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%I
%S 1,1,1,2,3,1,6,11,6,1,24,50,35,10,1,120,274,225,85,15,1,720,1764,1624,
%T 735,175,21,1,5040,13068,13132,6769,1960,322,28,1,40320,109584,118124,
%U 67284,22449,4536,546,36,1,362880,1026576,1172700,723680,269325,63273
%N Triangle T(n,k), 0<=k<=n, read by rows, giving coefficients of the polynomial (x+1)(x+2)...(x+n), expanded in increasing powers of x. T(n,k) is also the unsigned Stirling number |s(n+1,k+1)|, denoting the number of permutations on n+1 elements that contain exactly k+1 cycles.
%C This triangle is an unsigned version of the triangle of Stirling numbers of the first kind, A008275, which is the main entry for these numbers. - _N. J. A. Sloane_, Jan 25 2011
%C Or, triangle T(n,k), 0<=k<=n, read by rows given by [1,1,2,2,3,3,4,4,5,5,6,6,...] DELTA [1,0,1,0,1,0,1,0,1,0,1,0,...] where DELTA is the operator defined in A084938 .
%C Reversal of A094638 .
%C Equals A132393*A007318, as infinite lower triangular matrices . - _Philippe DELEHAM_, Nov 13 2007
%C Contribution from _Johannes W. Meijer_, Oct 07 2009: (Start)
%C The higher order exponential integrals E(x,m,n) are defined in A163931. The asymptotic expansion of the exponential integrals E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + n*(n+1)/x^2 - n*(n+1)*(n+2)/x^3 + .. ), see Abramowitz and Stegun. This formula follows from the general formula for the asymptotic expansion, see A163932. We rewrite E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + (n^2+n)/x^2 - (2*n+3*n^2+n^3)/x^3 + (6*n+11*n^2+6*n^3+n^4)/x^3 - .. ) and observe that the T(n,m) are the polynomials coefficients in the denominators. Looking at the a(n,m) formula of A028421, A163932 and A163934, and shifting the offset given above to 1, we can write T(n-1,m-1) = a(n,m) = (-1)^(n+m)*stirling1(n,m), see the Maple program.
%C The asymptotic expansion leads for values of n from one to eleven to known sequences, see the cross-references. With these sequences one can form the triangles A008279 (right hand columns) and A094587 (left hand columns).
%C See A163936 for information about the o.g.f.s. of the right hand columns of this triangle.
%C (End)
%C The number of elements greater than i to the left of i in a permutation gives the i-th element of the inversion vector. (Skiena-Pemmaraju 2003 p.69). T(n,k) is the number of n-permutations that have exactly k 0's in their inversion vector. See evidence in Mathematica code below. [From _Geoffrey Critzer_, May 07 2010]
%D Sriram Pemmaraju and Steven Skiena,Computational Discrete Mathematics,Cambridge University Press,2003,pp. 69-71 [From _Geoffrey Critzer_, May 07 2010]
%H T. D. Noe, <a href="/A130534/b130534.txt">Rows n=0..50 of triangle, flattened</a>
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.nrbook.com/abramowitz_and_stegun/">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, Chapter 5, pp. 227-251. [From _Johannes W. Meijer_, Oct 07 2009]
%H Dennis Walsh, <a href="http://frank.mtsu.edu/~dwalsh/STIRLIN1.pdf">A short note on unsigned Stirling numbers</a>
%F T(0,0)=1, T(n,k)=0 if k>n or if n<0, T(n,k)=T(n-1,k-1)+n*T(n-1,k). T(n,0)=n!=A000142(n). T(2*n,n)=A129505(n+1). Sum_{k, 0<=k<=n}T(n,k)=(n+1)!=A000142(n+1). Sum_{k, 0<=k<=n}T(n,k)^2=A047796(n+1). T(n,k)=|Stirling1(n+1,k+1)|, see A008275 . (x+1)(x+2)...(x+n)=Sum_{k, 0<=k<=n}T(n,k)*x^k. [Corrected by Arie Bos, Jul 11 2008]
%F Sum_{k, 0<=k<=n}T(n,k)*x^k = A000007(n), A000142(n), A000142(n+1), A001710(n+2), A001715(n+3), A001720(n+4), A001725(n+5), A001730(n+6), A049388(n), A049389(n), A049398(n), A051431(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively . - _Philippe DELEHAM_, Nov 13 2007
%F For 1<=k<=n, let A={a1,a2,...,ak} denote a size-k subset of {1,2,...,n}. Then T(n,n-k)=sum(product(ai)) where the sum is over all subsets A and the product is over i=1,..,k. For example, T(4,1)=50 since (1)(2)(3)+(1)(2)(4)+(1)(3)(4)+(2)(3)(4)=50. - _Dennis P. Walsh_, Jan 25 2011.
