 0

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Author

Paul D. Hanna, Sep 09 2008

Keywords

nonn

Comments

Row sums of A186492 - Peter Bala, Feb 22 2011.

Examples

E.g.f.: A(x) = 1 + x + 5*x^2/2! + 29*x^3/3! + 265*x^4/4! + 3001*x^5/5! +... log(A(x)) = x + 4*x^2/2! + 16*x^3/3! + 128*x^4/4! + 1280*x^5/5! +... A(x)^2/A(-x)^2 = 1 + 4*x + 16*x^2/2! + 128*x^3/3! +...+ 4^n*A000111(n)*x^n/n! +... O.g.f.: 1/(1-x - 4*1*1*x^2/(1-5*x - 4*2*3*x^2/(1-9*x - 4*3*5*x^2/(1-13*x - 4*4*7*x^2/(1-17*x - 4*5*9*x^2/(1-...)))))) [continued fraction by Sergei Gladkovskii].

Crossrefs

Cf. A000111, A001586, A007788, A186492, A230134, A227544, A230114, A235329.

Cf. A007696, A136630.

Programs

Mathematica

CoefficientList[Series[1/(1-Sin[4*x])^(1/4), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 26 2013 *)

PARI

{a(n)=local(X=x+x*O(x^n)); n!*polcoeff((cos(2*X)-sin(2*X))^(-1/2), n)} for(n=0, 20, print1(a(n), ", "))

PARI

{a(n)=local(A=1+x+x*O(x^n));for(i=0,n,A=exp(intformal(A^2/subst(A^2,x,-x))));n!*polcoeff(A,n)} for(n=0, 20, print1(a(n), ", "))

PARI

/* From A'(x) = A(x)^3 / A(-x)^2: */ {a(n)=local(A=1); for(i=0, n, A=1+intformal(A^3/subst(A, x, -x)^2 +x*O(x^n) )); n!*polcoeff(A, n)} for(n=0, 20, print1(a(n), ", "))

PARI

/* 1/sqrt(1-2*Series_Reversion(Integral 1/sqrt(1+4*x-4*x^2) dx)): */ {a(n)=local(A=1);A=1/sqrt(1-2*serreverse(intformal(1/sqrt(1+4*x-4*x^2 +x*O(x^n)))));n!*polcoeff(A, n)} for(n=0,20,print1(a(n),", "))

PARI

a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j)); a007696(n) = prod(k=0, n-1, 4*k+1); a(n) = sum(k=0, n, a007696(k)*(4*I)^(n-k)*a136630(n, k)); \\ Seiichi Manyama, Jun 24 2025

Formula

E.g.f. A(x) satisfies:

(1) A(x) = (cos(2*x) - sin(2*x))^(-1/2).

(2) A(x)^2/A(-x)^2 = 1/cos(4*x) + tan(4*x).

(3) A(x) = exp( Integral A(x)^2/A(-x)^2 dx).

(4) A'(x) = A(x)^3/A(-x)^2 with A(0) = 1.

(5) A(x) = 1/sqrt(1 - 2*Series_Reversion( Integral 1/sqrt(1+4*x-4*x^2) dx )).

G.f.: 1/G(0) where G(k) = 1 - x*(4*k+1) - 4*x^2*(k+1)*(2*k+1)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 11 2013.

a(n) ~ 2^(3*n+5/4)*n^n/(exp(n)*Pi^(n+1/2)). - Vaclav Kotesovec, Jun 26 2013

a(n) = Sum_{k=0..n} A007696(k) * (4*i)^(n-k) * A136630(n,k), where i is the imaginary unit. - Seiichi Manyama, Jun 24 2025

A162660 Triangle read by rows: coefficients of the complementary Swiss-Knife polynomials.

Original entry on oeis.org

0, 1, 0, 0, 2, 0, -2, 0, 3, 0, 0, -8, 0, 4, 0, 16, 0, -20, 0, 5, 0, 0, 96, 0, -40, 0, 6, 0, -272, 0, 336, 0, -70, 0, 7, 0, 0, -2176, 0, 896, 0, -112, 0, 8, 0, 7936, 0, -9792, 0, 2016, 0, -168, 0, 9, 0, 0, 79360, 0, -32640, 0, 4032, 0, -240, 0, 10, 0

Offset: 0

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Peter Luschny, Jul 09 2009

Keywords

sign
tabl

Comments

Definition. V_n(x) = (skp(n, x+1) - skp(n, x-1))/2 where skp(n,x) are the Swiss-Knife polynomials A153641. - Peter Luschny, Jul 23 2012

Equivalently, let the polynomials V_n(x) (n>=0) defined by V_n(x) = Sum_{k=0..n} Sum_{v=0..k} (-1)^v*C(k,v)*L(k)*(x+v+1)^n; the sequence L(k) = -1 - H(k-1)*(-1)^floor((k-1)/4) / 2^floor(k/2) if k > 0 and L(0)=0; H(k) = 1 if k mod 4 <> 0, otherwise 0.

(1) V_n(0) = 2^n * Euler(n,1) for n > 0, A155585.

(2) V_n(1) = 1 - Euler(n).

(3) V_{n-1}(0) n / (4^n - 2^n) = B_n gives for n > 1 the Bernoulli numbers A027641/A027642.

(4) V_{n-1}(0) n (2/2^n-2)/(2^n-1) = G_n the Genocchi number A036968 for n > 1.

(5) V_n(1/2)2^{n} - 1 is a signed version of the generalized Euler (Springer) numbers, see A001586.

The Swiss-Knife polynomials (A153641) are complementary to the polynomials defined here. Adding both gives polynomials with e.g.f. exp(x*t)*(sech(t)+tanh(t)), the coefficients of which are a signed variant of A109449.

