A130534 -id:A130534 - cp's OEIS frontend

A076620 Coefficient of x^a(n) in (x+1)(x+2)...*(x+n) is the largest one.

0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 1

Benoit Cloitre, Nov 10 2002

nonn

			In (x+1)(x+2)(x+3) = x^3 + 6*x^2 + 11*x + 6, the largest coefficient (11) appears at x^1, hence a(3)=1.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A065048, A130534, A309237.

a(n) = my(p=prod(j=1, n, x+j), m=vecmax(Vec(p))); for (i=0, poldegree(p), if (polcoef(p, i)==m, return(i))); \\ Michel Marcus, Feb 19 2021

first(n) = {res = vector(n); my(r = 1); v = [1]; for(i = 1, n, v = concat(0, v) + concat(v, 0)*i; for(j = r + 1, #v, if(v[j] > v[j - 1], r++ , next ); ); res[i] = r-1 ); res } \\ David A. Corneth, Feb 21 2021

from sympy import prod, Poly
from sympy.abc import x
def A076620(n):
    y = Poly(prod(x+i for i in range(1,n+1))).all_coeffs()[::-1]
    return y.index(max(y)) # Chai Wah Wu, Mar 07 2021

Is a(n) - floor(log(n)) bounded?

A112002 Seventh diagonal of triangle A008275 (Stirling1) and seventh column of |A008276|.

Original entry on oeis.org

720, 13068, 118124, 723680, 3416930, 13339535, 44990231, 135036473, 368411615, 928095740, 2185031420, 4853222764, 10246937272, 20692933630, 40171771630, 75289668850, 136717357942, 241276443496, 414908513800, 696829576300

Offset: 1

Wolfdieter Lang, Sep 12 2005

T. D. Noe, Table of n, a(n) for n = 1..1000

Sixth diagonal A053567; A130534.

[StirlingFirst(n+6, n): n in [1..20]]; // Vincenzo Librandi, Aug 09 2015

A112002 := proc(n) combinat[stirling1](n+6,n) ; end proc: # R. J. Mathar, Jun 08 2011

Table[StirlingS1[n+6, n], {n, 1, 20}] (* Jean-François Alcover, Mar 05 2014 *)

[stirling_number1(n,n-6) for n in range(7, 27)] # Zerinvary Lajos, May 16 2009

a(n)= Stirling1(n+6, n), n>=1, with Stirling1(n, k)= A008275(n, k).
E.g.f. with offset 6: exp(x)*sum(A112486(6, m)*(x^(6+m))/(6+m)!, m=0..6).
a(n)= (f(n+5, 6)/12!)*sum(A112486(6, m)*f(12, 6-m)*f(n-1, m), m=0..min(6, n-1)), with the falling factorials f(n, k):=n*(n-1)*...*(n-(k-1)). From the e.g.f.
a(n)=(binomial(n+6, 7)/r(8, 5))*sum(A112007(5, m)*r(n+7, 5-m)*f(n-1, m), m=0..5), with rising factorials r(n, k):=n*(n+1)*...*(n+(k-1)) and falling factorials f(n, m). From the g.f.
G.f.: x*(720+3708*x+4400*x^2+1452*x^3+114*x^4+x^5)/(1-x)^13. See row k=5 of triangles A112007 or A008517 for the coefficients.
Explicit formula: a(n) = n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)(n + 6)(63n^5 + 1575n^4 + 15435n^3 + 73801n^2 + 171150n + 152696)/2903040. - Vaclav Kotesovec, Jan 30 2010

A193651 a(n) = ((2*n + 1)!! + 1)/2.

