A000392 -id:A000392 - cp's OEIS frontend

A049435 Stirling numbers of second kind: 10th column of Stirling2 triangle A008277.

1, 55, 1705, 39325, 752752, 12662650, 193754990, 2758334150, 37112163803, 477297033785, 5917584964655, 71187132291275, 835143799377954, 9593401297313460, 108254081784931500, 1203163392175387500, 13199555372846848005, 143197070509423605675

Offset: 10

Wolfdieter Lang

See A000771.

Index entries for linear recurrences with constant coefficients, signature (55,-1320,18150,-157773,902055,-3416930,8409500,-12753576,10628640,-3628800).

Cf. A000225, A000392, A000453, A000481, A000770, A000771, A049434, A049447. a(n) = A008277(n, 10).

lst={};Do[f=StirlingS2[n, 10];AppendTo[lst, f], {n, 10, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Sep 27 2008 *)
CoefficientList[Series[1/((1 - x) (1 - 2 x) (1 - 3 x) (1 - 4 x) (1 - 5 x) (1 - 6 x) (1 - 7 x) (1 - 8 x) (1 - 9 x) (1 - 10 x)), {x, 0, 25}], x] (* Vladimir Joseph Stephan Orlovsky, Jun 20 2011 *)
StirlingS2[Range[10,35],10] (* Harvey P. Dale, Nov 07 2020 *)

G.f.: x^10/Product_{k=1..10} (1-k*x).
E.g.f.: ((exp(x)-1)^10)/10!.
a(n) = det(|s(i+10,j+9)|, 1 <= i,j <= n-10), where s(n,k) are Stirling numbers of the first kind. - Mircea Merca, Apr 06 2013

A049447 Stirling numbers of second kind: 9th column of Stirling2 triangle A008277.

Original entry on oeis.org

1, 45, 1155, 22275, 359502, 5135130, 67128490, 820784250, 9528822303, 106175395755, 1144614626805, 12011282644725, 123272476465204, 1241963303533920, 12320068811796900, 120622574326072500, 1167921451092973005, 11201516780955125625, 106563273280541795575

Offset: 9

Wolfdieter Lang

See A000771.

Cf. A000225, A000392, A000453, A000481, A000770, A000771, A049434.

lst={};Do[f=StirlingS2[n, 9];AppendTo[lst, f], {n, 9, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Sep 27 2008 *)
CoefficientList[Series[1/((1 - x) (1 - 2 x) (1 - 3 x) (1 - 4 x) (1 - 5 x) (1 - 6 x) (1 - 7 x) (1 - 8 x) (1 - 9 x)), {x, 0, 25}], x] (* Vladimir Joseph Stephan Orlovsky, Jun 20 2011 *)
StirlingS2[Range[9,30],9] (* Harvey P. Dale, Dec 12 2022 *)

a(n)= A008277(n, 9).
G.f.: x^9/product_{k=1..9} (1-k*x).
E.g.f.: ((exp(x)-1)^9)/9!.
a(n) = det(|s(i+9,j+8)|, 1 <= i,j <= n-9), where s(n,k) are Stirling numbers of the first kind. - Mircea Merca, Apr 06 2013

A140945 Triangle read by rows: counts series-parallel networks by the number of series connections.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 25, 25, 1, 1, 90, 290, 90, 1, 1, 301, 2450, 2450, 301, 1, 1, 966, 17451, 41580, 17451, 966, 1, 1, 3025, 112035, 544971, 544971, 112035, 3025, 1, 1, 9330, 671980, 6076350, 12122502, 6076350, 671980, 9330, 1, 1, 28501, 3846700, 60738700, 217523922, 217523922, 60738700, 3846700, 28501, 1

Offset: 1

Brian Drake, Jul 24 2008

T(n,k) is the number of series-parallel matroids on [n+1] of rank k. - Andrew Howroyd, Mar 08 2023

			Triangle begins:
  1;
  1,   1;
  1,   6,     1;
  1,  25,    25,     1;
  1,  90,   290,    90,     1;
  1, 301,  2450,  2450,   301,   1;
  1, 966, 17451, 41580, 17451, 966, 1;
  ...

