%I #26 Sep 08 2022 08:45:57
%S 2,1,6,1,10,1,5,1,6,3,22,3,26,1,30,0,34,2,38,5,42,3,9,3,1,7,6,7,58,1,
%T 62,1,66,3,70,3,74,5,78,5,82,11,29,11,30,3,13,1,14,3,102,1,106,1,110,
%U 7,114,15,118,15,41,1,42,1,130,17,134,17,138,3,142,3,29,19,30,19,154
%N [Squarefree part of (ABC)]/C for A=2, C=A+B, as a function of B, rounded to nearest integer.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Abc_conjecture">abc Conjecture</a>
%e For B=10, we have C=12 so SQP(ABC)=SQP(240)=2*3*5=30, so SQP(ABC)/C=30/12=2.5, which rounds off to 3.
%e For B=16, we have C=18 so SQP(ABC)=SQP(576)=2*3=6, so SQP(ABC)/C=6/18=0.33, which rounds off to 0.
%o (Magma) SQP:=func< n | &*[ f[j, 1]: j in [1..#f] ] where f is Factorization(n) >; A191093:=func< n | Round(SQP(a*n*c)/c) where c is a+n where a is 2 >; [ A191093(n): n in [1..80] ]; // _Klaus Brockhaus_, May 27 2011
%o (PARI) rad(n)=my(f=factor(n)[,1]); prod(i=1,#f,f[i])
%o a(n)=rad(2*n^2+4*n)\/(n+2) \\ _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(2*n**2 + 4*n)/(n + 2))) # _Indranil Ghosh_, May 24 2017
%Y Cf. A190846, A191100, A120498.
%K nonn,easy
%O 1,1
%A _Darrell Minor_, May 25 2011