A068566 -id:A068566 - cp's OEIS frontend

A145656 a(n) = numerator of polynomial of genus 1 and level n for m = 2.

0, 2, 5, 32, 131, 661, 1327, 18608, 148969, 447047, 89422, 1967410, 7869871, 102309709, 204620705, 2046213056, 32739453941, 556571077357, 556571247527, 10574855234543, 42299423848079, 42299425233749, 84598851790183

Offset: 1

Artur Jasinski, Oct 16 2008

For the numerators of the polynomials of genus 1 and level n for m = 1 see A001008.
Definition: The polynomial A[1,2*n+1](m) = A[genus 1,level n] is here defined as Sum_{d = 1..n-1} m^(n-d)/d.
First few A[1,n](m):
  n = 1: A[1,1](m) = 0
  n = 2: A[1,2](m) = m
  n = 3: A[1,3](m) = m/2 + m^2
  n = 4: A[1,4](m) = m/3 + m^2/2 + m^3
  n = 5: A[1,5](m) = m/4 + m^2/3 + m^3/2 + m^4
The general formula which uses these polynomials is the following:
(1/(n+1))*Hypergeometric2F1[1, n, n+1, 1/m] = Sum_{k >= 0} m^(-k)/(k + n) = m^n * ArcTanh[(2*m-1)/(2*m^2-2*m+1)] - A[1,n](m) = (m^n)*Log[m/(m-1)] - A[1,n](m).
Conjecture: a(n) = numerator( (2^n)*log(2) - 2^(n+1)*Integral_{x = 0..1} x^(2*n-1)/(1 + x^2)^n ). - Peter Bala, Jun 10 2024
a(n) appears to be a multiple of A068566(n): The sequence {a(n)/A068566(n) : n >= 2} begins [2, 1, 16, 1, 1, 1, 16, 1, 1, 2, 2, 1, 1, 1, 64, 1, 1, 1, 1, 1, 1, 1, ...]. - Peter Bala, Aug 07 2025

Cf. A145609-A145640.

A145656 := proc(n) add( 2^(n-d)/d, d = 1..n-1) end: seq(numer(A145656(n)), n = 1..20); # R. J. Mathar, Feb 01 2011

m = 2; aa = {}; Do[k = 0; Do[k = k + m^(r - d)/d, {d, 1, r - 1}]; AppendTo[aa, Numerator[k]], {r, 1, 30}]; aa
(* or *)
a[n_]:=2Integrate[(2-x^n)/(2-x),{x,0,1}]+4(2^(n-1)-1)Log[2]
Table[a[n] // Simplify // Numerator,{n,0,22}] (* Gerry Martens, Jun 04 2016 *)

A104709 Triangle read by rows: T(n,k) = Sum_{j=0..n} 2^(n-j)binomial(j,k) for n >= 0 and 0 <= k <= n; also, Riordan array (1/((1-x)(1-2*x)), x/(1-x)).

Original entry on oeis.org

1, 3, 1, 7, 4, 1, 15, 11, 5, 1, 31, 26, 16, 6, 1, 63, 57, 42, 22, 7, 1, 127, 120, 99, 64, 29, 8, 1, 255, 247, 219, 163, 93, 37, 9, 1, 511, 502, 466, 382, 256, 130, 46, 10, 1, 1023, 1013, 968, 848, 638, 386, 176, 56, 11, 1, 2047, 2036, 1981, 1816, 1486, 1024, 562, 232, 67

Offset: 0

Gary W. Adamson, Mar 19 2005

This array (A104709) is the mirror of the fission, A054143, of the polynomial sequence ((x+1)^n: n >= 0) by the polynomial sequence (q(n,x): n >= 0) given by q(n,x) = x^n + x^(n-1) + ... + x + 1.  See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
The elements of the matrix inverse appear to be T^(-1)(n,k) = (-1)^(n+k)*A110813(n,k) assuming the same offset in both triangles. - R. J. Mathar, Mar 15 2013
From Paul Curtz, Jun 12 2019: (Start)
Numerators of the triangle [Curtz, page 15, triangle (E)]:
     1/2;
     3/4,  1/4;
     7/8,  4/8,    1/8;
   15/16, 11/16,  5/16,  1/16;
   31/32, 26/31, 16/32,  6/32, 1/32;
   63/64, 57/64, 42/64, 22/64, 7/64, 1/64;
   ...
Denominators - Numerators: Triangle A054143.
  1;
  1, 3;
  1, 4, 7;
  1, 5, 11, 15;
  ...
(E) is a transform which accelerates the convergence of series.
For log(2) = 1 - 1/2 + 1/3 - 1/4 ... = 0.6931..., we have
  1*(1/2) = 1/2,
  1*(3/4) - (1/2)*(1/4) = 5/8,
  1*(7/8) - (1/2)*(4/8) + (1/3)*(1/8) = 2/3,
  1*(15/16) - (1/2)*(11/16) + (1/3)*(5/16) - (1/4)*1/16 = 131/192,
  ...
This is A068566/A068565. (End)

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
   1;
   3,  1;
   7,  4,  1;
  15, 11,  5,  1;
  31, 26, 16,  6,  1;
  63, 57, 42, 22,  7,  1;
  ...

