A051379 Generalized Stirling number triangle of first kind.
1, -8, 1, 72, -17, 1, -720, 242, -27, 1, 7920, -3382, 539, -38, 1, -95040, 48504, -9850, 995, -50, 1, 1235520, -725592, 176554, -22785, 1645, -63, 1, -17297280, 11393808, -3197348, 495544, -45815, 2527, -77, 1, 259459200, -188204400, 59354028, -10630508, 1182769, -83720, 3682, -92, 1
Offset: 0
Examples
{1}; {-8,1}; {72,-17,1}; {-720,242,-27,1}; ... s(2,x)=72-17*x+x^2; S1(2,x)= -x+x^2 (Stirling1).
Links
- Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
- D. S. Mitrinovic, M. S. Mitrinovic, Tableaux d'une classe de nombres reliés aux nombres de Stirling, Univ. Beograd. Pubi. Elektrotehn. Fak. Ser. Mat. Fiz. 77 (1962).
Crossrefs
Programs
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Haskell
a051379 n k = a051379_tabl !! n !! k a051379_row n = a051379_tabl !! n a051379_tabl = map fst $ iterate (\(row, i) -> (zipWith (-) ([0] ++ row) $ map (* i) (row ++ [0]), i + 1)) ([1], 8) -- Reinhard Zumkeller, Mar 12 2014
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Mathematica
a[n_, m_] := Pochhammer[m + 1, n - m] SeriesCoefficient[Log[1 + x]^m/(1 + x)^8, {x, 0, n}]; Table[a[n, m], {n, 0, 8}, {m, 0, n}] // Flatten (* Jean-François Alcover, Oct 29 2019 *)
Formula
a(n, m)= a(n-1, m-1) - (n+7)*a(n-1, m), n >= m >= 0; a(n, m) := 0, n
E.g.f. for m-th column of signed triangle: ((log(1+x))^m)/(m!*(1+x)^8).
Triangle (signed) = [ -8, -1, -9, -2, -10, -3, -11, -4, -12, ...] DELTA A000035; triangle (unsigned) = [8, 1, 9, 2, 10, 3, 11, 4, 12, 5, ...] DELTA A000035; where DELTA is Deléham's operator defined in A084938.
If we define f(n,i,a)=sum(binomial(n,k)*stirling1(n-k,i)*product(-a-j,j=0..k-1),k=0..n-i), then T(n,i) = f(n,i,8), for n=1,2,...;i=0...n. - Milan Janjic, Dec 21 2008
Extensions
Typo fixed in data by Reinhard Zumkeller, Mar 12 2014
A051380 Generalized Stirling number triangle of first kind.
1, -9, 1, 90, -19, 1, -990, 299, -30, 1, 11880, -4578, 659, -42, 1, -154440, 71394, -13145, 1205, -55, 1, 2162160, -1153956, 255424, -30015, 1975, -69, 1, -32432400, 19471500, -4985316, 705649, -59640, 3010, -84, 1, 518918400, -343976400, 99236556, -16275700, 1659889, -107800, 4354, -100, 1
Offset: 0
Comments
a(n,m)= ^9P_n^m in the notation of the given reference with a(0,0) := 1. The monic row polynomials s(n,x) := sum(a(n,m)*x^m,m=0..n) which are s(n,x)= product(x-(9+k),k=0..n-1), n >= 1 and s(0,x)=1 satisfy s(n,x+y) = sum(binomial(n,k)*s(k,x)*S1(n-k,y),k=0..n), with the Stirling1 polynomials S1(n,x)=sum(A008275(n,m)*x^m, m=1..n) and S1(0,x)=1. In the umbral calculus (see the S. Roman reference given in A048854) the s(n,x) polynomials are called Sheffer for (exp(9*t),exp(t)-1).
Examples
{1}; {-9,1}; {90,-19,1}; {-990,299,-30,1}; ... s(2,x)= 90-19*x+x^2; S1(2,x)= -x+x^2 (Stirling1).
Links
- Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
- D. S. Mitrinovic, M. S. Mitrinovic, Tableaux d'une classe de nombres reliés aux nombres de Stirling, Univ. Beograd. Pubi. Elektrotehn. Fak. Ser. Mat. Fiz. 77 (1962).
