%I #19 Mar 04 2020 20:18:32
%S 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,
%T 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,
%U 1035,648,331,152,72,31,15,6,3,1,1,2295,1457,727,347,155,73,31,15,6,3,1,1
%N 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.
%C 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).
%F 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:
%F 1, 1;
%F 1, 0, 1;
%F 1, 1, 0, 1;
%F 1, 0, 0, 0, 1;
%F 1, 1, 1, 0, 0, 1;
%F 1, 0, 0, 0, 0, 0, 1;
%F ...
%F Example: top row of Q^3 = (4, 3, 1, 1). - _Gary W. Adamson_, Jul 07 2011
%e The linear substitution sequence is:
%e 1
%e 1,2
%e 1,2,1,3
%e 1,2,1,3,1,2,1,2,4
%e 1,2,1,3,1,2,1,2,4,1,2,1,3,1,2,1,3,1,5
%e (* limiting sequence is eigenfunction of the operator *)
%e Counting each of the integers results in:
%e {1},
%e {1,1},
%e {2,1,1},
%e {4,3,1,1},
%e {9,5,3,1,1}
%t (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}]
%Y Cf. A008683.
%K tabl,nonn
%O 1,4
%A _Wouter Meeussen_ and _Paul D. Hanna_, Dec 31 2005; revised Jan 23 2006