login
This site is supported by donations to The OEIS Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A239605 First column of A316273. Recurrence similar to series reversion giving Catalan numbers. 3
1, 1, 2, 4, 10, 24, 66, 178, 508, 1464, 4320, 12886, 38992, 119030, 366740, 1138036, 3554962, 11167292, 35259290, 111825840, 356100044, 1138107490, 3649507278, 11738028470, 37857909164, 122411024822 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

Setting second term in coefficients equal to -1 gives alternating powers of 2.

The algorithm is as follows:

Step 1. Start with a lower triangular matrix "T" with the first column equal to any number sequence. Let the columns from the second column onwards in "T" be defined by the divisor recurrence: If n >= k: Sum from i = 1 to k - 1 of T[n - i, k - 1] - T[n - i, k], else T[n,k] = 0.

Step 2. Downshift matrix "T" one step to get matrix "X".

Step 3. Calculate matrix powers X^0, X^1, X^2, X^3, X^4, X^5, ... and multiply by coefficient list c0, c1, c2, c3, c4, c5, ... to get matrix "B": B = c0*X^0 + c1*X^1 + c2*X^2 + c3*X^3 + c4*X^4 + c5*X^5 + ...

Step 4. Take first column in matrix "B" and put it into the first column of a matrix "T".

Repeat steps 1 to 4 until first column in matrix "T" has converged. This sequence a(n) is the converged first column in matrix "T", when coefficients c0, c1, c2, c3, c4, c5, ... = 1,1,1,1,1,1,...

From Mats Granvik, Jul 25 2014: (Start)

In the program, changing the line

If[n >= k, Sum[t[n - i, k - 1], {i, 1, k - 1}] - Sum[t[n - i, k], {i, 1, k - 1}], 0];

to

If[n >= k, Sum[t[n - i, k - 1], {i, 1, n - 1}] - Sum[t[n - i, k], {i, 1, n - 1}], 0];

that is, summing to (n-1) instead of (k-1), gives Catalan numbers A000108 and series reversion of x/Sum[x^n, {n, 0, nn}]. (End)

From Mats Granvik, Dec 30 2017, Feb 01 2018: (Start)

Let x be an arbitrary real or complex number and let "coeff" in the Mathematica program be the coefficients of the series expansion of the expression:

1/(1 + x*y) + (2*x*y)/(1 + x*y) - (x*y^2)/(1 + x*y) + (x^2*y^2)/(1 + x*y) at y=0.

and the output will be the series expansion of (1 + y)/(1 + (1 - x) y) at y=0.

This is equivalent to setting the coefficients "coeff" in the Mathematica program equal to:

coeff = Join[{1, x}, (-Sign[x]*Abs[x])^Range[10]]

which gives an output "cc" that is:

Join[{1}, x*(x - 1)^Range[0, Length[coeff] - 2]]

for all values of "x" (with the special case 0^0=1 for x=1).

(End)

LINKS

Table of n, a(n) for n=1..26.

MATHEMATICA

Clear[t, n, k, i, nn, x];

coeff = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};

mp[m_, e_] :=

If[e == 0, IdentityMatrix@Length@m, MatrixPower[m, e]]; nn =

Length[coeff]; cc = Range[nn]*0 + 1; Monitor[

Do[Clear[t]; t[n_, 1] := t[n, 1] = cc[[n]];

  t[n_, k_] :=

   t[n, k] =

    If[n >= k,

     Sum[t[n - i, k - 1], {i, 1, k - 1}] -

      Sum[t[n - i, k], {i, 1, k - 1}], 0];

  A4 = Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}];

  A5 = A4[[1 ;; nn - 1]]; A5 = Prepend[A5, ConstantArray[0, nn]];

  cc = Total[

    Table[coeff[[n]]*mp[A5, n - 1][[All, 1]], {n, 1, nn}]]; , {i, 1,

   nn}], i]; cc

(* Mats Granvik, Aug 26 2015 *)

CROSSREFS

Cf. A051731, A000079.

Sequence in context: A049146 A000682 A001997 * A000084 A057734 A151516

Adjacent sequences:  A239602 A239603 A239604 * A239606 A239607 A239608

KEYWORD

nonn

AUTHOR

Mats Granvik, Mar 22 2014

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified October 16 08:03 EDT 2018. Contains 316259 sequences. (Running on oeis4.)