Original entry on oeis.org

1, 2, 8, 53, 473, 5198, 67568, 1013513, 17229713, 327364538, 6874655288, 158117071613, 3952926790313, 106729023338438, 3095141676814688, 95949391981255313, 3166329935381425313, 110821547738349885938, 4100397266318945779688, 159915493386438885407813

Offset: 0

Clark Kimberling, Aug 02 2011

Previous name was: Q-residue of the triangle A130534, where Q is the triangular array (t(i,j)) given by t(i,j)=1. For the definition of Q-residue, see A193649.

a(404) has 1002 decimal digits. - Michael De Vlieger, Apr 25 2016

Michael De Vlieger, Table of n, a(n) for n = 0..403

Cf. A001147, A193649, A130534.

seq((1+doublefactorial(2*n+1))/2,n=0..18); # Peter Luschny, Aug 20 2014

q[n_, k_] := 1;
r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}]
u[0, x_] := 1; u[n_, x_] := (x + n)*u[n - 1, x]
p[n_, k_] := Coefficient[u[n, x], x, k]
v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
Table[v[n], {n, 0, 18}]    (* A193651 *)
TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
Table[r[k], {k, 0, 8}]  (* 2^k *)
TableForm[Table[p[n, k], {n, 0, 6}, {k, 0, n}]]  (* A130534 *)
Table[((2 n + 1)!! + 1)/2, {n, 0, 18}] (* or *)
Table[(2^n Gamma[n + 3/2])/Sqrt[Pi] + 1/2, {n, 0, 18}] (* or *)
Table[2^n Pochhammer[1/2, n + 1] + 1/2, {n, 0, 18}] (* Michael De Vlieger, Apr 25 2016 *)

def A():
    n, a, b = 1, 1, 2
    yield a
    while True:
        yield b
        n += 1
        a, b = b, ((2*(b-a)*n + a)*n - b)/(n-1)
A193651 = A()
[next(A193651) for i in range(19)] # Peter Luschny, Aug 20 2014

From Peter Luschny, Aug 20 2014: (Start)
a(n) = (2^n*Gamma(n+3/2))/sqrt(Pi) + 1/2.
a(n) = 2^n*Pochhammer(1/2, n+1) + 1/2.
a(n) = ((2*a(n-1) - 2*a(n-2))*n^2 + a(n-2)*n - a(n-1))/(n-1) for n>1, a(0)=1, a(1)=2. (End)
(-n+1)*a(n) +(2*n^2-1)*a(n-1) -n*(2*n-1)*a(n-2)=0. - R. J. Mathar, Feb 19 2015
E.g.f.: (exp(x) + 1/(1-2*x)^(3/2))/2. - Vladimir Reshetnikov, Apr 25 2016

New name from Peter Luschny, Aug 20 2014

A213167 a(n) = n! - (n-2)!.

Original entry on oeis.org

1, 5, 22, 114, 696, 4920, 39600, 357840, 3588480, 39553920, 475372800, 6187104000, 86699289600, 1301447347200, 20835611596800, 354379753728000, 6381450915840000, 121289412980736000, 2426499634470912000

Offset: 2

Olivier Gérard, Nov 02 2012

Row sums of A134433 starting from k=3.
a(n) = sum( (-1)^k*k*A008276(n,k), k=1..n-1).
a(n) = sum( (-1)^k*k*A054654(n,k), k=1..n-2).
For n >= 3, a(n) = number whose factorial base representation (A007623) begins with digits {n-1} and {n-2} followed by n-3 zeros. Viewed in that base, this sequence looks like this: 1, 21, 320, 4300, 54000, 650000, 7600000, 87000000, 980000000, A900000000, BA000000000, ... (where "digits" A and B stand for placeholder values 10 and 11 respectively). - Antti Karttunen, May 07 2015.

Index entries for sequences related to factorial base representation

Column 4 of A257503 (apart from initial 1. Equally, row 4 of A257505).
Cf. A000142, A007623, A134433.
Cf. A008276, A094638, A008275, A130534.
Cf. A054654, A048994, A132393.
Cf. A067318.

Table[n! - (n - 2)!, {n, 2, 20}]
#[[3]]-#[[1]]&/@Partition[Range[0,20]!,3,1] (* Harvey P. Dale, Aug 10 2023 *)

A213167(n):=n!-(n-2)!$
makelist(A213167(n),n,2,30); /* Martin Ettl, Nov 03 2012 */

(define (A213167 n) (- (A000142 n) (A000142 (- n 2)))) ;; Antti Karttunen, May 07 2015

a(n) = n! - (n-2)!.
G.f.: (1/G(0) - 1 - x)/x^2 where G(k) = 1 - x/(x - 1/(x - (k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Dec 13 2012
G.f.: (1+x)/x^2*(1/Q(0)-1), where Q(k)= 1 - 2*k*x - x^2*(k + 1)^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 08 2013
G.f.: 2*Q(0), where Q(k)= 1 - 1/( (k+1)*(k+2) - x*(k+1)^2*(k+2)^2*(k+3)/(x*(k+1)*(k+2)*(k+3) - 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 08 2013

A330797 Evaluation of the Stirling cycle polynomials at -1/2 and normalized with (-2)^n.

