

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
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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_{i=1..k1} T(ni, k1)  T(ni, k), otherwise 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 (n1) instead of (k1), 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 * A309508 A000084 A057734
Adjacent sequences: A239602 A239603 A239604 * A239606 A239607 A239608


KEYWORD

nonn


AUTHOR

Mats Granvik, Mar 22 2014


STATUS

approved



