A354578 Number of ways to choose a divisor of each part of the n-th composition in standard order such that no adjacent divisors are equal.
1, 1, 2, 0, 2, 1, 1, 0, 3, 1, 2, 0, 1, 1, 0, 0, 2, 2, 3, 0, 3, 1, 1, 0, 2, 1, 1, 0, 0, 0, 0, 0, 4, 1, 4, 0, 2, 2, 1, 0, 4, 2, 2, 0, 1, 1, 0, 0, 1, 2, 2, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 3, 3, 0, 5, 2, 2, 0, 5, 1, 3, 0, 1, 1, 0, 0, 3, 3, 5, 0, 3, 1, 1
Offset: 0
Examples
The terms 2^(n - 1) through 2^n - 1 sum to 2^n. As a triangle: 1 1 2 0 2 1 1 0 3 1 2 0 1 1 0 0 2 2 3 0 3 1 1 0 2 1 1 0 0 0 0 0 The a(n) compositions for selected n: n=1: n=2: n=8: n=32: n=68: n=130: ---------------------------------------------------------------------- (1) (2) (4) (6) (4,3) (6,2) (1,1) (2,2) (3,3) (2,2,3) (3,3,2) (1,1,1,1) (2,2,2) (4,1,1,1) (6,1,1) (1,1,1,1,1,1) (1,1,1,1,3) (3,3,1,1) (2,2,1,1,1) (2,2,2,1,1) (1,1,1,1,1,1,2)
Crossrefs
First column is 1 followed by A000005.
Row-sums are A011782.
Standard compositions are listed by A066099.
Positions of 0's are A354904.
Positions of first appearances are A354905.
A005811 counts runs in binary expansion.
A353860 counts collapsible compositions.
A353863 counts run-sum-complete partitions.
Programs
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Mathematica
stc[n_]:=Differences[Prepend[Join@@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse; antirunQ[y_]:=Length[Split[y]]==Length[y]; Table[Length[Select[Tuples[Divisors/@stc[n]],antirunQ]],{n,0,30}]
Comments