Revisions to A special type of Langford pairing

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edited Nov 26, 2019 at 4:10

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The answer to the first question seems to be yes: there are such Langford pairings for all suitable $n$. The number of them is given by the following sequence, not yet in the OEIS: $$0,0,1,1,0,0,3,10,0,0,76,140,0,0,\dots$$

$$0,0,1,1,0,0,3,10,0,0,76,140,0,0,2478,5454,0,0\dots$$

The 10 sequences for $n=8$ are the following:

5 8 4 1 7 1 5 4 6 3 8 2 7 3 2 6
3 6 2 7 3 2 8 5 6 4 1 7 1 5 4 8
5 8 2 3 7 2 5 3 6 4 8 1 7 1 4 6
4 6 1 7 1 4 8 5 6 2 3 7 2 5 3 8
4 2 5 7 2 4 8 6 5 3 1 7 1 3 6 8
5 2 4 7 2 8 5 4 6 3 1 7 1 3 8 6
3 1 7 1 3 8 6 4 2 5 7 2 4 6 8 5
3 1 7 1 3 8 4 5 6 2 7 4 2 5 8 6
3 1 7 1 3 5 8 6 4 2 7 5 2 4 6 8
3 1 7 1 3 6 8 5 2 4 7 2 6 5 4 8

The answer to the first question seems to be yes: there are such Langford pairings for all suitable $n$. The number of them is given by the following sequence, not yet in the OEIS: $$0,0,1,1,0,0,3,10,0,0,76,140,0,0,\dots$$

The 10 sequences for $n=8$ are the following:

5 8 4 1 7 1 5 4 6 3 8 2 7 3 2 6
3 6 2 7 3 2 8 5 6 4 1 7 1 5 4 8
5 8 2 3 7 2 5 3 6 4 8 1 7 1 4 6
4 6 1 7 1 4 8 5 6 2 3 7 2 5 3 8
4 2 5 7 2 4 8 6 5 3 1 7 1 3 6 8
5 2 4 7 2 8 5 4 6 3 1 7 1 3 8 6
3 1 7 1 3 8 6 4 2 5 7 2 4 6 8 5
3 1 7 1 3 8 4 5 6 2 7 4 2 5 8 6
3 1 7 1 3 5 8 6 4 2 7 5 2 4 6 8
3 1 7 1 3 6 8 5 2 4 7 2 6 5 4 8

The answer to the first question seems to be yes: there are such Langford pairings for all suitable $n$. The number of them is given by the following sequence, not yet in the OEIS:

$$0,0,1,1,0,0,3,10,0,0,76,140,0,0,2478,5454,0,0\dots$$

The 10 sequences for $n=8$ are the following:

5 8 4 1 7 1 5 4 6 3 8 2 7 3 2 6
3 6 2 7 3 2 8 5 6 4 1 7 1 5 4 8
5 8 2 3 7 2 5 3 6 4 8 1 7 1 4 6
4 6 1 7 1 4 8 5 6 2 3 7 2 5 3 8
4 2 5 7 2 4 8 6 5 3 1 7 1 3 6 8
5 2 4 7 2 8 5 4 6 3 1 7 1 3 8 6
3 1 7 1 3 8 6 4 2 5 7 2 4 6 8 5
3 1 7 1 3 8 4 5 6 2 7 4 2 5 8 6
3 1 7 1 3 5 8 6 4 2 7 5 2 4 6 8
3 1 7 1 3 6 8 5 2 4 7 2 6 5 4 8

added 2 characters in body

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edited Nov 24, 2019 at 0:44

Freddy Barrera

588
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The answer to the first question seems to be yes: there are such Langford pairings for all suitable $n$. The number of them is given by the following sequence, not yet in the OEIS: $$0,0,1,1,0,0,3,10,0,0,76,140,0,0 ...$$$$0,0,1,1,0,0,3,10,0,0,76,140,0,0,\dots$$

The 10 sequences for $n=8$ are the following:

5 8 4 1 7 1 5 4 6 3 8 2 7 3 2 6
3 6 2 7 3 2 8 5 6 4 1 7 1 5 4 8
5 8 2 3 7 2 5 3 6 4 8 1 7 1 4 6
4 6 1 7 1 4 8 5 6 2 3 7 2 5 3 8
4 2 5 7 2 4 8 6 5 3 1 7 1 3 6 8
5 2 4 7 2 8 5 4 6 3 1 7 1 3 8 6
3 1 7 1 3 8 6 4 2 5 7 2 4 6 8 5
3 1 7 1 3 8 4 5 6 2 7 4 2 5 8 6
3 1 7 1 3 5 8 6 4 2 7 5 2 4 6 8
3 1 7 1 3 6 8 5 2 4 7 2 6 5 4 8

The answer to the first question seems to be yes: there are such Langford pairings for all suitable $n$. The number of them is given by the following sequence, not yet in the OEIS: $$0,0,1,1,0,0,3,10,0,0,76,140,0,0 ...$$

The 10 sequences for $n=8$ are the following:

