A001711 -id:A001711 - cp's OEIS frontend

A051564 Second unsigned column of triangle A051523.

0, 1, 21, 362, 6026, 101524, 1763100, 31813200, 598482000, 11752855200, 240947474400, 5154170774400, 114942011990400, 2669517204076800, 64496340380102400, 1619153396908185600, 42188624389562112000

Offset: 0

Wolfdieter Lang

The asymptotic expansion of the higher order exponential integral E(x,m=2,n=10) ~ exp(-x)/x^2*(1 - 21/x + 362/x^2 - 6026/x^3 + 101524/x^4 - 1763100/x^5 + 31813200/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

Mitrinovic, D. S. and Mitrinovic, R. S. see reference given for triangle A051523.

G. C. Greubel, Table of n, a(n) for n = 0..440

Cf. A049398 (first unsigned column).

Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k=2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564. - Gary Detlefs Jan 04 2011

f[n_] := n!*Sum[(-1)^k*Binomial[-10, k]/(n - k), {k, 0, n - 1}]; Array[f, 17, 0]
Range[0, 16]! CoefficientList[ Series[-Log[(1 - x)]/(1 - x)^10, {x, 0, 16}], x]
(* Or, using elementary symmetric functions: *)
f[k_] := k + 9; t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[n - 1, t[n]]
Table[a[n], {n, 1, 16}]
(* Clark Kimberling, Dec 29 2011 *)

a(n) = A051523(n, 2)*(-1)^(n-1).
E.g.f.: -log(1-x)/(1-x)^10.
a(n) = n!*Sum_{k=0..n-1}((-1)^k*binomial(-10,k)/(n-k)), for n>=1. - Milan Janjic, Dec 14 2008
a(n) = n!*[9]h(n), where [k]h(n) denotes the k-th successive summation of the harmonic numbers from 0 to n. - Gary Detlefs, Jan 04 2011

A052752 a(n) = (3*n+1)^(n-1).

Original entry on oeis.org

1, 1, 7, 100, 2197, 65536, 2476099, 113379904, 6103515625, 377801998336, 26439622160671, 2064377754059776, 177917621779460413, 16777216000000000000, 1718264124282290785243, 189937030341242876870656, 22539340290692258087863249, 2857942574656970690381479936

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

G. C. Greubel, Table of n, a(n) for n = 0..333
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 708
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.

Cf. A000169, A052750, A052774, A052782, A000272, A001711.

[(3*n+1)^(n-1): n in [0..50]]; // G. C. Greubel, Nov 16 2017

spec := [S,{B=Prod(S,S,S,Z),S=Set(B)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);

Table[(3n+1)^(n-1),{n,0,20}] (* Harvey P. Dale, Aug 14 2015 *)
With[{nmax = 50}, CoefficientList[Series[Exp[-LambertW[-3*x]/3], {x, 0, nmax}], x]*Range[0, nmax]!] (* G. C. Greubel, Nov 16 2017 *)

for(n=0,50, print1((3*n+1)^(n-1), ", ")) \\ G. C. Greubel, Nov 16 2017

x='x+O('x^50); Vec(serlaplace(exp(-lambertw(-3*x)/3))) \\ G. C. Greubel, Nov 16 2017

E.g.f.: exp(-(1/3)*LambertW(-3*x)).
From Peter Bala, Dec 19 2013: (Start)
The e.g.f. A(x) = 1 + x + 7*x^2/2! + 100*x^3/3! + 2197*x^4/4! + ... satisfies:
1) A(x*exp(-3*x)) = exp(x) = 1/A(-x*exp(3*x));
2) A^3(x) = 1/x*series reversion(x*exp(-3*x));
3) A(x^3) = 1/x*series reversion(x*exp(-x^3));
4) A(x) = exp(x*A(x)^3);
5) A(x) = 1/A(-x*A(x)^6). (End)
E.g.f.: (-LambertW(-3*x)/(3*x))^(1/3). - Vaclav Kotesovec, Dec 07 2014
Related to A001711 by Sum_{n >= 1} a(n)*x^n/n! = series reversion( 1/(1 + x)^3*log(1 + x) ) = series reversion(x - 7*x^2/2! + 47*x^3/3! - 342*x^4/4! + ...). Cf. A000272, A052750. - Peter Bala, Jun 15 2016
a(n) = Sum_{k=1..n} (-1)^(n-k)*(2n+k)^(n-1)*binomial(n,k-1), a(0)=1. - Vladimir Kruchinin, Aug 14 2025

Better description from Vladeta Jovovic, Sep 02 2003

A165674 Triangle generated by the asymptotic expansions of the E(x,m=2,n).

Original entry on oeis.org

1, 3, 1, 11, 5, 1, 50, 26, 7, 1, 274, 154, 47, 9, 1, 1764, 1044, 342, 74, 11, 1, 13068, 8028, 2754, 638, 107, 13, 1, 109584, 69264, 24552, 5944, 1066, 146, 15, 1, 1026576, 663696, 241128, 60216, 11274, 1650, 191, 17, 1

