A112682 Triangle read by rows: T(n,k) counts the occurrences of integer k in the sequence generated by replacing integer i with the sorted sequence of divisors of (i+1), starting on 1 and iterating n times.

1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 9, 5, 3, 1, 1, 19, 13, 6, 3, 1, 1, 43, 26, 14, 6, 3, 1, 1, 94, 61, 29, 15, 6, 3, 1, 1, 210, 130, 68, 30, 15, 6, 3, 1, 1, 464, 297, 146, 71, 31, 15, 6, 3, 1, 1, 1035, 648, 331, 152, 72, 31, 15, 6, 3, 1, 1, 2295, 1457, 727, 347, 155, 73, 31, 15, 6, 3, 1, 1

Offset: 1

Author

Wouter Meeussen and Paul D. Hanna, Dec 31 2005; revised Jan 23 2006

Keywords

• tabl
• nonn

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).

Examples

			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}
		

Crossrefs

Cf. A008683.

Programs

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}]

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