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50; first 17 rows from Brian Drake, Jul 24 2008)
Brian Drake, An inversion theorem for labeled trees and some limits of areas under lattice paths (Example 1.5.1), A dissertation presented to the Faculty of the Graduate School of Arts and Sciences of Brandeis University.
Luis Ferroni and Matt Larson, Kazhdan-Lusztig polynomials of braid matroids, arXiv:2303.02253 [math.CO], 2023.
Nicholas Proudfoot, Yuan Xu, and Ben Young, On the enumeration of series-parallel matroids, arXiv:2406.04502 [math.CO], 2024.

Row sums are A006351.
Second column is A000392.
Cf. A359985.

N:=6: 1/a*log(1+a*y)+1*log(1+b*y)/b-y=x: solve(%, y):series(%, x, N): simplify(%, symbolic): convert(%, polynom): subs(b=1, %): R:= [seq(i!*coeff(%, x, i), i=1..N-1)]: seq( seq(coeff(R[i], a, j), j=0..i-1), i=1..N-1);

T(n) = {[Vecrev(p) | p<-Vec(serlaplace(intformal(serreverse(log(1 + x*y + O(x*x^n))/y + log(1 + x + O(x*x^n)) - x))))]}
{ my(A=T(10)); for(i=1, #A, print(A[i])) }  \\ Andrew Howroyd, Mar 08 2023

E.g.f. is reversion of log(1+ax)/a+log(1+bx)/b-x.

Let f(x,t) = (1+x)*(1+x*t)/(1-x^2*t) and let D be the operator f(x,t)*d/dx. Then the n-th row polynomial equals (D^n)(f(x,t)) evaluated at x = 0. - Peter Bala, Sep 29 2011

A249992 Expansion of 1/((1+x)(1+2x)(1-3x)).

Original entry on oeis.org

1, 0, 7, 6, 49, 84, 379, 882, 3157, 8448, 27391, 78078, 242425, 710892, 2165443, 6430794, 19423453, 58008216, 174548935, 522598230, 1569891841, 4705481220, 14124832267, 42357719586, 127106713189, 381253030704, 1143893309839, 3431411494062, 10294771353097

Offset: 0

Alex Ratushnyak, Dec 27 2014

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,7,6).

Cf. A249993.
Cf. A000392: expansion of x^3/((1-x)*(1-2*x)*(1-3*x)).
Cf. A091002: expansion of x^2/((1-x)*(1+2*x)*(1-3*x)).
Cf. A094705: expansion of   x/((1+x)*(1-2*x)*(1-3*x)).

[(3^(n+2) + (-1)^n*(2^(n+4) - 5))/20: n in [0..50]]; // G. C. Greubel, Jul 21 2022

seq((9/20)*3^n+(4/5)*(-2)^n-(1/4)*(-1)^n, n=0 .. 100); # Robert Israel, Dec 28 2014

LinearRecurrence[{0, 7, 6}, {1, 0, 7}, 29] (* Jean-François Alcover, Oct 05 2017 *)
CoefficientList[Series[1/((1+x)(1+2x)(1-3x)),{x,0,30}],x] (* Harvey P. Dale, May 26 2020 *)

Vec(1/((1+x)*(1+2*x)*(1-3*x)) + O(x^50)) \\ Michel Marcus, Dec 28 2014

[(3^(n+2) +(-1)^n*(2^(n+4) -5))/20 for n in (0..50)] # G. C. Greubel, Jul 21 2022

G.f.: 1/((1+x)*(1+2*x)*(1-3*x)).
a(n) = ( 3^(n+2) + (2^(n+4) - 5)*(-1)^n )/20. - Colin Barker, Dec 28 2014
a(n) = 7*a(n-2) + 6*a(n-3). - Colin Barker, Dec 28 2014
E.g.f.: (9/20)*exp(3*x) + (4/5)*exp(-2*x) - (1/4)*exp(-x). - Robert Israel, Dec 28 2014

A006230 Bitriangular permutations.

