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 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 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

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Last modified May 5 19:31 EDT 2021. Contains 343573 sequences. (Running on oeis4.)