Crossrefs
Programs
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Haskell
a051380 n k = a051380_tabl !! n !! k a051380_row n = a051380_tabl !! n a051380_tabl = map fst $ iterate (\(row, i) -> (zipWith (-) ([0] ++ row) $ map (* i) (row ++ [0]), i + 1)) ([1], 9) -- Reinhard Zumkeller, Mar 12 2014
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Mathematica
a[n_, m_] := Pochhammer[m + 1, n - m] SeriesCoefficient[Log[1 + x]^m/(1 + x)^9, {x, 0, n}]; Table[a[n, m], {n, 0, 8}, {m, 0, n}] // Flatten (* Jean-François Alcover, Oct 29 2019 *)
Formula
a(n, m)= a(n-1, m-1) - (n+8)*a(n-1, m), n >= m >= 0; a(n, m) := 0, n
E.g.f. for m-th column of signed triangle: ((log(1+x))^m)/(m!*(1+x)^9).
Triangle (signed) = [ -9, -1, -10, -2, -11, -3, -12, -4, -13, ...] DELTA A000035; triangle (unsigned) = [9, 1, 10, 2, 11, 3, 12, 4, 13, 5, ...] DELTA A000035; where DELTA is Deléham's operator defined in A084938.
If we define f(n,i,a)=sum(binomial(n,k)*stirling1(n-k,i)*product(-a-j,j=0..k-1),k=0..n-i), then T(n,i) = f(n,i,9), for n=1,2,...;i=0...n. - Milan Janjic, Dec 21 2008
A167569 The lower left triangle of the ED2 array A167560.
1, 2, 4, 6, 16, 32, 24, 80, 192, 384, 120, 480, 1344, 3072, 6144, 720, 3360, 10752, 27648, 61440, 122880, 5040, 26880, 96768, 276480, 675840, 1474560, 2949120, 40320, 241920, 967680, 3041280, 8110080, 19169280, 41287680, 82575360
Offset: 1
Comments
We discovered that the numbers that appear in the lower left triangle of the ED2 array A167560 (m <= n) behave in a regular way, see the formula below. This rather simple regularity doesn't show up in the upper right triangle of the ED2 array (m > n).
Examples
The first few triangle rows are: [1] [2, 4] [6, 16, 32] [24, 80, 192, 384] [120, 480, 1344, 3072, 6144] [720, 3360, 10752, 27648, 61440, 122880]
Links
- G. C. Greubel, Table of n, a(n) for the first 50 rows
Crossrefs
Programs
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Maple
a := proc(n, m): 4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)! end: seq(seq(a(n, m), m=1..n), n=1..8); # Johannes W. Meijer, revised Nov 23 2012
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Mathematica
Flatten[Table[4^(m - 1)*(m - 1)!*(n + m - 1)!/(2*m - 1)!, {n, 1, 50}, {m, n}]] (* G. C. Greubel, Jun 16 2016 *)
Formula
a(n,m) = 4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)!.
A257606 Triangle read by rows: T(n,k) = t(n-k, k); t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = x + 4.
1, 4, 4, 16, 40, 16, 64, 296, 296, 64, 256, 1928, 3552, 1928, 256, 1024, 11688, 34808, 34808, 11688, 1024, 4096, 67656, 302352, 487312, 302352, 67656, 4096, 16384, 379240, 2423016, 5830000, 5830000, 2423016, 379240, 16384, 65536, 2076424, 18330496, 62617144, 93280000, 62617144, 18330496, 2076424, 65536
Offset: 0
Examples
Triangle begins as: 1; 4, 4; 16, 40, 16; 64, 296, 296, 64; 256, 1928, 3552, 1928, 256; 1024, 11688, 34808, 34808, 11688, 1024; 4096, 67656, 302352, 487312, 302352, 67656, 4096; 16384, 379240, 2423016, 5830000, 5830000, 2423016, 379240, 16384;
Links
- G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Crossrefs
Programs
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Mathematica
T[n_, k_, a_, b_]:= T[n, k, a, b]= If[k<0 || k>n, 0, If[n==0, 1, (a*(n-k)+b)*T[n-1, k-1, a, b] + (a*k+b)*T[n-1, k, a, b]]]; Table[T[n,k,1,4], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 24 2022 *)
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Sage
def T(n,k,a,b): # A257606 if (k<0 or k>n): return 0 elif (n==0): return 1 else: return (a*k+b)*T(n-1,k,a,b) + (a*(n-k)+b)*T(n-1,k-1,a,b) flatten([[T(n,k,1,4) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 24 2022
Formula
T(n,k) = t(n-k, k); t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = x + 4.
