cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-9 of 9 results.

A196837 Coefficient table of numerator polynomials of o.g.f.s for partial sums of powers of positive integers.

Original entry on oeis.org

1, 2, -3, 3, -12, 11, 4, -30, 70, -50, 5, -60, 255, -450, 274, 6, -105, 700, -2205, 3248, -1764, 7, -168, 1610, -7840, 20307, -26264, 13068, 8, -252, 3276, -22680, 89796, -201852, 236248, -109584, 9, -360, 6090, -56700, 316365, -1077300, 2171040, -2345400, 1026576, 10, -495, 10560, -127050, 946638, -4510275, 13667720, -25228500, 25507152, -10628640
Offset: 1

Views

Author

Wolfdieter Lang, Oct 10 2011

Keywords

Comments

The k-th power of the positive integers has partial sums Sum_{j=1..n} j^k given as column number n >= 1, in the array A103438 (not in the triangle; see the example array given there; note that 0^0 has been set to 0 there).
The o.g.f. of column number n >= 1 of the array A103438 is obtained via Laplace transformation from the e.g.f. which is given there as
exp(x)*(exp(n*x)-1)/(exp(x)-1) = Sum_{j=1..n} exp(j*x)
(it is trivial that the sum is the e.g.f.).
The o.g.f. is, therefore, Sum_{j=1..n} 1/(1-j*x), which is rewritten as P(n,x)/Product_{j=1..n} (1-j*x). This defines the row polynomials P(n,x) of the present triangle. See the link for details.
This e.g.f. - o.g.f. connection proves some conjectures by Simon Plouffe. See the o.g.f. Maple programs under, e.g., A001551(n=4) and A001552 (n=5).
This triangle organizes the sum of powers of the first n positive integers in terms of the column no. n of the Stirling2 numbers A048993 (see the formula and example given below, as well as the link).
From Wolfdieter Lang, Oct 12 2011: (Start)
With the formulas given below one finds for n >= 1, k >= 0, Sum_{j=1..n} j^k =
Sum_{m=0..min(k,n-1)} ((n-m)*S1(n+1, n-m+1)*S2(k+n-m, n)),
with the Stirling numbers S1 from A048994 and S2 from A048993 (this formula I did not (yet) find in the literature). See the link for the proof.
For two other formulas expressing these sums of k-th powers of the first n positive integers in terms of the row no. k of Stirling2 numbers and binomials in n see the D. E. Knuth reference given under A093556, p. 285.
See also the given link below, eqs. (11) and (12). (End)

Examples

			n\m  0    1    2     3     4      5...
1    1
2    2   -3
3    3  -12   11
4    4  -30   70   -50
5    5  -60  255  -450   274
6    6 -105  700 -2205  3248  -1764
...
n=4 (A001551=2*A196836): the row polynomial factorizes into 2*(2-5*x)*(1-5*x+5*x^2).
n=5: 1^k + 2^k + 3^k + 4^k + 5^k, k>=0, (A001552) has as e.g.f. Sum_{j=1..5} exp(j*x). The o.g.f. is
  Sum_{j=1..5} 1/(1-j*x), and this is
  (5 - 60*x + 255*x^2 - 450*x^3 + 274*x^4)/Product_{j=1..5} (1-j*x).
n=6 (A001553): the row polynomial factorizes into
     (2 - 7*x)*(3 - 42*x + 203*x^2 - 392*x^3 + 252*x^4).
Sums of powers of the first n positive integers in terms of S2:
n=4: A001551(k) = 4*S2(k+4,4) - 30*S2(k+3,4) + 70*S2(k+2,4) - 50*S2(k+1,4), k >= 0. E.g., k=3: 4*350 - 30*65 + 70*10 - 50*1 = 100 = A001551(3).
From _Wolfdieter Lang_, Oct 12 2011: (Start)
Row polynomial for n=3: P(3,x) = (1-2*x)*(1-3*x) + (1-1*x)*(1-3*x) + (1-1*x)*(1-2*x) = 3 - 12*x + 11*x^2.
a(3,2) = +(sigma_2(2,3) + sigma_2(1,3) + sigma_2(1,2)) =
  2*3 + 1*3 + 1*2 = 11 = +1*sigma_2(1,2,3) = +1*|S1(4,4-2)|.
S1,S2 formula for sums of powers with n=4, k=3:
A001551(3) = Sum_{j=1..n} j^3 = 1*4*350 - 3*10*65 + 2*35*10 - 1*50*1 = 100. (End)
		

Crossrefs

Cf. A103438, A093556/A093557 (for sums of powers).