Offset: 1

Johannes W. Meijer, Oct 05 2009

The higher order exponential integrals E(x,m,n) are defined in A163931. The asymptotic expansion of the E(x,m=2,n) ~ (exp(-x)/x^2)*(1 - (1+2*n)/x + (2+6*n+3*n^2)/x^2 - (6+22*n+18*n^2+ 4*n^3)/x^3 + ... ) is discussed in A028421. The formula for the asymptotic expansion leads for n = 1, 2, 3, .., to the left hand columns of the triangle given above.
The recurrence relations of the right hand columns of this triangle lead to Pascal's triangle A007318, their a(n) formulas lead to Wiggen's triangle A028421 and their o.g.f.s lead to Wood's polynomials A126671; cf. A080663, A165676, A165677, A165678 and A165679.
The row sums of this triangle lead to A093344. Surprisingly the e.g.f. of the row sums Egf(x) = (exp(1)*Ei(1,1-x) - exp(1)*Ei(1,1))/(1-x) leads to the exponential integrals in view of the fact that E(x,m=1,n=1) = Ei(n=1,x). We point out that exp(1)*Ei(1,1) = A073003.
The Maple programs generate the coefficients of the triangle given above. The first one makes use of a relation between the triangle coefficients, see the formulas, and the second one makes use of the asymptotic expansions of the E(x,m=2,n).
Amarnath Murthy discovered triangle A093905 which is the reversal of our triangle.
A165675 is an extended version of this triangle. Its reversal is A105954.
Triangle A094587 is generated by the asymptotic expansions of E(x,m=1,n).

A093905 is the reversal of this triangle.
A000254, A001705, A001711, A001716, A001721, A051524, A051545, A051560, A051562, A051564 are the first ten left hand columns.
A080663, n>=2, is the third right hand column.
A165676, A165677, A165678 and A165679 are the next right hand columns, A093344 gives the row sums.
A073003 is Gompertz's constant.
A094587 is generated by the asymptotic expansions of E(x, m=1, n).
Cf. A165675, A105954 (Quet) and A067176 (Bottomley).
Cf. A007318 (Pascal), A028421 (Wiggen), A126671 (Wood).

nmax:=9; for n from 1 to nmax do a(n, n) := 1 od: for n from 2 to nmax do a(n, 1) := n*a(n-1, 1) + (n-1)! od: for n from 3 to nmax do for m from 2 to n-1 do a(n, m) := (n-m+1)*a(n-1, m) + a(n-1, m-1) od: od: seq(seq(a(n, m), m = 1..n), n = 1..nmax);
# End program 1
nmax := nmax+1: m:=2; with(combinat): EA := proc(x, m, n) local E, i; E:=0: for i from m-1 to nmax+2 do E := E + sum((-1)^(m+k1+1) * binomial(k1, m-1) * n^(k1-m+1) * stirling1(i, k1), k1=m-1..i) / x^(i-m+1) od: E:= exp(-x)/x^(m) * E: return(E); end: for n1 from 1 to nmax do f(n1-1) := simplify(exp(x) * x^(nmax+3) * EA(x, m, n1)); for m1 from 0 to nmax+2 do b(n1-1, m1) := coeff(f(n1-1), x, nmax+2-m1) od: od: for n1 from 0 to nmax-1 do for m1 from 0 to n1-m+1 do a(n1-m+2, m1+1) := abs(b(m1, n1-m1)) od: od: seq(seq(a(n, m), m = 1..n),n = 1..nmax-1);
# End program 2
# Maple programs revised by Johannes W. Meijer, Sep 22 2012

a(n,m) = (n-m+1)*a(n-1,m) + a(n-1,m-1), for 2 <= m <= n-1, with a(n,n) = 1 and a(n,1) = n*a(n-1,1) + (n-1)!.

a(n,m) = product(i, i= m..n)*sum(1/i, i = m..n).

A165675 Triangle read by rows. T(n, k) = (n - k + 1)! * H(k, n - k), where H are the hyperharmonic numbers. For 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 11, 5, 1, 24, 50, 26, 7, 1, 120, 274, 154, 47, 9, 1, 720, 1764, 1044, 342, 74, 11, 1, 5040, 13068, 8028, 2754, 638, 107, 13, 1, 40320, 109584, 69264, 24552, 5944, 1066, 146, 15, 1, 362880, 1026576, 663696, 241128, 60216, 11274, 1650, 191, 17, 1

Offset: 0

Johannes W. Meijer, Oct 05 2009

Previous name: Extended triangle related to the asymptotic expansions of the E(x, m = 2, n).
For the definition of the hyperharmonic numbers see the formula section.
This triangle is the same as triangle A165674 except for the extra left-hand column T(n, 0) = n!. The T(n) formulas for the right-hand columns generate the coefficients of this extra left-hand column.
Leroy Quet discovered triangle A105954 which is the reversal of our triangle.
In square format, row k gives the (n-1)-st elementary symmetric function of {k, k+1, k+2,..., k+n}, as in the Mathematica section. - Clark Kimberling, Dec 29 2011

			Triangle T(n, k) begins:
  [0]    1;
  [1]    1,     1;
  [2]    2,     3,    1;
  [3]    6,    11,    5,    1;
  [4]   24,    50,   26,    7,   1;
  [5]  120,   274,  154,   47,   9,   1;
  [6]  720,  1764, 1044,  342,  74,  11,  1;
  [7] 5040, 13068, 8028, 2754, 638, 107, 13, 1;
Seen as an array (the triangle arises when read by descending antidiagonals):
  [0] 1,  1,   2,    6,    24,    120,     720,     5040, ...
  [1] 1,  3,  11,   50,   274,   1764,   13068,   109584, ...
  [2] 1,  5,  26,  154,  1044,   8028,   69264,   663696, ...
  [3] 1,  7,  47,  342,  2754,  24552,  241128,  2592720, ...
  [4] 1,  9,  74,  638,  5944,  60216,  662640,  7893840, ...
  [5] 1, 11, 107, 1066, 11274, 127860, 1557660, 20355120, ...
  [6] 1, 13, 146, 1650, 19524, 245004, 3272688, 46536624, ...
  [7] 1, 15, 191, 2414, 31594, 434568, 6314664, 97053936, ...

