A191093 -id:A191093 - cp's OEIS frontend

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

Darrell Minor, May 25 2011

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)]

			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.

Eric Weisstein's World of Mathematics, abc Conjecture

Cf. A190846, A191093, A120498.

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

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

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

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

Original entry on oeis.org

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

Darrell Minor, May 25 2011

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.

			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.

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

Cf. A191093, A191100, A120498.

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

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

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

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

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

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

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