A117713 a(1)=1, a(2)=3, a(3)=8; for n>=4, a(n) = 10*a(n-3) + 8 (if a(n-3) is odd) or + 9 (if a(n-3) is even).

1, 3, 8, 18, 38, 89, 189, 389, 898, 1898, 3898, 8989, 18989, 38989, 89898, 189898, 389898, 898989, 1898989, 3898989, 8989898, 18989898, 38989898, 89898989, 189898989, 389898989, 898989898, 1898989898, 3898989898, 8989898989, 18989898989, 38989898989, 89898989898

Offset: 1

Author

Louis Ciotti (lciotti(AT)twcny.rr.com), Apr 13 2006

Keywords

• nonn
• easy

Comments

From a puzzle (1,3,8,18,?,89,189) given on a civil service test.

Another possibility is that 1,3,8,18,?,89,189,... is an erroneous version of A117727. - Hugo van der Sanden, Apr 14 2006

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,0,9,-9,0,10,-10).

Crossrefs

Cf. A010892, A117727.

Programs

Magma

I:=[1,3,8,18,38,89,189]; [n le 7 select I[n] else Self(n-1) +9*Self(n-3) -9*Self(n-4) +10*Self(n-6) -10*Self(n-7): n in [1..40]]; // G. C. Greubel, Jul 23 2023

Maple

f:=proc(n) option remember; local t1; if n=1 then RETURN(1); fi; if n=2 then RETURN(3); fi; if n=3 then RETURN(8); fi; t1:=10*f(n-3)+8; if f(n-3) mod 2 = 0 then t1:=t1+1; fi; RETURN(t1); end;

Mathematica

a[n_]:= a[n]= If[n<4, Fibonacci[2*n], 10*a[n-3] +If[Mod[a[n-3], 2]==1, 8, 9]];
Table[a[n], {n, 40}] (* G. C. Greubel, Jul 23 2023 *)

SageMath

@CachedFunction
def a(n): # a = A117713
    if (n<4): return fibonacci(2*n)
    elif (a(n-3)%2)==1: return 10*a(n-3) + 8
    else: return 10*a(n-3) + 9
[a(n) for n in range(1,41)] # G. C. Greubel, Jul 23 2023

Formula

From G. C. Greubel, Jul 23 2023: (Start)

a(n) = (1/198)*(2*(89*b(n) + 188*b(n-1) + 386*b(n-2)) + 6*(A010892(n) + A010892(n-1)) - 187 + 3*(-1)^n), where b(n) = 10^floor(n/3)*floor((n-1 mod 3)/2).

G.f.: x*(1 + 2*x + 5*x^2 + x^3 + 2*x^4 + 6*x^5)/((1-x^2)*(1-x+x^2)*(1-10*x^3)). (End)

Extensions

Solution proposed by Mohammed BOUAYOUN (mohammed.bouayoun(AT)yahoo.fr), Apr 14 2006