Original entry on oeis.org

1, 1, -1, 3, -15, 105, -945, 10395, -135135, 2027025, -34459425, 654729075, -13749310575, 316234143225, -7905853580625, 213458046676875, -6190283353629375, 191898783962510625, -6332659870762850625, 221643095476699771875, -8200794532637891559375, 319830986772877770815625

Offset: 0

Peter Luschny, Jan 06 2020

sign

G. C. Greubel, Table of n, a(n) for n = 0..400
Y. Alp and E. G. Kocer, Exponential Almost-Riordan Arrays, Results Math 79, 173 (2024). See page 25.

The equivalent for Stirling2 is A009235.
Cf. A001147, A101485, A103639, A130534.
Cf. A330799, A330800, A330802, A330803.

m:=30;
R:=PowerSeriesRing(Rationals(), m+2);
A330797:= func< n | Coefficient(R!(Laplace( Sqrt(1+2*x) )), n) >;
[A330797(n): n in [0..m]]; // G. C. Greubel, Sep 14 2023

a := n -> ((-2)^(n-1)*GAMMA(n-1/2))/sqrt(Pi): seq(a(n), n=1..9);
# Alternative:
arec := proc(n) option remember: if n = 0 then 1 else
(3 - 2*n)*arec(n-1) fi end: seq(arec(n), n=0..20);
# Or:
gf := (1+2*x)^(1/2); ser := series(gf, x, 24);
seq(n!*coeff(ser, x, n), n=0..20);

a[n_]:= (-2)^n*Sum[Abs[StirlingS1[n, k]]*(-1/2)^k, {k, 0, n}];
Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Nov 19 2021 *)
Table[(-2)^(n-1)*Pochhammer[1/2, n-1], {n,0,30}] (* G. C. Greubel, Sep 14 2023 *)

def A330797(n): return (-2)^(n-1)*rising_factorial(1/2, n-1)
[A330797(n) for n in (0..20)]

a(n) = (-2)^n*Sum_{k=0..n} |Stirling1(n,k)|*(-1/2)^k.
a(n) = (-2)^(n-1)*RisingFactorial(1/2, n-1).
a(n) = ((-2)^(n-1)*Gamma(n - 1/2))/sqrt(Pi).
a(n) = n!*[x^n] (1+2*x)^(1/2).
D-finite with recurrence a(n) = (3 - 2*n)*a(n-1).
a(n) = (-1)^(n-1)*(2*n-3)!! = (-1)^(n-1)*A001147(n-1).
a(2*n) = -2^(2*n-1)*RisingFactorial(1/2, 2*n-1) = -A103639(n-1).
a(2*n+1) = 4^n*RisingFactorial(1/2, 2*n) = A101485(n).
a(n) ~ -((-2*n)^n/exp(n))/(sqrt(2)*n).
Sum_{n>=0} 1/a(n) = 2 - sqrt(Pi/(2*e))*erfi(1/sqrt(2)), where erfi is the imaginary error function. - Amiram Eldar, Jan 08 2023
O.g.f.: 1+x*2F0(1/2,1;;-2*x). - R. J. Mathar, Aug 10 2025

A136662 Number of cycles of the permutations of [1,2,...,n].

