OFFSET
1,1
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)
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).
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).
EXAMPLE
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.
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Jun 23 2022
STATUS
approved