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First differences of the Beatty sequence A276854 for 1 + sqrt(5).
4

%I #23 Mar 16 2021 20:32:39

%S 3,3,3,3,4,3,3,3,4,3,3,3,4,3,3,3,4,3,3,3,3,4,3,3,3,4,3,3,3,4,3,3,3,4,

%T 3,3,3,3,4,3,3,3,4,3,3,3,4,3,3,3,4,3,3,3,3,4,3,3,3,4,3,3,3,4,3,3,3,4,

%U 3,3,3,3,4,3,3,3,4,3,3,3,4,3,3,3,4,3

%N First differences of the Beatty sequence A276854 for 1 + sqrt(5).

%H Clark Kimberling, <a href="/A276863/b276863.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = floor(n*r) - floor(n*r - r), where r = 1 + sqrt(5), n >= 1.

%F a(n) = A188187(n) + 3, as follows right from the definitions. - _Michel Dekking_, Sep 02 2019

%F a(n) = 1+floor(n*sqrt(5))-floor((n-1)*sqrt(5)). - _Chai Wah Wu_, Mar 16 2021

%t z = 500; r = 1+Sqrt[5]; b = Table[Floor[k*r], {k, 0, z}]; (* A276854 *)

%t Differences[b] (* A276863 *)

%o (Python)

%o from sympy import integer_nthroot

%o def A276863(n): return 1+integer_nthroot(5*n**2,2)[0]-integer_nthroot(5*(n-1)**2,2)[0] # _Chai Wah Wu_, Mar 16 2021

%Y Cf. A188187, A276854, A276881.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Sep 24 2016