A024395 -id:A024395 - cp's OEIS frontend

A225470 Triangle read by rows, s_3(n, k) where s_m(n, k) are the Stirling-Frobenius cycle numbers of order m; n >= 0, k >= 0.

1, 2, 1, 10, 7, 1, 80, 66, 15, 1, 880, 806, 231, 26, 1, 12320, 12164, 4040, 595, 40, 1, 209440, 219108, 80844, 14155, 1275, 57, 1, 4188800, 4591600, 1835988, 363944, 39655, 2415, 77, 1, 96342400, 109795600, 46819324, 10206700, 1276009, 95200, 4186, 100, 1

Offset: 0

Peter Luschny, May 16 2013

The Stirling-Frobenius subset numbers S_{m}(n,k), for m >= 1 fixed, regarded as an infinite lower triangular matrix, can be inverted by Sum_{k} S_{m}(n,k)*s_{m}(k,j)*(-1)^(n-k) = [j = n]. The inverse numbers s_{m}(k,j), which are unsigned, are the Stirling-Frobenius cycle numbers. For m = 1 this gives the classical Stirling cycle numbers A132393. The Stirling-Frobenius subset numbers are defined in A225468.

Triangle T(n,k), read by rows, given by (2, 3, 5, 6, 8, 9, 11, 12, 14, 15, ... (A007494)) DELTA (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, May 14 2015

			Triangle starts:
  [n\k][    0,      1,     2,     3,    4,  5,  6]
  [0]       1,
  [1]       2,      1,
  [2]      10,      7,     1,
  [3]      80,     66,    15,     1,
  [4]     880,    806,   231,    26,    1,
  [5]   12320,  12164,  4040,   595,   40,  1,
  [6]  209440, 219108, 80844, 14155, 1275, 57,  1.
  ...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence (see Maple program): T(4, 2) = T(3, 1) + (3*4 - 1)*T(3, 2) = 66 + 11*15 = 231.
Boas-Buck type recurrence for column k = 2 and n = 4: T(4, 2) = (4!/2)*(3*(2 + 6*(5/12))*T(2, 2)/2! + 1*(2 + 6*(1/2))*T(3,2)/3!) = (4!/2)*(3*9/4 + 5*15/3!) = 231. (End)

Peter Bala, A 3 parameter family of generalized Stirling numbers
Peter Luschny, Generalized Eulerian polynomials
Peter Luschny, The Stirling-Frobenius numbers

Cf. A225468; A132393 (m=1), A028338 (m=2), A225471(m=4).
Column k=0..4 give A008544, A024395(n-1), A286722(n-2), A383221, A383222.
T(n, n) ~ A000012; T(n, n-1) ~ A005449; T(n, n-2) ~ A024391; T(n, n-3) ~ A024392.
row sums ~ A032031; alternating row sums ~ A007559.
Cf. A132393.

SF_C := proc(n, k, m) option remember;
if n = 0 and k = 0 then return(1) fi;
if k > n or  k < 0 then return(0) fi;
SF_C(n-1, k-1, m) + (m*n-1)*SF_C(n-1, k, m) end:
seq(print(seq(SF_C(n, k, 3), k = 0..n)), n = 0..8);

SFC[0, 0, ] = 1; SFC[n, k_, ] /; (k > n || k < 0) = 0; SFC[n, k_, m_] := SFC[n, k, m] = SFC[n-1, k-1, m] + (m*n-1)*SFC[n-1, k, m]; Table[SFC[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 26 2013, after Maple *)

For a recurrence see the Maple program.
From Wolfdieter Lang, May 18 2017: (Start)
This is the Sheffer triangle (1/(1 - 3*x)^{-2/3}, -(1/3)*log(1-3*x)). See the P. Bala link where this is called exponential Riordan array, and the signed version is  denoted by s_{(3,0,2)}.
E.g.f. of row polynomials in the variable x (i.e., of the triangle): (1 - 3*z)^{-(2+x)/3}.
E.g.f. of column k: (1-3*x)^(-2/3)*(-(1/3)*log(1-3*x))^k/k!, k >= 0.
Recurrence for row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k: R(n, x) = (x+2)*R(n-1,x+3), with R(0, x) = 1.
R(n, x) = risefac(3,2;x,n) := Product_{j=0..(n-1)} (x + (2 + 3*j)). (See the P. Bala link, eq. (16) for the signed s_{3,0,2} row polynomials.)
T(n, k) = Sum_{j=0..(n-m)} binomial(n-j, k)* S1p(n, n-j)*2^(n-k-j)*3^j, with S1p(n, m) = A132393(n, m). (End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 3^(n-1-p)*(2 + 3*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/A002209(k+1), beginning {1/2, 5/12, 3/8, 251/720, ...}. See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017

A136212 Triple factorial array, read by antidiagonals, where row n+1 is generated from row n by first removing terms in row n at positions {[m*(m+5)/6], m >= 0} and then taking partial sums, starting with all 1's in row 0.

