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 1-3 of 3 results.

A354045 Row sums of A354043.

Original entry on oeis.org

1, 1, 2, 9, 83, 1391, 38498, 1638505, 101604805, 8802059341, 1030071680506, 158430831805049, 31302182958422223, 7792558134438194339, 2404183465743291972650, 906172339589098358635601, 412063269903446575653003241, 223575203315439598127658946681, 143324925086774266814499800353234
Offset: 0

Views

Author

Peter Luschny, May 17 2022

Keywords

Crossrefs

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)).

A354042 Triangle read by rows. The Faulhaber numbers. F(0, k) = 1 and otherwise F(n, k) = (n + 1)!*(-1)^(k+1)*Sum_{j=0..floor((k-1)/2)} C(2*k-2*j, k+1)*C(2*n+1, 2*j+1) * Bernoulli(2*n-2*j) / (k - j).

Original entry on oeis.org

1, 0, 1, 0, -1, 2, 0, 4, -8, 6, 0, -36, 72, -60, 24, 0, 600, -1200, 1020, -480, 120, 0, -16584, 33168, -28320, 13776, -4200, 720, 0, 705600, -1411200, 1206240, -591360, 188160, -40320, 5040, 0, -43751232, 87502464, -74813760, 36747648, -11813760, 2661120, -423360, 40320
Offset: 0

Views

Author

Peter Luschny, May 17 2022

Keywords

Comments

I. Gessel and X. Viennot call the rational numbers F(n, k)/(n + 1)! 'Faulhaber numbers'. However, for our purposes it is more convenient to define the integers F(n, k). For the Faulhaber polynomials see A335951/A335952.
Let S(r, m) = Sum_{k=0..m} k^r, with 0^0 = 1 and S(0, m) = m + 1. Faulhaber's theorem (the sums of powers formula) is:
S(2*n+1, m) = (1/(n+1)!)*(1/2)*Sum_{k=0..n} F(n, k)*(m*(m + 1))^(k + 1).
Gessel and Viennot give two combinatorial interpretations for the Faulhaber numbers, for this see A354043.

Examples

			Triangle starts:
0: 1
1: 0,         1
2: 0,        -1,        2
3: 0,         4,       -8,         6
4: 0,       -36,       72,       -60,       24
5: 0,       600,    -1200,      1020,     -480,       120
6: 0,    -16584,    33168,    -28320,    13776,     -4200,     720
7: 0,    705600, -1411200,   1206240,  -591360,    188160,  -40320,    5040
8: 0, -43751232, 87502464, -74813760, 36747648, -11813760, 2661120, -423360, 40320
.
Let n = 4 and m = 3, then S(2*n + 1, m) = S(9, 3) = 20196. Faulhaber's formula gives this as (0*12 + (-36)*144 + 72*1728 + (-60)*20736 + 24*248832) / (2*120).
		

Crossrefs

Programs

  • Maple
    F := (n, k) -> ifelse(n = 0, 1, (n + 1)!*(-1)^(k + 1)*add(binomial(2*k - 2*j, k + 1)*binomial(2*n + 1, 2*j + 1)*bernoulli(2*n - 2*j) / (k - j), j = 0..(k - 1)/2)): for n from 0 to 8 do seq(F(n, k), k = 0..n) od;

Formula

F(n,1) = (2*n +1)*Bernoulli(2*n)*(n+1)! for n >= 1.
F(n,2) = -(4*n+2)*Bernoulli(2*n)*(n+1)! for n >= 2.
F(n,3) = ((10*n+5)*Bernoulli(2*n) + binomial(2*n+1,3)*Bernoulli(2*n-2)/2)*(n+1)! for n >= 3.
Showing 1-3 of 3 results.