Original entry on oeis.org

1, 13, 73, 301, 1081, 3613, 11593, 36301, 111961, 342013, 1038313, 3139501, 9467641, 28501213, 85700233, 257493901, 773268121, 2321377213, 6967277353, 20908123501, 62736953401, 188236026013, 564758409673, 1694375892301, 5083329003481, 15250389663613

Offset: 4

N. J. A. Sloane

Prepending the term 0 and setting the offset to 0 makes this sequence row 3 of A371761. In this form it can be generated by the Akiyama-Tanigawa algorithm for powers (see the Python script). - Peter Luschny, Apr 12 2024

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Colin Barker, Table of n, a(n) for n = 4..1000
Irving Kaplansky and John Riordan, The problem of the rooks and its applications, Duke Mathematical Journal 13.2 (1946): 259-268.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
J. Riordan, Letter to N. J. A. Sloane, Dec. 1976
Index entries for linear recurrences with constant coefficients, signature (6,-11,6).

Cf. A136301 (row 4), A371761 (row 3).

A006230:=-(z+1)*(6*z+1)/(z-1)/(3*z-1)/(2*z-1); # Conjectured by Simon Plouffe in his 1992 dissertation.

12*StirlingS2[n+1, 3]+1; (* Brian Parsonnet, Feb 25 2011 *)
Sum[ StirlingS2[n,i] * StirlingS2[ 3,i ] * i!^2, {i,3} ]; (* alternative, Brian Parsonnet, Feb 25 2011 *)

Vec(x^4*(1 + x)*(1 + 6*x) / ((1 - x)*(1 - 2*x)*(1 - 3*x)) + O(x^40))
\\ Colin Barker, Dec 27 2017

# Using the Akiyama-Tanigawa algorithm for powers from A371761.
print(ATPowList(3, 27))  # Peter Luschny, Apr 12 2024

a(n) = 12*S(n-2) + 1, with S(n)=A000392(n) the Stirling numbers of second kind, 3rd column. - Ralf Stephan, Jul 07 2003
a(n+3) = Sum_{i=1..3} A008277(n,i) * A008277(3,i) * i!^2. - Brian Parsonnet, Feb 25 2011
From Colin Barker, Dec 27 2017: (Start)
G.f.: x^4*(1 + x)*(1 + 6*x) / ((1 - x)*(1 - 2*x)*(1 - 3*x)).
a(n) = 12*(3 - 3*2^(n-2) + 3^(n-2))/6 + 1.
a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) for n>6. (End)

A056509 Number of periodic palindromic structures of length n using exactly three different symbols.

Original entry on oeis.org

0, 0, 0, 1, 1, 5, 6, 19, 25, 64, 90, 208, 301, 656, 966, 2035, 3025, 6250, 9330, 19035, 28501, 57740, 86526, 174436, 261625, 525994, 788970, 1583119, 2375101, 4760516, 7141686, 14303011, 21457825, 42954850, 64439010, 128953341, 193448101, 387046700, 580606446, 1161504423

Offset: 1

Marks R. Nester

nonn

			For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome. Permuting the symbols will not change the structure.

M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Andrew Howroyd, Table of n, a(n) for n = 1..200

Column 3 of A285012.

Cf. A000392, A056503, A056504, A056519.

a(n) = A056504(n) - A056503(n).

Inverse Moebius transform of A056519. - Andrew Howroyd, Oct 01 2019

a(17)-a(35) from Andrew Howroyd, Apr 07 2017

Terms a(36) and beyond from Andrew Howroyd, Oct 01 2019

A306550 Array read by antidiagonals where A(n,k) is the number of labeled k-antichains covering n vertices.

