A068565 -id:A068565 - cp's OEIS frontend

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)).

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

A068566 Numerator of Sum_{k=1..n} 1/(k * 2^k).

Original entry on oeis.org

1, 5, 2, 131, 661, 1327, 1163, 148969, 447047, 44711, 983705, 7869871, 102309709, 204620705, 31972079, 32739453941, 556571077357, 556571247527, 10574855234543, 42299423848079, 42299425233749, 84598851790183

Offset: 1

Benoit Cloitre, Mar 25 2002

Sum_{k>=1} 1/(k * 2^k) = log(2).
From Paul Curtz, Jun 11 2019: (Start)
(Link) page 9:
   T0 =           1/2   =    1/2
   T1 =   1/2   + 1/8   =    5/8
   T2 =   5/8   + 1/24  =    2/3
   T3 =   2/3   + 1/64  =  131/192
   T4 = 131/192 + 1/160 =  661/960
  (T5 = 661/960 + 1/384 = 1327/1920)
... .
a(n)/A068565(n) is the first and the third column.
The denominators of the second column are essentially A036289, A097064 and A134401. (End)

Robert Israel, Table of n, a(n) for n = 1..1228
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.

Cf. A068565.

Cf. A036289, A097064, A134401.

List([1..30], n-> NumeratorRat( Sum([1..n], k-> 1/(2^k*k)) ) ) # G. C. Greubel, Jun 30 2019

[Numerator( (&+[1/(2^k*k): k in [1..n]]) ): n in [1..30]]; // G. C. Greubel, Jun 30 2019

map(numer, ListTools:-PartialSums([seq(1/k/2^k,k=1..100)])); # Robert Israel, Jul 10 2015

Numerator[Accumulate[Table[1/(k 2^k),{k,30}]]] (* Harvey P. Dale, May 11 2013 *)
a[n_]:=Log[2]-Hypergeometric2F1[1+n,1+n,2+n,-1]/(1+n);
Numerator[Table[Simplify[a[n]],{n,1,30}]] (* Gerry Martens, Aug 06 2015 *)

vector(30, n, numerator(sum(k=1,n, 1/(k * 2^k)))) \\ Michel Marcus, Aug 07 2015

[numerator( sum(1/(2^k*k) for k in (1..n)) ) for n in (1..30)] # G. C. Greubel, Jun 30 2019

From Peter Bala, Feb 05 2024: (Start)
Integral_{x = 0..1} x^n/(1 + x)^(n+1) dx = log(2) - Sum_{k = 1..n} 1/(k * 2^k).
Hence a(n) = the numerator of Integral_{x = 0..1} ((1 + x)^n - x^n)/(1 + x)^(n+1) dx.
Integral_{x = 0..1/2} x^n/(1 - x) dx = Integral_{x >= 2} 1/(x^(n+2) - x^(n+1)) dx = log(2) - a(n)/A068565(n). (End)

cp's OEIS Frontend

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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A068566 Numerator of Sum_{k=1..n} 1/(k * 2^k).

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cp's OEIS Frontend

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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Author

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Examples

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Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A068566 Numerator of Sum_{k=1..n} 1/(k * 2^k).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

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)).