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A348536
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Number of partitions of n into 3 parts that divide n.
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2
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0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 0, 3, 0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 0, 3, 0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 0, 3, 0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 0, 3, 0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 0, 3, 0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 0, 3, 0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 0, 3, 0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 0
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OFFSET
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1,6
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COMMENTS
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Proof of formula: suppose we have a partition d_1 + d_2 + d_3 = n where 0 < d_1 <= d_2 <= d_3 and d_1 | n, d_2 | n and d_3 | n.
Then let m_i * d_i = n for i in (1, 2, 3).
Dividing d_1 + d_2 + d_3 = n by n gives d_1/n + d_2/n + d_3/n = n/n, i.e., 1/m_1 + 1/m_2 + 1/m_3 = 1. There are 3 solutions to that equation namely (m_1, m_2, m_3) in {(2, 4, 4), (2, 3, 6), (3, 3, 3)}. The lcm of these numbers is 12 so the sequence is periodic with a period of 12 and then calculating the first 12 terms defines the sequence.
Alternative name: number of divisors of n from {3, 4, 6}. (End)
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LINKS
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FORMULA
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a(n) = Sum_{j=1..floor(n/3)} Sum_{i=j..floor((n-j)/2)} c(n/j) * c(n/i) * c(n/(n-i-j)), where c(n) = 1 - ceiling(n) + floor(n).
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EXAMPLE
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a(12) = 3 via 2 + 4 + 6 = 3 + 3 + 6 = 4 + 4 + 4. - David A. Corneth, Oct 08 2022
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MATHEMATICA
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Block[{c}, c[n_] := 1 - Ceiling[n] + Floor[n]; Array[Sum[Sum[c[#/j]*c[#/i]*c[#/(# - i - j)], {i, j, Floor[(# - j)/2]} ], {j, Floor[#/3]} ] &, 105]] (* Michael De Vlieger, Oct 21 2021 *)
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PROG
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(PARI) a(n) = [0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 0, 3][(n-1)%12 + 1] \\ David A. Corneth, Oct 08 2022
(PARI) a(n) = { my(d = divisors(n), res = 0); d = d[^#d]; forvec(x = vector(2, i, [1, #d]), s = d[x[1]] + d[x[2]]; if(n - s >= d[x[2]], if(n % (n - s) == 0, print([d[x[1]], d[x[2]], n-s]); res++ ) ) , 1 ); res } \\ 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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