A105954 is the reversal of this triangle.
A165674, A138771 and A165680 are related triangles.
A080663 equals the third right hand column.
A000142 equals the first left hand column.
A093345 are the row sums.
Columns include A165676, A165677, A165678 and A165679.

nmax := 8; for n from 0 to nmax do a(n, 0) := n! od: for n from 0 to nmax do a(n, n) := 1 od: for n from 2 to nmax do for m from 1 to n-1 do a(n, m) := (n-m+1)*a(n-1, m) + a(n-1, m-1) od: od: seq(seq(a(n, m), m=0..n), n=0..nmax);
# Johannes W. Meijer, revised Nov 27 2012
# Shows the array format, using hyperharmonic numbers.
H := proc(n, k) option remember; if n = 0 then 1/(k+1)
else add(H(n - 1, j), j = 0..k) fi end:
seq(lprint(seq((k + 1)!*H(n, k), k = 0..7)), n = 0..7);
# Shows the array format, using the hypergeometric formula.
A := (n, k) -> (k+1)*((n + k)! / n!)*hypergeom([-k, 1, 1], [2, n + 1], 1):
seq(lprint(seq(simplify(A(n, k)), k = 0..7)), n = 0..7);
# Peter Luschny, Jul 03 2022

a[n_] := SymmetricPolynomial[n - 1, t[n]]; z = 10;
t[n_] := Table[k - 1, {k, 1, n}]; t1 = Table[a[n], {n, 1, z}]  (* A000142 *)
t[n_] := Table[k,     {k, 1, n}]; t2 = Table[a[n], {n, 1, z}]  (* A000254 *)
t[n_] := Table[k + 1, {k, 1, n}]; t3 = Table[a[n], {n, 1, z}]  (* A001705 *)
t[n_] := Table[k + 2, {k, 1, n}]; t4 = Table[a[n], {n, 1, z}]  (* A001711 *)
t[n_] := Table[k + 3, {k, 1, n}]; t5 = Table[a[n], {n, 1, z}]  (* A001716 *)
t[n_] := Table[k + 4, {k, 1, n}]; t6 = Table[a[n], {n, 1, z}]  (* A001721 *)
t[n_] := Table[k + 5, {k, 1, n}]; t7 = Table[a[n], {n, 1, z}]  (* A051524 *)
t[n_] := Table[k + 6, {k, 1, n}]; t8 = Table[a[n], {n, 1, z}]  (* A051545 *)
t[n_] := Table[k + 7, {k, 1, n}]; t9 = Table[a[n], {n, 1, z}]  (* A051560 *)
t[n_] := Table[k + 8, {k, 1, n}]; t10 = Table[a[n], {n, 1, z}] (* A051562 *)
t[n_] := Table[k + 9, {k, 1, n}]; t11 = Table[a[n], {n, 1, z}] (* A051564 *)
t[n_] := Table[k + 10, {k, 1, n}];t12 = Table[a[n], {n, 1, z}] (* A203147 *)
t = {t1, t2, t3, t4, t5, t6, t7, t8, t9, t10};
TableForm[t]  (* A165675 in square format *)
m[i_, j_] := t[[i]][[j]];
(* A165675 as a sequence *)
Flatten[Table[m[i, n + 1 - i], {n, 1, 10}, {i, 1, n}]]
(* Clark Kimberling, Dec 29 2011 *)
A[n_, k_] := (k + 1)*((n + k)! / n!)*HypergeometricPFQ[{-k, 1, 1}, {2, n + 1}, 1];
Table[A[n, k], {n, 0, 7}, {k, 0, 7}] // TableForm (* Peter Luschny, Jul 03 2022 *)

from functools import cache
@cache
def Trow(n: int) -> list[int]:
    if n == 0:
        return [1]
    row = Trow(n - 1) + [1]
    for m in range(n - 1, 0, -1):
        row[m] = (n - m + 1) * row[m] + row[m - 1]
    row[0] *= n
    return row
for n in range(9): print(Trow(n))  # Peter Luschny, Feb 27 2025

The hyperharmonic numbers are H(n, k) = Sum_{j=0..k} H(n - 1, j), with base condition H(0, k) = 1/(k + 1).
T(n, k) = (n - k + 1)*T(n - 1, k) + T(n - 1, k - 1), 1 <= k <= n-1, with T(n, 0) = n! and T(n, n) = 1.
From Peter Luschny, Jul 03 2022: (Start)
The rectangular array is given by:
A(n, k) = (k + 1)!*H(n, k).
A(n, k) = (k + 1)*((n + k)! / n!)*hypergeom([-k, 1, 1], [2, n + 1], 1). (End)
From Werner Schulte, Feb 26 2025: (Start)
T(n, k) = n * T(n-1, k) + (n-1)! / (k-1)! for 0 < k < n.
T(n, k) = (Sum_{i=k..n} 1/i) * n! / (k-1)! for 0 < k <= n.
Matrix inverse M = T^(-1) is given by: M(n, n) = 1, M(n, n-1) = 1 - 2 * n for n > 0, M(n, n-2) = (n-1)^2 for n > 1, and M(i, j) = 0 otherwise. (End)

