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

1,1

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