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 6, 0, 0, 0, 0, 1, 25, 2, 0, 0, 0, 0, 1, 90, 56, 0, 0, 0, 0, 0, 1, 301, 790, 25, 0, 0, 0, 0, 0, 1, 966, 8380, 1895, 6, 0, 0, 0, 0, 0, 1, 3025, 76482, 70370, 2116, 1, 0, 0, 0, 0

Offset: 0

Gus Wiseman, Feb 23 2019

			Array begins:
    n=0: n=1: n=2: n=3: n=4: n=5:
---------------------------------
k=0:  1    0    0    0    0    0
k=1:  1    1    1    1    1    1
k=2:  0    0    1    6   25   90
k=3:  0    0    0    2   56  790
k=4:  0    0    0    0   25 1895
k=5:  0    0    0    0    6 2116
Column n = 3 counts the following antichains:
  {{123}}  {{1}{23}}   {{1}{2}{3}}
           {{2}{13}}   {{12}{13}{23}}
           {{3}{12}}
           {{12}{13}}
           {{12}{23}}
           {{13}{23}}

Column sums are A006126. Row k = 2 is A000392. Rows k = 3-9 are A056046-A056049, A056052, A056101, A056104.

Cf. A000372, A006126, A006602, A014466, A261005, A305000, A305844, A317674.

nn=8;
stableSets[u_,Q_,k_]:=If[k==0,{{}},If[Length[u]==0,{},With[{w=First[u]},Join[stableSets[DeleteCases[u,w],Q,k],Prepend[#,w]&/@stableSets[DeleteCases[u,r_/;r==w||Q[r,w]||Q[w,r]],Q,k-1]]]]];
ae[n_,k_]:=Length[Select[stableSets[Subsets[Range[n]],SubsetQ,k],Union@@#==Range[n]&]];
Table[ae[k,n-k],{n,0,nn},{k,0,n}]

A321163 Square array A(n,k), n >= 1, k >= 1, read by antidiagonals, where A(n,k) is the sum of distinct products Product_{j=1..k} b_j with 1 <= b_j<= n.

Original entry on oeis.org

1, 1, 3, 1, 7, 6, 1, 15, 25, 10, 1, 31, 90, 61, 15, 1, 63, 301, 310, 136, 21, 1, 127, 966, 1441, 990, 244, 28, 1, 255, 3025, 6370, 6391, 2220, 440, 36, 1, 511, 9330, 27301, 38325, 17731, 5300, 680, 45, 1, 1023, 28501, 114670, 218926, 130851, 54831, 9660, 1022, 55

Offset: 1

Seiichi Manyama, Jan 10 2019

			In case of (n,k) = (3,2):
  | 1  2  3
--+--------
1 | 1, 2, 3
2 | 2, 4, 6
3 | 3, 6, 9
Distinct products are 1,2,3,4,6,9. So A(3,2) = 1+2+3+4+6+9 = 25.
Square array begins:
    1,    1,     1,      1,       1,        1, ...
    3,    7,    15,     31,      63,      127, ...
    6,   25,    90,    301,     966,     3025, ...
   10,   61,   310,   1441,    6370,    27301, ...
   15,  136,   990,   6391,   38325,   218926, ...
   21,  244,  2220,  17731,  130851,   916714, ...
   28,  440,  5300,  54831,  514668,  4519390, ...
   36,  680,  9660, 116991, 1280916, 13092430, ...
   45, 1022, 17130, 242091, 3070935, 36184072, ...
   55, 1472, 28670, 467391, 6807045, 91765822, ...

