A284255 Largest divisor of n such that all its prime factors are less than the square of the smallest prime factor of n, a(1) = 1.
1, 2, 3, 4, 5, 6, 7, 8, 9, 2, 11, 12, 13, 2, 15, 16, 17, 18, 19, 4, 21, 2, 23, 24, 25, 2, 27, 4, 29, 6, 31, 32, 3, 2, 35, 36, 37, 2, 3, 8, 41, 6, 43, 4, 45, 2, 47, 48, 49, 2, 3, 4, 53, 54, 55, 8, 3, 2, 59, 12, 61, 2, 63, 64, 65, 6, 67, 4, 3, 2, 71, 72, 73, 2, 75, 4, 77, 6, 79, 16, 81, 2, 83, 12, 85, 2, 3, 8, 89, 18, 91, 4, 3, 2, 95, 96, 97, 2, 9, 4, 101, 6
Offset: 1
Keywords
Examples
For n = 50 = 2*5*5, only prime less than 2^2 is 2, thus a(50) = 2. For n = 90 = 2*3*3*5, the primes less than 2^2 are 2, 3 and 3, thus a(90) = 2*3*3 = 18.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10001
Crossrefs
Programs
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Mathematica
Table[If[n == 1, 1, Function[d, First[Select[Reverse@ First@ d, Times @@ Boole@ Map[# < Last[d]^2 &, FactorInteger[#][[All, 1]]] == 1 &] /. {} -> {1}]]@ {#, First@ Select[#, PrimeQ]} &@ Divisors@ n], {n, 102}] (* Michael De Vlieger, Mar 24 2017 *)
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PARI
A(n) = if(n<2, return(1), my(f=factor(n)[, 1]); for(i=2, #f, if(f[i]>f[1]^2, return(f[i]))); return(1)); a(n) = if(A(n)==1, 1, A(n)*a(n/A(n))); for(n=1, 150, print1(n/a(n),", ")) \\ Indranil Ghosh, after David A. Corneth, Mar 24 2017
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Python
from sympy import primefactors def A(n): pf = primefactors(n) if pf: min_pf2 = min(pf)**2 for i in pf: if i > min_pf2: return i return 1 def a(n): return 1 if A(n)==1 else A(n)*a(n//A(n)) print([n//a(n) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 2017
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Scheme
(define (A284255 n) (/ n (A284254 n)))