

A191093


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


3



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


LINKS

Table of n, a(n) for n=1..77.
Wikipedia, abc Conjecture


EXAMPLE

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.


PROG

(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


CROSSREFS

Cf. A190846, A191100, A120498.
Sequence in context: A181569 A274085 A302129 * A266303 A267354 A139625
Adjacent sequences: A191090 A191091 A191092 * A191094 A191095 A191096


KEYWORD

nonn,easy


AUTHOR

Darrell Minor, May 25 2011


STATUS

approved



