A268099 a(n) = 2^(n mod 2)*5*10^floor(n/2) - 1.

4, 9, 49, 99, 499, 999, 4999, 9999, 49999, 99999, 499999, 999999, 4999999, 9999999, 49999999, 99999999, 499999999, 999999999, 4999999999, 9999999999, 49999999999, 99999999999, 499999999999, 999999999999, 4999999999999, 9999999999999, 49999999999999

Offset

0,1

Comments

These numbers x also have the property that the Fibonacci sequence starting (1, x, ...) contains the concatenation of 1 and x, but there are other solutions, e.g., x = 14285.

Links

Table of n, a(n) for n=0..26.

R. Israel, in reply to E. Angelini, Fibonacci concatenated, SeqFan list, Jan. 26, 2016

Index entries for linear recurrences with constant coefficients, signature (1,10,-10).

Formula

G.f.: (-4/(1 - x) + (2*Sqrt(10))/(1 - Sqrt(10)*x) - 10/(-1 + Sqrt(10)*x) + 10/(1 + Sqrt(10)*x) - (2*Sqrt(10))/(1 + Sqrt(10)*x))/4. - Michael De Vlieger, Jan 26 2016

G.f.: ( 4+5*x ) / ( (x-1)*(10*x^2-1) ). - R. J. Mathar, Mar 12 2017

Mathematica

Table[2^Mod[n, 2]*5*10^Floor[n/2] - 1, {n, 0, 30}] (* or *) LinearRecurrence[{0, 10}, {5, 10}, 30] - 1 (* or *) CoefficientList[Series[(-4/(1 - x) + (2 Sqrt[10])/(1 - Sqrt[10] x) - 10/(-1 + Sqrt[10] x) + 10/(1 + Sqrt[10] x) - (2 Sqrt[10])/(1 + Sqrt[10] x))/4, {x, 0, 30}], x] (* Michael De Vlieger, Jan 26 2016 *)

Prog

(PARI) a(n)=2^bittest(n, 0)*5*10^(n\2)-1

Crossrefs

Sequence in context: A053961 A055812 A368337 * A061867 A019544 A053059

Adjacent sequences: A268096 A268097 A268098 * A268100 A268101 A268102

Keyword

nonn,easy,changed

Author

M. F. Hasler, Jan 26 2016

Status

approved