Programs

  • Mathematica
    a[n_, m_] := (n-m)*StirlingS1[n+1, n+1-m]; Flatten[ Table[ a[n, m], {n, 1, 10}, {m, 0, n-1}] ] (* Jean-François Alcover, Dec 02 2011, after Wolfdieter Lang *)
  • Python
    from itertools import count, islice
    from sympy.functions.combinatorial.numbers import stirling
    def A196837_T(n,m): return (n-m)*stirling(n+1,n+1-m,kind=1,signed=True)
    def A196837_gen(): # generator of terms
        return (A196837_T(n,m) for n in count(1) for m in range(n))
    A196837_list = list(islice(A196837_gen(),40)) # Chai Wah Wu, Oct 24 2024

Formula

a(n,m) = [x^m] P(n,x), m=0..n-1, with the row polynomials defined by
(Sum_{j=1..n} 1/(1-j*x))*Product_{j=1..n} (1-j*x) (see the comment given above).
Sum_{j=1..n} j^k = Sum_{m=0..n-1} a(n,m)*S2(k+n-m,n), n >= 1, k >= 0, with the Stirling2 triangle A048993.
From Wolfdieter Lang, Oct 12 2011: (Start)
The row polynomial P(n,x) is therefore
Sum_{j=1..n} (Product_{k=1..n omitting k=j} (1-k*x)), n >= 1. This leads to:
a(n,m) = (n-m)*S1(n+1, n+1-m), n-1 >= m >= 0, with the (signed) Stirling1 numbers A048994. For the proof see the link.
(End)
A similar polynomial occurs in the expansion of 1/(n+x)^2 as a series with factorials in the denominator: 1/(n+x)^2 = -Sum_{k>=1} n!/(n+k+1)! * P(k,1/x) x^(k-1). - Matt Majic, Nov 01 2019

A093557 Triangle of denominators of coefficients of Faulhaber polynomials in Knuth's version.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 6, 15, 3, 15, 30, 1, 1, 1, 3, 3, 3, 1, 1, 1, 1, 2, 1, 1, 5, 2, 5, 10, 1, 1, 3, 2, 7, 1, 3, 42, 21, 21, 1, 1, 2, 3, 2, 1, 6, 15, 3, 5, 10, 1, 1, 1, 5, 3, 10, 5, 15, 5, 5, 1, 1, 1, 1, 6, 3, 2, 3, 3, 7, 1, 1, 14, 21, 42, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1
Offset: 1

Views

Author

Wolfdieter Lang, Apr 02 2004

Keywords

Comments

The companion triangle with the numerators is A093556, where more information can be found.

Examples

			Triangle begins:
  [1];
  [1,1];
  [1,2,1];
  [1,3,3,1];
  ...
Denominators of [1]; [1,0]; [1,-1/2,0]; [1,-4/3,2/3,0]; ... (see W. Lang link in A093556.)
		

Crossrefs

Cf. A093556 (numerators).

Programs

  • Mathematica
    a[m_, k_] := (-1)^(m-k)* Sum[ Binomial[2*m, m-k-j]*Binomial[m-k+j, j]*((m-k-j)/(m-k+j))*BernoulliB[m+k+j], {j, 0, m-k}]; Flatten[ Table[ Denominator[a[m, k]], {m, 1, 14}, {k, 0, m-1}]] (* Jean-François Alcover, Oct 25 2011 *)
  • PARI
    T(n,k) = denominator((-1)^(n-k)*sum(j=0, n-k, binomial(2*n, n-k-j)*binomial(n-k+j,j)*(n-k-j)/(n-k+j) * bernfrac(n+k+j))); \\ Michel Marcus, Aug 03 2025

Formula

a(m, k) = denominator(A(m, k)) with recursion: A(m, 0)=1, A(m, k)=-(sum(binomial(m-j, 2*k+1-2*j)*A(m, j), j=0..k-1))/(m-k) if 0<= k <= m-1, else 0. From the 1993 Knuth reference, given in A093556, p. 288, eq.(*) with A^{(m)}_k = A(m, k).

