

A191100


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


4



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
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OFFSET

1,1


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., OesterleMasser 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.


LINKS

Table of n, a(n) for n=1..76.
Eric Weisstein's World of Mathematics, abc Conjecture


EXAMPLE

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.


PROG

(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


CROSSREFS

Cf. A190846, A191093, A120498.
Sequence in context: A280580 A288872 A329207 * A322944 A019576 A141906
Adjacent sequences: A191097 A191098 A191099 * A191101 A191102 A191103


KEYWORD

nonn,easy


AUTHOR

Darrell Minor, May 25 2011


STATUS

approved



