Factor-perfect numbers: Difference between revisions
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OEIS A163272: |
OEIS A163272: |
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0, 1, 48, 1280, 2496, 28672, 29808, 454656, 2342912, |
0, 1, 48, 1280, 2496, 28672, 29808, 454656, 2342912, |
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</pre> |
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=={{header|Python}}== |
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<syntaxheader lang=python>''' Rosetta Code task Factor-perfect_numbers ''' |
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from sympy import divisors |
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def more_multiples(to_seq, from_seq): |
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''' Uses the first definition and recursion to generate the sequences ''' |
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onemores = [to_seq + [i] |
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for i in from_seq if i > to_seq[-1] and i % to_seq[-1] == 0] |
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if len(onemores) == 0: |
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return [] |
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for i in range(len(onemores)): |
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for arr in more_multiples(onemores[i], from_seq): |
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onemores.append(arr) |
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return onemores |
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listing = sorted(more_multiples([1], divisors(48)[1:-1]) + [[1, 48]]) |
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print('48 sequences using first definition:') |
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for j, seq in enumerate(listing): |
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print(f'{str(seq):20}', end='\n' if (j + 1) % 4 == 0 else '') |
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# Derive second definition's sequences |
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print('\n48 sequences using second definition:') |
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for k, seq in enumerate(listing): |
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if seq[-1] != 48: |
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seq.append(48) |
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seq2 = [seq[i] // seq[i - 1] for i in range(1, len(seq))] |
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print(f'{str(seq2):20}', end='\n' if (k + 1) % 4 == 0 else '') |
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def count_multiple_sequences(number): |
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''' Counts using the first definition, plus one extra for [1, n] ''' |
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return len(more_multiples([1], divisors(number)[1:-1])) + 1 |
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print("\nOEIS A163272: ", end='') |
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for num in range(500_000): |
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if num == 0 or count_multiple_sequences(num) == num: |
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print(num, end=', ') |
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</syntaxhighlight>{{out}} |
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<pre> |
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48 sequences using first definition: |
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[1, 2] [1, 2, 4] [1, 2, 4, 8] [1, 2, 4, 8, 16] |
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[1, 2, 4, 8, 24] [1, 2, 4, 12] [1, 2, 4, 12, 24] [1, 2, 4, 16] |
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[1, 2, 4, 24] [1, 2, 6] [1, 2, 6, 12] [1, 2, 6, 12, 24] |
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[1, 2, 6, 24] [1, 2, 8] [1, 2, 8, 16] [1, 2, 8, 24] |
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[1, 2, 12] [1, 2, 12, 24] [1, 2, 16] [1, 2, 24] |
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[1, 3] [1, 3, 6] [1, 3, 6, 12] [1, 3, 6, 12, 24] |
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[1, 3, 6, 24] [1, 3, 12] [1, 3, 12, 24] [1, 3, 24] |
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[1, 4] [1, 4, 8] [1, 4, 8, 16] [1, 4, 8, 24] |
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[1, 4, 12] [1, 4, 12, 24] [1, 4, 16] [1, 4, 24] |
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[1, 6] [1, 6, 12] [1, 6, 12, 24] [1, 6, 24] |
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[1, 8] [1, 8, 16] [1, 8, 24] [1, 12] |
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[1, 12, 24] [1, 16] [1, 24] [1, 48] |
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48 sequences using second definition: |
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[2, 24] [2, 2, 12] [2, 2, 2, 6] [2, 2, 2, 2, 3] |
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[2, 2, 2, 3, 2] [2, 2, 3, 4] [2, 2, 3, 2, 2] [2, 2, 4, 3] |
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[2, 2, 6, 2] [2, 3, 8] [2, 3, 2, 4] [2, 3, 2, 2, 2] |
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[2, 3, 4, 2] [2, 4, 6] [2, 4, 2, 3] [2, 4, 3, 2] |
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[2, 6, 4] [2, 6, 2, 2] [2, 8, 3] [2, 12, 2] |
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[3, 16] [3, 2, 8] [3, 2, 2, 4] [3, 2, 2, 2, 2] |
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[3, 2, 4, 2] [3, 4, 4] [3, 4, 2, 2] [3, 8, 2] |
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[4, 12] [4, 2, 6] [4, 2, 2, 3] [4, 2, 3, 2] |
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[4, 3, 4] [4, 3, 2, 2] [4, 4, 3] [4, 6, 2] |
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[6, 8] [6, 2, 4] [6, 2, 2, 2] [6, 4, 2] |
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[8, 6] [8, 2, 3] [8, 3, 2] [12, 4] |
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[12, 2, 2] [16, 3] [24, 2] [48] |
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OEIS A163272: 0, 1, 48, 1280, 2496, 28672, 29808, 454656, |
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</pre> |
</pre> |
Revision as of 17:35, 6 October 2022
Consider the list of factors (divisors) of an integer, such as 12. The factors of 12 are [1, 2, 3, 4, 6, 12]. Consider all sorted sequences of the factors of n such that each succeeding number in such a sequnce is a multiple of its predecessor. So, for 6, we have the factors (divisors) [1, 2, 3, 6]. The 3 unique lists of sequential multiples starting with 1 and ending with 6 that can be derived from these factors are [1, 6], [1, 2, 6], and [1, 3, 6].