Extensions

More terms from Michel Marcus, Aug 03 2025

A093558 Triangle of numerators of coefficients of Faulhaber polynomials used for sums of even powers.

Original entry on oeis.org

1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1, -5, 17, -5, 5, 1, -5, 41, -236, 691, -691, 1, -7, 14, -22, 359, -7, 7, 1, -14, 77, -293, 1519, -1237, 3617, -3617, 1, -6, 217, -1129, 8487, -6583, 750167, -43867, 43867, 1, -5, 23, -470, 689, -28399, 1540967, -1254146, 174611, -174611
Offset: 2

Views

Author

Wolfdieter Lang, Apr 02 2004

Keywords

Comments

The companion triangle with the denominators is A093559.
Sum_{k=1..n} k^(2*(m-1)) = (2*n+1)*Sum_{j=0..m-1} Fe(m,k)*(n*(n+1))^(m-1-j), m >= 2. Sums of even powers of the first n integers >0 as polynomials in u := n*(n+1) (falling powers of u). See bottom of p. 288 of the 1993 Knuth reference.

Examples

			Triangle begins:
  [1];
  [1,-1];
  [1,-1,1];
  [1,-1,1,-1];
  [1,-5,17,-5,5]
   ...
Numerators of:
  [1/6];
  [1/10,-1/30];
  [1/14,-1/14,1/42];
  [1/18,-1/9,1/10,-1/30];
  [1/22,-5/33,17/66,-5/22,5/66];
  ... (see Lang link)
		

References

  • Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser Verlag, Basel, Boston, Berlin, 1993, ch. 7, pp. 131-159.

Crossrefs

Cf. A093556/A093557, A093559 (denominators).

Programs

  • Mathematica
    a[m_, k_] := (-1)^(m-k)*Sum[Binomial[2*m, m-k-j]*Binomial[m-k+j, j]*((m-k-j)/(m-k+j))*BernoulliB[m+k+j], {j, 0, m-k}]; t[m_, k_] := (m-k)*a[m, k]/(2*m*(2*m-1)); Table[t[m, k] // Numerator, {m, 2, 12}, {k, 0, m-2}] // Flatten (* Jean-François Alcover, Mar 03 2014 *)

Formula

a(n, m) = numerator(Fe(m, k), with Fe(m, k):=(m-k)*A(m, k)/(2*m*(2*m-1)) with Faulhaber numbers A(m, k):=A093556(m, k)/A093557(m, k) in Knuth's version. From the bottom of p. 288 of the 1993 Knuth reference.

A093559 Triangle of denominators of coefficients of Faulhaber polynomials used for sums of even powers.

Original entry on oeis.org

6, 10, 30, 14, 14, 42, 18, 9, 10, 30, 22, 33, 66, 22, 66, 26, 26, 78, 273, 910, 2730, 30, 30, 15, 9, 90, 2, 6, 34, 51, 51, 51, 102, 51, 170, 510, 38, 19, 95, 95, 190, 57, 3990, 266, 798, 42, 14, 7, 21, 6, 66, 1386, 693, 110, 330, 46, 138, 46, 23, 230, 690, 345, 23, 230, 46
Offset: 2

Views

Author

Wolfdieter Lang, Apr 02 2004

Keywords

Comments

The companion triangle with the numerators is A093558. See comment there.

Examples

			Triangle begins:
  [6];
  [10,30];
  [14,14,42];
  [18,9,10,30]; ...
Denominators of:
  [1/6];
  [1/10,-1/30];
  [1/14,-1/14,1/42];
  [1/18,-1/9,1/10,-1/30];
  ... (see W. Lang link in A093558.)
		

References

  • Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser Verlag, Basel, Boston, Berlin, 1993, ch. 7, pp. 131-159.

