A247453 -id:A247453 - cp's OEIS frontend

A109449 Triangle read by rows, T(n,k) = binomial(n,k)*A000111(n-k), 0 <= k <= n.

1, 1, 1, 1, 2, 1, 2, 3, 3, 1, 5, 8, 6, 4, 1, 16, 25, 20, 10, 5, 1, 61, 96, 75, 40, 15, 6, 1, 272, 427, 336, 175, 70, 21, 7, 1, 1385, 2176, 1708, 896, 350, 112, 28, 8, 1, 7936, 12465, 9792, 5124, 2016, 630, 168, 36, 9, 1, 50521, 79360, 62325, 32640, 12810, 4032, 1050, 240, 45, 10, 1

Offset: 0

Philippe Deléham, Aug 27 2005

The boustrophedon transform {t} of a sequence {s} is given by t_n = Sum_{k=0..n} T(n,k)*s(k). Triangle may be called the boustrophedon triangle.
The 'signed version' of the triangle is the exponential Riordan array [sech(x) + tanh(x), x]. - Peter Luschny, Jan 24 2009
Up to signs, the matrix is self-inverse: T^(-1)(n,k) = (-1)^(n+k)*T(n,k). - R. J. Mathar, Mar 15 2013

			Triangle starts:
      1;
      1,     1;
      1,     2,     1;
      2,     3,     3,     1;
      5,     8,     6,     4,     1;
     16,    25,    20,    10,     5,    1;
     61,    96,    75,    40,    15,    6,    1;
    272,   427,   336,   175,    70,   21,    7,   1;
   1385,  2176,  1708,   896,   350,  112,   28,   8,  1;
   7936, 12465,  9792,  5124,  2016,  630,  168,  36,  9,  1;
  50521, 79360, 62325, 32640, 12810, 4032, 1050, 240, 45, 10, 1; ...

Reinhard Zumkeller, Rows n = 0..125 of table, flattened
Peter Luschny, The Swiss-Knife polynomials.
J. Millar, N. J. A. Sloane and N. E. Young, A new operation on sequences: the Boustrophedon transform, J. Combin. Theory, 17A 44-54 1996 (Abstract, pdf, ps).
Wikipedia, Boustrophedon transform
Index entries for sequences related to boustrophedon transform

Cf. A000111, A000667, A000795, A002084, A003719, A007318, A009747.
See also : A000182, A000964, A009739, A062161, A062272.
Cf. A153641, A162660.
Cf. A000667 (row sums), A247453.

a109449 n k = a109449_row n !! k
a109449_row n = zipWith (*)
                (a007318_row n) (reverse $ take (n + 1) a000111_list)
a109449_tabl = map a109449_row [0..]
-- Reinhard Zumkeller, Nov 02 2013

f:= func< n,x | Evaluate(BernoulliPolynomial(n+1), x) >;
A109449:= func< n,k | k eq n select 1 else 2^(2*n-2*k+1)*Binomial(n,k)*Abs(f(n-k,3/4) - f(n-k,1/4) + f(n-k,1) - f(n-k,1/2))/(n-k+1) >;
[A109449(n,k): k in [0..n], n in [0..13]]; // G. C. Greubel, Jul 10 2025

From Peter Luschny, Jul 10 2009, edited Jun 06 2022: (Start)
A109449 := (n,k) -> binomial(n, k)*A000111(n-k):
seq(print(seq(A109449(n, k), k=0..n)), n=0..9);
B109449 := (n,k) -> 2^(n-k)*binomial(n, k)*abs(euler(n-k, 1/2)+euler(n-k, 1)) -`if`(n-k=0, 1, 0): seq(print(seq(B109449(n, k), k=0..n)), n=0..9);
R109449 := proc(n, k) option remember; if k = 0 then A000111(n) else R109449(n-1, k-1)*n/k fi end: seq(print(seq(R109449(n, k), k=0..n)), n=0..9);
E109449 := proc(n) add(binomial(n, k)*euler(k)*((x+1)^(n-k)+ x^(n-k)), k=0..n) -x^n end: seq(print(seq(abs(coeff(E109449(n), x, k)), k=0..n)), n=0..9);
sigma := n -> ifelse(n=0, 1, [1,1,0,-1,-1,-1,0,1][n mod 8 + 1]/2^iquo(n-1,2)-1):
L109449 := proc(n) add(add((-1)^v*binomial(k, v)*(x+v+1)^n*sigma(k), v=0..k), k=0..n) end: seq(print(seq(abs(coeff(L109449(n), x, k)), k=0..n)), n=0..9);
X109449 := n -> n!*coeff(series(exp(x*t)*(sech(t)+tanh(t)), t, 24), t, n): seq(print(seq(abs(coeff(X109449(n), x, k)), k=0..n)), n=0..9);
(End)

