A343149 Floor-powerfree numbers: positive integers not expressible as a (nontrivially) nested floor function using the same positive real slope throughout the nesting.

2, 3, 6, 7, 15, 23, 24, 44, 47, 48, 56, 57, 58, 59, 60, 61, 62, 63, 79, 97, 98, 113, 143, 167, 184, 185, 186, 210, 211, 212, 213, 214, 215, 222, 223, 247, 287, 320, 321, 356, 381, 462, 463, 474, 475, 481, 482, 483, 507, 508, 520, 521, 522, 553, 559, 604, 623

Offset

1,1

Comments

Any of these integers can be expressed by a composition of floor functions f(n) = [mu*n] and g(n) = [nu*n], provided that the composition applies at least one f(n) and one g(n), for an irrational slope 1 < mu < 2 and its conjugate nu = 1/(1-1/mu). This follows from the Rayleigh-Beatty theorem. See reference in link. A064801 gives "floor squares."

Links

Table of n, a(n) for n=1..57.

J. Parker Shectman, A Quit after Fibonacci, Part 2 of 3: Cohorts, Free Monoids, and Numeration

Example

Example (of calculation by sieve, see reference in link, p. 221): The first term, 2, while given by the (un-nested) floor [mu] of a real slope 2 <= mu < 3, is too big to result from a twice-nested floor [[mu]mu], thrice-nested floor [mu[mu[mu]]], etc. for mu < 2. Yet for mu >= 2, the integer 2 is too small to result from a twice-nested, thrice-nested, etc. floor. Sequence A064801 = 1,4,5,9,... gives the "floor squares" - positive integers that are expressible as the twice-nested floor [mu[mu]] for a positive real slope mu. Thus 2,3,6,7 and 8 are not "floor squares". Besides 0 and 1, the next smallest integer obtainable from nesting a floor function with real positive slope t times is 2^t. Thus, the sequence of positive "floor cubes" starts with 1 and continues 8,9,12,13,14,27,... Hence, the first level of the sieve catches the floor squares 1,4,5,9,..., the second level of the sieve catches the floor cubes 1,8,... So, 2,3,6, and 7 are the initial floor-powerfree numbers passing the sieve for all t >= 2.

Mathematica

(*Define the nested floor function.*)
NestedFloor[slope_, t_] := Nest[Function[Floor[#*slope]], 1, t]
(*Specify an upper bound on a(n) in DATA.*)
aMax = 1017;
(*Calculate the number of floor powers that must be sifted out.*)
tMax = Ceiling[Log[2, aMax]];
(*Initialize slopes for each floor power.*)
slopes = Table[{1}, {tMax}]; slopes[[1]] = Table[n, {n, 1, aMax}];
(*Initialize "floor-powerful" numbers for each floor power.*)
powerfuls = Table[{1}, {tMax}]; powerfuls[[1]] = Table[n, {n, 1, aMax}];
Do[n = 2; While[Last[powerfuls[[t]]] < aMax,
  (*Include slopes from previously sifted power as coarse slopes.*) coarseSlope = slopes[[t - 1]][[n]]; AppendTo[slopes[[t]], coarseSlope]; AppendTo[powerfuls[[t]], NestedFloor[coarseSlope, t]];
  (*Generate fine slopes between the coarse slopes; use floor-powerful numbers from previously sifted floor power as denominators q, start with a numerator p that gives the least fine slope exceeding the current coarse one*) q = powerfuls[[t - 1]][[n]]; p = Floor[coarseSlope*q] + 1; fineSlope = p/q;
  (*Insert fine slope(s) (if any) between the current coarse slope and the next smallest one.*) nextCoarse = slopes[[t - 1]][[n + 1]]; While[fineSlope < nextCoarse, AppendTo[slopes[[t]], fineSlope]; AppendTo[powerfuls[[t]], NestedFloor[fineSlope, t]]; p++; fineSlope = p/q; ]; n++], {t, 2, tMax}]
(*Sift out all floor-powerful numbers to output the floor-powerfree numbers, a(n)*)
Complement[Table[n, {n, 1, aMax}], Union[Flatten[Rest[powerfuls]]]]

Crossrefs

Cf. A064801.

Sequence in context: A125167 A137604 A034901 * A275390 A109976 A011768

Adjacent sequences: A343146 A343147 A343148 * A343150 A343151 A343152

Keyword

nonn,easy

Author

J. Parker Shectman, Apr 06 2021

Status

approved