A191093 [Squarefree part of (ABC)]/C for A=2, C=A+B, as a function of B, rounded to nearest integer.

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, 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, 7, 114, 15, 118, 15, 41, 1, 42, 1, 130, 17, 134, 17, 138, 3, 142, 3, 29, 19, 30, 19, 154

Offset: 1

Author

Darrell Minor, May 25 2011

Keywords

• nonn
• easy

Examples

			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.
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.
		

Links

• Wikipedia, abc Conjecture

Crossrefs

Cf. A190846, A191100, A120498.

Programs

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

PARI

rad(n)=my(f=factor(n)[,1]); prod(i=1,#f,f[i])
a(n)=rad(2*n^2+4*n)\/(n+2) \\ 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(2*n**2 + 4*n)/(n + 2))) # Indranil Ghosh, May 24 2017