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A159335 Triangle read by rows: numerator of n/binomial(n,m).

%I A159335 #15 Sep 08 2022 08:45:43
%S A159335 0,1,1,2,1,2,3,1,1,3,4,1,2,1,4,5,1,1,1,1,5,6,1,2,3,2,1,6,7,1,1,1,1,1,
%T A159335 1,7,8,1,2,1,4,1,2,1,8,9,1,1,3,1,1,3,1,1,9,10,1,2,1,1,5,1,1,2,1,10,11,
%U A159335 1,1,1,1,1,1,1,1,1,1,11,12,1,2,3,4,1,1,1,4,3,2,1,12,13,1,1,1,1,1,1,1,1,1,1,1,1,13
%N A159335 Triangle read by rows: numerator of n/binomial(n,m).
%C A159335 This triangle first differs from A109004 (read as a triangle) at T(10, 4) and T(10,6).
%C A159335 T(n,m) is the smallest positive integer such that binomial(n,m)*T(n,m) is a multiple of n.
%H A159335 G. C. Greubel, <a href="/A159335/b159335.txt">Rows n=0..100 of triangle, flattened</a>
%F A159335 T(n,m) = n/gcd(n,binomial(n,m)).
%e A159335 Row 10 of Pascal's triangle is: 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1. {a(10,m)} of this sequence (A159335) is: 10, 1, 2, 1, 1, 5, 1, 1, 2, 1,10. Multiplying the corresponding integers, we get multiples of 10: 1*10=10,10*1=10, 45*2=90, 120*1=120, 210*1=210, 252*5=1260, 210*1=210, 120*1=120, 45*2=90, 10*1=10, 1*10=10.
%t A159335 Table[n/GCD[n, Binomial[n, k]], {n, 0, 10}, {k, 0, n}] // Flatten (* _G. C. Greubel_, Jun 25 2018 *)
%o A159335 (PARI) for(n=0, 10, for(k=0,n, print1(n/gcd(n, binomial(n,k)), ", "))) \\ _G. C. Greubel_, Jun 25 2018
%o A159335 (Magma) /* As triangle */ [[n/GCD(n,Binomial(n, k)): k in [0..n]]: n in [0..10]]; // _G. C. Greubel_, Jun 25 2018
%Y A159335 Cf. A165661 (denominators), A007318, A020475, A109004.
%K A159335 frac,nonn,tabl
%O A159335 0,4
%A A159335 _Leroy Quet_, Apr 10 2009
%E A159335 Extended by _Ray Chandler_, Jun 19 2009
%E A159335 Edited by _Franklin T. Adams-Watters_, Sep 24 2009