5 8 4 1 7 1 5 4 6 3 8 2 7 3 2 6
3 6 2 7 3 2 8 5 6 4 1 7 1 5 4 8
5 8 2 3 7 2 5 3 6 4 8 1 7 1 4 6
4 6 1 7 1 4 8 5 6 2 3 7 2 5 3 8
4 2 5 7 2 4 8 6 5 3 1 7 1 3 6 8
5 2 4 7 2 8 5 4 6 3 1 7 1 3 8 6
3 1 7 1 3 8 6 4 2 5 7 2 4 6 8 5
3 1 7 1 3 8 4 5 6 2 7 4 2 5 8 6
3 1 7 1 3 5 8 6 4 2 7 5 2 4 6 8
3 1 7 1 3 6 8 5 2 4 7 2 6 5 4 8

The answer to the first question seems to be yes: there are such Langford pairings for all suitable $n$. The number of them is given by the following sequence, not yet in the OEIS: $$0,0,1,1,0,0,3,10,0,0,76,140,0,0,\dots$$

The 10 sequences for $n=8$ are the following:

5 8 4 1 7 1 5 4 6 3 8 2 7 3 2 6
3 6 2 7 3 2 8 5 6 4 1 7 1 5 4 8
5 8 2 3 7 2 5 3 6 4 8 1 7 1 4 6
4 6 1 7 1 4 8 5 6 2 3 7 2 5 3 8
4 2 5 7 2 4 8 6 5 3 1 7 1 3 6 8
5 2 4 7 2 8 5 4 6 3 1 7 1 3 8 6
3 1 7 1 3 8 6 4 2 5 7 2 4 6 8 5
3 1 7 1 3 8 4 5 6 2 7 4 2 5 8 6
3 1 7 1 3 5 8 6 4 2 7 5 2 4 6 8
3 1 7 1 3 6 8 5 2 4 7 2 6 5 4 8

added 5 characters in body

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edited Nov 24, 2019 at 0:37

Bernardo Recamán Santos

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The answer to the first question seems to be yes: there are such Langford pairings for all suitable $n$. The number of them is given by the following sequence, not yet in the OEIS: $$0,0,1,1,0,0,3,10,0,0,76,140,0,0$$$$0,0,1,1,0,0,3,10,0,0,76,140,0,0 ...$$

The 10 sequences for $n=8$ are the following:

5 8 4 1 7 1 5 4 6 3 8 2 7 3 2 6
3 6 2 7 3 2 8 5 6 4 1 7 1 5 4 8
5 8 2 3 7 2 5 3 6 4 8 1 7 1 4 6
4 6 1 7 1 4 8 5 6 2 3 7 2 5 3 8
4 2 5 7 2 4 8 6 5 3 1 7 1 3 6 8
5 2 4 7 2 8 5 4 6 3 1 7 1 3 8 6
3 1 7 1 3 8 6 4 2 5 7 2 4 6 8 5
3 1 7 1 3 8 4 5 6 2 7 4 2 5 8 6
3 1 7 1 3 5 8 6 4 2 7 5 2 4 6 8
3 1 7 1 3 6 8 5 2 4 7 2 6 5 4 8

The answer to the first question seems to be yes: there are such Langford pairings for all suitable $n$. The number of them is given by the following sequence, not yet in the OEIS: $$0,0,1,1,0,0,3,10,0,0,76,140,0,0$$

The 10 sequences for $n=8$ are the following:

5 8 4 1 7 1 5 4 6 3 8 2 7 3 2 6
3 6 2 7 3 2 8 5 6 4 1 7 1 5 4 8
5 8 2 3 7 2 5 3 6 4 8 1 7 1 4 6
4 6 1 7 1 4 8 5 6 2 3 7 2 5 3 8
4 2 5 7 2 4 8 6 5 3 1 7 1 3 6 8
5 2 4 7 2 8 5 4 6 3 1 7 1 3 8 6
3 1 7 1 3 8 6 4 2 5 7 2 4 6 8 5
3 1 7 1 3 8 4 5 6 2 7 4 2 5 8 6
3 1 7 1 3 5 8 6 4 2 7 5 2 4 6 8
3 1 7 1 3 6 8 5 2 4 7 2 6 5 4 8

The answer to the first question seems to be yes: there are such Langford pairings for all suitable $n$. The number of them is given by the following sequence, not yet in the OEIS: $$0,0,1,1,0,0,3,10,0,0,76,140,0,0 ...$$

The 10 sequences for $n=8$ are the following:

5 8 4 1 7 1 5 4 6 3 8 2 7 3 2 6
3 6 2 7 3 2 8 5 6 4 1 7 1 5 4 8
5 8 2 3 7 2 5 3 6 4 8 1 7 1 4 6
4 6 1 7 1 4 8 5 6 2 3 7 2 5 3 8
4 2 5 7 2 4 8 6 5 3 1 7 1 3 6 8
5 2 4 7 2 8 5 4 6 3 1 7 1 3 8 6
3 1 7 1 3 8 6 4 2 5 7 2 4 6 8 5
3 1 7 1 3 8 4 5 6 2 7 4 2 5 8 6
3 1 7 1 3 5 8 6 4 2 7 5 2 4 6 8
3 1 7 1 3 6 8 5 2 4 7 2 6 5 4 8

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created Nov 24, 2019 at 0:30

Freddy Barrera

588
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3
11

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