A181718 a(n) = (1/9)*(10^(2*n) + 10^n - 2).

0, 12, 1122, 111222, 11112222, 1111122222, 111111222222, 11111112222222, 1111111122222222, 111111111222222222, 11111111112222222222, 1111111111122222222222, 111111111111222222222222, 11111111111112222222222222, 1111111111111122222222222222

Offset

0,2

Comments

In decimal, n times 1 followed by n times 2.

a(n) = 3 + 3*3, 33 + 33*33, 333 + 333*333, written with 3,6,9,12,... = A008585(n+1) 3's.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Formula

a(n) = A002277(n)*A093137(n).

G.f.: 6*x*(2-35*x) / ( (1-x)*(1-10*x)*(1-100*x) ). - R. J. Mathar, Feb 28 2011

a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3), a(0)=0, a(1)=12, a(2)=1122. - Harvey P. Dale, Jul 31 2013

a(n) = (A074992(n) - 1)/3. - Michel Marcus, Sep 14 2013

E.g.f.: (1/9)*(-2*exp(x) + exp(10*x) + exp(100*x)). - G. C. Greubel, Mar 25 2024

Mathematica

Table[FromDigits[Join[PadRight[{}, n, 1], PadRight[{}, n, 2]]], {n, 0, 20}] (* or *) LinearRecurrence[{111, -1110, 1000}, {0, 12, 1122}, 20] (* Harvey P. Dale, Jul 31 2013 *)

Prog

(Magma) [(1/9)*(10^(2*n) + 10^n - 2): n in [0..20]]; // Vincenzo Librandi, Aug 04 2011
(PARI) vector(30, n, n--; (10^(2*n) + 10^n - 2)/9) \\ G. C. Greubel, Nov 02 2018
(Python)
for n in range(30):
    print((10**(2*n)+10**n-2)//9, end=', ')
# Stefano Spezia, Nov 02 2018
(SageMath) [(100^n +10^n -2)//9 for n in range(31)] # G. C. Greubel, Mar 25 2024

Crossrefs

Cf. A002277, A008585, A074992, A093137.

Sequence in context: A260030 A160010 A325906 * A004073 A109229 A112457

Adjacent sequences: A181715 A181716 A181717 * A181719 A181720 A181721

Keyword

nonn,easy

Author

Paul Curtz, Nov 17 2010

Status

approved