|
%I #22 Sep 08 2022 08:45:57
%S 2,6,1,6,4,1,21,6,1,30,33,2,5,42,2,6,26,2,57,30,2,13,69,0,8,78,1,42,5,
%T 10,93,6,2,102,105,2,28,114,13,30,62,5,129,66,1,20,28,2,11,30,2,78,40,
%U 2,165,42,10,174,177,3,6,186,7,6,98,22,201,102,2,210,213,0,110,222,5,114
%N [Squarefree part of (ABC)]/C for A=3, C=A+B, as a function of B, rounded to nearest integer.
%C Given A,B natural numbers, and C=A+B. The ratio [squarefree part of (ABC)]/C (notation: SQP(ABC)/C) can get arbitrarily small, while the unsolved ABC conjecture (i.e., Oesterle-Masser conjecture) is equivalent to the statement that [SQP(ABC)]^n/C has a minimum value if n>1 (because there are conjectured to be finitely many instances of [SQP(ABC)^(1+epsilon)]<C). Here, B plays the role of the OEIS index n.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/abcConjecture.html">abc Conjecture</a>
%e For B=5, we have C=8 so SQP(ABC)=SQP(120)=2*3*5=30, so SQP(ABC)/C=30/8=3.75, which rounds off to 4.
%e For B=15, we have C=18 so SQP(ABC)=SQP(810)=2*3*5=30, so SQP(ABC)/C=30/18=1.67, which rounds off to 2.
%o (Magma) SQP:=func< n | &*[ f[j, 1]: j in [1..#f] ] where f is Factorization(n) >; A191100:=func< n | Round(SQP(a*n*c)/c) where c is a+n where a is 3 >; [ A191100(n): n in [1..80] ]; // _Klaus Brockhaus_, May 26 2011
%o (PARI) rad(n)=my(f=factor(n)[,1]); prod(i=1,#f,f[i])
%o a(n)=rad(3*n^2+9*n)\/(n+3) \\ _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(3*n**2 + 9*n)/(n + 3))) # _Indranil Ghosh_, May 24 2017
%Y Cf. A190846, A191093, A120498.
%K nonn,easy
%O 1,1
%A _Darrell Minor_, May 25 2011
|