Programs

  • Mathematica
    a[m_, k_] := (-1)^(m-k)*Sum[Binomial[2*m, m-k-j]*Binomial[m-k+j, j]*((m-k-j)/(m-k+j))*BernoulliB[m+k+j], {j, 0, m-k}]; t[m_, k_] := (m-k)*a[m, k]/(2*m*(2*m-1)); Table[t[m, k] // Denominator, {m, 2, 12}, {k, 0, m-2}] // Flatten (* Jean-François Alcover, Mar 03 2014 *)

Formula

a(n, m) = denominator(Fe(m, k), with Fe(m, k):=(m-k)*A(m, k)/(2*m*(2*m-1)) with Faulhaber numbers A(m, k):=A093556(m, k)/A093557(m, k) in Knuth's version. From the bottom of p. 288 of the 1993 Knuth reference.

A335951 Triangle read by rows. The numerators of the coefficients of the Faulhaber polynomials. T(n,k) for n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 0, -1, 4, 0, 0, 1, -4, 6, 0, 0, -3, 12, -20, 16, 0, 0, 5, -20, 34, -32, 16, 0, 0, -691, 2764, -4720, 4592, -2800, 960, 0, 0, 105, -420, 718, -704, 448, -192, 48, 0, 0, -10851, 43404, -74220, 72912, -46880, 21120, -6720, 1280
Offset: 0

Views

Author

Peter Luschny, Jul 16 2020

Keywords

Comments

There are many versions of Faulhaber's triangle: search the OEIS for his name.
Faulhaber's claim (in 1631) is: S_{2*m-1} = 1^(2*m-1) + 2^(2*m-1) + ... + n^(2*m-1) = F_m((n^2+2)/2). The first proof was given by Jacobi in 1834.
For the Faulhaber numbers see A354042 and A354043.

Examples

			The first few polynomials are:
  [0] 1;
  [1] x;
  [2] x^2;
  [3] (4*x - 1)*x^2*(1/3);
  [4] (6*x^2 - 4*x + 1)*x^2*(1/3);
  [5] (16*x^3 - 20*x^2 + 12*x - 3)*x^2*(1/5);
  [6] (16*x^4 - 32*x^3 + 34*x^2 - 20*x + 5)*x^2*(1/3);
  [7] (960*x^5 - 2800*x^4 + 4592*x^3 - 4720*x^2 + 2764*x - 691)*x^2*(1/105);
  [8] (48*x^6 - 192*x^5 + 448*x^4 - 704*x^3 + 718*x^2 - 420*x + 105)*x^2*(1/3);
  [9] (1280*x^7-6720*x^6+21120*x^5-46880*x^4+72912*x^3-74220*x^2+43404*x-10851)*x^2*(1/45);
Triangle starts:
  [0] 1;
  [1] 0, 1;
  [2] 0, 0,  1;
  [3] 0, 0, -1,     4;
  [4] 0, 0,  1,    -4,      6;
  [5] 0, 0, -3,     12,    -20,    16;
  [6] 0, 0,  5,    -20,     34,   -32,     16;
  [7] 0, 0, -691,   2764,  -4720,  4592,  -2800,  960;
  [8] 0, 0,  105,  -420,    718,  -704,    448,  -192,    48;
  [9] 0, 0, -10851, 43404, -74220, 72912, -46880, 21120, -6720, 1280;
		

References

  • Johann Faulhaber, Academia Algebra. Darinnen die miraculosische Inventiones zu den höchsten Cossen weiters continuirt und profitiert werden. Johann Ulrich Schönigs, Augsburg, 1631.

Crossrefs

Cf. A335952 (polynomial denominators), A000012 (row sums of the polynomial coefficients).
Other representations of the Faulhaber polynomials include A093556/A093557, A162298/A162299, A220962/A220963.
Cf. A354042 (Faulhaber numbers), A354043.

