A078140 Convolutory inverse of signed lower Wythoff sequence.

1, 3, 5, 9, 17, 30, 52, 90, 154, 262, 446, 758, 1285, 2176, 3683, 6230, 10533, 17803, 30085, 50831, 85873, 145063, 245037, 413891, 699082, 1180761, 1994293, 3368302, 5688920, 9608292, 16227841, 27407792, 46289925, 78180465, 132041227

Offset

1,2

Comments

Suppose that r is a real number in the interval [3/2, 5/3). Let C(r) = (c(k)) be the sequence of coefficients in the Maclaurin series for 1/(Sum_{k>=0} floor((k+1)*r))(-x)^k). It appears that c(k) > 0 for all k >= 0. Indeed, it appears that C(r) is strictly increasing and that the limit L(r) of c(k+1)/c(k) as k -> oo exists. Following is a guide for selected numbers r.

** r ** C(r) L(r)

sqrt(7/3) A188135 A288238

Pi/2 A288229 A288239

sqrt(5/2) A288230 A288240

4^(1/3) A288231 A288241

(1 + sqrt(5))/2 A078140 A281112

3e/5 A288232 A288242

sqrt(8/3) A288233 A288935

-1 + sqrt(7) A288234 A289003

sqrt(e) A288235 A289005

-4/5 + sqrt(6) A288236 A289032

sqrt(11/4) A288237 A289033

Links

Clark Kimberling, Table of n, a(n) for n = 1..1000

Clark Kimberling, Another question about the golden ratio and other numbers, MathOverflow, Jan 17 2017.

Formula

a(n) = d*[w(n)*a(1)-w(n-1)*a(2)+...+d*w(2)*a(n-1)], where d=(-1)^n, with a(1)=1 and w=floor(n*tau), tau=(1+sqrt(5))/2.

Example

a(5) = 17 = -[w(5)*a(1)-w(4)*a(2)+w(3)*a(3)-w(2)*a(4)] = -8*1+6*3-4*5+3*9. (a(1),a(2),...,a(n))(*)(w(1),-w(2),w(3),...,-d*w(n)) = (1,0,0,...,0), where (*) denotes convolution, w = lower Wythoff sequence, A000201.

Mathematica

CoefficientList[Series[1/Sum[Floor[GoldenRatio*(k + 1)] (-x)^k, {k, 0, 50}],
{x, 0, 50}], x]  (* Clark Kimberling, Dec 12 2016 *)

Crossrefs

Cf. A000201, A077607, A281112, A279676.

Sequence in context: A298338 A018162 A077879 * A279780 A289260 A279595

Adjacent sequences: A078137 A078138 A078139 * A078141 A078142 A078143

Keyword

nonn

Author

Clark Kimberling, Nov 23 2002

Extensions

Comments added by Clark Kimberling, Jul 10 2017

Status

approved