%F The preceding formula means T(n,k) = sigma_{n-k}(1,2,3,..,n) with the (n-k)-th elementary symmetric function sigma with the indeterminates chosen as 1,2,...,n. See the Oct 24 2011 comment in A094638 with sigma called there a. - _Wolfdieter Lang_, Feb 06 2013.
%F n-th row of the triangle = top row of M^n, where M is the production matrix:
%F 1, 1
%F 1, 2, 1
%F 1, 3, 3, 1
%F 1, 4, 6, 4, 1
%F ...
%F - _Gary W. Adamson_, Jul 08 2011
%e Triangle T(n,k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 ...
%e n=0: 1
%e n=1: 1 1
%e n=2: 2 3 1
%e n=3: 6 11 6 1
%e n=4: 24 50 35 10 1
%e n=5: 120 274 225 85 15 1
%e n=6: 720 1764 1624 735 175 21 1
%e n=7: 5040 13068 13132 6769 1960 322 28 1
%e n=8: 40320 109584 118124 67284 22449 4536 546 36 1
%e n=9: 362880, 1026576, 1172700, 723680, 269325, 63273, 9450, 870, 45, 1;
%e n=10: 3628800, 10628640, 12753576, 8409500, 3416930, 902055, 157773, 18150, 1320, 55, 1.
%e Reformatted and extended by _Wolfdieter Lang_, Feb 05 2013
%e T(3,2) = 6 because there are 6 permutations of {1,2,3,4} that have exactly 2 0's in their inversion vector:{1, 2, 4, 3}, {1, 3, 2, 4}, {1, 3, 4, 2}, {2, 1, 3, 4},{2, 3, 1, 4}, {2, 3, 4, 1}. The respective inversion vectors are: {0, 0, 1},{0, 1, 0}, {0, 2, 0}, {1, 0, 0}, {2, 0, 0}, {3, 0, 0}. [From _Geoffrey Critzer_, May 07 2010]
%e T(3,1)=11 since there are exactly 11 permutations of {1,2,3,4} with exactly 2 cycles, namely, (1)(234),(1)(243),(2)(134),(2)(143),(3)(124),(3)(142),
%e (4)(123),(4)(143),(12)(34),(13)(24), and(14)(23).
%e [From _Dennis P. Walsh_, Jan 25 2011]
%p with(combinat): A130534 := proc(n,m): (-1)^(n+m)*stirling1(n+1,m+1) end: seq(seq(A130534(n,m), m=0..n), n=0..10); # _Johannes W. Meijer_, Oct 07 2009, revised Sep 11 2012
%t Table[Table[ Length[Select[Map[ToInversionVector, Permutations[m]], Count[ #, 0] == n &]], {n, 0, m - 1}], {m, 0, 8}] // Grid [From _Geoffrey Critzer_, May 07 2010]
%o (Haskell)
%o a130534 n k = a130534_tabl !! n !! k
%o a130534_row n = a130534_tabl !! n
%o a130534_tabl = map (map abs) a008275_tabl
%o -- _Reinhard Zumkeller_, Mar 18 2013
%Y See A008275, which is the main entry for these numbers.
%Y Diagonals : A000012 A000217 A000914 A001303 A000915 A053567 A112002. Columns A000142 A000254 A000399 A000454 A000482 A001233 A001234.
%Y Contribution from _Johannes W. Meijer_, Oct 07 2009: (Start)
%Y Row sums equal A000142.
%Y The asymptotic expansions lead to A000142 (n=1), A000142(n=2; minus a(0)), A001710 (n=3), A001715 (n=4), A001720 (n=5), A001725 (n=6), A001730 (n=7), A049388 (n=8), A049389 (n=9), A049398 (n=10), A051431 (n=11), A008279 and A094587.
%Y Cf. A163931 (E(x,m,n)), A028421 (m=2), A163932 (m=3), A163934 (m=4), A163936.
%Y (End)
%K nonn,tabl
%O 0,4
%A _Philippe DELEHAM_, Aug 09 2007
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