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A356226
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Irregular triangle giving the lengths of maximal gapless submultisets of the prime indices of n.
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17
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1, 1, 2, 1, 2, 1, 3, 2, 1, 1, 1, 3, 1, 1, 1, 2, 4, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 4, 2, 1, 1, 3, 2, 1, 1, 3, 1, 5, 1, 1, 1, 1, 2, 4, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1, 3, 1, 1, 1, 5, 2, 1, 2, 1, 1, 2, 1, 1, 4, 1, 1, 3, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 2, 1, 6
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OFFSET
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1,3
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COMMENTS
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A sequence is gapless if it covers an unbroken interval of positive integers. For example, the multiset {2,3,5,5,6,9} has three maximal gapless submultisets: {2,3}, {5,5,6}, {9}.
A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.
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LINKS
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EXAMPLE
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Triangle begins: {}, {1}, {1}, {2}, {1}, {2}, {1}, {3}, {2}, {1,1}, {1}, {3}, {1}, {1,1}, {2}, {4}, {1}, {3}, {1}, {2,1}, ... For example, the prime indices of 20 are {1,1,3}, which separates into maximal gapless submultisets {{1,1},{3}}, so row 20 is (2,1).
The prime indices of 18564 are {1,1,2,4,6,7}, which separates into {1,1,2}, {4}, {6,7}, so row 18564 is (3,1,2). This corresponds to the factorization 18564 = 12 * 7 * 221.
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MATHEMATICA
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primeMS[n_]:=If[n==1, {}, Flatten[Cases[FactorInteger[n], {p_, k_}:>Table[PrimePi[p], {k}]]]];
Table[Length/@Split[primeMS[n], #1>=#2-1&], {n, 100}]
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CROSSREFS
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Standard composition numbers of rows are A356230.
Positions of first appearances are A356232.
A003963 multiplies together the prime indices of n.
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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