lim = 10; s = CoefficientList[Series[(1 + Sin[x])/Cos[x], {x, 0, lim}], x] Table[k!, {k, 0, lim}]; Table[Binomial[n, k] s[[n - k + 1]], {n, 0, lim}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 24 2015, after Jean-François Alcover at A000111 *)
T[n_, k_] := (n!/k!) SeriesCoefficient[(1 + Sin[x])/Cos[x], {x, 0, n - k}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 27 2019 *)

A109449(n,k)=binomial(n,k)*if(n>k,2*abs(polylog(k-n,I)),1) \\ M. F. Hasler, Oct 05 2017

R = PolynomialRing(ZZ, 'x')
@CachedFunction
def skp(n, x) :
    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 A109449_row(n):
    x = R.gen()
    return [abs(c) for c in list(skp(n,x)-skp(n,x-1)+x^n)]
for n in (0..10) : print(A109449_row(n)) # Peter Luschny, Jul 22 2012

Sum_{k>=0} T(n, k) = A000667(n).
Sum_{k>=0} T(2n, 2k) = A000795(n).
Sum_{k>=0} T(2n, 2k+1) = A009747(n).
Sum_{k>=0} T(2n+1, 2k) = A003719(n).
Sum_{k>=0} T(2n+1, 2k+1) = A002084(n).
Sum_{k>=0} T(n, 2k) = A062272(n).
Sum_{k>=0} T(n, 2k+1) = A062161(n).
Sum_{k>=0} (-1)^(k)*T(n, k) = A062162(n). - Johannes W. Meijer, Apr 20 2011
E.g.f.: exp(x*y)*(sec(x)+tan(x)). - Vladeta Jovovic, May 20 2007
T(n,k) = 2^(n-k)*C(n,k)*|E(n-k,1/2) + E(n-k,1)| - [n=k] where C(n,k) is the binomial coefficient, E(m,x) are the Euler polynomials and [] the Iverson bracket. - Peter Luschny, Jan 24 2009
From Reikku Kulon, Feb 26 2009: (Start)
A109449(n, 0) = A000111(n), approx. round(2^(n + 2) * n! / Pi^(n + 1)).
A109449(n, n - 1) = n.
A109449(n, n) = 1.
For n > 0, k > 0: A109449(n, k) = A109449(n - 1, k - 1) * n / k. (End)
From Peter Luschny, Jul 10 2009: (Start)
Let p_n(x) = Sum_{k=0..n} Sum_{v=0..k} (-1)^v C(k,v)*F(k)*(x+v+1)^n, where F(0)=1 and for k>0 F(k)=-1 + s_k 2^floor((k-1)/2), s_k is 0 if k mod 8 in {2,6}, 1 if k mod 8 in {0,1,7} and otherwise -1. T(n,k) are the absolute values of the coefficients of these polynomials.
Another way to express the polynomials p_n(x) is
p_n(x) = -x^n + Sum_{k=0..n} binomial(n,k)*Euler(k)((x+1)^(n-k) + x^(n-k)). (End)
From Peter Bala, Jan 26 2011: (Start)
An explicit formula for the n-th row polynomial is
x^n + i*Sum_{k=1..n}((1+i)/2)^(k-1)*Sum_{j=0..k} (-1)^j*binomial(k,j)*(x+i*j)^n, where i = sqrt(-1). This is the triangle of connection constants between the polynomial sequences {Z(n,x+1)} and {Z(n,x)}, where Z(n,x) denotes the zigzag polynomials described in A147309.
Denote the present array by M. The first column of the array (I-x*M)^-1 is a sequence of rational functions in x whose numerator polynomials are the row polynomials of A145876 - the generalized Eulerian numbers associated with the zigzag numbers. (End)
Let skp{n}(x) denote the Swiss-Knife polynomials A153641. Then
T(n,k) = [x^(n-k)] |skp{n}(x) - skp{n}(x-1) + x^n|. - Peter Luschny, Jul 22 2012
T(n,k) = A007318(n,k) * A000111(n - k), k = 0..n. - Reinhard Zumkeller, Nov 02 2013
T(n,k) = abs(A247453(n,k)). - Reinhard Zumkeller, Sep 17 2014

Edited, formula corrected, typo T(9,4)=2016 (before 2816) fixed by Peter Luschny, Jul 10 2009

A062162 Boustrophedon transform of (-1)^n.