Sum_{k=0..n} T(n, k) = A049388(n).
T(n,0) = T(n,n) = 4^n. - Georg Fischer, Oct 02 2021
From G. C. Greubel, Mar 24 2022: (Start)
T(n, k) = (a*k + b)*T(n-1, k) + (a*(n-k) + b)*T(n-1, k-1), with T(n, 0) = 1, a = 1, and b = 4.
T(n, n-k) = T(n, k).
T(n, 1) = 8*5^n - 4^n*(8+n).
T(n, 2) = 2*((56 +15*n +n^2)*4^(n-1) - 4*(8+n)*5^n + 3*6^(n+1)). (End)
Extensions
a(3) corrected by Georg Fischer, Oct 02 2021
A143499 Triangle of unsigned 4-Lah numbers.
1, 8, 1, 72, 18, 1, 720, 270, 30, 1, 7920, 3960, 660, 44, 1, 95040, 59400, 13200, 1320, 60, 1, 1235520, 926640, 257400, 34320, 2340, 78, 1, 17297280, 15135120, 5045040, 840840, 76440, 3822, 98, 1, 259459200, 259459200, 100900800, 20180160, 2293200
Offset: 4
Comments
This is the case r = 4 of the unsigned r-Lah numbers L(r;n,k). The unsigned 4-Lah numbers count the partitions of the set {1,2,...,n} into k ordered lists with the restriction that the elements 1, 2, 3 and 4 belong to different lists. For other cases see A105278 (r = 1), A143497 (r = 2 and comments on the general case) and A143498 (r = 3).
The unsigned 4-Lah numbers are related to the 4-Stirling numbers: the lower triangular array of unsigned 4-Lah numbers may be expressed as the matrix product St1(4) * St2(4), where St1(4) = A143493 and St2(4) = A143496 are the arrays of 4-restricted Stirling numbers of the first and second kind respectively. An alternative factorization for the array is as St1 * P^6 * St2, where P denotes Pascal's triangle, A007318, St1 is the triangle of unsigned Stirling numbers of the first kind, abs(A008275) and St2 denotes the triangle of Stirling numbers of the second kind, A008277.
Examples
Triangle begins n\k|......4......5......6......7......8......9 ============================================== 4..|......1 5..|......8......1 6..|.....72.....18......1 7..|....720....270.....30......1 8..|...7920...3960....660.....44......1 9..|..95040..59400..13200...1320.....60......1 ... T(5,4) = 8. The partitions of {1,2,3,4,5} into 4 ordered lists, such that the elements 1, 2, 3 and 4 lie in different lists, are: {1}{2}{3}{4,5} and {1}{2}{3}{5,4}, {1}{2}{4}{3,5} and {1}{2}{4}{5,3}, {1}{3}{4}{2,5} and {1}{3}{4}{5,2}, {2}{3}{4}{1,5} and {2}{3}{4}{5,1}.
Links
- Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 No. 1-3, 33-51 (2001).
- G. Nyul, G. Rácz, The r-Lah numbers, Discrete Mathematics, 338 (2015), 1660-1666.
- Michael J. Schlosser and Meesue Yoo, Elliptic Rook and File Numbers, Electronic Journal of Combinatorics, 24(1) (2017), #P1.31.
- M. Shattuck, Generalized r-Lah numbers, arXiv:1412.8721 [math.CO], 2014
Crossrefs
Programs
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Maple
with combinat: T := (n, k) -> (n-4)!/(k-4)!*binomial(n+3,k+3): for n from 4 to 13 do seq(T(n, k), k = 4..n) end do;
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Mathematica
T[n_, k_] := (n-4)!/(k-4)!*Binomial[n+3, k+3]; Table[T[n, k], {n, 4, 10}, {k, 4, n}] // Flatten (* Amiram Eldar, Nov 26 2018 *)
Formula
T(n,k) = (n-4)!/(k-4)!*binomial(n+3,k+3), n,k >= 4.
Recurrence: T(n,k) = (n+k-1)*T(n-1,k) + T(n-1,k-1) for n,k >= 4, with the boundary conditions: T(n,k) = 0 if n < 4 or k < 4; T(4,4) = 1.
E.g.f. for column k: Sum_{n >= k} T(n,k)*t^n/(n-4)! = 1/(k-4)!*t^k/(1-t)^(k+4) for k >= 4.
