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A190846
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(Squarefree part of (ABC))/C for A=1, C=A+B, as a function of B, rounded to the nearest integer.
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3
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1, 2, 2, 2, 5, 6, 2, 1, 3, 10, 6, 6, 13, 14, 2, 2, 6, 6, 10, 10, 21, 22, 6, 1, 5, 3, 2, 14, 29, 30, 2, 2, 33, 34, 6, 6, 37, 38, 10, 10, 41, 42, 22, 7, 15, 46, 6, 1, 1, 10, 26, 26, 6, 6, 14, 14, 57, 58, 30, 30, 61, 21, 1, 2, 65, 66, 34, 34, 69, 70, 6, 6, 73, 15, 8, 38, 77, 78, 10, 0, 3, 82, 42
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OFFSET
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1,2
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COMMENTS
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Given A, B natural numbers, and C=A+B, the ABC conjecture deals with the ratio of the squarefree part of the product A*B*C, divided by C. Here, B plays the role of the OEIS index n.
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LINKS
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EXAMPLE
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For B=14, we have C=15, so SQP(ABC)=SQP(210)=2*3*5*7=210, so SQP(ABC)/C=210/15=14.
For B=19, we have C=20, so SQP(ABC)=SQP(380)=2*5*19=190, so SQP(ABC)/C=190/20=9.5, which rounds to 10.
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MAPLE
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MATHEMATICA
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Array[Round[SelectFirst[Reverse@ Divisors[#1 #2], SquareFreeQ]/#2] & @@ {#, # + 1} &, 83] (* Michael De Vlieger, Feb 19 2019 *)
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PROG
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(Magma) SQP:=func< n | &*[ f[j, 1]: j in [1..#f] ] where f is Factorization(n) >; A190846:=func< n | Round(SQP(a*n*c)/c) where c is a+n where a is 1 >; [ A190846(n): n in [1..85] ]; // Klaus Brockhaus, May 27 2011
(PARI) rad(n)=my(f=factor(n)[, 1]); prod(i=1, #f, f[i])
(Python)
from operator import mul
from sympy import primefactors
def rad(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a(n): return int(round(rad(n**2 + n)/(n + 1))) # Indranil Ghosh, May 24 2017
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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