Original entry on oeis.org

1, 0, 0, 1, 0, 5, 10, 61, 280, 1665, 10470, 73621, 561660, 4650425, 41441530, 395757181, 4031082640, 43626778785, 499925138190, 6046986040741, 76992601769220, 1029315335116745, 14416214547400450, 211085887742964301, 3225154787165157400, 51329932704636904305

Offset: 0

Frank Ellermann, Jun 10 2001

Inverse binomial transform of Euler numbers A000111. - Paul Barry, Jan 21 2005

a(n) = abs(sum of row n in A247453). - Reinhard Zumkeller, Sep 17 2014

Alois P. Heinz, Table of n, a(n) for n = 0..200
Peter Luschny, An old operation on sequences: the Seidel transform.
Ludwig 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. [USA access only through the HATHI TRUST Digital Library]
Ludwig 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. [Access through ZOBODAT]
Wikipedia, Boustrophedon transform.
Index entries for sequences related to boustrophedon transform

Cf. A000111 (binomial transform).
Cf. A000667.
Cf. A247453.

a062162 = abs . sum . a247453_row -- Reinhard Zumkeller, Sep 17 2014

CoefficientList[Series[E^(-x)*(Tan[x]+1/Cos[x]), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Oct 05 2013 *)
t[n_, 0] := (-1)^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 *)

from itertools import islice, accumulate
def A062162_gen(): # generator of terms
    blist, m = tuple(), -1
    while True:
        yield (blist := tuple(accumulate(reversed(blist),initial=(m:=-m))))[-1]
A062162_list = list(islice(A062162_gen(),20)) # Chai Wah Wu, Jun 10 2022

# Generalized algorithm of L. Seidel (1877)
def A062162_list(n) :
    R = []; A = {-1:0, 0:0}
    k = 0; e = 1
    for i in range(n) :
        Am = (-1)^i
        A[k + e] = 0
        e = -e
        for j in (0..i) :
            Am += A[k]
            A[k] = Am
            k += e
        R.append(A[e*i//2])
    return R
A062162_list(22) # Peter Luschny, Jun 02 2012

E.g.f.: exp(-x)*(tan(x) + sec(x)). - Vladeta Jovovic, Feb 11 2003
a(n) ~ 4*(2*n/Pi)^(n+1/2)/exp(n+Pi/2). - Vaclav Kotesovec, Oct 05 2013
G.f.: E(0)*x/(1+x) + 1/(1+x), where E(k) = 1 - x^2*(k+1)*(k+2)/(x^2*(k+1)*(k+2) - 2*(x*k-1)*(x*(k+1)-1)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Jan 16 2014

A294033 Triangle read by rows, expansion of exp(xz)z*(tanh(z) + sech(z)), T(n, k) for n >= 1 and 0 <= k <= n-1.

Original entry on oeis.org

1, 2, 2, -3, 6, 3, -8, -12, 12, 4, 25, -40, -30, 20, 5, 96, 150, -120, -60, 30, 6, -427, 672, 525, -280, -105, 42, 7, -2176, -3416, 2688, 1400, -560, -168, 56, 8, 12465, -19584, -15372, 8064, 3150, -1008, -252, 72, 9, 79360, 124650, -97920, -51240, 20160, 6300, -1680, -360, 90, 10, -555731, 872960, 685575, -359040, -140910, 44352, 11550, -2640, -495, 110, 11

Offset: 1

Peter Luschny, Oct 24 2017

			Triangle starts:
  [1][   1]
  [2][   2,   2]
  [3][  -3,   6,    3]
  [4][  -8, -12,   12,    4]
  [5][  25, -40,  -30,   20,    5]
  [6][  96, 150, -120,  -60,   30,  6]
  [7][-427, 672,  525, -280, -105, 42, 7]

T(n, 0) = signed A065619. Row sums of abs(T(n,k)) = A231179.
Diagonals A000027, A002378, A027480, A162668.
A003506 (m=1), this seq. (m=2), A294034 (m=3).
Cf. A000111, A247453.

