A105954 -id:A105954 - cp's OEIS frontend

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

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

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

A274266 Expansion of e.g.f. (1 + x)^3*log(1 + x).

Original entry on oeis.org

1, 5, 11, 6, -6, 12, -36, 144, -720, 4320, -30240, 241920, -2177280, 21772800, -239500800, 2874009600, -37362124800, 523069747200, -7846046208000, 125536739328000, -2134124568576000, 38414242234368000, -729870602452992000, 14597412049059840000

Offset: 1

Peter Bala, Jun 19 2016

First four terms [1, 5, 11, 6] form row 3 of A105954 read as a triangular array.

			E.g.f.= x + 5*x^2/2 + 11*x^3/3! + 6*x^4/4! - 6*x^5/5! + ....

Cf. A000169, A105954, A274265, A274268, A274270.

[1,5,11] cat [(-1)^n*6*Factorial(n-4): n in [4..25]]; // Vincenzo Librandi, Jun 20 2016

CoefficientList[Series[(1+t)^3 * Log[1+t], {t, 1, 20}], t]*Range[1, 20]! (* G. C. Greubel, Jun 19 2016 *)

a(n) = (-1)^n*6*(n - 4)! for n >= 4.
E.g.f.: A(x) = (1 + x)^3*log(1 + x).
Series reversion(A(x)) = exp(-1/3*T(-3*x)) - 1 = x - 5*x^2/2! + 8^2*x^3/3! - 11^3*x^4/4! + 14^4*x^5/5! - ... is the e.g.f. for a signed version of A274265, where T(x) = Sum_{n >= 1} n^(n-1)*x^n/n! is Euler's tree function - see A000169.
Sum_{n>=1} 1/a(n) = 71/55 + 1/(6*e). - Amiram Eldar, Feb 02 2023

A274270 Expansion of e.g.f. (1 + x)^5*log(1 + x).

Original entry on oeis.org

1, 9, 47, 154, 274, 120, -120, 240, -720, 2880, -14400, 86400, -604800, 4838400, -43545600, 435456000, -4790016000, 57480192000, -747242496000, 10461394944000, -156920924160000, 2510734786560000, -42682491371520000, 768284844687360000, -14597412049059840000

Offset: 1

Peter Bala, Jun 19 2016

The first six terms [1, 9, 47, 154, 274, 120] form row 5 of A105954 read as a triangular array.

			E.g.f.= x + 9*x^2/2 + 47*x^3/3! + 154*x^4/4! + 274*x^5/5! + 120*x^6/6! - 120*x^7/7! + ....

Cf. A000169, A105954, A274265, A274266, A274267, A274268, A274269, A274270.

[1,9,47,154,274] cat [(-1)^n*120*Factorial(n - 6): n in [6..25]]; // Vincenzo Librandi, Jun 20 2016

CoefficientList[Series[(1+t)^5 * Log[1+t], {t,1,20}],t]*Range[1,20]! (* G. C. Greubel, Jun 19 2016 *)

a(n) = (-1)^n*120*(n - 6)! for n >= 6.
E.g.f.: A(x) = (1 + x)^5*log(1 + x).
Series reversion(A(x)) = exp(-1/5*T(-5*x)) - 1 = x - 9*x^2/2! + 14^2*x^3/3! - 19^3*x^4/4! + 24^4*x^5/5! - ... is the e.g.f. for a signed version of A274269, where T(x) = Sum_{n >= 1} n^(n-1)*x^n/n! is Euler's tree function - see A000169.
Sum_{n>=1} 1/a(n) = 5098232/4462227 + 1/(120*e). - Amiram Eldar, Feb 02 2023

A274268 Expansion of e.g.f. (1 + x)^4*log(1 + x).

Original entry on oeis.org

1, 7, 26, 50, 24, -24, 48, -144, 576, -2880, 17280, -120960, 967680, -8709120, 87091200, -958003200, 11496038400, -149448499200, 2092278988800, -31384184832000, 502146957312000, -8536498274304000, 153656968937472000, -2919482409811968000, 58389648196239360000

Offset: 1

Peter Bala, Jun 19 2016

First five terms [1, 7, 26, 50, 24] form row 4 of A105954 read as a triangular array.

