A386686 Classification of the interval [4*n+1, 4*n+2, 4*n+3] according to positions of squarefree composite numbers.

0, 2, 2, 3, 0, 6, 2, 2, 7, 3, 2, 2, 1, 1, 6, 2, 6, 6, 2, 6, 2, 7, 1, 7, 0, 2, 6, 3, 3, 3, 3, 0, 6, 6, 2, 7, 6, 0, 3, 3, 4, 6, 2, 2, 6, 3, 7, 2, 3, 0, 7, 6, 6, 7, 7, 6, 2, 3, 1, 6, 0, 3, 4, 7, 3, 2, 7, 0, 6, 2, 2, 7, 3, 1, 3, 7, 4, 6, 2, 3, 7, 3, 6, 3, 1, 4, 6

Offset

0,2

Comments

The maximum run of squarefree numbers is 3, since 4*n is nonsquarefree. Because of this, we may classify the n-th interval according to which numbers, 4*n+1, 4*n+2, or 4*n+3, are squarefree. Since a number is either squarefree (in A005117) or not (in A013929), there 2^3 possibilities, hence 8 classes.

The sequence that employs this classification methodology is A063736; this sequence borrows Labos' classification in A063736 but ignores primes.

Links

Michael De Vlieger, Table of n, a(n) for n = 0..14641

Michael De Vlieger, Plot a(n) at (x,y) = (n mod p^2, -floor(n/p^2)) for prime p = {3, 5, 7, 11, 13} and y <= 120 using color function described in the example.

Formula

Let f(x) = A008966(x) - A010051(x) = A354819(x) for x > 1, with f(1) = 0.

Thus f(k) = 1 for squarefree composite k (in A120944), else f(k) = 0 for k that are neither squarefree nor composite (in A363597).

a(n) = 4*f(4*n+1) + 2*f(4*n+2) + f(4*n+3).

Example

Table of possible configurations of squarefree numbers that come between multiples of 4, along with a simple algorithm of representing same with RGB color:
    R          G          B
f(4*n+1)   f(4*n+2)   f(4*n+3) -> binary -> a(n)   RGB color
------------------------------------------------------------
    0          0          0        000        0    black
    0          0          1        001        1    blue
    0          1          0        010        2    green
    0          1          1        011        3    cyan
    1          0          0        100        4    red
    1          0          1        101        5    magenta
    1          1          0        110        6    yellow
    1          1          1        111        7    white
a(0) = 0 since, for n = 0, 4*n+1 = 1, 4*n+2 = 2, and 4*n+3 = 3, none are both squarefree and composite, therefore we have the binary number 000_2 = 0.
a(1) = 2 since, for n = 1, 4*n+1 = 5, 4*n+2 = 6, and 4*n+3 = 7, only 6 is squarefree and composite, thus we have the binary number 010_2 = 2.
a(2) = 2 since, for n = 2, 4*n+1 = 9, 4*n+2 = 10, and 4*n+3 = 11, only 10 is squarefree and composite, thus we have the binary number 010_2 = 2.
a(5) = 6 since, for n = 5, 4*n+1 = 21, 4*n+2 = 22, and 4*n+3 = 23, only the first 2 are both squarefree and composite, thus we have the binary number 110_2 = 6.
a(8) = 7 since, for n = 8, 4*n+1 = 33, 4*n+2 = 34, and 4*n+3 = 35, all are both squarefree and composite, thus we have the binary number 111_2 = 7, etc.

Mathematica

Table[FromDigits[#, 2] &@ Map[Boole[And[CompositeQ[#], SquareFreeQ[#] ] ] &, 4*n + Range[3] ], {n, 0, 120}]

Crossrefs

Cf. A008966, A010051, A063736, A120944, A354819, A363597, A389685.

Sequence in context: A199784 A376931 A349384 * A127466 A342314 A099118

Adjacent sequences: A386683 A386684 A386685 * A386687 A386688 A386689

Keyword

nonn,base,easy

Author

Michael De Vlieger, Oct 25 2025

Status

approved