Original entry on oeis.org

1, 2, 1, 3, 2, 2, 1, 1, 2, 4, 3, 3, 2, 2, 3, 3, 2, 2, 1, 1, 2, 2, 1, 3, 2, 2, 1, 1, 2, 2, 3, 1, 2, 5, 4, 4, 3, 3, 4, 4, 3, 3, 2, 2, 3, 3, 2, 4, 3, 3, 2, 2, 3, 3, 4, 2, 3, 4, 3, 3, 2, 2, 3, 3, 2, 2, 1, 1, 2, 2, 1, 3, 2, 2, 1, 1, 2, 2, 3, 1, 2, 3, 2, 2, 1, 1, 2, 4, 3, 3, 2, 2, 3, 3, 2, 2, 1, 1, 2, 2, 3, 1, 2, 2, 1

Offset: 1

Wolfdieter Lang, Feb 22 2008, May 21 2008

The row lengths sequence is A000142(n), n>=1, (factorials).

The permutations of [1,2,...,n] are ordered in the standard way (lexicographic or numerically increasing). E.g., in Maple as permute(n) list for not too large n (around 10).

			Triangle begins:
  [1];
  [2,1];
  [3,2,2,1,1,2];
  [4,3,3,2,2,3,3,2,2,1,1,2,2,1,3,2,2,1,1,2,2,3,1,2];
  ...
Row n=3: permutations of [1,2,3] in the order [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]. Cycle decomposition: [[[1], [2], [3]], [[1], [2, 3]], [[1, 2], [3]], [[1, 2, 3]], [[1, 3, 2]], [[1, 3], [2]]]. Number of cycles: [3,2,2,1,1,2], the entries of row n=3.

Alois P. Heinz, Rows n = 1..8, flattened
FindStat - Combinatorial Statistic Finder, The number of cycles in the cycle decomposition of a permutation
Wolfdieter Lang, First rows and cycle decompositions.

Row sums (total cycle numbers) A000254.

Cf. A130534.

a(n,k) = number of cycles of the k-th permutation of [1,2,...,n] in standard (increasing) order.

A193246 Ordered unsigned Stirling numbers |S1(n,k)|.

Original entry on oeis.org

0, 1, 2, 3, 6, 10, 11, 15, 21, 24, 28, 35, 36, 45, 50, 55, 66, 78, 85, 91, 105, 120, 136, 153, 171, 175, 190, 210, 225, 231, 253, 274, 276, 300, 322, 325, 351, 378, 406, 435, 465, 496, 528, 546, 561, 595, 630, 666, 703, 720, 735, 741, 780, 820, 861, 870, 903

Offset: 1

Vaclav Kotesovec, Jan 01 2013

nonn

List all unsigned Stirling numbers of the first kind, sort and remove duplicates.

The only numbers below 10^8 that appear in more than one place in the table A130534 are 1, 6, and 120. - Pontus von Brömssen, Jul 20 2024

Pontus von Brömssen, Table of n, a(n) for n = 1..10000

Cf. A193245, A024412, A130534.

s1 = Union[Flatten[Table[Table[Abs[StirlingS1[n, k]], {k, 1, 100}], {n, 1, 100}]]]; Table[s1[[j]], {j, 1, 100}]

A221914 Array of products of the list entries of the nonempty combinations of n, ordered in a standard way.

Original entry on oeis.org

1, 1, 2, 2, 1, 2, 3, 2, 3, 6, 6, 1, 2, 3, 4, 2, 3, 4, 6, 8, 12, 6, 8, 12, 24, 24, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 24, 30, 40, 60, 120, 120, 1, 2, 3, 4, 5, 6, 2, 3, 4, 5, 6, 6, 8, 10, 12, 12, 15, 18, 20, 24, 30, 6, 8, 10, 12, 12, 15, 18, 20, 24, 30, 24, 30, 36, 40, 48, 60, 60, 72, 90, 120, 24, 30, 36, 40, 48

Offset: 1

Wolfdieter Lang, Feb 06 2013

The sequence of the row lengths is 2^n-1 = A000225(n), n >= 1.
The combination choose(n,0) is the empty list [], appearing first in the list of combinations of n each taken as a list, called choose(n). The empty combination is not considered in this array. The list of lists choose(n) is ordered according to increasing m from choose(n,m), m = 0, ..., n, and for each m from 1, ..., n the lists are ordered lexicographically. E.g., choose(3) = [[], [1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]]. The corresponding list of the products of these list entries is then (without the empty list) the row n=3 of the present array [1, 2, 3, 2, 3, 6, 6].
The row sums are (n+1)! - 1 = A033312(n+1).
One could add a column k=0 (for the empty combination) and a row n=0 by defining a(n,0) = 1, n >= 0. This enlarged array produces then after summation of the entries deriving from the choose(n,m) list the triangle T(n,m), n >= m >= 0, which coincides with the unsigned Stirling1 triangle if the rows are read backwards: T(n,m) = |S1(n+1,n+1-m)|. See A130534 for |Strirling1(n+1,m+1)|. Proof by showing the recurrence.

