A191100 [Squarefree part of (ABC)]/C for A=3, C=A+B, as a function of B, rounded to nearest integer.

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, 10, 93, 6, 2, 102, 105, 2, 28, 114, 13, 30, 62, 5, 129, 66, 1, 20, 28, 2, 11, 30, 2, 78, 40, 2, 165, 42, 10, 174, 177, 3, 6, 186, 7, 6, 98, 22, 201, 102, 2, 210, 213, 0, 110, 222, 5, 114

Offset: 1

Author

Darrell Minor, May 25 2011

Keywords

• nonn
• easy

Comments

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)]

Examples

			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.
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.
		

Links

• Eric Weisstein's World of Mathematics, abc Conjecture

Crossrefs

Cf. A190846, A191093, A120498.

Programs

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

PARI

rad(n)=my(f=factor(n)[,1]); prod(i=1,#f,f[i])
a(n)=rad(3*n^2+9*n)\/(n+3) \\ Charles R Greathouse IV, Mar 11 2014

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(3*n**2 + 9*n)/(n + 3))) # Indranil Ghosh, May 24 2017

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