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%I #23 Feb 15 2021 22:32:38
%S 2,4,12,14,16,24,26,28,70,72,74,82,84,86,94,96,98,140,142,144,152,154,
%T 156,164,166,168,408,410,412,420,422,424,432,434,436,478,480,482,490,
%U 492,494,502,504,506,548,550,552,560,562,564,572,574,576,816,818
%N Numbers m such that Sum_{k=1..m} (<1/2 + k*r> - <k*r>) = 0, where r=sqrt(2) and < > denotes fractional part.
%C Suppose that r and c are real numbers, 0 < c < 1, and
%C ...
%C s(m) = Sum_{k=1..m} (<c+k*r> - <k*r>)
%C ...
%C where < > denotes fractional part. The inequalities s(m) < 0, s(m) = 0, s(m) > 0 yield up to three sequences that partition the set of positive integers, as in the examples cited below. Of particular interest are choices of r and c for which s(m) >= 0 for every m >= 1.
%C .
%C Note that s(m) = m*c - Sum_{k=1..m} floor(c + <k*r>). This shows that if c is a rational number p/q, then the range of s(m) is a set of rational numbers having denominator q. In this case, it is easy to prove that if s(m)=0, then m is an integer multiple of q, yielding a sequence of quotients denoted by [[m/q>]] in the following list:
%C .
%C r..........p/q....s(m)<0....s(m)=0....[[m/q]]...s(m)>0
%C sqrt(2)....1/2....(empty)...A194368...A194369...A194370
%C sqrt(3)....1/2....A194371...A194372.............A194373
%C sqrt(5)....1/2....(empty)...A194374.............A194375
%C sqrt(6)....1/2....(empty)...A194376.............A194377
%C sqrt(7)....1/2....A194378...A194379.............A194380
%C sqrt(8)....1/2....A194381...A194382...A194383...A194384
%C sqrt(10)...1/2....(empty)...A194385.............A194386
%C sqrt(11)...1/2....A194387...A194388.............A194389
%C sqrt(12)...1/2....(empty)...A194390.............A194391
%C sqrt(13)...1/2....A194392...A194393.............A194394
%C sqrt(14)...1/2....A194395...A194396.............A194397
%C sqrt(15)...1/2....A194398...A194399.............A194400
%C tau........1/2....A194401...A194402...A194403...A194404
%C e..........1/2....A194405...A194406.............A194407
%C Pi.........1/2....A194408...A194409.............A194410
%C sqrt(2)....1/3....A194411...A194412...A194413...A194414
%C sqrt(3)....1/3....A194415...A194416...A194417...A194418
%C sqrt(5)....1/3....A194419...A194420.............A194421
%C sqrt(2)....2/3....A194422...A194423...A194424...A194425
%C tau.....<tau>/2...A194461.......................A194462
%C tau.....<tau/2>...A194463.......................A194464
%C sqrt(2)....1/r.......A194465....................A194466
%C sqrt(3)....1/r.......A194467....................A194468
%C .
%C Next, suppose that r and c are chosen so that s(m)=0 for all m. Then the sets X={m : s(m)<0} and Y={m : s(m)>0} represent a pair of "generalized Beatty sequences" in this sense: if c=1/<r>, the sets X and Y represent the Beatty sequences of 1/<r> and 1<-r>. Examples:
%C ...
%C r..........c.........X.........Y......
%C sqrt(2)....r-1.......A003151...A003152
%C sqrt(3)....r-1.......A003511...A003512
%C tau........r-1.......A000201...A001950
%C sqrt(1/2)..r.........A001951...A001952
%C e..........e-2.......A000062...A098005
%D Ivan Niven, Diophantine Approximations, Interscience Publishers, 1963.
%H Ronald L. Graham, Shen Lin, Chio-Shih Lin, <a href="http://www.jstor.org/stable/2689998">Spectra of numbers</a>, Math. Mag. 51 (1978), 174-176.
%t r = Sqrt[2]; c = 1/2;
%t x[n_] := Sum[FractionalPart[k*r], {k, 1, n}]
%t y[n_] := Sum[FractionalPart[c + k*r], {k, 1, n}]
%t t1 = Table[If[y[n] < x[n], 1, 0], {n, 1, 100}];
%t Flatten[Position[t1, 1]] (* empty *)
%t t2 = Table[If[y[n] == x[n], 1, 0], {n, 1, 800}];
%t Flatten[Position[t2, 1]] (* A194368 *)
%t %/2 (* A194369 *)
%t t3 = Table[If[y[n] > x[n], 1, 0], {n, 1, 100}];
%t Flatten[Position[t3, 1]] (* A194370 *)
%Y Cf. A184369=(1/2)*A184368, A194285, A194469, A194470.
%K nonn
%O 1,1
%A _Clark Kimberling_, Aug 23 2011
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