gf := exp(x*z)*z*(tanh(z)+sech(z)):
s := n -> n!*coeff(series(gf,z,n+2),z,n):
C := n -> PolynomialTools:-CoefficientList(s(n),x):
ListTools:-FlattenOnce([seq(C(n), n=1..7)]);
# Alternatively:
T := (n, k) -> `if`(n = k+1, n,
(k+1)*binomial(n,k+1)*2^(n-k-1)*(euler(n-k-1, 1/2)+euler(n-k-1, 1))):
for n from 1 to 7 do seq(T(n,k), k=0..n-1) od;

L[0] := 1; L[n_] := (-1)^Binomial[n, 2] 2 Abs[PolyLog[-n, -I]];
p[n_] := n Sum[Binomial[n - 1, k - 1] L[k - 1] x^(n - k), {k, 0, n}];
Table[CoefficientList[p[n], x], {n, 1, 11}] // Flatten

T(n, k) = (k+1)*binomial(n,k+1)*2^(n-k-1)*(Euler(n-k-1, 1/2) + Euler(n-k-1, 1)) for 0 <= k <= n-2.

T(n, k) is the coefficient of x^k of the polynomial p(n) = n*Sum_{k=1..n} binomial(n-1, k-1)*L(k-1)*x^(n-k) and L(n) = (-1)^binomial(n,2)*A000111(n). In particular n divides T(n, k).

A333303 T(n, k) = [x^k] (-2)^n*(B(n, x/2) - B(n, (x+1)/2)) where B(n, x) are the Bernoulli polynomials. Triangle read by rows, for 0 <= k <= n.

Original entry on oeis.org

0, 1, 1, -2, 0, -3, 3, -1, 0, 6, -4, 0, 5, 0, -10, 5, 3, 0, -15, 0, 15, -6, 0, -21, 0, 35, 0, -21, 7, -17, 0, 84, 0, -70, 0, 28, -8, 0, 153, 0, -252, 0, 126, 0, -36, 9, 155, 0, -765, 0, 630, 0, -210, 0, 45, -10, 0, -1705, 0, 2805, 0, -1386, 0, 330, 0, -55, 11

Offset: 0

Peter Luschny, May 07 2020

Can be seen as the Bernoulli counterpart of the Euler triangles A247453 and A109449.

			B*(8, z) = 1024*(Zeta(-7, (z+1)/2) - Zeta(-7, z/2))
         = -17 + 84*z^2 - 70*z^4 + 28*z^6 - 8*z^7.
Triangle starts:
[ 0] [  0]
[ 1] [  1]
[ 2] [  1,    -2]
[ 3] [  0,    -3,    3]
[ 4] [ -1,     0,    6,   -4]
[ 5] [  0,     5,    0,  -10,   5]
[ 6] [  3,     0,  -15,    0,  15,    -6]
[ 7] [  0,   -21,    0,   35,   0,   -21,    7]
[ 8] [-17,     0,   84,    0, -70,     0,   28,  -8]
[ 9] [  0,   153,    0, -252,   0,   126,    0, -36,  9]
[10] [155,     0, -765,    0, 630,     0, -210,   0, 45, -10]
[11] [  0, -1705,    0, 2805,   0, -1386,    0, 330,  0, -55, 11]

Michael De Vlieger, Table of n, a(n) for n = 0..11325 (rows 0 <= n <= 150, flattened)
Digital Library of Mathematical Functions, Lerch’s Transcendent.
Peter H. N. Luschny, An introduction to the Bernoulli function, arXiv:2009.06743 [math.HO], 2020.
Eric Weisstein's World of Mathematics, Lerch Transcendent.

Row sums are (-1)^n*A226158(n). Alternating row sums are A239977(n).

Cf. A181983, A247453, A109449, (A053382/A053383) Bernoulli polynomials.