Another way to see these sequences is as an set of all the ordered factorizations of a number taken so that their product is that number (excluding 1 from the sequence). So, for 6, we would have [6], [2, 3], and [3, 2]. In this description of the sequences, we are looking at the numbers needed to multiply by, in order to generate the next element in the sequences previously listed in our first definition of the sequence type, as we described it in the preceding paragraph, above.
For example, for the factorization of 6, if the first type of sequence is [1, 6], this is generated by [6] since 1 * 6 = 6. Similarly, the first type of sequence [1, 2, 6] is generated by the second type of sequence [2, 3] because 1 * 2 = 2 and 2 * 3 = 6. Similarly, [1, 3, 6] is generated by [3, 2] because 1 * 3 = 3 and 3 * 2 = 6.
If we count the number of such sorted sequences of multiples, or ordered factorizations, and using that count find all integers `n` for which the count of such sequences equals `n`, we have re-created the sequence of the "factor-perfect" numbers (OEIS 163272).
By some convention, on its OEIS page, the factor-perfect number sequence starts with 0 rather than 1. As might be expected
with a sequence involving factorization and combinations, finding factor-perfect numbers becomes
more demanding on CPU time as the numbers become large.
- Task
- Show all 48 ordered sequences for each of the two methods for n = 48, which is the first non-trivial factor-perfect number.
- Write a program to calculate and show the first 7 numbers of the factor-perfect numbers.
- Stretch task
- Calculate and show more of the subsequent numbers in the sequence.
- see also
[OEIS A163272]
[On the maximal order of numbers in the “factorisatio numerorum” problem]
Julia
using Primes
""" Return the factors of n, including 1, n """
function factors(n::T)::Vector{T} where T <: Integer
sort(vec(map(prod, Iterators.product((p.^(0:m) for (p, m) in eachfactor(n))...))))
end
""" Uses the first definition and recursion to generate the sequences """
function more_multiples(to_seq, from_seq)
onemores = [[to_seq; i] for i in from_seq if i > to_seq[end] && i % to_seq[end] == 0]
isempty(onemores) && return Int[]
return append!(onemores, mapreduce(seq -> more_multiples(seq, from_seq), append!, onemores))
end
listing = sort!(push!(more_multiples([1], factors(48)[begin:end-1]), [1, 48]))
println("48 sequences using first definition:")
for (i, seq) in enumerate(listing)
print(rpad(seq, 20), i % 4 == 0 ? "\n" : "")
end
println("\n48 sequences using second definition:")
for (i, seq) in enumerate(listing)
seq[end] != 48 && push!(seq, 48)
seq2 = [seq[i] ÷ seq[i - 1] for i in 2:length(seq)]
print(rpad(seq2, 20), i % 4 == 0 ? "\n" : "")
end
""" Get factorization sequence count """
count_multiple_sequences(n) = length(more_multiples([1], factors(n)[begin:end-1])) + 1
println("\nOEIS A163272: ")
for n in 0:2_400_000
if n == 0 || count_multiple_sequences(n) == n
print(n, ", ")
end
end
- Output:
