A056609 a(n) = rad(n!)/rad(A001142(n)) where rad(n) is the squarefree kernel of n, A007947(n).
1, 1, 2, 1, 3, 1, 2, 3, 5, 1, 1, 1, 7, 5, 2, 1, 3, 1, 5, 7, 11, 1, 1, 5, 13, 3, 7, 1, 1, 1, 2, 11, 17, 7, 1, 1, 19, 13, 1, 1, 7, 1, 11, 1, 23, 1, 1, 7, 5, 17, 13, 1, 3, 11, 1, 19, 29, 1, 1, 1, 31, 1, 2, 13, 11, 1, 17, 23, 1, 1, 1, 1, 37, 5, 19, 11, 13, 1, 1, 3, 41, 1, 1, 17, 43, 29, 11, 1, 1, 13
Offset: 1
Keywords
Examples
From _Luc Rousseau_, Jan 02 2019: (Start) In Pascal's triangle, - row n=3 (1 3 3 1) contains no number with prime factor 2, so a(3) = 2; - row n=4 (1 4 6 4 1) contains, for all p prime <= 4, a multiple of p, so a(4) = 1; - row n=5 (1 5 10 10 5 1) contains no number with prime factor 3, so a(5) = 3; etc. (End)
Links
- Luc Rousseau, Table of n, a(n) for n = 1..1000 (first 90 terms from Labos Elemer)
Programs
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Mathematica
L[n_] := Table[Binomial[n, k], {k, 1, Floor[n/2]}] c[n_] := Complement[Prime /@ Range[PrimePi[n]], First /@ FactorInteger[Times @@ L[n]]] a[n_] := Module[{x = c[n]}, If[x == {}, 1, First[x]]] Table[a[n], {n, 1, 100}] (* Luc Rousseau, Jan 01 2019 *)
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PARI
rad(n) = factorback(factorint(n)[, 1]); \\ A007947 b(n) = prod(m=1, n, binomial(n, m)); \\ A001142 a(n) = rad(n!)/rad(b(n)); \\ Michel Marcus, Jan 02 2019
Formula
Extensions
Definition and example changed by Luc Rousseau, Jan 02 2019
Comments