A190846 (Squarefree part of (ABC))/C for A=1, C=A+B, as a function of B, rounded to the nearest integer.

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, 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, 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

Offset: 1

Author

Darrell Minor, May 25 2011

Keywords

• nonn
• easy

Comments

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.

Examples

			For B=14, we have C=15, so SQP(ABC)=SQP(210)=2*3*5*7=210, so SQP(ABC)/C=210/15=14.
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.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Abderrahmane Nitaj, The ABC conjecture homepage
• Eric Weisstein's World of Mathematics, abc Conjecture
• Wikipedia, ABC conjecture

Crossrefs

Cf. A191093, A191100, A120498.

Programs

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

Maple

A190846 := proc(n) c := 1+n ; round(A007947(n*c)/c) ; end proc:
seq(A190846(n),n=1..80) ; # R. J. Mathar, Jun 10 2011

Mathematica

Array[Round[SelectFirst[Reverse@ Divisors[#1 #2], SquareFreeQ]/#2] & @@ {#, # + 1} &, 83] (* Michael De Vlieger, Feb 19 2019 *)

PARI

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