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%I #21 Apr 29 2023 20:46:39
%S 0,0,1,1,2,4,8,5,9,3,9,2,5,7,5,9,2,8,0,0,8,9,9,8,8,7,5,1,4,0,6,0,7,4,
%T 2,4,0,7,1,9,9,1,0,0,1,1,2,4,8,5,9,3,9,2,5,7,5,9,2,8,0,0,8,9,9,8,8,7,
%U 5,1,4,0,6,0,7,4,2,4,0,7,1,9,9,1,0,0,1,1,2,4,8,5,9,3,9,2,5,7,5
%N Decimal expansion of 1/889.
%C Sum_{j>=0} tribonacci(j) / 10^(j+1).
%C From _Daniel Forgues_, May 04 2013: (Start)
%C Generalization (since tribonacci(j+3) =
%C tribonacci(j+2) + tribonacci(j+1) + tribonacci(j)):
%C 1/889 = Sum_{j>=0} tribonacci(j) / 10^(j+1), (this sequence)
%C 1/989899 = Sum_{j>=0} tribonacci(j) / 100^(j+1),
%C 1/998998999 = Sum_{j>=0} tribonacci(j) / 1000^(j+1),
%C 1/999899989999 = Sum_{j>=0} tribonacci(j) / 10000^(j+1),
%C ...
%C 1 / ((10^k)^3 - (10^k)^2 - (10^k)^1 - (10^k)^0) = 1 / (10^(3k) - 10^(2k) - 10^k - 1) = Sum_{j>=0} tribonacci(j) / (10^k)^(j+1), k >= 1.
%C Sum_{j>=0} 111^j / 1000^(j+1).
%C Generalization (since 111^(j+1) = 111*111^j):
%C 1/889 = Sum_{j>=0} 111^j / 1000^(j+1), (this sequence)
%C 1/9889 = Sum_{j>=0} 111^j / 10000^(j+1),
%C 1/99889 = Sum_{j>=0} 111^j / 100000^(j+1),
%C 1/999889 = Sum_{j>=0} 111^j / 1000000^(j+1),
%C ...
%C 1 / ((10^k)^1 - 111*(10^k)^0) = 1 / (10^k - 111) = Sum_{j>=0} 111^j / (10^k)^(j+1), k >= 3. (End)
%H <a href="/index/Rec#order_22">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1,1).
%F From _Chai Wah Wu_, Feb 03 2021: (Start)
%F a(n) = a(n-1) - a(n-21) + a(n-22) for n > 21.
%F G.f.: (-x^21 - 8*x^20 + 8*x^18 - 6*x^17 + 7*x^16 - 4*x^15 + 2*x^14 - 2*x^13 - 3*x^12 + 7*x^11 - 6*x^10 + 6*x^9 - 4*x^8 + 3*x^7 - 4*x^6 - 2*x^5 - x^4 - x^2)/(x^22 - x^21 + x - 1). (End)
%t Join[{0,0},RealDigits[1/889,10,120][[1]]] (* or *) PadRight[{},120,{0,0,1,1,2,4,8,5,9,3,9,2,5,7,5,9,2,8,0,0,8,9,9,8,8,7,5,1,4,0,6,0,7,4,2,4,0,7,1,9,9,1}] (* _Harvey P. Dale_, Dec 02 2018 *)
%Y Cf. A000073.
%K nonn,cons
%O 0,5
%A _N. J. A. Sloane_
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