

A358151


Earliest infinite sequence of distinct integers on a square spiral such that every number equals the sum of its eight adjacent neighbors. See the Comments.


4



0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 6, 11, 5, 6, 20, 15, 8, 7, 17, 12, 9, 18, 32, 21, 13, 8, 16, 38, 14, 30, 7, 11, 37, 57, 60, 23, 9, 10, 24, 34, 24, 60, 10, 13, 31, 72, 109, 82, 20, 12, 14, 108, 182, 142, 28, 188, 15, 16, 160, 168, 82, 67, 128, 120, 21, 22, 43, 22
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OFFSET

0,4


COMMENTS

We define earliest as the number with the smallest magnitude, and where a positive number is considered earlier than its negative value.
Numerous conditions must be met when choosing the next earliest available number so the sequence is infinite. The main consideration is that the first two numbers chosen in any new row or column totally determine the remaining values in that row/column. Usually these two terms will be the earliest unused numbers, although sometimes that is not possible. See the examples below and the attached text file explaining the selection criteria. Once these two numbers are chosen the other values tend to be larger in magnitude at the start and end of the row/column than in its middle, although all of these values are generally much much larger than the initial two terms. See the first linked image.
The values that are positive and negative make a pattern around the spiral axes; see the second linked image. There is no occurrence in the first 250000 terms where a positive number is surrounded by eight negative numbers, or vice versa. It is unknown is this can occur.
The sequence is conjectured to be a permutation of the integers. The values grow rapidly in size, e.g., in the first 1000 terms the largest magnitude number is a(899) = 3585008350, while the largest magnitude number in the first 100000 terms is a(99854) = 128...904, a number containing 103 digits.
The author thanks Eric Angelini whose sequence A358254 was the inspiration for this one.


LINKS



EXAMPLE

The square spiral begins:
. .
.
8...15...20...6....5 30
  
7 2....2...1 11 14
    
17 3 0....1 6 38
   
12 3....4...4...5 16
 
9...18...32...21...13...8
.
.
See the attached text file for a larger 31 by 31 example.
a(0) = 0. The earliest available integer.
a(1)..a(8) = 1,1,2,2,3,3,4,4. These are the earliest available eight numbers that sum to a(0) = 0 as required.
a(9) = 5. Despite this being a term where a seemingly free choice can be made, the earliest available number, 5, cannot be chosen; it is not immediately obvious as to why 5 fails since the addition of this number does not complete a new 3 by 3 block of numbers. See the attached text file for an explanation.
a(10) = 6. Given that a(9) = 5 the next number cannot be either 5 or 6 since those choices would force a(11) to equal 0 or 1 so that the eight numbers surrounding a(1) = 1 would sum to 1. But both 0 and 1 have already appeared thus a(10) cannot be 5 or 6.
a(11) = 11. Given a(9) = 5 and a(10) = 6, the seven terms around a(1) = 1 currently sum to 2  1 + 0 + 4  4  5  6 = 10, thus a(11) = 11 so that 11  10 = 1.
a(14) = 20. The earliest available numbers, 5 and 6, were able to be chosen for the start of this row, so the current sum of numbers around a(2) = 1 is 6 + 5 + 2 + 11 + 0 + 1  6 = 19. Therefore a(14) = 20 so that 20 + 19 = 1.


CROSSREFS



KEYWORD

sign


AUTHOR



STATUS

approved