48 sequences using first definition: [1, 2] [1, 2, 4] [1, 2, 4, 8] [1, 2, 4, 8, 16] [1, 2, 4, 8, 24] [1, 2, 4, 12] [1, 2, 4, 12, 24] [1, 2, 4, 16] [1, 2, 4, 24] [1, 2, 6] [1, 2, 6, 12] [1, 2, 6, 12, 24] [1, 2, 6, 24] [1, 2, 8] [1, 2, 8, 16] [1, 2, 8, 24] [1, 2, 12] [1, 2, 12, 24] [1, 2, 16] [1, 2, 24] [1, 3] [1, 3, 6] [1, 3, 6, 12] [1, 3, 6, 12, 24] [1, 3, 6, 24] [1, 3, 12] [1, 3, 12, 24] [1, 3, 24] [1, 4] [1, 4, 8] [1, 4, 8, 16] [1, 4, 8, 24] [1, 4, 12] [1, 4, 12, 24] [1, 4, 16] [1, 4, 24] [1, 6] [1, 6, 12] [1, 6, 12, 24] [1, 6, 24] [1, 8] [1, 8, 16] [1, 8, 24] [1, 12] [1, 12, 24] [1, 16] [1, 24] [1, 48] 48 sequences using second definition: [2, 24] [2, 2, 12] [2, 2, 2, 6] [2, 2, 2, 2, 3] [2, 2, 2, 3, 2] [2, 2, 3, 4] [2, 2, 3, 2, 2] [2, 2, 4, 3] [2, 2, 6, 2] [2, 3, 8] [2, 3, 2, 4] [2, 3, 2, 2, 2] [2, 3, 4, 2] [2, 4, 6] [2, 4, 2, 3] [2, 4, 3, 2] [2, 6, 4] [2, 6, 2, 2] [2, 8, 3] [2, 12, 2] [3, 16] [3, 2, 8] [3, 2, 2, 4] [3, 2, 2, 2, 2] [3, 2, 4, 2] [3, 4, 4] [3, 4, 2, 2] [3, 8, 2] [4, 12] [4, 2, 6] [4, 2, 2, 3] [4, 2, 3, 2] [4, 3, 4] [4, 3, 2, 2] [4, 4, 3] [4, 6, 2] [6, 8] [6, 2, 4] [6, 2, 2, 2] [6, 4, 2] [8, 6] [8, 2, 3] [8, 3, 2] [12, 4] [12, 2, 2] [16, 3] [24, 2] [48] OEIS A163272: 0, 1, 48, 1280, 2496, 28672, 29808, 454656, 2342912,
Python
<syntaxheader lang=python> Rosetta Code task Factor-perfect_numbers
from sympy import divisors
def more_multiples(to_seq, from_seq):
Uses the first definition and recursion to generate the sequences onemores = [to_seq + [i] for i in from_seq if i > to_seq[-1] and i % to_seq[-1] == 0] if len(onemores) == 0: return [] for i in range(len(onemores)): for arr in more_multiples(onemores[i], from_seq): onemores.append(arr) return onemores
listing = sorted(more_multiples([1], divisors(48)[1:-1]) + 1, 48)
print('48 sequences using first definition:')
for j, seq in enumerate(listing):
print(f'{str(seq):20}', end='\n' if (j + 1) % 4 == 0 else )
- Derive second definition's sequences
print('\n48 sequences using second definition:') for k, seq in enumerate(listing):
if seq[-1] != 48: seq.append(48) seq2 = [seq[i] // seq[i - 1] for i in range(1, len(seq))] print(f'{str(seq2):20}', end='\n' if (k + 1) % 4 == 0 else )
def count_multiple_sequences(number):
Counts using the first definition, plus one extra for [1, n] return len(more_multiples([1], divisors(number)[1:-1])) + 1
print("\nOEIS A163272: ", end=)
for num in range(500_000):
if num == 0 or count_multiple_sequences(num) == num: print(num, end=', ')
</syntaxhighlight>
- Output:
48 sequences using first definition: [1, 2] [1, 2, 4] [1, 2, 4, 8] [1, 2, 4, 8, 16] [1, 2, 4, 8, 24] [1, 2, 4, 12] [1, 2, 4, 12, 24] [1, 2, 4, 16] [1, 2, 4, 24] [1, 2, 6] [1, 2, 6, 12] [1, 2, 6, 12, 24] [1, 2, 6, 24] [1, 2, 8] [1, 2, 8, 16] [1, 2, 8, 24] [1, 2, 12] [1, 2, 12, 24] [1, 2, 16] [1, 2, 24] [1, 3] [1, 3, 6] [1, 3, 6, 12] [1, 3, 6, 12, 24] [1, 3, 6, 24] [1, 3, 12] [1, 3, 12, 24] [1, 3, 24] [1, 4] [1, 4, 8] [1, 4, 8, 16] [1, 4, 8, 24] [1, 4, 12] [1, 4, 12, 24] [1, 4, 16] [1, 4, 24] [1, 6] [1, 6, 12] [1, 6, 12, 24] [1, 6, 24] [1, 8] [1, 8, 16] [1, 8, 24] [1, 12] [1, 12, 24] [1, 16] [1, 24] [1, 48] 48 sequences using second definition: [2, 24] [2, 2, 12] [2, 2, 2, 6] [2, 2, 2, 2, 3] [2, 2, 2, 3, 2] [2, 2, 3, 4] [2, 2, 3, 2, 2] [2, 2, 4, 3] [2, 2, 6, 2] [2, 3, 8] [2, 3, 2, 4] [2, 3, 2, 2, 2] [2, 3, 4, 2] [2, 4, 6] [2, 4, 2, 3] [2, 4, 3, 2] [2, 6, 4] [2, 6, 2, 2] [2, 8, 3] [2, 12, 2] [3, 16] [3, 2, 8] [3, 2, 2, 4] [3, 2, 2, 2, 2] [3, 2, 4, 2] [3, 4, 4] [3, 4, 2, 2] [3, 8, 2] [4, 12] [4, 2, 6] [4, 2, 2, 3] [4, 2, 3, 2] [4, 3, 4] [4, 3, 2, 2] [4, 4, 3] [4, 6, 2] [6, 8] [6, 2, 4] [6, 2, 2, 2] [6, 4, 2] [8, 6] [8, 2, 3] [8, 3, 2] [12, 4] [12, 2, 2] [16, 3] [24, 2] [48] OEIS A163272: 0, 1, 48, 1280, 2496, 28672, 29808, 454656,