Original entry on oeis.org

1, 1, 1, 4, 2, 1, 28, 10, 3, 1, 280, 80, 18, 4, 1, 3640, 880, 162, 28, 5, 1, 58240, 12320, 1944, 280, 39, 6, 1, 1106560, 209440, 29160, 3640, 418, 52, 7, 1, 24344320, 4188800, 524880, 58240, 5714, 600, 66, 8, 1, 608608000, 96342400, 11022480, 1106560, 95064

Offset: 0

Paul D. Hanna, Dec 22 2007

This is the triple factorial variant of Moessner's factorial array described by A125714 and also of the double factorial array A135876. Another very interesting variant is A136217.

			Square array begins:
(1),(1),(1),1,(1),1,(1),1,(1),1,1,(1),1,1,(1),1,1,(1),1,1,1,(1),1,1,1,...;
(1),(2),(3),4,(5),6,(7),8,(9),10,11,(12),13,14,(15),16,17,(18),19,20,21,..;
(4),(10),(18),28,(39),52,(66),82,(99),118,138,(159),182,206,(231),258,286,..;
(28),(80),(162),280,(418),600,(806),1064,(1350),1696,2074,(2485),2966,3484,..;
(280),(880),(1944),3640,(5714),8680,(12164),16840,(22194),29080,36824,(45474),.;
(3640),(12320),(29160),58240,(95064),151200,(219108),315440,(428652),581680,...;
(58240),(209440),(524880),1106560,(1864456),3082240,...;
where terms in parenthesis are at positions {[m*(m+5)/6], m>=0}
and are removed before taking partial sums to obtain the next row.
To generate the array, start with all 1's in row 0; from then on,
obtain row n+1 from row n by first removing terms in row n at
positions {[m*(m+5)/6], m>=0} and then taking partial sums.
For example, to generate row 2 from row 1:
[(1),(2),(3),4,(5),6,(7),8,(9),10,11,(12),13,14,(15),16,17,(18),...],
remove terms at positions [0,1,2,4,6,8,11,14,17,...] to get:
[4, 6, 8, 10,11, 13,14, 16,17, 19,20,21, 23,24,25, 27,28,29, ...]
then take partial sums to obtain row 2:
[4, 10, 18, 28,39, 52,66, 82,99, 118,138,159, 182,206,231, ...].
Continuing in this way will generate all the rows of this array.

Cf. columns: A007559, A008544, A032031, A024216, A024395; related tables: A136213, A125714, A135876, A127054, A125781; A136217.

t[n_, k_] := t[n, k] = Module[{a = 0, m = 0, c = 0, d = 0}, If[n == 0, a = 1, While[d <= k, If[c == Quotient[(m*(m + 5)), 6], m += 1, a += t[n - 1, c]; d += 1]; c += 1]]; a]; Table[t[n - k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 06 2013, translated from Pari *)

{T(n, k)=local(A=0, m=0, c=0, d=0); if(n==0, A=1, until(d>k, if(c==(m*(m+5))\6, m+=1, A+=T(n-1, c); d+=1); c+=1)); A}

Columns 0, 1 and 2 form the triple factorials A007559, A008544 and A032031, respectively. Column 4 equals A024216; column 6 equals A024395.

A024216 a(n) = n-th elementary symmetric function of the first n+1 positive integers congruent to 1 mod 3.

Original entry on oeis.org

1, 5, 39, 418, 5714, 95064, 1864456, 42124592, 1077459120, 30777463360, 971142388160, 33547112941440, 1259204418129280, 51032742579123200, 2220990565060377600, 103308619261574809600, 5114702794181847910400

Offset: 0

Clark Kimberling

Comment by R. J. Mathar, Oct 01 2016: (Start)
The k-th elementary symmetric functions of the integers 1+j*3, j=0..n-1, form a triangle T(n,k), 0 <= k <= n, n >= 0:
1
1 1
1 5 4
1 12 39 28
1 22 159 418 280
1 35 445 2485 5714 3640
1 51 1005 9605 45474 95064 58240
1 70 1974 28700 227969 959070 1864456 1106560
1 92 3514 72128 859369 5974388 22963996 42124592 24344320
This here is the first subdiagonal. The diagonal seems to be A007559. The first columns are A000012, A000326, A024212, A024213, A024214. (End)

			From _Gheorghe Coserea_, Dec 24 2015: (Start)
For n = 1 we have a(1) = 1*4*(1/1 + 1/4) = 5.
For n = 2 we have a(2) = 1*4*7*(1/1 + 1/4 + 1/7) = 39.
For n = 3 we have a(3) = 1*4*7*10*(1/1 + 1/4 + 1/7 + 1/10) = 418.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A024395, A024382, A286718 (first column).