The Swiss-Knife polynomials as well as the complementary Swiss-Knife polynomials are closely related to the Bernoulli and Euler polynomials. Let F be a sequence and

P_{F}[n](x) = Sum_{k=0..n} Sum_{v=0..k} (-1)^v*C(k,v)*F(k)*(x+v+1)^n.

V_n(x) = P_{F}[n](x) with F(k)=L(k) defined above, are the Co-Swiss-Knife polynomials,

W_n(x) = P_{F}[n](x) with F(k)=c(k) the Chen sequence defined in A153641 are the Swiss-Knife polynomials.

B_n(x) = P_{F}[n](x-1) with F(k)=1/(k+1) are the Bernoulli polynomials,

E_n(x) = P_{F}[n](x-1) with F(k)=2^(-k) are the Euler polynomials.

The most striking formal difference between the Swiss-Knife-type polynomials and the Bernoulli-Euler type polynomials is: The SK-type polynomials have integer coefficients whereas the BE-type polynomials have rational coefficients.

Let R be the exponential Riordan array (exp(x)*sech(x), x) = P * A119879 = 2*P(I + P^2)^(-1) where P denotes Pascal's triangle A007318. Then T = R - I. - Peter Bala, Mar 07 2024

Examples

Triangle begins: [0] 0; [1] 1, 0; [2] 0, 2, 0; [3] -2, 0, 3, 0; [4] 0, -8, 0, 4, 0; [5] 16, 0, -20, 0, 5, 0; [6] 0, 96, 0, -40, 0, 6, 0; [7] -272, 0, 336, 0, -70, 0, 7, 0; [8] 0, -2176, 0, 896, 0, -112, 0, 8, 0; [9] 7936, 0, -9792, 0, 2016, 0, -168, 0, 9, 0;

Links

Ayse Yilmaz Ceylan and Yilmaz Simsek, Formulae for Generalization of Touchard Polynomials with Their Generating Functions, Symmetry (2025) Vol. 17, Issue 7, Art. No. 1126. See Eq. 28 and after.

Leonhard Euler (1735), De summis serierum reciprocarum, Opera Omnia I.14, E 41, 73-86; On the sums of series of reciprocals, arXiv:math/0506415 [math.HO], 2005-2008.

Peter Luschny, The Swiss-Knife polynomials.

Peter Luschny, Swiss-Knife Polynomials and Euler Numbers.

Wikipedia, Bernoulli number.

J. Worpitzky, Studien über die Bernoullischen und Eulerschen Zahlen, Journal für die reine und angewandte Mathematik, 94 (1883), 203-232.

Crossrefs

V_n(k), n=0, 1, ..., k=0: A155585, k=1: A009832,

V_n(k), k=0, 1, ..., V_0: A000004, V_1: A000012, V_2: A005843, V_3: A100536.

Cf. A153641, A154341, A154342, A154343, A154344, A154345.

Programs

Maple

# Polynomials V_n(x): V := proc(n,x) local k,pow; pow := (n,k) -> `if`(n=0 and k=0,1,n^k); add(binomial(n,k)*euler(k)*pow(x+1,n-k),k=0..n) - pow(x,n) end: # Coefficients a(n): seq(print(seq(coeff(n!*coeff(series(exp(x*t)*tanh(t),t,16),t,n),x,k),k=0..n)),n=0..8);

Mathematica

skp[n_, x_] := Sum[Binomial[n, k]*EulerE[k]*x^(n-k), {k, 0, n}]; v[n_, x_] := (skp[n, x+1]-skp[n, x-1])/2; t[n_, k_] := Coefficient[v[n, x], x, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 09 2014 *)

Sage

R = PolynomialRing(QQ, 'x') @CachedFunction def skp(n, x) : # Swiss-Knife polynomials A153641. if n == 0 : return 1 return add(skp(k, 0)*binomial(n, k)*(x^(n-k)-(n+1)%2) for k in range(n)[::2]) def A162660(n,k) : return 0 if k > n else R((skp(n, x+1)-skp(n, x-1))/2)[k] matrix(ZZ, 9, A162660) # Peter Luschny, Jul 23 2012

Formula

T(n, k) = [x^(n-k)](skp(n,x+1)-skp(n,x-1))/2 where skp(n,x) are the Swiss-Knife polynomials A153641. - Peter Luschny, Jul 23 2012

E.g.f. exp(x*t)*tanh(t) = 0*(t^0/0!) + 1*(t^1/1!) + (2*x)*(t^2/2!) + (3*x^2-2)*(t^3/3!) + ...

V_n(x) = -x^n + Sum_{k=0..n} C(n,k)*Euler(k)*(x+1)^(n-k).

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A104035 Triangle T(n,k), 0 <= k <= n, read by rows, defined by T(0,0) = 1; T(0,k) = 0 if k>0 or if k<0; T(n,k) = k*T(n-1,k-1) + (k+1)*T(n-1,k+1).

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A000281 Expansion of cos(x)/cos(2x).

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A188458 Expansion of e.g.f. exp(x)/cosh(2*x).

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A007788 Number of augmented Andre 3-signed permutations: E.g.f. (1-sin(3*x))^(-1/3).

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A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4*k+1).

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A212435 Expansion of e.g.f.: exp(-x) / cosh(2*x).

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A104035 Triangle T(n,k), 0 <= k <= n, read by rows, defined by T(0,0) = 1; T(0,k) = 0 if k>0 or if k<0; T(n,k) = kT(n-1,k-1) + (k+1)T(n-1,k+1).

A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+1).