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Worpitzky form polynomials for the {1,16,1} A142462 sequence: p(x,n) = Sum_{k=1..n} A(n, k)*binomial(x + k - 1, n - 1).
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%I #5 Mar 09 2024 10:29:39

%S 1,1,2,2,18,18,6,156,432,288,24,792,7416,13248,6624,120,-11280,64800,

%T 374400,496800,198720,720,-62640,-1254960,4968000,20865600,22057920,

%U 7352640,5040,24012000,-11854080,-125677440,389491200,1288103040

%N Worpitzky form polynomials for the {1,16,1} A142462 sequence: p(x,n) = Sum_{k=1..n} A(n, k)*binomial(x + k - 1, n - 1).

%C Row sums: {1, 3, 38, 882, 28104, 1123560, 53927280, 3019902480, 193273557120, 13915694298240, ...}.

%C In Comtet there is this function:

%C x^n = Sum_{k=1..n} Eulerian(n, k*binomial(x+k-1, n)

%C In the OEIS I was looking for an Umbral Calculus expansion for the MacMahon and found this "Worpitzky form":

%C Sum_{k=1..n} MacMahon(n,k)*binomial(x+k-1, n-1) = (2*x+1)^(n+1)

%C They use the infinite sums k, 2*k+1 type polynomials and are pretty much alike except for a sliding offset in n.

%C Conjecture: "Worpitzky forms"

%C Some general polynomial form: general Pascal recursion Pascal(n,k,m)

%C p(x,n,m) = Sum_{k=1..n} Pascal(n,k,m)*binomial(x+k-1,n-1)

%C where p(x,n,m) are the inverse z transform polynomials.

%F p(x,n) = Sum_{k=1..n} A(n, k)*binomial(x + k - 1, n - 1).

%e {1},

%e {1, 2},

%e {2, 18, 18},

%e {6, 156, 432, 288},

%e {24, 792, 7416, 13248, 6624},

%e {120, -11280, 64800, 374400, 496800, 198720},

%e {720, -62640, -1254960, 4968000, 20865600, 22057920, 7352640},

%e {5040, 24012000, -11854080, -125677440, 389491200, 1288103040, 1132306560, 323516160},

%e {40320, 192378240, 5004581760, -1669248000, -12569437440, 32116331520, 87702289920, 65997296640, 16499324160},

%e {362880, -119545632000, 57161064960, 868954106880, -218287560960, -1293900894720, 2812649495040, 6545378949120, 4306323605760, 956960801280}

%t (* Worpitzky form polynomials for A142462 *)

%t Clear[A, m, n, k, a, p]

%t m = 7;

%t A[n_, 1] := 1 A[n_, n_] := 1

%t A[n_, k_] := (m*n - m*k + 1)A[n - 1, k - 1] + (m*k - (m - 1))A[n - 1, k];

%t a = Table[A[n, k], {n, 10}, {k, n}];

%t p[x_, n_] = Sum[a[[n, k]]*Binomial[x + k - 1, n - 1], {k, 1, n}];

%t Table[CoefficientList[Expand[(n - 1)!*p[x, n]], x], {n, 1, 10}];

%t Flatten[%]

%Y Cf. A142462.

%K uned,sign

%O 1,3

%A _Roger L. Bagula_, Nov 22 2009