E.g.f: Sum_{n = 4..inf} Sum_{k = 4..n} T(n,k)*x^k*t^n/(n-4)! = (x*t)^4/(1-t)^8*exp(x*t/(1-t)) = (x*t)^4*(1 + (8+x)*t +(72+18*x+x^2)*t^2/2! + ...).
Generalized Lah identity: (x+7)*(x+8)*...*(x+n+2) = Sum_{k = 4..n} T(n,k)*(x-1)*(x-2)*...*(x-k+4).
The polynomials 1/n!*Sum_{k = 4..n+4} T(n+4,k)*(-x)^(k-4) for n >= 0 are the generalized Laguerre polynomials Laguerre(n,7,x).
A161742 Third left hand column of the RSEG2 triangle A161739.
1, 4, 13, 30, -14, -504, 736, 44640, -104544, -10644480, 33246720, 5425056000, -20843695872, -5185511654400, 23457840537600, 8506857655296000, -44092609863966720, -22430879475779174400, 130748316971139072000
Offset: 2
Crossrefs
Programs
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Maple
nmax:=21; for n from 0 to nmax do A008955(n,0):=1 end do: for n from 0 to nmax do A008955(n,n):=(n!)^2 end do: for n from 1 to nmax do for m from 1 to n-1 do A008955(n,m):= A008955(n-1,m-1)*n^2+A008955(n-1,m) end do: end do: for n from 1 to nmax do A028246(n,1):=1 od: for n from 1 to nmax do A028246(n,n):=(n-1)! od: for n from 3 to nmax do for m from 2 to n-1 do A028246(n,m):=m*A028246(n-1,m)+(m-1)*A028246(n-1,m-1) od: od: for n from 2 to nmax do a(n):=sum(((-1)^k/((k+1)!*(k+2)!)) *(n!)*A028246(n,k+2)* A008955(k+1,k),k=0..n-2) od: seq(a(n),n=2..nmax);
A161743 Fourth left hand column of the RSEG2 triangle A161739.
1, 10, 73, 425, 1561, -2856, -73520, 380160, 15376416, -117209664, -7506967104, 72162155520, 7045087741056, -80246202992640, -11448278791372800, 149576169325363200, 30017051616972275712, -440857664887810867200
Offset: 3
Crossrefs
Programs
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Maple
nmax:=21; for n from 0 to nmax do A008955(n,0):=1 end do: for n from 0 to nmax do A008955(n,n):=(n!)^2 end do: for n from 1 to nmax do for m from 1 to n-1 do A008955(n,m):= A008955(n-1,m-1)*n^2+A008955(n-1,m) end do: end do: for n from 1 to nmax do A028246(n,1):=1 od: for n from 1 to nmax do A028246(n,n):=(n-1)! od: for n from 3 to nmax do for m from 2 to n-1 do A028246(n,m):=m*A028246(n-1,m)+(m-1)*A028246(n-1,m-1) od: od: for n from 3 to nmax do a(n) := sum(((-1)^k/((k+2)!*(k+3)!))*(n!)*A028246(n,k+3)* A008955(k+2,k), k=0..n-3) od: seq(a(n),n=3..nmax);
A162995 A scaled version of triangle A162990.
1, 3, 1, 12, 4, 1, 60, 20, 5, 1, 360, 120, 30, 6, 1, 2520, 840, 210, 42, 7, 1, 20160, 6720, 1680, 336, 56, 8, 1, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 1814400, 604800, 151200, 30240, 5040, 720, 90, 10, 1
Offset: 1
Comments
We get this scaled version of triangle A162990 by dividing the coefficients in the left hand columns by their 'top-values' and then taking the square root.
T(n,k) = A173333(n+1,k+1), 1 <= k <= n. - Reinhard Zumkeller, Feb 19 2010
T(n,k) = A094587(n+1,k+1), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012
Examples
The first few rows of the triangle are: [1] [3, 1] [12, 4, 1] [60, 20, 5, 1]
Links
- Reinhard Zumkeller, Rows n = 1..150 of triangle, flattened
Crossrefs
Programs
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Haskell
a162995 n k = a162995_tabl !! (n-1) !! (k-1) a162995_row n = a162995_tabl !! (n-1) a162995_tabl = map fst $ iterate f ([1], 3) where f (row, i) = (map (* i) row ++ [1], i + 1) -- Reinhard Zumkeller, Jul 04 2012
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Maple
a := proc(n, m): (n+1)!/(m+1)! end: seq(seq(a(n, m), m=1..n), n=1..9); # Johannes W. Meijer, revised Nov 23 2012
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Mathematica
Table[(n+1)!/(m+1)!, {n, 10}, {m, n}] (* Paolo Xausa, Mar 31 2024 *)
Formula
a(n,m) = (n+1)!/(m+1)! for n = 1, 2, 3, ..., and m = 1, 2, ..., n.
