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A151775
Triangle read by rows: T(n,k) = value of (d^2n/dx^2n) (tan^(2k)(x)/cos(x)) at the point x = 0.
1
1, 1, 2, 5, 28, 24, 61, 662, 1320, 720, 1385, 24568, 83664, 100800, 40320, 50521, 1326122, 6749040, 13335840, 11491200, 3628800, 2702765, 98329108, 692699304, 1979524800, 2739623040, 1836172800, 479001600, 199360981, 9596075582
OFFSET
0,3
COMMENTS
From Emeric Deutsch, Jun 27 2009: (Start)
T(n,0) = A000364(n), the Euler (or secant) numbers.
Sum of entries in row n = A000281(n).
(End)
EXAMPLE
Triangle begins:
1;
1, 2;
5, 28, 24;
61, 662, 1320, 720;
1385, 24568, 83664, 100800, 40320;
50521, 1326122, 6749040, 13335840, 11491200, 3628800;
MAPLE
A151775 := proc(n, k) if n= 0 then 1 ; else taylor( (tan(x))^(2*k)/cos(x), x=0, 2*n+1) ; diff(%, x$(2*n)) ; coeftayl(%, x=0, 0) ; fi; end: for n from 0 to 10 do for k from 0 to n do printf("%d ", A151775(n, k)) ; od: printf("\n") ; od: # R. J. Mathar, Jun 24 2009
T := proc (n, k) if n = 0 and k = 0 then 1 elif n = 0 then 0 else simplify(subs(x = 0, diff(tan(x)^(2*k)/cos(x), `$`(x, 2*n)))) end if end proc: for n from 0 to 7 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form; Emeric Deutsch, Jun 27 2009
# alternative Maple program:
T:= (n, k)-> (2*n)!*coeff(series(tan(x)^(2*k)/cos(x), x, 2*n+1), x, 2*n):
seq(seq(T(n, k), k=0..n), n=0..10); # Alois P. Heinz, Aug 06 2017
MATHEMATICA
T[n_, k_] := (2n)! SeriesCoefficient[Tan[x]^(2k)/Cos[x], {x, 0, 2n}];
T[0, 0] = 1;
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 10 2019, after Alois P. Heinz *)
CROSSREFS
A subtriangle of A008294.
Cf. A000364, A000281. [Emeric Deutsch, Jun 27 2009]
Sequence in context: A208227 A127357 A025170 * A286879 A326230 A095159
KEYWORD
nonn,tabl
AUTHOR
N. J. A. Sloane, Jun 24 2009, at the suggestion of Alexander R. Povolotsky
EXTENSIONS
More values from R. J. Mathar and Emeric Deutsch, Jun 24 2009
STATUS
approved