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A328166
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Heinz number of the run-lengths of the divisors of n.
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21
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2, 3, 4, 6, 4, 10, 4, 12, 8, 12, 4, 28, 4, 12, 16, 24, 4, 40, 4, 36, 16, 12, 4, 112, 8, 12, 16, 48, 4, 120, 4, 48, 16, 12, 16, 224, 4, 12, 16, 144, 4, 120, 4, 48, 64, 12, 4, 448, 8, 48, 16, 48, 4, 160, 16, 144, 16, 12, 4, 832, 4, 12, 64, 96, 16, 160, 4, 48, 16
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OFFSET
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1,1
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COMMENTS
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The Heinz number of an integer partition or multiset {y_1,...,y_k} is prime(y_1)*...*prime(y_k).
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LINKS
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FORMULA
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EXAMPLE
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Splitting the divisors of 30 into runs gives {{1, 2, 3}, {5, 6}, {10}, {15}, {30}}, and the Heinz number of {1, 1, 1, 2, 3} is 120, so a(30) = 120.
Splitting the divisors of 1 into runs gives {1}, and the Heinz number of that is 2.
Splitting the divisors of 2 into runs gives {1, 2}, and the Heinz number of that is 3. [one run of length 2, therefore a(2) = prime(2)^1].
Splitting the divisors of 3 into runs gives {1} and {3}, and the Heinz number of that is 4. [two runs of length 1, therefore a(3) = prime(1)^2].
Splitting the divisors of 4 into runs gives {1, 2} and {4}, and the Heinz number of that is 6. [one run of length 1, and other run of length 2, therefore a(4) = prime(1)*prime(2)].
Splitting the divisors of 5 into runs gives {1} and {5}, and the Heinz number of that is 4. [two runs of length 1, therefore a(5) = prime(1)^2].
(End)
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MATHEMATICA
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Table[Times@@Prime/@Length/@Split[Divisors[n], #2==#1+1&], {n, 30}]
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PROG
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(PARI) A328166(n) = { my(rl=0, pd=0, v=vector(numdiv(n)), m=1); fordiv(n, d, if(d>(1+pd), v[rl]++; rl=0); pd=d; rl++); v[rl]++; for(i=1, #v, m *= prime(i)^v[i]); (m); }; \\ Antti Karttunen, Dec 09 2021
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CROSSREFS
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The longest run of divisors of n has length A055874(n).
Numbers whose divisors > 1 have no non-singleton runs are A088725.
The number of successive pairs of divisors of n is A129308(n).
The Heinz number of the set of divisors of n is A275700(n).
Numbers whose divisors do not have weakly decreasing run-lengths are A328165.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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