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A374673
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a(n) is the start of the least run of exactly n consecutive positive numbers with an equal value of A177329, or -1 if no such run exists.
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3
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OFFSET
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1,1
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COMMENTS
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For n > 1, a(n)! is the start of the least run of successive factorials of positive numbers (i.e., ignoring 0!) with an equal number of infinitary divisors (A037445).
a(9) > 320000, if it exists.
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LINKS
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EXAMPLE
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n | a(n) | A177329(k), k = a(n), a(n)+1, ..., a(n)+n-1
--|--------|------------------------------------------------
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MATHEMATICA
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s[n_] := Module[{e = FactorInteger[n!][[;; , 2]]}, Sum[DigitCount[e[[k]], 2, 1], {k, 1, Length[e]}]]; seq[len_] := Module[{v = Table[0, {len}], w = {0}, c = 0, k = 3, m, s1}, While[c < len, s1 = s[k]; m = Length[w]; If[s1 == w[[m]], AppendTo[w, s1], If[m <= len && v[[m]] == 0, v[[m]] = k-m; c++]; w = {s1}]; k++]; v]; seq[5]
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PROG
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(PARI) s(n) = {my(e = factor(n!)[, 2]); sum(k=1, #e, hammingweight(e[k])); }
lista(len) = {my(v = vector(len), w = [0], c = 0, k = 3, m, s1); while(c < len, s1 = s(k); m = #w; if(s1 == w[m], w = concat(w, s1), if(m < = len && v[m] == 0, v[m] = k-m; c++); w = [s1]); k++); v; }
(Python)
from itertools import count
from collections import Counter
from sympy import factorint
if n==1: return 2
c, a, l = Counter(), 0, 0
for m in count(2):
c += Counter(factorint(m))
b = sum(map(int.bit_count, c.values()))
if b==a:
l += 1
else:
if l==n-1:
return m-n
l = 0
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CROSSREFS
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KEYWORD
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nonn,hard,more,new
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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