			The array a(n,k) begins:
n\k  1  2  3  4  5  6  7  8  9  10 11 12  13  14  15 ...
1:   1
2:   1  2  2
3:   1  2  3  2  3  6  6
4:   1  2  3  4  2  3  4  6  8  12  6  8  12  24  24
...
Row n=5:  1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 24, 30, 40, 60, 120;
Row n=6: 1, 2, 3, 4, 5, 6, 2, 3, 4, 5, 6, 6, 8, 10, 12, 12, 15, 18, 20, 24, 30, 6, 8, 10, 12, 12, 15, 18, 20, 24, 30, 24, 30, 36, 40, 48, 60, 60, 72, 90, 120, 24, 30, 36, 40, 48, 60, 60, 72, 90, 120, 120, 144, 180, 240, 360, 120, 144, 180, 240, 360, 720, 720;
Row n=3: from the combinations list choose(3) (without the first entry []) [[1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]] leading to [1, 2, 3, 2, 3, 6, 6].
a(3,4) = 2 is the product of the entries of the combination list choose(3,1,1) = [1, 2], the first combination from choose(3,1).
|Stirling1| connection from like m summation: T(3,0) := 1 = |Stirling1(4,4)|, T(3,1) = 1 + 2 + 3 = 6 = |Stirling1(4,3)|,
  T(3,2) = 2 + 3 + 6 = 11 = |Stirling1(4,2)|, T(3,3) = 6 =
  |Stirling1(4,1)|.

a(n,k) = product(choose(n,m,j)[l],l=1..m) if the k-th entry of the choose(n) list (without the leading empty set and ordered as explained in a comment above), is choose(n,m,j), the j-th list member of the list choose(n,m), for n >= 1, 1 <= m <= n, 1 <= j <= binomial(n,m).

A237653 O.g.f.: Sum_{n>=0} x^n*Product_{k=1..n} (k + x).

Original entry on oeis.org

1, 1, 3, 9, 36, 176, 1030, 7039, 55098, 486346, 4780445, 51787405, 613045468, 7873065045, 109021348618, 1619197654575, 25675094145535, 432908683794379, 7733991639921585, 145933532935469016, 2900112108790279902, 60543749629794205640, 1324677739541613767983, 30312375400027348522996

Offset: 0

Paul D. Hanna, May 11 2014

nonn

Equals antidiagonal sums of the triangle of unsigned Stirling numbers of the first kind (A008275).

			G.f.: A(x) = 1 + x + 3*x^2 + 9*x^3 + 36*x^4 + 176*x^5 + 1030*x^6 + 7039*x^7 +...
where
A(x) = 1 + x*(1+x) + x^2*(1+x)*(2+x) + x^3*(1+x)*(2+x)*(3+x) + x^4*(1+x)*(2+x)*(3+x)*(4+x) +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..400

Cf. A124380, A008275, A130534.

{a(n)=polcoeff(sum(k=0, n, x^k*prod(j=1, k, j+x +x*O(x^n))), n)}
for(n=0,30,print1(a(n),", "))

A191685 Eighth diagonal a(n) = s(n,n-7) of the unsigned Stirling numbers of the first kind with n>7.

