A117442 Number triangle read by rows, related to exp(x)/(cos(x) + sin(x)).
1, -1, 1, 3, -2, 1, -11, 9, -3, 1, 57, -44, 18, -4, 1, -361, 285, -110, 30, -5, 1, 2763, -2166, 855, -220, 45, -6, 1, -24611, 19341, -7581, 1995, -385, 63, -7, 1, 250737, -196888, 77364, -20216, 3990, -616, 84, -8, 1, -2873041, 2256633, -885996, 232092, -45486, 7182, -924, 108, -9, 1
Offset: 0
Examples
Triangle begins
1;
-1, 1;
3, -2, 1;
-11, 9, -3, 1;
57, -44, 18, -4, 1;
-361, 285, -110, 30, -5, 1;
2763, -2166, 855, -220, 45, -6, 1;
-24611, 19341, -7581, 1995, -385, 63, -7, 1;
Links
- G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Crossrefs
Programs
-
Maple
A117442_row := proc(n) 2^n*add(binomial(n,k)*euler(k)*((x+1)/2)^(n-k), k=0..n); seq((-1)^(n-j)*abs(coeff(%,x,j)),j=0..n) end: seq(print(A117442_row(n)),n=0..5); # Peter Luschny, Jun 08 2013
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Mathematica
row[n_] := row[n] = 2^n Sum[Binomial[n, k] EulerE[k] ((x+1)/2)^(n-k), {k, 0, n}]; T[n_, k_] := (-1)^(n-k) Abs[Coefficient[row[n], x, k]]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Jean-François Alcover, Jun 13 2019, from Maple *) Table[(-1)^(n-k)*Binomial[n, k]*Abs[Numerator[EulerE[n-k, 1/4]]], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Jun 02 2021 *) -
PARI
E(n) = 2^n*2^(n+1)*(subst(bernpol(n+1, x), x, 3/4) - subst(bernpol(n+1, x), x, 1/4))/(n+1); \\ A122045 p(n) = 2^n*sum(k=0, n, binomial(n,k)*E(k)*((x+1)/2)^(n-k)); row(n) = my(rp=p(n)); vector(n+1, k, k--; (-1)^(n-k)*abs(polcoeff(rp, k))); \\ Michel Marcus, Nov 16 2020
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Sage
def f(n): return (1/4)^n*sum( binomial(n, j)*2^j*euler_number(j) for j in (0..n)) # f(n) = Euler(n, 1/4) def A117442(n,k): return (-1)^(n+k)*binomial(n,k)*abs(numerator(f(n-k))) flatten([[A117442(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 02 2021
Formula
T(n, 0) = (-1)^n*A001586(n).
Sum_{k=0..n} T(n, k) = A117443(n).
Column k has e.g.f. (x^k/k!)/(cos(x) + sin(x)).
Apart from signs the T(n,k) are the coefficients of the polynomials p(n, x) = 2^n*Sum_{k=0..n} binomial(n,k)*euler(k)*((x+1)/2)^(n-k). - Peter Luschny, Jun 08 2013
From G. C. Greubel, Jun 02 2021: (Start)
T(n, k) = (-1)^(n+k) * binomial(n, k) * abs(numerator( Euler(n-k, 1/4) )), where Euler(n, x) is the Euler number polynomial.
T(n, n) = 1.
T(n, n-1) = -A000027(n) = -binomial(n+1, 1).
T(n, n-2) = A045943(n+1) = 3*binomial(n+2, 2).
T(n, n-3) = -A111080(n) = -11*binomial(n+3, 3).
T(j, k) = (-1)^k * binomial(j+k, k) * abs(numerator( Euler(k, 1/4) )) (columns).
T(n, n-j) = (-1)^n * binomial(n+j, j) * abs(numerator( Euler(n, 1/4) )) (downward diagonals). (End)
The pair of triangles P*((I + P^4)/2)^(-1) and P^3*((I + P^4)/2)^(-1), where P denotes Pascal's triangle A007318, give the present triangle but with a different pattern of signs. - Peter Bala, Mar 07 2024