Paul Curtz, Accélération de la convergence de certaines séries alternées à l'aide des fonctions de sommation, Thèse de 3ème Cycle d'Analyse Numérique, Faculté des Sciences de l'Université de Paris, 4 mai 1965.
Clark Kimberling, Fusion, Fission, and Factors, Fib. Q., 52(3) (2014), 195-202.

Cf. A000124, A000295, A001787 (row sums), A007318, A054143, A055248, A068565, A068566, A104709, A140513.

A104709_row := proc(n) add(add(binomial(n,n-i)*x^(n-k-1),i=0..k),k=0..n-1);
coeffs(sort(%)) end; seq(print(A104709_row(n)),n=1..6); # Peter Luschny, Sep 29 2011

z = 10;
p[n_, x_] := (x + 1)^n;
q[0, x_] := 1; q[n_, x_] := x*q[n - 1, x] + 1;
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]] (* A054143 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]] (* A104709 *)
(* Clark Kimberling, Aug 07 2011 *)

Begin with A055248 as a triangle, delete leftmost column.
The Riordan array factors as (1/(1-2*x), x)*(1/(1-x), x/(1-x)) - the sequence array for 2^n times Pascal's triangle. - Paul Barry, Aug 05 2005
T(n,k) = Sum_{j=0..n-k} C(n-j, k)*2^j. - Paul Barry, Jan 12 2006
T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 2*T(n-2,k) - 2*T(n-2,k-1), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 30 2013
Working with an offset of 0, we have exp(x) * (e.g.f. for row n) = (e.g.f. for diagonal n). For example, for n = 3 we have exp(x)*(15 + 11*x + 5*x^2/2! + x^3/3!) = 15 + 26*x + 42*x^2/2! + 64*x^3/3! + 93*x^4/4! + .... The same property holds more generally for Riordan arrays of the form (f(x), x/(1 - x)). - Peter Bala, Dec 21 2014
From Petros Hadjicostas, Jun 05 2020: (Start)
Bivariate o.g.f.: A(x,y) = Sum_{n,k >= 0} T(n,k)*x^n*y^k = 1/(1 - 3*x - x*y + 2*x^2 + 2*x^2*y) = 1/((1 - 2*x)*(1 - x*(y+1))).
The o.g.f. of the n-th row is (2^(n+1) - (1 + y)^(n+1))/(1 - y).
Let B(x,y) be the bivariate o.g.f. of triangular array A054143. Because A054143 is the mirror image of the current array, we have A(x,y) = B(x*y, 1/y) and B(x,y) = A(x*y, 1/y). This makes it easy to identify lower diagonals of the array.
For example, if we want to identify the second lower diagonal of the array (i.e., 7, 11, 16, 22, ...), we take the 2nd derivative of B(x,y) with respect to y, set y = 0, and divide by 2!. (Note that columns in A054143 start at k = 0.) We get the g.f. x^2*(7 - 10*x + 4*x^2)/(1 - x)^3.
It is then easy to derive that T(n,n-2) = A000124(n+1) = (n+1)*(n+2)/2 + 1 for n >= 2 (by ignoring the first three terms of A000124). Of course, in the current case, it is much easier to use the formula for T(n,k) to find T(n,n-2). (End)
T(n,0) = 2^(n+1) - 1 for n >= 0; T(n,k) = T(n-1,k) + T(n-1,k-1) for 1 <= k <= n. - Peter Bala, Jan 30 2023
T(n,1) = 2^(n+1) - n - 2 = A000295(n+1) for n >= 1. - Bernard Schott, Feb 22 2023

Name edited and offset changed by Petros Hadjicostas, Jun 04 2020

cp's OEIS Frontend

A145656 a(n) = numerator of polynomial of genus 1 and level n for m = 2.

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A104709 Triangle read by rows: T(n,k) = Sum_{j=0..n} 2^(n-j)binomial(j,k) for n >= 0 and 0 <= k <= n; also, Riordan array (1/((1-x)(1-2*x)), x/(1-x)).

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A068565 Denominators of Sum_{k=1..n} 1/(k * 2^k).

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cp's OEIS Frontend

A145656 a(n) = numerator of polynomial of genus 1 and level n for m = 2.

Views

Author

Keywords

Comments

Crossrefs

Programs

Maple

Mathematica

A104709 Triangle read by rows: T(n,k) = Sum_{j=0..n} 2^(n-j)*binomial(j,k) for n >= 0 and 0 <= k <= n; also, Riordan array (1/((1-x)*(1-2*x)), x/(1-x)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A068565 Denominators of Sum_{k=1..n} 1/(k * 2^k).

Views

Author

Keywords

Comments

Crossrefs

Programs

PARI

A104709 Triangle read by rows: T(n,k) = Sum_{j=0..n} 2^(n-j)binomial(j,k) for n >= 0 and 0 <= k <= n; also, Riordan array (1/((1-x)(1-2*x)), x/(1-x)).