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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
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
1,2
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
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)
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 *)
PROG
(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
(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
CROSSREFS
Sequence in context: A172265 A258272 A117727 * A128552 A238361 A178420
KEYWORD
nonn,easy
AUTHOR
Louis Ciotti (lciotti(AT)twcny.rr.com), Apr 13 2006
EXTENSIONS
Solution proposed by Mohammed BOUAYOUN (mohammed.bouayoun(AT)yahoo.fr), Apr 14 2006
STATUS
approved