Seiichi Manyama, Antidiagonals n = 1..25, flattened

Columns 1-3 give A000217, A321165, A323334.
Rows 1-2 give A000012, A000225(n+1).
Main diagonal gives A321164.
Cf. A322967.
Row 3 gives A000392(n+4). - Fred Daniel Kline, Jan 11 2019

A[n_, k_] := Module[{b, bb}, bb = Array[b, k]; Table[Times @@ bb, Evaluate[Sequence @@ ({#, n}& /@ bb)]] // Flatten // Union // Total];
Table[A[n-k+1, k], {n, 1, 10}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Nov 25 2020 *)

A341091 Triangle read by rows: Coefficients for calculation of the sum of all the finite differences from order zero to order k. Sum_{n=0..k} T(n, k)*b(n) = b(0) + b(1) + ... + b(k) + (b(1) - b(0)) + ... + (b(k) - b(k-1)) + ((b(2) - b(1)) - (b(1) - b(0))) + ... .

Original entry on oeis.org

1, 0, 2, 1, -1, 3, 0, 3, -3, 4, 1, -2, 7, -6, 5, 0, 4, -8, 14, -10, 6, 1, -3, 13, -21, 25, -15, 7, 0, 5, -15, 35, -45, 41, -21, 8, 1, -4, 21, -49, 81, -85, 63, -28, 9, 0, 6, -24, 71, -129, 167, -147, 92, -36, 10, 1, -5, 31, -94, 201, -295, 315, -238, 129, -45, 11

Offset: 0

Thomas Scheuerle, Feb 13 2022

If we want to calculate the sum of finite differences for a sequence b(n):
  b(0)*T(0, n) + ... + b(n)*T(n, n) = b(0) + b(1) + ... + b(n) + (b(1) - b(0)) + ... + (b(n) - b(n-1)) + ((b(2) - b(1)) - (b(1) - b(0))) + ... This sum includes the sequence b(n) itself. This defines an invertible linear sequence transformation with a deep connection to Bernoulli numbers and other interesting sequences of rational numbers.
From Thomas Scheuerle, Apr 29 2024: (Start)
These are the coefficients of the polynomials defined by the recurrence: P(k, x) = P(k - 1, x) + (x^2 - x)*P(k - 2, x) + 1, with P(-1, x) = 0 and P(0, x) = 1. This can also be expressed as P(k, x) = Sum_{m=1..k+1} binomial(k+2 - m, m)*(x^2 - x)^(m - 1) = Sum_{n=0..k} T(n, k)*x^(k-n). If we would evaluate P(k, t) as sequence for some fixed t then we get the expansion of 1/((1 - x)*(1+(t-1)*x)*(1 - t*x)).
We may replace (x^2 - x) by (x^(-2) - x^(-1)) to get the coefficients in reverse order: x^k*Sum_{m=1..k+1} binomial(k+2 - m, m)*(x^(-2) - x^(-1))^(m - 1) = Sum_{n=0..k} T(n, k)*x^n = F(k, x). If we would evaluate F(k, t) as sequence for some fixed t then we get the expansion of 1/((1 - x)*(1 - (t-1)*x)*(1 - t*x)). (End)

			Triangle begins with T(n, k):
   n=   0,  1,   2,   3,   4,   5,   6,   7,   8
  k=0   1
  k=1   0,  2
  k=2   1, -1,   3
  k=3   0,  3,  -3,   4
  k=4   1, -2,   7,  -6,   5
  k=5   0,  4,  -8,  14, -10,   6
  k=6   1, -3,  13, -21,  25, -15,   7
  k=7   0,  5, -15,  35, -45,  41, -21,   8
  k=8   1, -4,  21, -49,  81, -85,  63, -28,   9
  ...

Cf. A000027, A000032, A000045, A000110, A000217.
Cf. A001045, A004006, A032346, A051744, A052875.
Cf. A084416, A130595, A145766.
Cf. A027642, A164555 (Numerators and denominators of Bernoulli numbers).
Cf. A001008, A002805 (Numerators and denominators of harmonic numbers).
Cf. A344037, A372245.
Sequences below will be obtained by evaluation of the associated polynomials:
Cf. A000295, A000392, A000975, A016208.
Cf. A016218, A016228, A016241, A016249.
Cf. A016256, A016261, A249997.