Original entry on oeis.org

5040, 109584, 1172700, 8409500, 45995730, 206070150, 790943153, 2681453775, 8207628000, 23057159840, 60202693980, 147560703732, 342252511900, 756111184500, 1599718388730, 3256091103430, 6400590336096, 12191224980000, 22563937825000, 40681506808800

Offset: 8

Paul Weisenhorn, Jun 11 2011

The Maple programs under I generate the sequence. The Maple program under II generates explicit formulas for a(n+1) = s(n+1,n+1-c) with c>=1 and n>=c.

			c=1; a(n+1) = binomial(n+1,2)
c=2; a(n+1) = binomial(n+1,3)*(2+3*n)/4
c=3; a(n+1) = binomial(n+1,4)*(n+n^2)/2
c=4; a(n+1) = binomial(n+1,5)*(-8-10*n+15*n^2 +15*n^3)/48
c=5; a(n+1) = binomial(n+1,6)*(-6*n-7*n^2+2*n^3+ 3*n^4)/16
c=6; a(n+1) = binomial(n+1,7)*(96+140*n-224*n^2-315*n^3+63*n^5)/576
c=7; a(n+1) = binomial(n+1,8)*(80*n+114*n^2-23*n^3-75*n^4-9*n^5+9*n^6)/144
c=8; a(n+1) = binomial(n+1,9)*(-1152-1936*n+2820*n^2+
  5320*n^3+735*n^4-1575*n^5-315*n^6+135*n^7)/3840
c=9; a(n+1) = binomial(n+1,10)*(-1008*n-1676*n^2 +100*n^3+1295*n^4+392*n^5-210*n^6-60*n^7 +15*n^8)/768

K. Seidel, Variation der Binomialkoeffizienten, Bild
der Wissenschaft, 6 (1980), pp. 127-128.

T. D. Noe, Table of n, a(n) for n = 8..1000

Cf. A130534, A000012 (c=0; 1st diagonal), A000217 (c=1; 2nd diagonal), A000914 (c=2; 3rd diagonal), A001303 (c=3; 4th diagonal), A000915 (c=4; 5th diagonal), A053567 (c=5; 6th diagonal), A112002 (c=6; 7th diagonal), A191685 (c=7; 8th diagonal).

I: programs generate the sequence:
with(combinat): c:=7; a:= proc(n) a(n):=abs(stirling1(n,n-c)); end: seq(a(n), n=c+1..28);
for n from 7 to 27 do a(n+1) := binomial(n+1,8)*(80*n+ 114*n^2- 23*n^3- 75*n^4- 9*n^5+ 9*n^6)/144 end do: seq(a(n),n=8..28);
II: program generates explicit formulas for a(n+1) =  s(n+1,n+1-c):
k[1,0]:=1: v:=1:
for c from 2 to 10 do
  c1:=c-1: c2:=c-2: p0:=0:
  for j from 0 to c2 do p0:=p0+k[c1,j]*m^j: end do:
  f:=expand(2*c*(m+1)*p0/v):
  p1:=0: p2:=0:
  for j from 0 to c1 do
    p1:=p1+k[c,j]*(m+1)^j:
    p2:=p2+k[c,j]*m^j:
  end do:
  g:=collect((m+2)*p1-(m-c1)*p2-f,m):
  kh[0]:=rem(g,m,m): Mk:=[kh[0]]: Mv:=[k[c,0]]:
  for j from 1 to c1 do
    kh[j]:=coeff(g,m^j):
    Mk:=[op(Mk),kh[j]]: Mv:=[k[c,j],op(Mv)]:
  end do:
  sol:=solve(Mk,Mv):
  v:=1:
  for j from 1 to c do
    k[c,c-j]:=eval(k[c,c-j],sol[1,j]):
    nen[j]:=denom(k[c,c-j]):
    v:=ilcm(v,nen[j]):
  end do:
  for j from 0 to c1 do k[c,j]:=k[c,j]*v:
    printf("%8d",k[c,j]): end do:
  p3:=0:
  for j from 0 to c1 do p3:=p3+k[c,j]*n^j: end do:
  s[n+1,n+1-c]:=binomial(n+1,c+1)*(c+1)*p3/(2^c*k[c,c1]):
end do:
for c from 2 to 10 do print("%a\n",s[n+1,n+1-c]):
end do:

a(n+1) = A130534(T(n,n-7)) = s(n+1,n+1-7)

a(n+1) = binomial(n+1,8)*(80*n+114*n^2-23*n^3-75*n^4-9*n^5+9*n^6)/144

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A076620 Coefficient of x^a(n) in (x+1)(x+2)...*(x+n) is the largest one.