			E.g.f.= x + 7*x^2/2 + 26*x^3/3! + 50*x^4/4! + 24*x^5/5! - 24*x^6/6! + ...

Cf. A000169, A105954, A274265, A274266, A274267, A274269, A274270.

[1,7,26,50] cat [(-1)^(n-1)*24*Factorial(n-5): n in [5..25]]; // Vincenzo Librandi, Jun 20 2016

CoefficientList[Series[(1+t)^4 * Log[1+t], {t, 1, 20}], t]*Range[1, 20]! (* G. C. Greubel, Jun 19 2016 *)

a(n) = (-1)^(n-1)*24*(n - 5)! for n >= 5.
E.g.f.: A(x) = (1 + x)^4*log(1 + x).
Series reversion(A(x)) = exp(-1/4*T(-4*x)) - 1 = x - 7*x^2/2! + 11^2*x^3/3! - 15^3*x^4/4! + 19^4*x^5/5! - ... is the e.g.f. for a signed version of A274267, where T(x) = Sum_{n >= 1} n^(n-1)*x^n/n! is Euler's tree function - see A000169.
Sum_{n>=1} 1/a(n) = 2733/2275 + 1/(24*e). - Amiram Eldar, Feb 02 2023

A136205 Square array read by antidiagonals: T(m,n) = H(n,2m)(2m)!/(2m+2*n-1). H(0,m) = 1/m, for all positive integers m. H(n,m) = Sum_{k=1..m} H(n-1,k).

Original entry on oeis.org

1, 1, 2, 1, 10, 24, 1, 22, 252, 720, 1, 38, 892, 12176, 40320, 1, 58, 2232, 60336, 966240, 3628800, 1, 82, 4632, 199440, 6202080, 114341760, 479001600, 1, 110, 8524, 526256, 25598016, 905049216, 18897709824, 87178291200, 1, 142, 14412, 1197360

Offset: 0

Leroy Quet, Mar 16 2008

In the array, the first m is 1; the first n is 0.

Every term of the array is a positive integer.

			Array: (The upper-leftmost term is T(1,0).)
  1, 2, 24, 720 (Row equals {(2*m-2)!}.)
  1, 10, 252 (Row equals {H(1,2*m)*(2*m)!/(2*m+1)}, where H(1,2*m) = the (2*m)th harmonic number.)
  1, 22 (Row equals {H(2,2*m)*(2*m)!/(2*m+3)}.)
  1 (Row equals {H(3,2*m)*(2*m)!/(2*m+5)}.)
The column {T(1,n)} consists entirely of 1's.

Cf. A105954.

Cf. A090288, A010050, A114450.

For n>=1, T(m,n) also equals (H(2*m+n-1) - H(n-1)) * (2*m+n-1)!/((2*m+2*n-1)*(n-1)!), where H(k) = H(1,k), the k-th harmonic number.

More terms from R. J. Mathar, Apr 01 2008

cp's OEIS Frontend

A108123 Duplicate of A105954.

Views

Author

Keywords

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

Views

Author

Keywords

Comments

Crossrefs

Programs

Maple

Formula

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.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

Python

Formula

Extensions

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A274266 Expansion of e.g.f. (1 + x)^3*log(1 + x).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Magma

Mathematica

Formula

A274270 Expansion of e.g.f. (1 + x)^5*log(1 + x).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Magma

Mathematica

Formula

A274268 Expansion of e.g.f. (1 + x)^4*log(1 + x).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Magma

Mathematica

Formula

A136205 Square array read by antidiagonals: T(m,n) = H(n,2*m)*(2*m)!/(2*m+2*n-1). H(0,m) = 1/m, for all positive integers m. H(n,m) = Sum_{k=1..m} H(n-1,k).

Views

Author

Keywords

Comments

Examples

Crossrefs

A136205 Square array read by antidiagonals: T(m,n) = H(n,2m)(2m)!/(2m+2*n-1). H(0,m) = 1/m, for all positive integers m. H(n,m) = Sum_{k=1..m} H(n-1,k).