New name from Peter Luschny, Jul 03 2022

A105954 Array read by descending antidiagonals: A(n, k) = (n + 1)! * H(k, n + 1), where H(n, k) is a higher-order harmonic number, H(0, k) = 1/k and H(n, k) = Sum_{j=1..k} H(n-1, j), for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 5, 11, 6, 1, 7, 26, 50, 24, 1, 9, 47, 154, 274, 120, 1, 11, 74, 342, 1044, 1764, 720, 1, 13, 107, 638, 2754, 8028, 13068, 5040, 1, 15, 146, 1066, 5944, 24552, 69264, 109584, 40320, 1, 17, 191, 1650, 11274, 60216, 241128, 663696, 1026576, 362880

Offset: 0

Leroy Quet, Jun 26 2005

Antidiagonal sums are A093345 (n! * (1 + Sum_{i=1..n}((1/i)*Sum_{j=0..i-1} 1/j!))). - Gerald McGarvey, Aug 27 2005
A recasting of A093905 and A067176. - R. J. Mathar, Mar 01 2009
The triangular array of this sequence is the reversal of A165675 which is related to the asymptotic expansion of the higher order exponential integral E(x,m=2,n); see also A165674. - Johannes W. Meijer, Oct 16 2009

			A(2, 2) = (1 + (1 + 1/2) + (1 + 1/2 + 1/3))*6 = 26.
Array A(n, k) begins:
  [n\k]  0       1       2        3        4        5          6
  -------------------------------------------------------------------
  [0]    1,      1,      1,       1,       1,       1,         1, ...
  [1]    1,      3,      5,       7,       9,       11,       13, ...
  [2]    2,     11,     26,      47,      74,      107,      146, ...
  [3]    6,     50,    154,     342,     638,     1066,     1650, ...
  [4]   24,    274,   1044,    2754,    5944,    11274,    19524, ...
  [5]  120,   1764,   8028,   24552,   60216,   127860,   245004, ...
  [6]  720,  13068,  69264,  241128,  662640,  1557660,  3272688, ...
  [7] 5040, 109584, 663696, 2592720, 7893840, 20355120, 46536624, ...

G. C. Greubel, Table of n, a(n) for the first 27 rows, flattened
Arthur T. Benjamin, David Gaebler and Robert Gaebler, A Combinatorial Approach to Hyperharmonic Numbers, INTEGERS, Electronic Journal of Combinatorial Number Theory, Volum 3, #A15, 2003.

Column 0 = A000142 (factorial numbers).
Column 1 = A000254 (Stirling numbers of first kind s(n, 2)) starting at n=1.
Column 2 = A001705 (Generalized Stirling numbers: a(n) = n!*Sum_{k=0..n-1}(k+1)/(n-k)), starting at n=1.
Column 3 = A001711 (Generalized Stirling numbers: a(n) = Sum_{k=0..n}(-1)^(n+k)*(k+1)*3^k*stirling1(n+1, k+1)).
Column 4 = A001716 (Generalized Stirling numbers: a(n) = Sum_{k=0..n}(-1)^(n+k)*(k+1)*4^k*stirling1(n+1, k+1)).
Column 5 = A001721 (Generalized Stirling numbers: a(n) = Sum_{k=0..n}(-1)^(n+k)*binomial(k+1, 1)*5^k*stirling1(n+1, k+1)).
Column 6 = A051524 (2nd unsigned column of A051338) starting at n=1.
Column 7 = A051545 (2nd unsigned column of A051339) starting at n=1.
Column 8 = A051560 (2nd unsigned column of A051379) starting at n=1.
Column 9 = A051562 (2nd unsigned column of A051380) starting at n=1.
Column 10= A051564 (2nd unsigned column of A051523) starting at n=1.
2nd row is A005408 (2n - 1, starting at n=1).
3rd row is A080663 (3n^2 - 1, starting at n=1).
Main diagonal gives A384024.
Cf. A000254, A165674 and A165675, A028421 and A126671.

H := proc(n, k) option remember; if n = 0 then 1/k else add(H(n - 1, j), j = 1..k) fi end: A := (n, k) -> (n + 1)!*H(k, n + 1):
# Alternative with standard harmonic number:
A := (n, k) -> if k = 0 then n! else (harmonic(n + k) - harmonic(k - 1))*(n + k)! / (k - 1)! fi:
for n from 0 to 7 do seq(A(n, k), k = 0..6) od;
# Alternative with hypergeometric formula:
A := (n, k) -> (n+1)*((n + k)! / k!)*hypergeom([-n, 1, 1], [2, k+1], 1):
seq(print(seq(simplify(A(n, k)), k = 0..6)), n=0..7); # Peter Luschny, Jul 01 2022

H[0, m_] := 1/m; H[n_, m_] := Sum[H[n - 1, k], {k, m}]; a[n_, m_] := m!H[n, m]; Flatten[ Table[ a[i, n - i], {n, 10}, {i, n - 1, 0, -1}]]
Table[ a[n, m], {m, 8}, {n, 0, m + 1}] // TableForm (* to view the table *)
(* Robert G. Wilson v, Jun 27 2005 *)