Programs

  • Maple
    FaulhaberPolynomial := proc(n) if n = 0 then return 1 fi;
    expand((bernoulli(2*n, x+1) - bernoulli(2*n,1))/(2*n));
    sort(simplify(expand(subs(x = (sqrt(8*x+1)-1)/2, %))), [x], ascending) end:
    Trow := n -> seq(coeff(numer(FaulhaberPolynomial(n)), x, k), k=0..n):
    seq(print(Trow(n)), n=0..9);
  • Python
    from math import lcm
    from itertools import count, islice
    from sympy import simplify,sqrt,bernoulli
    from sympy.abc import x
    def A335951_T(n,k):
        z = simplify((bernoulli(2*n,(sqrt(8*x+1)+1)/2)-bernoulli(2*n,1))/(2*n)).as_poly().all_coeffs()
        return z[n-k]*lcm(*(d.q for d in z))
    def A335951_gen(): # generator of terms
        yield from (A335951_T(n,k) for n in count(0) for k in range(n+1))
    A335951_list = list(islice(A335951_gen(),20)) # Chai Wah Wu, May 16 2022
    
  • SageMath
    def A335951Row(n):
        R. = PolynomialRing(QQ)
        if n == 0: return [1]
        b = expand((bernoulli_polynomial(x + 1, 2*n) -
                    bernoulli_polynomial(1, 2*n))/(2*n))
        s = expand(b.subs(x = (sqrt(8*x+1)-1)/2))
        return numerator(s).list()
    for n in range(10): print(A335951Row(n)) # Peter Luschny, May 17 2022

Formula

Let F_n(x) be the polynomial after substituting (sqrt(8*x + 1) - 1)/2 for x in b_n(x), where b_n(x) = (Bernoulli_{2*n}(x+1) - Bernoulli_{2*n}(1))/(2*n).
F_n(1) = 1 for all n >= 0.
T(n, k) = numerator([x^k] F_n(x)).

A065551 Triangle of Faulhaber numbers (numerators) read by rows.

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 1, -1, 1, 0, -3, 3, -1, 1, 0, 5, -5, 17, -2, 1, 0, -691, 691, -118, 41, -5, 1, 0, 35, -35, 359, -44, 14, -1, 1, 0, -3617, 3617, -1237, 1519, -293, 22, -7, 1, 0, 43867, -43867, 750167, -13166, 2829, -2258, 217, -4, 1, 0, -1222277, 1222277, -627073, 1540967, -198793, 689, -235, 46, -3, 1
Offset: 0

Views

Author

Wouter Meeussen, Dec 02 2001

Keywords

Comments

From Wolfdieter Lang, Jun 25 2011: (Start)
In the Gessel and Viennot reference f(n,k) = a(n,k)/A065553(n,k), n>=0, k>=0.
(n+1)*f(n,k) = A(n+1,n-k), with Knuth's A(m,k) =
A093556(m,k)/A093557(m,k). See the Knuth reference given in A093556, and the W. Lang link. (End)

Examples

			Triangle begins:
{1},
{0, 1},
{0, -1, 1},
{0, 1, -1, 1},
{0, -3, 3, -1, 1},
{0, 5, -5, 17, -2, 1}.
		

Crossrefs

Cf. A065553.
Cf. A103438.

Formula

sum(n>=0, k>=0, f(n, k)*t^k*x^(2*n+1)/(2*n+1)! ) is the expansion of (cosh(sqrt(1+4*t)*x/2)-cosh(x/2))/t/sinh(x/2).
a(n,k)=numerator(f(n,k)).

A385567 Triangle read by rows: T(n,k) is the numerator of A(n,k), such that A(n,k) satisfies the identity for sums of odd powers: Sum_{k=1..p} k^(2n-1) = 1/(2*n) * Sum_{k=0..n-1} A(n,k) * (p^2+p)^(n-k), for all integers p >= 1.

Original entry on oeis.org

1, 1, 1, 1, 0, -1, 1, -1, 0, 1, 1, -4, 2, 0, -1, 1, -5, 3, -3, 0, 5, 1, -4, 17, -10, 5, 0, -691, 1, -35, 287, -118, 691, -691, 0, 7, 1, -8, 112, -352, 718, -280, 140, 0, -3617, 1, -21, 66, -293, 4557, -3711, 10851, -10851, 0, 43867, 1, -40, 217, -4516, 2829, -26332, 750167, -438670, 219335, 0, -174611
Offset: 0

Views

Author

Kolosov Petro, Jul 31 2025

Keywords

Comments

The companion triangle with the denominators is A386728.
Extension of A093556 with k in the range 0 <= k <= n, and n >= 0.

