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A086764
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Triangle T(n, k), read by row, related to Euler's difference table A068106 (divide column k of A068106 by k!).
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20
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1, 0, 1, 1, 1, 1, 2, 3, 2, 1, 9, 11, 7, 3, 1, 44, 53, 32, 13, 4, 1, 265, 309, 181, 71, 21, 5, 1, 1854, 2119, 1214, 465, 134, 31, 6, 1, 14833, 16687, 9403, 3539, 1001, 227, 43, 7, 1, 133496, 148329, 82508, 30637, 8544, 1909, 356, 57, 8, 1
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OFFSET
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0,7
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COMMENTS
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The k-th column sequence, k >= 0, without leading zeros, enumerates the ways to distribute n beads, n >= 1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k+1 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords each contribute a factor 1, hence for n=0 one has 1. See A000255 for the description of a fixed cord with beads. This comment derives from a family of recurrences found by Malin Sjodahl for a combinatorial problem for certain quark and gluon diagrams (Feb 27 2010). - Wolfdieter Lang, Jun 02 2010
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LINKS
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FORMULA
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T(n, n) = 1; T(n+1, n) = n.
T(n+2, n) = A002061(n+1) = n^2 + n + 1; T(n+3, n) = n^3 + 3*n^2 + 5*n + 2.
T(n, k) = (k + 1)*T(n, k + 1) - T(n-1, k); T(n, n) = 1; T(n, k) = 0, if k > n.
T(n, k) = (n-1)*T(n-1, k) + (n-k-1)*T(n-2, k).
T(n, k) = (1/k!) * Sum_{j>=0} (-1)^j*binomial(n-k, j)*(n-j)!. - Philippe Deléham, Jun 13 2005
The following remarks all relate to the array read as a square array: e.g.f for column k: exp(-y)/(1-y)^(k+1); e.g.f. for array: exp(-y)/(1-x-y) = (1 + x + x^2 + x^3 + ...) + (x + 2*x^2 + 3*x^3 + 4*x^4 + ...)*y + (1 + 3*x + 7*x^2 + 13*x^3 + ...)*y^2/2! + ... .
This table is closely connected to the constant e. The row, column and diagonal entries of this table occur in series formulas for e.
Row n for n >= 2: e = n!*(1/T(n,0) + (-1)^n*[1/(1!*T(n,0)*T(n,1)) + 1/(2!*T(n,1)*T(n,2)) + 1/(3!*T(n,2)*T(n,3)) + ...]). For example, row 3 gives e = 6*(1/2 - 1/(1!*2*11) - 1/(2!*11*32) - 1/(3!*32*71) - ...). See A095000.
Column 0: e = 2 + Sum_{n>=2} (-1)^n*n!/(T(n,0)*T(n+1,0)) = 2 + 2!/(1*2) - 3 !/(2*9) + 4!/(9*44) - ... .
Column k, k >= 1: e = (1 + 1/1! + 1/2! + ... + 1/k!) + 1/k!*Sum_{n >= 0} (-1)^n*n!/(T(n,k)*T(n+1,k)). For example, column 3 gives e = 8/3 + 1/6*(1/(1*3) - 1/(3*13) + 2/(13*71) - 6/(71*465) + ...).
Main diagonal: e = 1 + 2*(1/(1*1) - 1/(1*7) + 1/(7*71) - 1/(71*1001) + ...).
First subdiagonal: e = 8/3 + 5/(3*32) - 7/(32*465) + 9/(465*8544) - ... .
Second subdiagonal: e = 2*(1 + 2^2/(1*11) - 3^2/(11*181) + 4^2/(181*3539) - ...). See A143413.
Third subdiagonal: e = 3 - (2*3*5)/(2*53) + (3*4*7)/(53*1214) - (4*5*9)/(1214*30637) + ... .
For the corresponding results for the constants 1/e, sqrt(e) and 1/sqrt(e) see A143409, A143410 and A143411 respectively. For other arrays similarly related to constants see A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)). (End)
G.f. for column k is hypergeom([1,k+1],[],x/(x+1))/(x+1). - Mark van Hoeij, Nov 07 2011
T(n, k) = (n!/k!)*hypergeom([k-n], [-n], -1). - Peter Luschny, Oct 05 2017
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EXAMPLE
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Formatted as a square array:
2 11 32 71 134 227 356 ... A094792;
9 53 181 465 1001 1909 ... A094793;
Formatted as a triangular array (mirror of A076731):
1;
0 1;
1 1 1;
2 3 2 1;
9 11 7 3 1;
44 53 32 13 4 1;
265 309 181 71 21 5 1;
1854 2119 1214 465 134 31 6 1;
14833 16687 9403 3539 1001 227 43 7 1;
133496 148329 82508 30637 8544 1909 356 57 8 1;
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MATHEMATICA
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T[n_, k_]:=(1/k!)*Sum[(-1)^j*Binomial[n-k, j]*(n-j)!, {j, 0, n}]; Flatten[Table[T[n, k], {n, 0, 11}, {k, 0, n}]] (* Indranil Ghosh, Feb 20 2017 *)
T[n_, k_] := (n!/k!) HypergeometricPFQ[{k-n}, {-n}, -1];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Peter Luschny, Oct 05 2017 *)
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PROG
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(Magma)
A086764:= func< n, k | (&+[(-1)^j*Binomial(n-k, j)*Factorial(n-j): j in [0..n]])/Factorial(k) >;
(SageMath)
def A086764(n, k): return sum((-1)^j*binomial(n-k, j)*factorial(n-j) for j in range(n+1))//factorial(k)
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CROSSREFS
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Columns: A000166, A000155, A000153, A000261, A001909, A001910, A176732, A176733, A176734, A176735, A176736.
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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