a(n, k) = polcoef(prod(j=0, n, 1+(j+k)*x), n); \\ Seiichi Manyama, May 19 2025

A(n, k) = (Harmonic(n + k) - Harmonic(k - 1))*(n + k)!/(k - 1)! if k > 0, otherwise n!.
From Gerald McGarvey, Aug 27 2005, edited by Peter Luschny, Jul 02 2022: (Start)
E.g.f. for column k: -log(1 - x)/(x*(1 - x)^k).
Row 3 is r(n) = 4*n^3 + 18*n^2 + 22*n + 6.
Row 4 is r(n) = 5*n^4 + 40*n^3 + 105*n^2 + 100*n + 24.
Row 5 is r(n) = 6*n^5 + 75*n^4 + 340*n^3 + 675*n^2 + 548*n + 120.
Row 6 is r(n) = 7*n^6 + 126*n^5 + 875*n^4 + 2940*n^3 + 4872*n^2 + 3528*n + 720.
Row 7 is r(n) = 8*n^7 + 196*n^6 + 1932*n^5 + 9800*n^4 + 27076*n^3 + 39396*n^2 + 26136*n + 5040.
The sum of the polynomial coefficients for the n-th row is |S1(n, 2)|, which are the unsigned Stirling1 numbers which appear in column 1.
A(m, n) = Sum_{k=1..m} n*A094645(m, n)*(n+1)^(k-1). (A094645 is Generalized Stirling number triangle of first kind, e.g.f.: (1-y)^(1-x).) (End)
In Gerard McGarvey's formulas for the row coefficients we find Wiggen's triangle A028421 and their o.g.f.s lead to Wood's polynomials A126671; see A165674. - Johannes W. Meijer, Oct 16 2009
A(n, k) = (n + 1)*((n + k)! / k!)*hypergeom([-n, 1, 1], [2, k + 1], 1). - Peter Luschny, Jul 01 2022
A(n,k) = [x^n] Product_{j=0..n} (1 + (j+k)*x). - Seiichi Manyama, May 19 2025

More terms from Robert G. Wilson v, Jun 27 2005

Edited by Peter Luschny, Jul 02 2022

A001712 Generalized Stirling numbers.

Original entry on oeis.org

1, 12, 119, 1175, 12154, 133938, 1580508, 19978308, 270074016, 3894932448, 59760168192, 972751628160, 16752851775360, 304473528961920, 5825460745532160, 117070467915075840, 2465958106403712000, 54336917746726272000, 1250216389189281024000

Offset: 0

N. J. A. Sloane

nonn

The asymptotic expansion of the higher order exponential integral E(x,m=3,n=3) ~ exp(-x)/x^3*(1 - 12/x + 119/x^2 - 1175/x^3 + 12154/x^4 - 133938/x^5 + ...) leads to the sequence given above. See A163931 and A163932 for more information. - Johannes W. Meijer, Oct 20 2009
From Petros Hadjicostas, Jun 11 2020: (Start)
For nonnegative integers n, m and complex numbers a, b (with b <> 0), the numbers R_n^m(a,b) were introduced by Mitrinovic (1961) using slightly different notation. They were further examined by Mitrinovic and Mitrinovic (1962).
These numbers are defined via the g.f. Product_{r=0..n-1} (x - (a + b*r)) = Sum_{m=0..n} R_n^m(a,b)*x^m for n >= 0.
As a result, R_n^m(a,b) = R_{n-1}^{m-1}(a,b) - (a + b*(n-1))*R_{n-1}^m(a,b) for n >= m >= 1 with R_1^0(a,b) = a, R_1^1(a,b) = 1, and R_n^m(a,b) = 0 for n < m. (Because an empty product is by definition 1, we may let R_0^0(a,b) = 1.)
With a = 0 and b = 1, we get the Stirling numbers of the first kind S1(n,m) = R_n^m(a=0, b=1) = A048994(n,m). (Array A008275 is the same as array A048994 but with no zero row and no zero column.)
We have R_n^m(a,b) = Sum_{k=0}^{n-m} (-1)^k * a^k * b^(n-m-k) * binomial(m+k, k) * S1(n, m+k) for n >= m >= 0.
For the current sequence, a(n) = R_{n+2}^2(a=-3, b=-1) for n >= 0. (End)

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..100
Matt Davis, Quadrant Marked Mesh Patterns and the r-Stirling Numbers, arXiv preprint arXiv:1412.0345 [math.CO], 2014.
Matt Davis, Quadrant Marked Mesh Patterns and the r-Stirling Numbers, J. Int. Seq. 18 (2015), #15.10.1.
Sergey Kitaev and Jeffrey Remmel, Simple marked mesh patterns, arXiv preprint arXiv:1201.1323 [math.CO], 2012.
Sergey Kitaev and Jeffrey Remmel, Quadrant Marked Mesh Patterns, J. Int. Seq. 15 (2012), #12.4.7.
D. S. Mitrinovic, Sur une classe de nombres reliés aux nombres de Stirling, Comptes rendus de l'Académie des sciences de Paris, t. 252 (1961), 2354-2356. [The numbers R_n^m(a,b) are introduced.]
D. S. Mitrinovic and M. S. Mitrinovic, Tableaux d'une classe de nombres reliés aux nombres de Stirling Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz. No. 77 (1962), 1-77.
D. S. Mitrinovic and R. S. Mitrinovic, Tableaux d'une classe de nombres reliés aux nombres de Stirling, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz., No. 77 (1962), 1-77 [jstor stable version].
Robert E. Moritz, On the sum of products of n consecutive integers, Univ. Washington Publications in Math., 1 (No. 3, 1926), 44-49. [Annotated scanned copy]

Cf. A001711, A008275, A048994, A163931, A163932.