A094646 Generalized Stirling number triangle of first kind.
1, -2, 1, 2, -3, 1, 0, 2, -3, 1, 0, 2, -1, -2, 1, 0, 4, 0, -5, 0, 1, 0, 12, 4, -15, -5, 3, 1, 0, 48, 28, -56, -35, 7, 7, 1, 0, 240, 188, -252, -231, 0, 42, 12, 1, 0, 1440, 1368, -1324, -1638, -231, 252, 114, 18, 1, 0, 10080, 11016, -7900, -12790, -3255, 1533, 1050, 240, 25, 1
Offset: 0
Comments
Triangle T(n,k), 0 <= k <= n, read by rows, given by [ -2, 1, -1, 2, 0, 3, 1, 4, 2, 5, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 23 2006
From Wolfdieter Lang, Jun 23 2011: (Start)
The row polynomials s(n,x):=Sum_{k=0..n} T(n,k)*x^k satisfy risefac(x-2,n)=s(n,x), with the rising factorials risefac(x-2,n):=Product_{j=0..n-1} (x-2+j), n >= 1, risefac(x-2,0)=1. Compare this with the formula risefac(x,n)=|S1|(n,x), with the row polynomials |S1|(n,x) of A132393 (unsigned Stirling1).
This is the third triangle of an a-family of Sheffer arrays, call them |S1|(a), with e.g.f. of the row polynomials |S1|(a;x;z) = ((1-z)^a)*exp(-x*log(1-z)). In the notation showing the column e.g.f.s this is Sheffer ((1-z)^a,-log(1-z)). In the umbral notation (see the Roman reference, given under A094645) this is called Sheffer for (exp(a*t),1-exp(-t)). For a=0 this becomes the unsigned Stirling1 triangle |S1|(0) = A132393 with row polynomials |S1|(0;n,x) =: s1(n,x).
E.g.f. column number k (with leading zeros): ((1-x)^a)*((-log(1-x))^k)/k!, k >= 0.
E.g.f. for row sums is (1-x)^(a-1), and the e.g.f. for the alternating row sums is (1-x)^(a+1).
Row polynomial recurrence:
|S1|(a;n,x)=(x+(n-1-a))*|S1|(a;n-1,m), |S1|(a;0,x)=1.
Meixner identity (see the reference under A060338):
|S1|(a;n,x) - |S1|(a;n,x-1) = n*|S1|(a;n-1,x), n >= 1,
Also (from the corollary 3.7.2 on p. 50 of the Roman reference): |S1|(a;n,x) = (x-a)*|S1|(a;n-1,x+1), n >= 1.
Recurrence: |S1|(a;n,k) = |S1|(a;n-1,k-1) + (n-(a+1))*|S1|(a;n-1,k); |S1|(a;n,k)=0 if n < m, |S1|(a;n,-1)=0, |S1|(a;0,0)=1.
Connection to |Stirling1|=|S1|(0):
|S1|(a;n,k) = Sum_{p=0..a} |S1|(a;a,p)*abs(Stirling1(n-a,k-p)), n >= a.
The exponential convolution identity is
|S1|(a;n,x+y) = Sum_{k=0..n} binomial(n,k)*|S1|(a;k,y)*s1(n-k,x), n >= 0, with symmetry x <-> y.
The Sheffer a- and z-sequences are (see the W. Lang link under A006232): Sha(a;n)=A164555(n)/A027642(n) (independent of a) with e.g.f. x/(1-exp(-x)), and the z-sequence has e.g.f. (exp(a*x)-1)/(exp(-x)-1).
The inverse Sheffer matrix has e.g.f. exp(a*z)*exp(x*(1-exp(-z))), in short notation (exp(a*z),1-exp(-z)),
(or in umbral notation ((1-t)^a,-log(1-t))).
(End)
Examples
Triangle begins 1; -2, 1; 2, -3, 1; 0, 2, -3, 1; 0, 2, -1, -2, 1; 0, 4, 0, -5, 0, 1; ... risefac(x-2,3) = (x-2)*(x-1)*x = 2*x-3*x^2+x^3. -1 = T(4,2) = T(3,1) + 1*T(3,2) = 2 + (-3). T(4,3) = 2*abs(S1(2,3)) - 3*abs(S1(2,2)) + 1*abs(S1(2,1)) = 2*0 - 3*1 + 1*1 = -2.