A341091(n, k) = sum(m=n, k,(-1)^(m+n)*binomial(m+1, n))

A341091(n, k) = (1/2)*(-1)^n*(2*(-1)^k*binomial(2+k, n)*hypergeom([1,k+3],k+3-n,-1)+(-1/2)^n*(2^(n+1)-1)) \\ Thomas Scheuerle, Apr 29 2024

b(0)*T(0, m) + b(1)*T(1, m) + ... + b(m)*T(m, m)
  = Sum_{j=0..m} Sum_{n=0..m-j} Sum_{k=0..n} (-1)^k*binomial(n, k)*b(j+n-k)
  = Sum_{n=0..m} b(n)*Sum_{j=n..m}(-1)^(j+n)*binomial(j+1, n).
T(n, k) = Sum_{m=n..k}(-1)^(m+n)*binomial(m+1, n).
T(n, k) = (1/2)*(-1)^n*(2*(-1)^k*binomial(2+k, n)*Hypergeometric2F1(1, k+3, k+3-n, -1)+(-1/2)^n*(2^(n+1) - 1)), where Hypergeometric2F1 is the Gaussian hypergeometric function 2F1 as defined in Mathematica. - Thomas Scheuerle, Apr 29 2024
T(k, k) = A000027(k+1) The positive integers.
|T(k-1, k)| = A000217(k) The triangular numbers.
T(k-2, k) = A004006(k).
|T(k-3, k)| = A051744(k).
T(0, k*2) = 1.
T(0, k*2 + 1) = 0.
T(1, k*2 + 1) = k + 2.
T(1, k*2 + 2) = -(k + 1).
T(n, k) with constant n and variable k, a linear recurrence relation with characteristic polynomial (x-1)*(x+1)^(n+1).
Sum_{n=0..k} T(n, k)*B_n = 1. B_n is the n-th Bernoulli number with B_1 = 1/2. B_n = A164555(n)/A027642(n).
Sum_{n=0..k} T(n, k)*(1 - B_n) = k.
Sum_{n=0..k} T(n, k)*(2*n - 3+3*B_n) = k^2.
Sum_{n=0..k} T(n, k)*A032346(n) = A032346(k+1).
From Thomas Scheuerle, Apr 29 2024: (Start)
Sum_{n=0..k} T(n, k)*A000110(n+1) = A000110(k+2) - 1.
Sum_{n=0..k} T(n, k)*(1/(1+n)) = H(1+floor(k/2)), where H(k) is the harmonic number A001008(k)/A002805(k). (End)
Sum_{n=0..k} T(n, k)*c(n) = c(k). C(k) = {-1, 0, 1/2, 1/2, 1/8, -7/20, ...} this sequence of rational numbers can be defined recursively: c(0) = -1, c(m) = (-c(m-1) + Sum_{k=0..m-1} A130595(m+1, k)*c(k))/m.
  c(m) is an eigensequence of this transformation, all eigensequences are c(m) multiplied by any factor.
Sum_{n=0..k} T(n, k)*A000045(n) = 2*(A000045(2*floor((k+1)/2) - 1) - 1). A000045 are the Fibonacci numbers.
Sum_{n=0..k} T(n, k)*A000032(n) = A000032(2*floor(k/2)+2) - 2. A000032 are the Lucas numbers.
Sum_{n=0..k} T(n, k)*A001045(n) = A145766(floor((k+1)/2)). A001045 is the Jacobsthal sequence.
This sequence acting as an operator onto a monomial n^w:
Sum_{n=0..k} T(n, k)*n^w = (1/(w+1))*k^(w+1) + Sum_{v=1..w} ((v+B_v)*(w)_v/v!)*k^(w+1-v) - A052875(w) + O_k(w) (w)_v is the falling factorial. If k > w-1 then O_k(w) = 0. If k <= w-1 then O_k(w) is A084416(w, 2+k), the sequence with the exponential generating function: (e^x-1)^(2+k)/(2-e^x).
From Thomas Scheuerle, Apr 29 2024: (Start)
This sequence acting by its inverse operator onto a monomial k^w:
Sum_{n=0..k} T(n, k)*( Sum_{m=0..k} ((-1)^(1+m+k)*binomial(k, m)*(2^(k-m) - 1)*n^m + A344037(m)*B_n) ) = k^w - A372245(w, k+3), note that A372245(w, k+3) = 0 if k+3 > w. B_n is the n-th Bernoulli number with B_1 = 1/2.
How this sequence will act as an operator onto a Dirichlet series may be developed by the formulas below:
Sum_{n=0..k} T(n, k)*2^n = A000295(k+2).
Sum_{n=0..k} T(n, k)*3^n = A000392(k+3).
Sum_{n=0..k} T(n, k)*4^n = A016208(k).
Sum_{n=0..k} T(n, k)*5^n = A016218(k).
Sum_{n=0..k} T(n, k)*6^n = A016228(k).
Sum_{n=0..k} T(n, k)*7^n = A016241(k).
Sum_{n=0..k} T(n, k)*8^n = A016249(k).
Sum_{n=0..k} T(n, k)*9^n = A016256(k).
Sum_{n=0..k} T(n, k)*10^n = A016261(k).
Sum_{n=0..k} T(n, k)*m^n = m^2*m^k/(m-1) - (m-1)^2*(m-1)^k/(m-2) + 1/((m-1)*(m-2)), for m > 2.
Sum_{n=0..k} T(n, k)*( m*B_n + (m-1)*Sum_{t=1..m} t^n )*(1/m^2) = m^k, for m > 0. B_n is the n-th Bernoulli number with B_1 = 1/2.
Sum_{n=0..k} T(n, k) zeta(-n) = Sum_{j=0..k} (-1)^(1+j)/(2+j) = (-1)^(k+1)*LerchPhi(-1, 1, k+3) - 1 + log(2).
Sum_{n=0..k} T(k - n, k)*2^n = A000975(k+1)
Sum_{n=0..k} T(k - n, k)*3^n = A091002(k+2)
Sum_{n=0..k} T(k - n, k)*4^n = A249997(k). (End)