Examples

			Triangle begins:
---------------------------------------------------------------------------------
k =   0    1     2     3     4       5       6        7       8      9      10
---------------------------------------------------------------------------------
n=0:  1;
n=1:  1,   1;
n=2:  1,   0,  -1;
n=3:  1,  -1,   0,     1;
n=4:  1,  -4,   2,     0,   -1;
n=5:  1,  -5,   3,    -3,    0,      5;
n=6:  1,  -4,  17,   -10,    5,      0,   -691;
n=7:  1, -35, 287,  -118,  691,   -691,      0,       7;
n=8:  1,  -8, 112,  -352,  718,   -280,    140,       0,  -3617;
n=9:  1, -21,  66,  -293, 4557,  -3711,  10851,  -10851,      0, 43867;
n=10: 1, -40, 217, -4516, 2829, -26332, 750167, -438670, 219335,     0, -174611;
...
		

Crossrefs

Programs

  • Mathematica
    FaulhaberCoefficient[n_, k_] := 0;
    FaulhaberCoefficient[n_, k_] := (-1)^(n - k) * Sum[Binomial[2 n, n - k - j]* Binomial[n - k + j, j] * (n - k - j)/(n - k + j) * BernoulliB[n + k + j], {j, 0, n - k}] /; 0 <= k < n;
    FaulhaberCoefficient[n_, k_] := BernoulliB[2 n] /; k == n;
    Flatten[Table[Numerator[FaulhaberCoefficient[n, k]], {n, 0, 10}, {k, 0, n}]]
  • PARI
    T(n,k) = numerator(if (k==n, bernfrac(2*n), if (kMichel Marcus, Aug 03 2025

Formula

A(n,k) = 0 if k>n or n<0
A(n,k) = (-1)^(n - k) * Sum_{j=0..n-k} binomial(2n, n - k - j) * binomial(n - k + j, j) * (n - k - j)/(n - k + j) * B_{n + k + j}, if 0 <= k < n;
A(n,k) = B_{2n}, if k = n;
T(n,k) = numerator(A(n,k)).

A386728 Triangle read by rows: T(n,k) is the denominator of A(n,k), such that A(n,k) satisfies the identity for sums of odd powers: Sum_{k=1..p} k^(2n-1) = 1/(2*n) * Sum_{k=0..n-1} A(n,k) * (p^2+p)^(n-k), for all integers p >= 1.

Original entry on oeis.org

1, 1, 6, 1, 1, 30, 1, 2, 1, 42, 1, 3, 3, 1, 30, 1, 2, 1, 2, 1, 66, 1, 1, 2, 1, 1, 1, 2730, 1, 6, 15, 3, 15, 30, 1, 6, 1, 1, 3, 3, 3, 1, 1, 1, 510, 1, 2, 1, 1, 5, 2, 5, 10, 1, 798, 1, 3, 2, 7, 1, 3, 42, 21, 21, 1, 330, 1, 2, 3, 2, 1, 6, 15, 3, 5, 10, 1, 138, 1
Offset: 0

Views

Author

Kolosov Petro, Jul 31 2025

Keywords

Comments

The companion triangle with the numerators is A385567.
Extension of A093557 with k in the range 0 <= k <= n.

Examples

			Triangle begins:
  ---------------------------------------------------------
  k =   0  1   2   3    4    5    6   7    8    9    10
  ---------------------------------------------------------
  n=0:  1;
  n=1:  1, 6;
  n=2:  1, 1, 30;
  n=3:  1, 2,  1, 42;
  n=4:  1, 3,  3,  1, 30;
  n=5:  1, 2,  1,  2,  1, 66;
  n=6:  1, 1,  2,  1,  1,  1, 2730;
  n=7:  1, 6, 15,  3, 15, 30,    1,  6;
  n=8:  1, 1,  3,  3,  3,  1,    1,  1, 510;
  n=9:  1, 2,  1,  1,  5,  2,    5, 10,   1, 798;
  n=10: 1, 3,  2,  7,  1,  3,   42, 21,  21,   1, 330;
  ...
		

