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%I #30 Mar 01 2019 04:47:09
%S 1,2,1,6,3,1,24,12,4,1,120,60,20,5,1,720,360,120,30,6,1,5040,2520,840,
%T 210,42,7,1,40320,20160,6720,1680,336,56,8,1,362880,181440,60480,
%U 15120,3024,504,72,9,1,3628800,1814400,604800,151200,30240,5040,720,90,10,1
%N Triangle read by rows: T(n,k) = n! / k!, 1<=k<=n.
%C Row sums give A002627;
%C central terms give A006963: T(2*n-1,n) = A006963(n+1);
%C T(2*n,n) = A001813(n); T(2*n,n+1) = A001761(n);
%C 1 < k <= n: T(n,k) = T(n,k-1) / k;
%C 1 <= k <= n: T(n+1,k) = A119741(n,n-k+1);
%C 1 <= k <= n: T(n+1,k+1) = A162995(n,k);
%C T(n,1) = A000142(n);
%C T(n,2) = A001710(n) for n>1;
%C T(n,3) = A001715(n) for n>2;
%C T(n,4) = A001720(n) for n>3;
%C T(n,5) = A001725(n) for n>4;
%C T(n,6) = A001730(n) for n>5;
%C T(n,7) = A049388(n-7) for n>6;
%C T(n,8) = A049389(n-8) for n>7;
%C T(n,9) = A049398(n-9) for n>8;
%C T(n,10) = A051431(n) for n>9;
%C T(n,n-7) = A159083(n+1) for n>7;
%C T(n,n-6) = A053625(n+1) for n>6;
%C T(n,n-5) = A052787(n) for n>5;
%C T(n,n-4) = A052762(n) for n>4;
%C T(n,n-3) = A007531(n) for n>3;
%C T(n,n-2) = A002378(n-1) for n>2;
%C T(n,n-1) = A000027(n) for n>1;
%C T(n,n) = A000012(n).
%C From _Wolfdieter Lang_, Jun 27 2012: (Start)
%C T(n-1,k), k=1,...,n-1, gives the number of representative necklaces with n beads (C_N symmetry) of n+1-k distinct colors, say c[1],c[2],...,c[n-k+1], corresponding to the color signature determined by the partition k,1^(n-k) of n. The representative necklaces have k beads of color c[1]. E.g., n=4, k=2: partition 2,1,1, color signature (parts as exponents) c[1]c[1]c[2]c[3], 3=T(3,2) necklaces (write j for color c[j]): cyclic(1123), cyclic(1132) and cyclic(1213). See A212359 for the numbers for general partitions or color signatures. (End)
%C T(n,k) = A094587(n,k), 1 <= k <= n. - _Reinhard Zumkeller_, Jul 05 2012
%H Reinhard Zumkeller, <a href="/A173333/b173333.txt">Rows n = 1..150 of triangle, flattened</a>
%H <a href="/index/Fa#factorial">Index entries for sequences related to factorial numbers</a>.
%F E.g.f.: (exp(x*y) - 1)/(x*(1 - y)). - _Olivier Gérard_, Jul 07 2011
%e Triangle starts:
%e n\k 1 2 3 4 5 6 7 8 9 10 ...
%e 1 1
%e 2 2 1
%e 3 6 3 1
%e 4 24 12 4 1
%e 5 120 60 20 5 1
%e 6 720 360 120 30 6 1
%e 7 5040 2520 840 210 42 7 1
%e 8 40320 20160 6720 1680 336 56 8 1
%e 9 362880 181440 60480 15120 3024 504 72 9 1
%e 10 3628800 1814400 604800 151200 30240 5040 720 90 10 1
%e ... - _Wolfdieter Lang_, Jun 27 2012
%t Table[n!/k!, {n, 1, 10}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Mar 01 2019 *)
%o (Haskell)
%o a173333 n k = a173333_tabl !! (n-1) !! (k-1)
%o a173333_row n = a173333_tabl !! (n-1)
%o a173333_tabl = map fst $ iterate f ([1], 2)
%o where f (row, i) = (map (* i) row ++ [1], i + 1)
%o -- _Reinhard Zumkeller_, Jul 04 2012
%Y Cf. A138533, A002627.
%K nonn,tabl
%O 1,2
%A _Reinhard Zumkeller_, Feb 19 2010
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