A001712 := proc(n)
    add((-1)^(n+k)*binomial(k+2, 2)*3^k*Stirling1(n+2, k+2), k=0..n) ;
end proc:
seq(A001712(n), n=0..10) ; # R. J. Mathar, Jun 09 2018

nn = 22; t = Range[0, nn]! CoefficientList[Series[Log[1 - x]^2/(2*(1 - x)^3), {x, 0, nn}], x]; Drop[t, 2] (* T. D. Noe, Aug 09 2012 *)

a(n) = sum(k=0, n, (-1)^(n+k)*binomial(k+2, 2)*3^k*stirling(n+2, k+2, 1)) \\ Michel Marcus, Jan 20 2016

b(n) = prod(r=0, n+1, r+3);
c(n) = sum(i=0, n+1, sum(j=i+1, n+1, 1/((3+i)*(3+j))));
for(n=0, 18, print1(b(n)*c(n),",")) \\ Petros Hadjicostas, Jun 11 2020

a(n) = Sum_{k=0..n} (-1)^(n+k)*binomial(k+2, 2)*3^k*Stirling1(n+2, k+2). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
E.g.f.: (1 - 7*log(1 - x) + 6*log(1 - x)^2)/(1 - x)^5. - Vladeta Jovovic, Mar 01 2004
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 a(n-2) = |f(n,2,3)|, for n >= 2. [Milan Janjic, Dec 21 2008]
Conjecture: a(n) + 3*(-n-3)*a(n-1) + (3*n^2 + 15*n + 19)*a(n-2) - (n+2)^3*a(n-3)=0. - R. J. Mathar, Jun 09 2018
From Petros Hadjicostas, Jun 11 2020: (Start)
a(n) = [x^2] Product_{r=0}^{n+1} (x + 3 + r) = (Product_{r=0}^{n+1} (r+3)) * Sum_{0 <= i < j <= n+1} 1/((3+i)*(3+j)).
Since a(n) = R_{n+2}^2(a=-3, b=-1) and A001711(n) = R_{n+1}^1(a=-3,b=-1), the equation R_{n+2}^2(a=-3,b=-1) = R_{n+1}^1(a=-3,b=-1) + (n+4)*R_{n+1}^2(a=-3,b=-1) implies the following:
(i) a(n) = A001711(n) + (n+4)*a(n-1) for n >= 1.
(ii) a(n) = (n+2)!/2 + (2*n+7)*a(n-1) - (n+3)^2*a(n-2) for n >= 2.
(iii) R. J. Mathar's recurrence above. (End)

More terms from Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004

A188386 a(n) = numerator(H(n+2)-H(n-1)), where H(n) = Sum_{k=1..n} 1/k is the n-th harmonic number.

Original entry on oeis.org

11, 13, 47, 37, 107, 73, 191, 121, 299, 181, 431, 253, 587, 337, 767, 433, 971, 541, 1199, 661, 1451, 793, 1727, 937, 2027, 1093, 2351, 1261, 2699, 1441, 3071, 1633, 3467, 1837, 3887, 2053, 4331, 2281, 4799, 2521, 5291, 2773, 5807, 3037, 6347, 3313, 6911, 3601

Offset: 1

Gary Detlefs, Mar 29 2011

Denominators are listed in A033931.
A027446 appears to be divisible by a(n).
The sequence lists also the largest odd divisors of 3*m^2-1 (A080663) for m>1. In fact, for m even, the largest odd divisor is 3*m^2-1 itself; for m odd, the largest odd divisor is (3*m^2-1)/2. From this follows the second formula given in Formula field. - Bruno Berselli, Aug 27 2013

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A033931 (denominators), A001008, A001711, A027446, A002805, A003154, A080663, A158463, A213998.

import Data.Ratio ((%), numerator)
a188386 n = a188386_list !! (n-1)
a188386_list = map numerator $ zipWith (-) (drop 3 hs) hs
   where hs = 0 : scanl1 (+) (map (1 %) [1..])
-- Reinhard Zumkeller, Jul 03 2012

[Numerator((3*n^2+6*n+2)/((n*(n+1)*(n+2)))): n in [1..50]]; // Vincenzo Librandi, Mar 30 2011

seq((3-(-1)^n)*(3*n^2+6*n+2)/4, n=1..100);

Table[(3 - (-1)^n)*(3*n^2 + 6*n + 2)/4, {n, 40}] (* Wesley Ivan Hurt, Jan 29 2017 *)
Numerator[#[[4]]-#[[1]]]&/@Partition[HarmonicNumber[Range[0,50]],4,1] (* or *) LinearRecurrence[{0,3,0,-3,0,1},{11,13,47,37,107,73},50] (* Harvey P. Dale, Dec 31 2017 *)

a(n) = numerator((3*n^2+6*n+2)/(n*(n+1)*(n+2))).
a(n) = (3-(-1)^n)*(3*n^2+6*n+2)/4.
a(2n+1) = A158463(n+1), a(2n) = A003154(n+1).
G.f.: -x*(11+13*x+14*x^2-2*x^3-x^4+x^5) / ( (x-1)^3*(1+x)^3 ). - R. J. Mathar, Apr 09 2011
a(n) = numerator of coefficient of x^3 in the Maclaurin expansion of sin(x)*exp((n+1)*x). - Francesco Daddi, Aug 04 2011
H(n+3) = 3/2 + 2*f(n)/((n+2)*(n+3)), where f(n) = Sum_{k=0..n}((-1)^k*binomial(-3,k)/(n+1-k)). - Gary Detlefs, Jul 17 2011
a(n) = A213998(n+2,2). - Reinhard Zumkeller, Jul 03 2012
Sum_{n>=1} 1/a(n) = c*(tan(c) - cot(c)/2) - 1/2, where c = Pi/(2*sqrt(3)). - Amiram Eldar, Sep 27 2022

A067176 A triangle of generalized Stirling numbers: sum of consecutive terms in the harmonic sequence multiplied by the product of their denominators.

