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A194368
Numbers m such that Sum_{k=1..m} (<1/2 + k*r> - <k*r>) = 0, where r=sqrt(2) and < > denotes fractional part.
68
2, 4, 12, 14, 16, 24, 26, 28, 70, 72, 74, 82, 84, 86, 94, 96, 98, 140, 142, 144, 152, 154, 156, 164, 166, 168, 408, 410, 412, 420, 422, 424, 432, 434, 436, 478, 480, 482, 490, 492, 494, 502, 504, 506, 548, 550, 552, 560, 562, 564, 572, 574, 576, 816, 818
OFFSET
1,1
COMMENTS
Suppose that r and c are real numbers, 0 < c < 1, and
...
s(m) = Sum_{k=1..m} (<c+k*r> - <k*r>)
...
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.
.
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:
.
r..........p/q....s(m)<0....s(m)=0....[[m/q]]...s(m)>0
sqrt(2)....1/2....(empty)...A194368...A194369...A194370
sqrt(3)....1/2....A194371...A194372.............A194373
sqrt(5)....1/2....(empty)...A194374.............A194375
sqrt(6)....1/2....(empty)...A194376.............A194377
sqrt(7)....1/2....A194378...A194379.............A194380
sqrt(8)....1/2....A194381...A194382...A194383...A194384
sqrt(10)...1/2....(empty)...A194385.............A194386
sqrt(11)...1/2....A194387...A194388.............A194389
sqrt(12)...1/2....(empty)...A194390.............A194391
sqrt(13)...1/2....A194392...A194393.............A194394
sqrt(14)...1/2....A194395...A194396.............A194397
sqrt(15)...1/2....A194398...A194399.............A194400
tau........1/2....A194401...A194402...A194403...A194404
e..........1/2....A194405...A194406.............A194407
Pi.........1/2....A194408...A194409.............A194410
sqrt(2)....1/3....A194411...A194412...A194413...A194414
sqrt(3)....1/3....A194415...A194416...A194417...A194418
sqrt(5)....1/3....A194419...A194420.............A194421
sqrt(2)....2/3....A194422...A194423...A194424...A194425
tau.....<tau>/2...A194461.......................A194462
tau.....<tau/2>...A194463.......................A194464
sqrt(2)....1/r.......A194465....................A194466
sqrt(3)....1/r.......A194467....................A194468
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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:
...
r..........c.........X.........Y......
sqrt(2)....r-1.......A003151...A003152
sqrt(3)....r-1.......A003511...A003512
tau........r-1.......A000201...A001950
sqrt(1/2)..r.........A001951...A001952
e..........e-2.......A000062...A098005
REFERENCES
Ivan Niven, Diophantine Approximations, Interscience Publishers, 1963.
LINKS
Ronald L. Graham, Shen Lin, Chio-Shih Lin, Spectra of numbers, Math. Mag. 51 (1978), 174-176.
MATHEMATICA
r = Sqrt[2]; c = 1/2;
x[n_] := Sum[FractionalPart[k*r], {k, 1, n}]
y[n_] := Sum[FractionalPart[c + k*r], {k, 1, n}]
t1 = Table[If[y[n] < x[n], 1, 0], {n, 1, 100}];
Flatten[Position[t1, 1]] (* empty *)
t2 = Table[If[y[n] == x[n], 1, 0], {n, 1, 800}];
Flatten[Position[t2, 1]] (* A194368 *)
%/2 (* A194369 *)
t3 = Table[If[y[n] > x[n], 1, 0], {n, 1, 100}];
Flatten[Position[t3, 1]] (* A194370 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Aug 23 2011
STATUS
approved