Crossrefs

Programs

  • Mathematica
    FaulhaberCoefficient[n_, k_] := 0;
    FaulhaberCoefficient[n_, k_] := (-1)^(n - k) * Sum[Binomial[2 n, n - k - j]* Binomial[n - k + j, j] * (n - k - j)/(n - k + j) * BernoulliB[n + k + j], {j, 0, n - k}] /; 0 <= k < n;
    FaulhaberCoefficient[n_, k_] := BernoulliB[2 n] /; k == n;
    Flatten[Table[Denominator[FaulhaberCoefficient[n, k]], {n, 0, 10}, {k, 0, n}]]
  • PARI
    T(n,k) = denominator(if (k==n, bernfrac(2*n), if (kMichel Marcus, Aug 03 2025

Formula

A(n,k) = 0 if k>n or n<0;
A(n,k) = (-1)^(n - k) * Sum_{j=0..n-k} binomial(2n, n - k - j) * binomial(n - k + j, j) * (n - k - j)/(n - k + j) * B_{n + k + j}, if 0 <= k < n;
A(n,k) = B_{2n}, if k = n;
T(n,k) = denominator(A(n,k)).

A251926 The Faulhaber-Knuth a(0,n) sequence.

Original entry on oeis.org

2, 1, 1, 1, 1, 0, 0, 1, 37, -60, -5, 37, 174, -955, -10545, 38610, 176297, -322740, -205420, 4512655, 56820585, -104019264, -25907081, 94854194, 1141847218, -2090335775, -414239903275, 6066664425833, 85621405759989, -156743813184120, -4337631088920, 47644406040193, 1265208493396175131, -2316168508680582540, -192288633159406495
Offset: 4

Views

Author

Roman Witula, Dec 11 2014

Keywords

Comments

a(n) is equal to the remainder when dividing the polynomial T_n(x) by x^2 + x - 1. T_n(x) (in Z[x]) is the positive integer multiplicity of the modified Faulhaber polynomial T*_n(x), coefficients of which have GCD equal to 1. We have T*_n(x) = S(n;x)/x^2(x+1)^2 if n is odd, and T*_n(x) = S(n;x)/x(x+1)(2x+1) if n is even, n >= 4, where S(n;x) denotes the n-th Faulhaber polynomial, i.e., S(n;x) = 1/(n+1) sum{taken over i=0,1,...,n} Bin(n+1,i)Bern(i)x^(n+1-i), and Bern(i) denotes the i-th Bernoulli number with Bern(1)=1/2.
We note that every T_n(x) is a polynomial in the variable (x^2 + x - 1), for example T_7(x) = 3(x^2 + x - 1)^2 + 2(x^2 + x - 1) + 1. Furthermore, every T_n(x) is a polynomial in (x^2 + x + a) for each complex a. But only for a = -1 is the element a(n) also equal to the remainder when dividing S(n;x) by x^2 + x + a if n is odd and S(n;x)/(2x+1) by x^2 + x + a if n is even.

Examples

			We have: T_4(x) = 3x^2 + 3x - 1, T_4(x) - T_5(x) = x^2 + x, T_6(x) - T_7(x) = x^2 + x - 1, T_9(x) = (x^2 + x - 1)(2x^4 + 4x^3 - x^2 - 3x + 3) and T_15(x) - T_12(x) is divisible by (x^2 + x - 1), which implies a(0)=2, a(1)=1, a(2)=a(3), a(5)=0 and a(8)=a(11).
		

Crossrefs

Programs

  • Mathematica
    coeffFaulh[n_] := Module[{t, tab = {}, s, p, x},
      If[n < 4, Return["Give n greater than 3."]];
      t = Table[1, {n + 2}];
      Do[t[[i + 1]] = BernoulliB[i], {i, 1, n + 1}];
      t[[2]] = 1/2;
      s[m_, x_] := (Sum[Binomial[m + 1, i]t[[ i + 1]] x^(m + 1 - i),{i,0,m}])/(m + 1);
      Do[If[Mod[i, 2] == 0,
        p = PolynomialRemainder[FactorList[Factor[s[i, x]] (i + 1)/(x (x + 1) (2 x + 1))][[2,1]], -1 + x + x^2, x],
        p = PolynomialRemainder[FactorList[Factor[s[i, x]] (i + 1)/(x^2 (x + 1)^2)][[2,1]], -1 + x + x^2, x]];
       tab = Append[tab, p], {i, 4, n}];
      tab]
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