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A081168 Differences of Beatty sequence for square root of 10.

%I A081168 #26 Jan 17 2024 01:52:40
%S A081168 3,3,3,3,3,3,4,3,3,3,3,3,4,3,3,3,3,3,4,3,3,3,3,3,4,3,3,3,3,3,4,3,3,3,
%T A081168 3,3,4,3,3,3,3,3,3,4,3,3,3,3,3,4,3,3,3,3,3,4,3,3,3,3,3,4,3,3,3,3,3,4,
%U A081168 3,3,3,3,3,4,3,3,3,3,3,3,4,3,3,3,3,3,4,3,3,3,3,3,4,3,3,3,3,3,4,3,3,3,3,3,4
%N A081168 Differences of Beatty sequence for square root of 10.
%C A081168 Let S(0) = 3; obtain S(k) from S(k-1) by applying the morphism 3 -> 333334, 4 -> 3333334; sequence is S(0), S(1), S(2), ...
%C A081168 More generally, for a(n,m) = floor((n+1)*sqrt(m^2+ 1)) - floor(n*sqrt(m^2+1)) start with m and apply the morphism: m -> m^(2m-1), m+1; m+1 -> m^(2m), m+1.
%H A081168 G. C. Greubel, <a href="/A081168/b081168.txt">Table of n, a(n) for n = 0..10000</a>
%F A081168 a(n) = floor((n+1)*sqrt(10)) - floor(n*sqrt(10)).
%t A081168 Differences[Floor[Sqrt[10]*Range[0, 120]]] (* _G. C. Greubel_, Jan 15 2024 *)
%o A081168 (PARI) a(n)=floor((n+1)*sqrt(10))-floor(n*sqrt(10))
%o A081168 (Magma)
%o A081168 A081168:= func< n | Floor((n+1)*Sqrt(10)) - Floor(n*Sqrt(10)) >;
%o A081168 [A081168(n): n in [0..120]]; // _G. C. Greubel_, Jan 15 2024
%o A081168 (SageMath)
%o A081168 def A081168(n): return floor((n+1)*sqrt(10)) - floor(n*sqrt(10))
%o A081168 [A081168(n) for n in range(121)] # _G. C. Greubel_, Jan 15 2024
%Y A081168 Cf. A010467, A006337, A177102.
%K A081168 nonn
%O A081168 0,1
%A A081168 _Benoit Cloitre_, Apr 16 2003