A346975 Expansion of e.g.f. log( 1 + (exp(x) - 1)^3 / 3! ).

Original entry on oeis.org

1, 6, 25, 80, 91, -1694, -22875, -193740, -1119569, -768394, 101162425, 1930987240, 23583202371, 181575384906, -306743537075, -45405986594980, -1070132302146089, -16439720013909794, -145808623945689375, 1048196472097011600, 84226169502099763051

Offset: 3

Ilya Gutkovskiy, Aug 09 2021

sign

Cf. A000392, A327504, A346974, A346976, A346977.

Cf. A346390, A346945.

nmax = 23; CoefficientList[Series[Log[1 + (Exp[x] - 1)^3/3!], {x, 0, nmax}], x] Range[0, nmax]! // Drop[#, 3] &
a[n_] := a[n] = StirlingS2[n, 3] - (1/n) Sum[Binomial[n, k] StirlingS2[n - k, 3] k a[k], {k, 1, n - 1}]; Table[a[n], {n, 3, 23}]

a(n) = Stirling2(n,3) - (1/n) * Sum_{k=1..n-1} binomial(n,k) * Stirling2(n-k,3) * k * a(k).
a(n) ~ -(n-1)! * 2^(n+1) * cos(n*arctan(2*arctan(3^(5/6)/(2^(2/3) + 3^(1/3))) / log(1 + 6^(1/3) + 6^(2/3)))) / (4*arctan(3^(5/6)/(2^(2/3) + 3^(1/3)))^2 + log(1 + 6^(1/3) + 6^(2/3))^2)^(n/2). - Vaclav Kotesovec, Aug 09 2021
a(n) = Sum_{k=1..floor(n/3)} (-1)^(k-1) * (3*k)! * Stirling2(n,3*k)/(k * 6^k). - Seiichi Manyama, Jan 23 2025

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