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A343943
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Number of distinct possible alternating sums of permutations of the multiset of prime factors of n.
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3
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1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 1, 2, 2, 2, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 4, 1, 2, 2, 1, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 1, 2, 1, 4, 2, 2, 2
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OFFSET
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1,6
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COMMENTS
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The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. Of course, the alternating sum of prime factors is also the reverse-alternating sum of reversed prime factors.
Also the number of distinct "sums of prime factors" of divisors d|n such that bigomega(d) = bigomega(n)/2 rounded up.
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LINKS
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EXAMPLE
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The divisors of 525 with 2 prime factors are: 15, 21, 25, 35, with prime factors {3,5}, {3,7}, {5,5}, {5,7}, with distinct sums {8,10,12}, so a(525) = 3.
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MATHEMATICA
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prifac[n_]:=If[n==1, {}, Flatten[ConstantArray@@@FactorInteger[n]]];
Table[Length[Union[Total/@Subsets[prifac[n], {Ceiling[PrimeOmega[n]/2]}]]], {n, 100}]
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PROG
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(Python)
from sympy import factorint
from sympy.utilities.iterables import multiset_combinations
fs = factorint(n)
return len(set(sum(d) for d in multiset_combinations(fs, (sum(fs.values())+1)//2))) # Chai Wah Wu, Aug 23 2021
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CROSSREFS
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The half-length submultisets are counted by A114921.
Including all multisets of prime factors gives A305611(n) + 1.
The strict rounded version appears to be counted by A342343.
The version for prime indices instead of prime factors is A345926.
A071321 gives the alternating sum of prime factors (reverse: A071322).
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
A334968 counts subsequence-sums of standard compositions.
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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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