A355200 Numbers k that can be written as the sum of 3 divisors of k (not necessarily distinct).

3, 4, 6, 8, 9, 12, 15, 16, 18, 20, 21, 24, 27, 28, 30, 32, 33, 36, 39, 40, 42, 44, 45, 48, 51, 52, 54, 56, 57, 60, 63, 64, 66, 68, 69, 72, 75, 76, 78, 80, 81, 84, 87, 88, 90, 92, 93, 96, 99, 100, 102, 104, 105, 108, 111, 112, 114, 116, 117, 120, 123, 124, 126, 128, 129, 132, 135

Offset: 1

Author

Wesley Ivan Hurt, Jun 23 2022

Keywords

• nonn
• easy

Comments

From Bernard Schott, Aug 06 2023: (Start)

Equivalently: positive numbers that are divisible by 3 or by 4.

Proof (similar to the proof proposed by Robert Israel in A355641).

If k is divisible by 3, then k is in the sequence because k = k/3 + k/3 + k/3.

If k is divisible by 4, then k is in the sequence because k = k/2 + k/4 + k/4.

Moreover, if k is positive and divisible by 6 (A008588), then k = k/3 + k/3 + k/3, but k is also in the sequence because k = k/2 + k/3 + k/6.

Conversely, to show that every term of this sequence is divisible by 3 or by 4, we consider all positive integer solutions of the equation 1 = 1/a + 1/b + 1/c. Without loss of generality, we may assume a <= b <= c, then 3/a >= 1/a + 1/b + 1/c = 1. So a <= 3. Similarly, given a, we have 2/b >= 1/b + 1/c = 1 - 1/a, so b <= 2/(1 - 1/a).

-> if a = 1, then 1 = 1 + 1/b + 1/c; this equation has clearly no solution.

-> if a = 2, then 1/2 = 1/b + 1/c with b <= 2/(1 - 1/2) = 4; in this case, there are two solutions: (a,b,c) = (2,3,6) or (a,b,c) = (2,4,4).

-> if a = 3, then 2/3 = 1/b + 1/c with b <= 2/(1 - 1/3) = 3; in this case, there is one solution: (a,b,c) = (3,3,3).

It turns out that there are only 3 solutions with a <= b <= c. Each corresponds to a possible pattern k = k/a + k/b + k/c for writing k as the sum of 3 of its divisors, which works when k is divisible by 3 or by 4. (End)

From David A. Corneth, Aug 07 2023: (Start)

Proof that a(n + 6) = a(n) + 12.

As k is in the sequence, k = k/d1 + k/d2 + k/d3 where d1, d2 and d3 | k and they are not necessarily distinct. By discussion above from Bernard Schott, Aug 06 2023, (d1, d2, d3) are in {(2, 3, 6), (2, 4, 4), (3, 3, 3)}. The lcm of these tuples are 6, 4 and 3 respectively. So any number k in the sequence is divisible by 3, 4 or 6.

This tells us that if k is in the sequence then k + 12 is in the sequence since k + 12 is divisible by one of 3, 4 or 6 since lcm(3, 4, 6) = 12.

So we can write a(n + m) = a(n) + 12 for some m. Inspection gives m = 6 so a(n + 6) = a(n) + 12. (End)

Examples

			6 is in the sequence since it can be written as the sum of 3 of its (not necessarily distinct) divisors: 6 = 1+2+3 = 2+2+2 with 1|6, 2|6, and 3|6.
		

Links

• Ray Chandler, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-2,2,-1).

Crossrefs

Cf. A002966, A008588, A348536.

Numbers k that can be written as the sum of j divisors of k (not necessarily distinct) for j=1..10: A000027 (j=1), A299174 (j=2), this sequence (j=3), A354591 (j=4), A355641 (j=5), A356609 (j=6), A356635 (j=7), A356657 (j=8), A356659 (j=9), A356660 (j=10).

Equals positive terms of A008585 union A008586.

Cf. A005843, A134667.

Programs

Mathematica

q[n_, k_] := AnyTrue[Tuples[Divisors[n], k], Total[#] == n &]; Select[Range[135], q[#, 3] &] (* Amiram Eldar, Aug 21 2022 *)
Table[2n + (Sin[Pi*n/3] + Sin[2*Pi*n/3])/Sqrt[3], {n, 100}] (* Wesley Ivan Hurt, Oct 30 2023 *)
CoefficientList[Series[(3 - 2*x + 4*x^2 - 2*x^3 + 3*x^4)/((x - 1)^2*(1 + x^2 + x^4)), {x, 0, 80}], x] (* Wesley Ivan Hurt, Jul 17 2025 *)

PARI

isok(k) = my(d=divisors(k)); forpart(p=k, if (setintersect(d, Set(p)) == Set(p), return(1)), , [3,3]); \\ Michel Marcus, Aug 21 2022

PARI

is(n) = my(v = [3,4,6]); sum(i = 1, 3, n%v[i] == 0) > 0 \\ David A. Corneth, Oct 08 2022

Formula

a(n + 6) = a(n) + 12. - David A. Corneth, Oct 08 2022

Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2)/3 - log(3)/4. - Amiram Eldar, Sep 10 2023

From Wesley Ivan Hurt, Oct 30 2023: (Start)

a(n) = 2*n + (sin(Pi*n/3) + sin(2*Pi*n/3))/sqrt(3).

a(n) = A005843(n) + A134667(n). (End)

From Wesley Ivan Hurt, Jul 17 2025: (Start)

G.f.: x*(3-2*x+4*x^2-2*x^3+3*x^4)/((x-1)^2*(1+x^2+x^4)).

a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - 2*a(n-4) + 2*a(n-5) - a(n-6). (End)