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A355200
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Numbers k that can be written as the sum of 3 divisors of k (not necessarily distinct).
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12
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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
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OFFSET
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1,1
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COMMENTS
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Equivalently: positive numbers that are divisible by 3 or by 4.
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)
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)
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LINKS
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FORMULA
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Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2)/3 - log(3)/4. - Amiram Eldar, Sep 10 2023
a(n) = 2*n + (sin(Pi*n/3) + sin(2*Pi*n/3))/sqrt(3).
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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