I:=[5,39]; [1] cat [n le 2 select I[n] else (6*n-1) * Self(n-1) - (3*n-2)^2 * Self(n-2) : n in [1..30]]; // Vincenzo Librandi, Aug 30 2015

f:= gfun:-rectoproc({-(3*n+1)^2*a(n-1)+(6*n+5)*a(n)-a(n+1), a(0) = 1, a(1) = 5, a(2) = 39}, a(n), remember):
map(f, [$0..30]); # Robert Israel, Aug 30 2015

Rest[CoefficientList[Series[-(1/3)*Log[1-3*x]/(1-3*x)^(1/3), {x, 0, 20}], x]* Range[0, 20]!] (* Vaclav Kotesovec, Oct 07 2013 *)

n = 33; a = vector(n); a[1] = 5; a[2] = 39;
for (k = 2, n-1, a[k+1] = (6*k+5) * a[k] - (3*k+1)^2 * a[k-1]);
print(concat(1,a));  \\ Gheorghe Coserea, Aug 29 2015

E.g.f. (for offset 1): -(1/3)*log(1-3*x)/(1-3*x)^(1/3). - Vladeta Jovovic, Sep 26 2003
For n >= 1, a(n-1) = 3^(n-1)*n!*Sum_{k=0..n-1} binomial(k-2/3, k)/(n-k). - Milan Janjic, Dec 14 2008, corrected by Peter Bala, Oct 08 2013
a(n) ~ (n+1)! * GAMMA(2/3) * 3^(n+3/2) * (log(n) + gamma + Pi*sqrt(3)/6 + 3*log(3)/2) / (6*Pi*n^(2/3)), where "GAMMA" is the Gamma function and "gamma" is the Euler-Mascheroni constant (A001620). - Vaclav Kotesovec, Oct 07 2013
a(n+1) = (6*n+5) * a(n) - (3*n+1)^2 * a(n-1). - Gheorghe Coserea, Aug 29 2015
E.g.f.: (3 - log(1-3*x))/(3*(1-3*x)^(4/3)). - Robert Israel, Aug 30 2015
a(n) = A286718(n+1, 1), n >= 0.
Boas-Buck type recurrence: a(0) = 1 and for n >= 1: a(n) = ((n+1)!/n) * Sum_{p=1..n} 3^(n-p)*(1 + 3*beta(n-p))*a(p-1)/p!, with beta(k) = A002208(k+1) / A002209(k+1). Proof from a(n) = A286718(n+1, 1). - Wolfdieter Lang, Aug 09 2017

More terms from Vladeta Jovovic, Sep 26 2003

A286722 Column k=2 of the triangle A225470; Sheffer ((1 - 3x)^(-2/3), (-1/3)log(1 - 3*x)).

Original entry on oeis.org

1, 15, 231, 4040, 80844, 1835988, 46819324, 1327098024, 41436870696, 1414064576672, 52383613213920, 2094099207620160, 89873259151044160, 4122137910567440640, 201246677825480820480, 10420702442559832716800, 570477791902749185318400, 32923432900388514432614400

Offset: 0

Wolfdieter Lang, May 29 2017

a(n) is, for n >= 1, the total volume of the binomial(n+2, n) rectangular polytopes (hyper-cuboids) built from n orthogonal vectors with lengths of the sides from the set {2 + 3*j | j=0..n+1}. See the formula a(n) = sigma[3,2]^{(n+2)}_n and an example below.

			a(2) = 231 because sigma[3,2]^{(4)}_2 = 2*(5 + 8 + 11) + 5*(8 + 11) + 8*11 = 231. There are six rectangles (2D rectangular polytopes) built from two orthogonal vectors of different lengths from the set of {2,5,8,11} of total area 231.

Cf. A008544 (k=0), A024395 (k=1), A225470.

a(n) = A225470(n+2, 2), n >= 0.
E.g.f.: (d^2/dx^2)((1 - 3*x)^(-2/3)*((-1/3)*log(1 - 3*x))^2/2!) = (5*(log(1-3*x))^2 - 21*log(1-3*x) + 9)/(3^2*(1-3*x)^(8/3)).
a(n) = sigma[3,2]^{(n+2)}_n, n >= 0, with the elementary symmetric function sigma[3,2]^{n+2}_n of degree n of the n+2 numbers 2, 5, 8, ..., (2 + 3*(n+1)).

cp's OEIS Frontend

A225470 Triangle read by rows, s_3(n, k) where s_m(n, k) are the Stirling-Frobenius cycle numbers of order m; n >= 0, k >= 0.

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A136212 Triple factorial array, read by antidiagonals, where row n+1 is generated from row n by first removing terms in row n at positions {[m*(m+5)/6], m >= 0} and then taking partial sums, starting with all 1's in row 0.

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A024216 a(n) = n-th elementary symmetric function of the first n+1 positive integers congruent to 1 mod 3.

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A286722 Column k=2 of the triangle A225470; Sheffer ((1 - 3*x)^(-2/3), (-1/3)*log(1 - 3*x)).

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A383261 Expansion of e.g.f. f(x) * exp(2 * f(x)), where f(x) = (exp(3*x) - 1)/3.

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A286722 Column k=2 of the triangle A225470; Sheffer ((1 - 3x)^(-2/3), (-1/3)log(1 - 3*x)).