OFFSET
1,4
COMMENTS
T(n+1,1) = sum of previous row (each integer produces the trivial divisor 1). The matrix inverse M (after padding to the right with zeros) consists of columns that, from the main diagonal down, are nested Shift-Moebius Transforms of {1,0,0,..0} (see Mathematica code).
FORMULA
n-th row of the triangle = top row terms in (n-1)-th power of the production matrix Q, where Q = the inverse Mobius transform with the first "1" deleted:
1, 1;
1, 0, 1;
1, 1, 0, 1;
1, 0, 0, 0, 1;
1, 1, 1, 0, 0, 1;
1, 0, 0, 0, 0, 0, 1;
...
Example: top row of Q^3 = (4, 3, 1, 1). - Gary W. Adamson, Jul 07 2011
EXAMPLE
The linear substitution sequence is:
1
1,2
1,2,1,3
1,2,1,3,1,2,1,2,4
1,2,1,3,1,2,1,2,4,1,2,1,3,1,2,1,3,1,5
(* limiting sequence is eigenfunction of the operator *)
Counting each of the integers results in:
{1},
{1,1},
{2,1,1},
{4,3,1,1},
{9,5,3,1,1}
MATHEMATICA
(Length/@ Split[Sort[ # ]])&/@ NestList[Flatten[ #/. k_:>Divisors[1+k]]&, {1}, 12]; or, more efficiently: Nest[Apply[Plus, Map[Last, Split[Sort[Apply[Sequence, Thread[w[Divisors[1 +Range[Length[ # ]]]& @ #, List/@# ]]/. w->(Outer[Sequence, ## ]&), {1}]], First[ #1]===First[ #2]&], {2}], {1}]&, {1}, 63]; or, using a Shift-Moebius Transform: upper=MapIndexed[Drop[ #1, -1+First@#2]&, IdentityMatrix[17], {1}]; tran=Rest/@ MapIndexed[Nest[ Prepend[moebius[ # ], 0]&, #1, First@#2]&, upper]; MapIndexed[Take[ #1, First@#2]&, Transpose[Inverse[tran]], {1}]
CROSSREFS
KEYWORD
tabl,nonn
AUTHOR
Wouter Meeussen and Paul D. Hanna, Dec 31 2005; revised Jan 23 2006
STATUS
approved