Original entry on oeis.org

0, 1, 0, 3, 1, 0, 11, 5, 1, 0, 50, 26, 7, 1, 0, 274, 154, 47, 9, 1, 0, 1764, 1044, 342, 74, 11, 1, 0, 13068, 8028, 2754, 638, 107, 13, 1, 0, 109584, 69264, 24552, 5944, 1066, 146, 15, 1, 0, 1026576, 663696, 241128, 60216, 11274, 1650, 191, 17, 1, 0, 10628640

Offset: 0

Henry Bottomley, Jan 09 2002

In the Coupon Collector's Problem with n types of coupon, the expected number of coupons required until there are only k types of coupon uncollected is a(n,k)*k!/(n-1)!.
If n+k is even, then a(n,k) is divisible by (n+k+1). For n>=k and k>= 0, a(n,k) = (n-k)!*H(k+1,n-k), where H(m,n) is a generalized harmonic number, i.e., H(0,n) = 1/n and H(m,n) = Sum_{j=1..n} H(m-1,j). - Leroy Quet, Dec 01 2006
This triangle is the same as triangle A165674, which is generated by the asymptotic expansion of the higher order exponential integral E(x,m=2,n), minus the first right hand column. - Johannes W. Meijer, Oct 16 2009

			Rows start 0; 1,0; 3,1,0; 11,5,1,0; 50,26,7,1,0; 274,154,47,9,1,0 etc. a(5,2) = 3*4*5*(1/3 + 1/4 + 1/5) = 4*5 + 3*5 + 3*4 = 20 + 15 + 12 = 47.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Columns are A000254, A001705, A001711, A001716, A001721, A051524, A051545, A051560, A051562, A051564, etc.

Cf. A093905 and A165674. - Johannes W. Meijer, Oct 16 2009

T[0, k_] := 1; T[n_, k_] := T[n, k] = Sum[ i*k^(i - 1)*Abs[StirlingS1[n - k, i]], {i, 1, n - k}]; Table[T[n,k], {n,1,10}, {k,1,n}] (* G. C. Greubel, Jan 21 2017 *)

a(n, k) = (n!/k!)*Sum_{j=k+1..n} 1/j = (A000254(n) - A000254(k)*A008279(n, n-k))/A000142(k) = a(n-1, k)*n + (n-1)!/k! = (a(n, k-1)-n!/k!)/k.

a(n, k) = Sum_{i=1..n-k} i*k^(i-1)*abs(stirling1(n-k, i)). - Vladeta Jovovic, Feb 02 2003

A093905 Triangle read by rows: for 0 <= k < n, a(n, k) is the sum of the products of all subsets of {n-k, n-k+1, ..., n} with k members.

Original entry on oeis.org

1, 1, 3, 1, 5, 11, 1, 7, 26, 50, 1, 9, 47, 154, 274, 1, 11, 74, 342, 1044, 1764, 1, 13, 107, 638, 2754, 8028, 13068, 1, 15, 146, 1066, 5944, 24552, 69264, 109584, 1, 17, 191, 1650, 11274, 60216, 241128, 663696, 1026576, 1, 19, 242, 2414, 19524, 127860

Offset: 1

Amarnath Murthy, Apr 24 2004

Triangle A165674, which is the reversal of this triangle, is generated by the asymptotic expansion of the higher order exponential integral E(x,m=2,n). - Johannes W. Meijer, Oct 16 2009

			Triangle begins:
1
1 3
1 5 11
1 7 26 50
1 9 47 154 274
...
a(5, 3) = 4*3*2+5*3*2+5*4*2+5*4*3 = 154.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

The leading diagonal is given by A000254, Stirling numbers of first kind. The next nine diagonals are A001705, A001711, A001716, A001721, A051524, A051545, A051560, A051562 and A051564, generalized Stirling numbers.
Cf. A001705, A001711, A067176, A093344, A105954.
A165674 is the reversal of this triangle. - Johannes W. Meijer, Oct 16 2009

T[n_, 0] := 1; T[n_, k_]:= Product[i, {i, n - k, n}]*Sum[1/i, {i, n - k, n}]; Table[T[n, k], {n, 1, 10}, {k, 0, n - 1}] (* G. C. Greubel, Jan 21 2017 *)

a(n, k) = prod(i=n-k, n, i)*sum(i=n-k,n,1/i);
tabl(nn) = for (n=1, nn, for (k=0, n-1, print1(a(n,k), ", ")); print()); \\ Michel Marcus, Jan 21 2017

a(n, k) = (Product_{i=n-k..n} i)*(Sum_{i=n-k..n} 1/i), where a(n, 0) = 1.

a(n, k) = A067176(n, n-k-1) = A105954(k+1, n-k). Row sums are given by A093344.

Edited and extended by David Wasserman, Apr 24 2007

A001713 Generalized Stirling numbers.

