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A093548
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a(n) is the smallest number m such that each of the numbers m and m+1 has n distinct prime divisors.
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14
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2, 14, 230, 7314, 254540, 11243154, 965009045, 65893166030, 5702759516090, 490005293940084, 76622240600506314
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OFFSET
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1,1
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COMMENTS
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Prime factors may be repeated in m and m+1. The difference between this sequence and A052215 is that in the latter, no prime factor may be repeated. So A052215 imposes more stringent conditions, hence a(n) <= A052215(n). - N. J. A. Sloane, Nov 21 2015
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REFERENCES
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J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 230, p. 65, Ellipses, Paris 2008.
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LINKS
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FORMULA
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a[n_] := (For[m=1, !(Length[FactorInteger[m]]==n && Length[FactorInteger[m+1]]==n), m++ ];m)
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EXAMPLE
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a(5) = 254540 because 254540=2^2*5*11*13*89; 254541=3*7*17*23*31
and 254540 is the smallest number m which each of the numbers m & m+1 has 5 distinct prime divisors.
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MATHEMATICA
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a[n_] := (For[m=1, !(Length[FactorInteger[m]]==n && Length[FactorInteger[m+1]]==n), m++ ]; m); Do[Print[a[n]], {n, 7}]
Flatten[Table[SequencePosition[PrimeNu[Range[260000]], {n, n}, 1], {n, 5}], 1][[;; , 1]] (* To generate more terms, increase the Range and n constants. *) (* Harvey P. Dale, Jun 08 2023 *)
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PROG
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(Python)
from sympy import primefactors, primorial
def a(n):
m = primorial(n)
while True:
if len(primefactors(m)) == n:
if len(primefactors(m+1)) == n: return m
else: m += 2
else: m += 1
for n in range(1, 6):
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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