

A104053


Triangle of coefficients in the numerators of rational functions in tanh(1) that express the (2n)th Du BoisReymond constants as C_0 = 0, C_2 = 4  1/(1tanh(1)), for n>1, C_2n = 3  (Sum_{k=0..n} a(n,k)*tanh(1)^k) / (2^n*n! * (1tanh(1))^n).


0



0, 1, 0, 1, 1, 1, 1, 0, 0, 3, 1, 5, 18, 13, 7, 11, 70, 135, 65, 10, 45, 111, 609, 1215, 1350, 1275, 621, 141, 1009, 6188, 16758, 27335, 26845, 12474, 2548, 1883, 10977, 81353, 270004, 511791, 584710, 420287, 216468, 70169, 3599, 146691, 1248210, 4715217, 10303461, 14439411
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OFFSET

0,10


COMMENTS

For n>0 the row sums = (1)^(n1) * (n1)! For n odd, the sum of the absolute values of the coefficients in the nth row = (2*(n1))!/n! (every other entry of A001761).
The sum of the (2n)th Du BoisReymond constants = 1/5 or is very close to 1/5.
For the 6th and 9th rows, the coefficients were adjusted from results of the residue evaluations so that double factorials ((2n)!! = 2^n*n! (A000165)) are in the denominators. For the 6th row they were multiplied by 3, for the 9th row they were multiplied by 9.
For n>1, Sum_{k=0..n} (nk+1)*a(n,k) = (1)^(n)*A001286(n1) [A001286 are Lah numbers: (n1)*n!/2].


LINKS

Table of n, a(n) for n=0..49.
Eric Weisstein. Du Bois Reymond Constants from MathWorld
Eric Weisstein. Double Factorial


FORMULA

For n>1, C_2n = 3  2 * Residue_{x=i} (x^2/((1+x^2)^n * (tan(x)  x))) (see MathWorld article).
For n>1, Sum_{k=0..n} (1)^(n+k)*a(n, k) = (2*(n1))!/n! (i.e., A001761(n1)).


MATHEMATICA

Table[2 Residue[x^2/((1+x^2)^n (Tan[x]x)), {x, I}], {n, 0, 9}]


CROSSREFS

Cf. A000142, A000165, A001761, A062545, A062546, A085466, A085467.
Sequence in context: A124740 A073597 A338172 * A187369 A039512 A140825
Adjacent sequences: A104050 A104051 A104052 * A104054 A104055 A104056


KEYWORD

hard,sign,tabl


AUTHOR

Gerald McGarvey, Mar 02 2005


EXTENSIONS

Added the keyword tabl Gerald McGarvey, Aug 20 2009


STATUS

approved



