OFFSET
1,1
COMMENTS
This sequence is a 2D square lattice version of the Enots Wolley sequence A336957. Like that sequence, for this sequence to be infinite no number can be a prime or prime power - the first term must therefore be 6. Likewise, for each of its orthogonal nearest neighbors, a new number must have at least one prime factor that is not a factor of that neighbor. When choosing each new number a further test must also be performed against the previous number that is two squares ahead of the current number and on the edge of the spiral - this number is outside the eight nearest neighbors of the number being determined. See the examples below.
Like A336957 is it conjectured that all nonprime powers eventually appear. In the first 10000 terms the largest value is a(6975) = 3005315.
LINKS
Scott R. Shannon, Image of the spiral up to n=10201 with even numbers shown in yellow. Zoom in to see the numbers and spiral line.
Scott R. Shannon, Image of the spiral up to n=10201 color coded with the minimum prime factor. The numbers with a minimum prime factor of 2, 3, 5, 7, 11, 13, 17, >=19 are shown as red, orange, yellow, green, blue, indigo, violet, dark grey respectively.
EXAMPLE
The spiral begins:
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105----91---539---147---385----33----55 99
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110 26----44----24----20----18 80 72
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847 273 77----21----35 63 175 189
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176 30 22 6----10 12 50 66
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1001 195 143----39----65---117 455 351
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28 36----88----42----40----48----14 54
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119----51---187---153----85---459---595----15
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a(5) = 22. The two orthogonal nearest neighbors to the square at (-1,0), relative to the starting central square 6, are 6 and 77. Therefore the new number at that position must share a factor with both of these numbers while containing a factor they do not have. Also the new number must have no factor in common with any diagonal next-nearest neighbors. Only one exists in this case, namely 21. The smallest unused number satisfying all of these conditions is 22.
a(6) = 143. For the square at (-1,-1) the orthogonal nearest neighbor is 22, while the diagonal next-nearest neighbor is 6. As 22 has 2 and 11 as factors, while 6 has 2 and 3, the new number can only be a multiple of 11, must have a factor other than 2 and 11 while not being a multiple of 2 or 3. The smallest unused number satisfying these conditions is 55. However this is where another check must be performed, namely against the number two squares ahead on the spiral from the current number - the number at (1,0) which in this case is 10. The reason this must be checked is that if 55 is chosen as the new number at (-1,-1) then that would force the next number at (0,-1) to be a multiple of 5 - only 5 and 11 are factors of 55 but 11 cannot be a factor of the new number at (0,-1) since it is an orthogonal neighbor to 22 with which it cannot share a factor. But forcing the number at (0,-1) to have 5 as a factor is not permitted since it would then share a factor with its diagonal neighbor 10 at (1,0). This is why the prime factors of the number two squares ahead on the spiral need to be checked when choosing the current number. Once a candidate number for the current square is chosen a list of the prime factors it has that are not factors of the previous number, 22 in this case, is created. This list is then compared against the factors of the number two squares ahead on the spiral, if such a number exists. If that number has as factors all the numbers in the list, then a new candidate number must be chosen else it would lead to the issue described above when the next new number is created. In this case choosing 55 leads to a list containing only 5, but that is a factor of 10, so 55 cannot be chosen. The next valid number that satisfies all the factor requirements and also has a factor not in either 22 or 10 is 11*13 = 143.
CROSSREFS
KEYWORD
nonn
AUTHOR
Scott R. Shannon, Nov 01 2022
STATUS
approved