Original entry on oeis.org

1, 18, 245, 3135, 40369, 537628, 7494416, 109911300, 1698920916, 27679825272, 474957547272, 8572072384512, 162478082312064, 3229079010579072, 67177961946534528, 1460629706845766400, 33139181950164806400, 783398920650352012800, 19268391564147377318400

Offset: 0

N. J. A. Sloane

nonn

The asymptotic expansion of the higher order exponential integral E(x,m=4,n=3) ~ exp(-x)/x^4*(1 - 18/x + 245/x^2 - 3135/x^3 + 40369/x^4 - 537628/x^5 + ...) leads to the sequence given above. See A163931 and A163934 for more information. - Johannes W. Meijer, Oct 20 2009
From Petros Hadjicostas, Jun 12 2020: (Start)
For nonnegative integers n, m and complex numbers a, b (with b <> 0), the numbers R_n^m(a,b) were introduced by Mitrinovic (1961) and Mitrinovic and Mitrinovic (1962) using slightly different notation.
These numbers are defined via the g.f. Product_{r=0..n-1} (x - (a + b*r)) = Sum_{m=0..n} R_n^m(a,b)*x^m for n >= 0.
As a result, R_n^m(a,b) = R_{n-1}^{m-1}(a,b) - (a + b*(n-1))*R_{n-1}^m(a,b) for n >= m >= 1 with R_0^0(a,b) = 1, R_1^0(a,b) = a, R_1^1(a,b) = 1, and R_n^m(a,b) = 0 for n < m.
With a = 0 and b = 1, we get the Stirling numbers of the first kind S1(n,m) = R_n^m(a=0, b=1) = A048994(n,m) for n, m >= 0.
We have R_n^m(a,b) = Sum_{k=0}^{n-m} (-1)^k * a^k * b^(n-m-k) * binomial(m+k, k) * S1(n, m+k) for n >= m >= 0.
For the current sequence, a(n) = R_{n+3}^3(a=-3, b=-1) for n >= 0. (End)

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..100
D. S. Mitrinovic, Sur une classe de nombres reliés aux nombres de Stirling, Comptes rendus de l'Académie des sciences de Paris, t. 252 (1961), 2354-2356. [The numbers R_n^m(a,b) are introduced.]
D. S. Mitrinovic and R. S. Mitrinovic, Tableaux d'une classe de nombres reliés aux nombres de Stirling, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz., No. 77 (1962), 1-77 [jstor stable version].
D. S. Mitrinovic and M. S. Mitrinovic, Tableaux d'une classe de nombres reliés aux nombres de Stirling, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz. No. 77 (1962), 1-77.

Cf. A000254, A001706, A001711, A001712, A001719, A048994.

nn = 23; t = Range[0, nn]! CoefficientList[Series[-Log[1 - x]^3/(6*(1 - x)^3), {x, 0, nn}], x]; Drop[t, 3] (* T. D. Noe, Aug 09 2012 *)

a(n) = sum(k=0, n, (-1)^(n+k)*binomial(k+3, 3)*3^k*stirling(n+3, k+3, 1)); \\ Michel Marcus, Jan 20 2016

b(n) = prod(r=0, n+2, r+3);
c(n) = sum(i=0, n+2, sum(j=i+1, n+2, sum(k=j+1, n+2, 1/((3+i)*(3+j)*(3+k)))));
for(n=0, 18, print1(b(n)*c(n), ", ")) \\ Petros Hadjicostas, Jun 12 2020

E.g.f.: Sum_{n>=0} a(n)*x^(n+3)/(n+3)! = (log(1 - x)/(x - 1))^3/6. - Vladeta Jovovic, May 05 2003 [Edited by Petros Hadjicostas, Jun 13 2020]
a(n) = Sum_{k=0..n} (-1)^(n+k) * binomial(k+3, 3) * 3^k * Stirling1(n+3, k+3). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
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 a(n-3) = |f(n,3,3)| for n >= 3. - Milan Janjic, Dec 21 2008
From Petros Hadjicostas, Jun 12 2020: (Start)
a(n) = [x^3] Product_{r=0}^{n+2} (x + 3 + r) = (Product_{r=0}^{n+2} (r+3)) * Sum_{0 <= i < j < k <= n+2} 1/((3+i)*(3+j)*(3+k)).
Since a(n) = R_{n+3}^3(a=-3, b=-1), A001712(n) = R_{n+2}^2(a=-3,b=-1), and A001711(n) = R_{n+1}^1(a=-3, b=-1), the equation R_{n+3}^3(a=-3,b=-1) = R_{n+2}^2(a=-3,b=-1) + (n+5)*R_{n+2}^3(a=-3,b=-1) implies the following:
(i) a(n) = A001712(n) + (n+5)*a(n-1) for n >= 1.
(ii) a(n) = A001711(n) + (2*n+9)*a(n-1) - (n+4)^2*a(n-2) for n >= 2.
(iii) a(n) = (n+2)!/2 + 3*(n+4)*a(n-1) - (3*n^2+21*n+37)*a(n-2) + (n+3)^3*a(n-3) for n >= 3.
(iv) a(n) = 2*(2*n+7)*a(n-1) - (6*n^2+36*n+55)*a(n-2) + (2*n^2+10*n+13)*(2*n+5)*a(n-3) - (n+2)^4*a(n-4) for n >= 4. (End)

More terms from Vladeta Jovovic, May 05 2003

cp's OEIS Frontend

A051564 Second unsigned column of triangle A051523.

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A052752 a(n) = (3*n+1)^(n-1).

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A165674 Triangle generated by the asymptotic expansions of the E(x,m=2,n).

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A165675 Triangle read by rows. T(n, k) = (n - k + 1)! * H(k, n - k), where H are the hyperharmonic numbers. For 0 <= k <= n.

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A105954 Array read by descending antidiagonals: A(n, k) = (n + 1)! * H(k, n + 1), where H(n, k) is a higher-order harmonic number, H(0, k) = 1/k and H(n, k) = Sum_{j=1..k} H(n-1, j), for 0 <= k <= n.

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A001712 Generalized Stirling numbers.

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A188386 a(n) = numerator(H(n+2)-H(n-1)), where H(n) = Sum_{k=1..n} 1/k is the n-th harmonic number.

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