OFFSET
0,3
COMMENTS
The denominators seem to be the natural numbers A000027.
The z-sequence gives the recurrence for column k=0 entries of the triangle T = A271703 (using also lower rows of T): T(n, 0) = Sum_{j=0..n-1} z(j)*T(n-1, j), n >=1, with T(0, 0) = 1. Because column k=0 has e.g.f. exp(x) all entries T(n, 0) = 1, and one obtains rational representations of 1 by the z-sequence recurrence. See the examples.
LINKS
FORMULA
E.g.f. for the rationals r(n): ((1+x)/x)*(1 - exp(-x/(1+x))).
a(n) = numerator(r(n)) (in lowest terms).
From G. C. Greubel, Jan 09 2022: (Start)
a(n) = (-1)^n * (n+1)! * Laguerre(n+1, -2, -1), with a(0) = 1.
a(n) = (-1)^n * (n+1)! * Sum_{j=0..n-1} binomial(n-1, j)/(j+2)!, with a(0) = 1. (End)
EXAMPLE
The rationals r(n) begin: 1, -1/2, 4/3, -21/4, 136/5, -1045/6, 9276/7, -93289/8, 1047376/9, -12975561/10, 175721140/11, -2581284541/12, 40864292184/13, -693347907421/14, 12548540320876/15, ...
Recurrence with T= P*Lah = A271703, rational representations of 1:
1 = T(2, 0) = 2*(1*T(1, 0) + (-1/2)*T(1, 1)) = 2*(1 - 1/2) = 1.
1 = T(3, 0) = 3*(1*1 + (-1/2)*4 + (4/3)*1) = 1.
1 = T(4, 0) = 4*(1*1 + (-1/2)*15 + (4/3)*9 + (-21/4)*1) = 1.
...
MATHEMATICA
Table[If[n==0, 1, (-1)^n*(n+1)!*LaguerreL[n+1, -2, -1]], {n, 0, 30}] (* G. C. Greubel, Jan 09 2022 *)
PROG
(Magma) [1] cat [(-1)^n*Factorial(n+1)*Evaluate(LaguerrePolynomial(n+1, -2), -1): n in [1..30]]; // G. C. Greubel, Jan 09 2022
(Sage) [1]+[(-1)^n*factorial(n+1)*gen_laguerre(n+1, -2, -1) for n in (1..30)] # G. C. Greubel, Jan 09 2022
(PARI) a(n) = if (n, (-1)^n*(n+1)!*subst(pollaguerre(n+1, -2), x, -1), 1); \\ Michel Marcus, Jan 09 2022
CROSSREFS
KEYWORD
sign,easy,frac
AUTHOR
Wolfdieter Lang, Jun 20 2017
STATUS
approved