B[n_, x_] := (-2)^n (BernoulliB[n, x/2] - BernoulliB[n, (x + 1)/2]);
Prepend[Table[CoefficientList[B[n, x], x], {n, 1, 11}], 0] // Flatten

def Bstar(n,x):
    return (-2)^n*(bernoulli_polynomial(x/2,n) - bernoulli_polynomial((x+1)/2,n))
print(flatten([expand(Bstar(n, x)).list() for n in (0..11)]))

Let B*(n, x) denote the alternating Bernoulli rational polynomial functions defined by Z*(s, x) = Phi(-1, s, x) and B*(s, x) = -s Z*(1 - s, x). Here Phi(z, s, x) is the Hurwitz-Lerch transcendent defined as an analytic continuation of Sum_{k>=0} z^k/(k+x)^s. Then T(n, k) = (-1)^n [x^k] 2 B*(n, x).
T(n, 0) = 2*(2^n - 1)*Bernoulli(n, 1) = n*Euler(n - 1, 1) = -A226158(n).
Main diagonal is (-1)^(n+1)*n = A181983(n).

A363394 Triangle read by rows. T(n, k) = A081658(n, k) + A363393(n, k) for k > 0 and T(n, 0) = 1.

Original entry on oeis.org

1, 1, 1, 1, 2, -1, 1, 3, -3, -2, 1, 4, -6, -8, 5, 1, 5, -10, -20, 25, 16, 1, 6, -15, -40, 75, 96, -61, 1, 7, -21, -70, 175, 336, -427, -272, 1, 8, -28, -112, 350, 896, -1708, -2176, 1385, 1, 9, -36, -168, 630, 2016, -5124, -9792, 12465, 7936

Offset: 0

Peter Luschny, Jun 06 2023

			The triangle T(n, k) begins:
  [0] 1;
  [1] 1, 1;
  [2] 1, 2,  -1;
  [3] 1, 3,  -3,   -2;
  [4] 1, 4,  -6,   -8,   5;
  [5] 1, 5, -10,  -20,  25,   16;
  [6] 1, 6, -15,  -40,  75,   96,   -61;
  [7] 1, 7, -21,  -70, 175,  336,  -427,  -272;
  [8] 1, 8, -28, -112, 350,  896, -1708, -2176,  1385;
  [9] 1, 9, -36, -168, 630, 2016, -5124, -9792, 12465, 7936;

Richard P. Stanley, A survey of alternating permutations, arXiv:0912.4240 [math.CO], 2009.
Index entries for sequences related to Euler numbers.

Variants (row reversed): A109449, A247453.

Cf. A081658 (signed secant part), A363393 (signed tangent part), A000111 (main diagonal), A122045, A155585 (aerated main diagonal), A000667, A062162 (row sums of signless variant).

# Variant, computes abs(T(n, k)):
P := n -> n!*coeff(series((sec(y) + tan(y))/exp(x*y), y, 24), y, n):
seq(print(seq((-1)^(n - k)*coeff(P(n), x, n - k), k = 0..n)), n = 0..9);

from functools import cache
@cache
def T(n: int, k: int) -> int:
    if k == 0: return 1
    if k == n:
        p = k % 2
        return p - sum(T(n, j) for j in range(p, n - 1, 2))
    return (T(n - 1, k) * n) // (n - k)
for n in range(10): print([T(n, k) for k in range(n + 1)])

|T(n, k)| = (-1)^(n - k) * n! * [x^(n - k)][y^n] (sec(y) + tan(y)) / exp(x*y).
T(n, k) = [x^(n - k)] -2^(k-(0^k))*(Euler(k, 0) + Euler(k, 1/2)) / (x-1)^(k + 1).
For a recursion see the Python program.
T(n, k) = [x^n] ((-1) + Sum_{j=0..n} binomial(n, j)*(Euler(j, 1) + Euler(j, 1/2))*(2*x)^j). - Peter Luschny, Nov 17 2024

cp's OEIS Frontend

A109449 Triangle read by rows, T(n,k) = binomial(n,k)*A000111(n-k), 0 <= k <= n.

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A062162 Boustrophedon transform of (-1)^n.

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A294033 Triangle read by rows, expansion of exp(x*z)*z*(tanh(z) + sech(z)), T(n, k) for n >= 1 and 0 <= k <= n-1.

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A333303 T(n, k) = [x^k] (-2)^n*(B(n, x/2) - B(n, (x+1)/2)) where B(n, x) are the Bernoulli polynomials. Triangle read by rows, for 0 <= k <= n.

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A363394 Triangle read by rows. T(n, k) = A081658(n, k) + A363393(n, k) for k > 0 and T(n, 0) = 1.

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A294033 Triangle read by rows, expansion of exp(xz)z*(tanh(z) + sech(z)), T(n, k) for n >= 1 and 0 <= k <= n-1.