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%I #36 Sep 12 2022 03:48:20
%S 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,
%T 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,
%U 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
%N (Squarefree part of (ABC))/C for A=1, C=A+B, as a function of B, rounded to the nearest integer.
%C 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.
%H Michael De Vlieger, <a href="/A190846/b190846.txt">Table of n, a(n) for n = 1..10000</a>
%H Abderrahmane Nitaj, <a href="https://nitaj.users.lmno.cnrs.fr/abc.html">The ABC conjecture homepage</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/abcConjecture.html">abc Conjecture</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/ABC_conjecture">ABC conjecture</a>
%e For B=14, we have C=15, so SQP(ABC)=SQP(210)=2*3*5*7=210, so SQP(ABC)/C=210/15=14.
%e 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.
%p A190846 := proc(n) c := 1+n ; round(A007947(n*c)/c) ; end proc:
%p seq(A190846(n),n=1..80) ; # _R. J. Mathar_, Jun 10 2011
%t Array[Round[SelectFirst[Reverse@ Divisors[#1 #2], SquareFreeQ]/#2] & @@ {#, # + 1} &, 83] (* _Michael De Vlieger_, Feb 19 2019 *)
%o (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
%o (PARI) rad(n)=my(f=factor(n)[,1]); prod(i=1,#f,f[i])
%o a(n)=rad(n^2+n)\/(n+1) \\ _Charles R Greathouse IV_, Mar 11 2014
%o (Python)
%o from operator import mul
%o from sympy import primefactors
%o def rad(n): return 1 if n<2 else reduce(mul, primefactors(n))
%o def a(n): return int(round(rad(n**2 + n)/(n + 1))) # _Indranil Ghosh_, May 24 2017
%Y Cf. A191093, A191100, A120498.
%K nonn,easy
%O 1,2
%A _Darrell Minor_, May 25 2011
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