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%I #25 Apr 25 2022 08:04:48
%S 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,
%T 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,
%U 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
%N a(n) = rad(n!)/rad(A001142(n)) where rad(n) is the squarefree kernel of n, A007947(n).
%C The previous name, which does not match the data as observed by _Luc Rousseau_, was: Quotient of squarefree kernels of A002944(n) and A001405.
%C a(n) is the unique prime p not greater than n missing in the prime factorization of A001142(n), if such a prime exists; a(n) is 1 otherwise. - _Luc Rousseau_, Jan 01 2019
%H Luc Rousseau, <a href="/A056609/b056609.txt">Table of n, a(n) for n = 1..1000</a> (first 90 terms from Labos Elemer)
%F a(n) = A034386(n) / A056606(n). - _Sean A. Irvine_, Apr 24 2022
%e From _Luc Rousseau_, Jan 02 2019: (Start)
%e In Pascal's triangle,
%e - row n=3 (1 3 3 1) contains no number with prime factor 2, so a(3) = 2;
%e - row n=4 (1 4 6 4 1) contains, for all p prime <= 4, a multiple of p, so a(4) = 1;
%e - row n=5 (1 5 10 10 5 1) contains no number with prime factor 3, so a(5) = 3;
%e etc.
%e (End)
%t L[n_] := Table[Binomial[n, k], {k, 1, Floor[n/2]}]
%t c[n_] := Complement[Prime /@ Range[PrimePi[n]], First /@ FactorInteger[Times @@ L[n]]]
%t a[n_] := Module[{x = c[n]}, If[x == {}, 1, First[x]]]
%t Table[a[n], {n, 1, 100}]
%t (* _Luc Rousseau_, Jan 01 2019 *)
%o (PARI) rad(n) = factorback(factorint(n)[, 1]); \\ A007947
%o b(n) = prod(m=1, n, binomial(n, m)); \\ A001142
%o a(n) = rad(n!)/rad(b(n)); \\ _Michel Marcus_, Jan 02 2019
%Y Cf. A002944, A001405, A003418, A002110, A001142, A034386, A056606.
%K nonn
%O 1,3
%A _Labos Elemer_, Aug 07 2000
%E Definition and example changed by _Luc Rousseau_, Jan 02 2019
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