|
%I #13 Aug 30 2019 22:05:35
%S 1,1,1,1,2,1,1,3,3,1,1,4,2,4,1,1,5,10,10,5,1,1,6,3,2,3,6,1,1,7,21,35,
%T 35,21,7,1,1,8,4,56,2,56,4,8,1,1,9,36,3,126,126,3,36,9,1,1,10,5,120,
%U 10,2,10,120,5,10,1,1,11,55,165,330,462,462,330,165,55,11,1,1,12,6,4,3,792,2,792,3,4,6,12,1
%N Modified Pascal-triangle, read by rows. All C(n,j) binomial coefficients are replaced by C(n/g, j/g), where g = gcd(n,j).
%H G. C. Greubel, <a href="/A082905/b082905.txt">Rows n = 0..100 of triangle, flattened</a>
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 2, 1;
%e 1, 3, 3, 1;
%e 1, 4, 2, 4, 1;
%e 1, 5, 10, 10, 5, 1;
%e 1, 6, 3, 2, 3, 6, 1;
%e 1, 7, 21, 35, 35, 21, 7, 1;
%e 1, 8, 4, 56, 2, 56, 4, 8, 1;
%e 1, 9, 36, 3, 126, 126, 3, 36, 9, 1;
%e 1, 10, 5, 120, 10, 2, 10, 120, 5, 10, 1;
%t Flatten[Table[Table[Binomial[n/GCD[n, j], j/GCD[n, j]], {j, 0, n}], {n, 1, 32}], 1]
%o (PARI) T(n,k) = my(g=gcd(n,k)); if (!g, g=1); binomial(n/g, k/g);
%o tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n,k), ", "))); \\ _Michel Marcus_, Aug 30 2019
%o (Sage)
%o def T(n,k):
%o if k==0 or k==n: return 1
%o else: return binomial(n/gcd(n,k), k/gcd(n,k))
%o [[T(n,k) for k in (0..n)] for n in (0..12)] # _G. C. Greubel_, Aug 30 2019
%o (GAP)
%o T:= function(n,k)
%o if k=0 or k=n then return 1;
%o else return Binomial(n/Gcd(n,k), k/Gcd(n,k));
%o fi;
%o end;
%o Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # _G. C. Greubel_, Aug 30 2019
%Y Cf. A000005, A007318, A056045.
%K nonn,tabl
%O 0,5
%A _Labos Elemer_, Apr 23 2003
%E More terms from _Michel Marcus_, Aug 30 2019
|