Programs
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Maple
A094646_row := n -> seq((-1)^(n-k)*coeff(expand(pochhammer(x-n+3, n)), x, k), k=0..n): seq(print(A094646_row(n)), n = 0..6); # Peter Luschny, May 16 2013
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Mathematica
Flatten[ Table[ CoefficientList[ Pochhammer[x-2, n], x], {n, 0, 10}]] (* Jean-François Alcover, Sep 26 2011 *)
Formula
E.g.f.: (1-y)^(2-x).
Sum_{k=0..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 = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 respectively. - Philippe Deléham, Nov 13 2007
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j), then |T(n,i)| = |f(n,i,-2)|, for n=1,2,...; i=0..n. - Milan Janjic, Dec 21 2008
From Wolfdieter Lang, Jun 23 2011: (Start)
risefac(x-2,n) = Sum_{k=0..n} T(n,k)*x^k, n >= 0, with the rising factorials (see a comment above).
Recurrence: T(n,k) = T(n-1,k-1) + (n-3)*T(n-1,k); T(n,k)=0 if n < m, T(n,-1)=0, T(0,0)=1.
T(n,k) = 2*abs(S1(n-2,k)) - 3*abs(S1(n-2,k-1)) + abs(S1(n-2,k-2)), n >= 2, with S1(n,k) = Stirling1(n,k) = A048994(n,k).
E.g.f. column number k (with leading zeros):
((1-x)^2)*((-log(1-x))^k)/k!, k >= 0.
E.g.f. for row sums is 1-x, i.e., [1,-1,0,0,...],
and the e.g.f. for the alternating row sums is (1-x)^3. i.e., [1,-3,3,1,0,0,...]. (End)
A176734 a(n) = (n+7)*a(n-1) + (n-1)*a(n-2), a(-1)=0, a(0)=1.
1, 8, 73, 746, 8425, 104084, 1395217, 20157542, 312129649, 5155334720, 90449857081, 1679650774658, 32908313146393, 678322072223756, 14672571587601985, 332293083938376254, 7862829504396683617, 194024597448534426872, 4984283037788104293289, 133083801736564331309210
Offset: 0
Comments
a(n) enumerates the possibilities for distributing n beads, n>=1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k=8 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords contribute a factor 1 in the counting, e.g., a(0):= 1*1 =1. See A000255 for the description of a fixed cord with beads. This produces for a(n) the exponential (aka binomial) convolution of the subfactorial sequence {A000166(n)} and the sequence {A049388(n) = (n+7)!/7!}. See the necklaces and cords problem comment in A000153. Therefore the recurrence with inputs holds. 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).
Examples
Necklaces and 8 cords problem. For n=4 one considers the following weak 2 part compositions of 4: (4,0), (3,1), (2,2), and (0,4), where (1,3) does not appear because there are no necklaces with 1 bead. These compositions contribute respectively !4*1,binomial(4,3)*!3*c8(1), (binomial(4,2)*!2)*c8(2), and 1*c8(4) with the subfactorials !n:=A000166(n) (see the necklace comment there) and the c8(n):=A049388(n) numbers for the pure 8-cord problem (see the remark on the e.g.f. for the k cords problem in A000153; here for k=8: 1/(1-x)^8). This adds up as 9 + 4*2*8 + (6*1)*72 + 7920 = 8425 = a(4).
Crossrefs
Cf. A176733 (necklaces and k=7 cords).
Programs
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Mathematica
nxt[{n_,a_,b_}]:={n+1,b,(n+8)b+n*a}; Transpose[NestList[nxt,{1,1,8},20]][[2]] (* Harvey P. Dale, Mar 19 2013 *) Table[(-1)^n HypergeometricPFQ[{9, -n}, {}, 1], {n, 0, 20}] (* Benedict W. J. Irwin, May 29 2016 *)
Formula
E.g.f. (exp(-x)/(1-x))*(1/(1-x)^8) = exp(-x)/(1-x)^9, equivalent to the given recurrence.
a(n) = A086764(n+8,8).
a(n) = (-1)^n*2F0(9,-n;;1). - Benedict W. J. Irwin, May 29 2016
Comments