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%I #31 Mar 10 2024 15:11:31
%S 1,-1,1,3,-2,1,-11,9,-3,1,57,-44,18,-4,1,-361,285,-110,30,-5,1,2763,
%T -2166,855,-220,45,-6,1,-24611,19341,-7581,1995,-385,63,-7,1,250737,
%U -196888,77364,-20216,3990,-616,84,-8,1,-2873041,2256633,-885996,232092,-45486,7182,-924,108,-9,1
%N Number triangle read by rows, related to exp(x)/(cos(x) + sin(x)).
%H G. C. Greubel, <a href="/A117442/b117442.txt">Rows n = 0..50 of the triangle, flattened</a>
%F T(n, 0) = (-1)^n*A001586(n).
%F Sum_{k=0..n} T(n, k) = A117443(n).
%F Column k has e.g.f. (x^k/k!)/(cos(x) + sin(x)).
%F 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
%F From _G. C. Greubel_, Jun 02 2021: (Start)
%F 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.
%F T(n, n) = 1.
%F T(n, n-1) = -A000027(n) = -binomial(n+1, 1).
%F T(n, n-2) = A045943(n+1) = 3*binomial(n+2, 2).
%F T(n, n-3) = -A111080(n) = -11*binomial(n+3, 3).
%F T(j, k) = (-1)^k * binomial(j+k, k) * abs(numerator( Euler(k, 1/4) )) (columns).
%F T(n, n-j) = (-1)^n * binomial(n+j, j) * abs(numerator( Euler(n, 1/4) )) (downward diagonals). (End)
%F 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
%e Triangle begins
%e 1;
%e -1, 1;
%e 3, -2, 1;
%e -11, 9, -3, 1;
%e 57, -44, 18, -4, 1;
%e -361, 285, -110, 30, -5, 1;
%e 2763, -2166, 855, -220, 45, -6, 1;
%e -24611, 19341, -7581, 1995, -385, 63, -7, 1;
%p A117442_row := proc(n) 2^n*add(binomial(n,k)*euler(k)*((x+1)/2)^(n-k), k=0..n);
%p seq((-1)^(n-j)*abs(coeff(%,x,j)),j=0..n) end:
%p seq(print(A117442_row(n)),n=0..5); # _Peter Luschny_, Jun 08 2013
%t row[n_] := row[n] = 2^n Sum[Binomial[n, k] EulerE[k] ((x+1)/2)^(n-k), {k, 0, n}];
%t T[n_, k_] := (-1)^(n-k) Abs[Coefficient[row[n], x, k]];
%t Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* _Jean-François Alcover_, Jun 13 2019, from Maple *)
%t 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 *)
%o (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
%o p(n) = 2^n*sum(k=0, n, binomial(n,k)*E(k)*((x+1)/2)^(n-k));
%o row(n) = my(rp=p(n)); vector(n+1, k, k--; (-1)^(n-k)*abs(polcoeff(rp, k))); \\ _Michel Marcus_, Nov 16 2020
%o (Sage)
%o 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)
%o def A117442(n,k): return (-1)^(n+k)*binomial(n,k)*abs(numerator(f(n-k)))
%o flatten([[A117442(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Jun 02 2021
%Y Inverse of A117440.
%Y Second column contains A161722 as subsequence.
%Y Cf. A000027, A001586, A045943, A111080, A122045, A117443 (row sums).
%K easy,sign,tabl
%O 0,4
%A _Paul Barry_, Mar 16 2006
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