The original name was: "The sequence is an antidiagonal of the decimal of a mapped 4ary Gray code matrix as a triangular sequence."
Gary W. Adamson's explanation of the sequence: Here's the conversion rules for the codons, 4Ary gray code, which "turns out" to be the most appropriate format for mapping the Codons on a gray code Karnaugh map. The "why" this is the appropriate format relates to a degree of trial and error to find the proper fit in terms of the numbers of hydrogen bonds per codon anticodon. (Antti Karttunen's comment: obscure definition. The "degree of trial and error" should be defined transparently.)
1) The "Hbond codonanticodon magic square" map by Gary Adamson, published on page 287 of Cliff Pickover's book "Zen of Magic Squares..." looks like this:
CCC CCU CUU CUC UUC UUU UCU UCC
CCA CCG CUG CUA UUA UUG UCG UCA
CAA CAG CGG CGA UGA UGG UAG UAA
CAC CAU CGU CGC UGC UGU UAU UAC
AAC AAU AGU AGC GGC GGU GAU GAC
AAA AAG AGG AGA GGA GGG GAG GAA
ACA ACG AUG AUA GUA GUG GCG GCA
ACC ACU AUU AUC GUC GUU GCU GCC
2) Using the conversion rules: 0 = C, 1 = A, 2 = G, 3 = U, we convert to 4ary gray code:
000 003 033 030 330 333 303 300
001 002 032 031 331 332 302 301
011 012 022 021 321 322 312 311
010 013 023 020 320 323 313 310
110 113 123 120 220 223 213 210
111 112 122 121 221 222 212 211
101 102 132 131 231 232 202 201
100 103 133 130 230 233 203 200
3) To convert back to decimal:
+0 +3 14 15 58 57 62 63
+1 +2 13 12 59 56 61 60
+6 +7 +8 11 54 55 50 49
+5 +4 +9 10 53 52 51 48
26 25 30 31 32 35 46 47
27 24 29 28 33 34 45 44
22 23 18 17 38 39 40 43
21 20 19 16 37 36 41 42
... and that's it! Notice how the 1,2,3... jumps around, somewhat like a Peano curve, from one 4unit cell to the next.
Antti Karttunen's notes: The steps 1 & 2 are clear, but the step 3 would not produce the array given here, but instead the array A163239. Furthermore, in Pickover's book the conversion rules C=0, A=1, U=2, G=3 are used, in which case we get the array A163235. Also, the path taken by the terms does not form a continuous Peano curve (Hamiltonian path), because there are discontinuities, e.g. when going from 3 to 4, or from 15 to 16. See A163357/A163359 & A163334/A163336 for examples of continuous Peano/Hilbert curves/paths in an N x N grid. However, this sequence is uniquely defined by the formula a(n) = A163485(A057300(A054238(n))). The 8 X 8 array given at the step 3 is the top left corner of the infinite square array whose antidiagonal gives this sequence.
From Gary W. Adamson, Aug 04 2009: (Start)
This entry was originally only an e mail to the coauthor; but given that the terms are correct, the complete set of rules for the system can be presented.
Using 3 bit terms, we write out the Gray code for (0  7) as row headings; doing the same as the left column, then each of the 64 entries places the left column term (of 3 bits) underneath the top row headings. Then reading 2 bits from top to down in each entry, we use (0,0) = C; (1,1) = G; (0,1) = A and (1,0) = U. This gives the Gray code Karnaugh map along with 64 codons:
.
000...001...011...010...110...111...101...100
000...000...000...000...000...000...000...000
CCC...CCU...CUU...CUC...UUC...UUU...UCU...UCC
000...001...011...010...110...111...101...100
001...001...001...001...001...001...001...001
CCA...CCG...CUG...CUA...UUA...UUG...UCG...UCA
000...001...011...010...110...111...101...100
011...011...011...011...011...011...011...011
CAA...CAG...CGG...CGA...UGA...UGG...UAG...UAA
000...001...011...010...110...111...101...100
010...010...010...010...010...010...010...010
CAC...CAU...CGU...CGC...UGC...UGU...UAU...UAC
000...001...011...010...110...111...101...100
110...110...110...110...110...110...110...110
AAC...AAU...AGU...AGC...GGC...GGU...GAU...GAC
000...001...011...010...110...111...101...100
111...111...111...111...111...111...111...111
AAA...AAG...AGG...AGA...GGA...GGG...GAG...GAA
000...001...011...010...110...111...101...100
101...101...101...101...101...101...101...101
ACA...ACG...AUG...AUA...GUA...GUG...GCG...GCA
000...001...011...010...110...111...101...100
100...100...100...100...100...100...100...100
ACC...ACU...AUU...AUC...GUC...GUU...GCU...GCC
.
Next, reading again from top 3 bits to bottom, we convert the base 2 Gray code to 4ary Gray code using the rules (0,0) = 0; (0,1) = 1; (1,1) = 2; and (1,0) = 3; giving the array given using numbers (0,1,2, and 3) = 4ary Gray code. The previous 2 maps have the unique Gray code property of having only a 1 bit (or 1 letter) change in any direction: up, down, right, left, including wraparounds.
Last part of this system, we need create a linear system of Codons with only 1 bit (letter) change from one term to the next, giving an ordered decimal term for each Codon. This is done by converting the array with the (0,1,2,3) terms to the corresponding decimal term. Thus Given the array: 000...003...033...030...330...333...etc; considered as 4ary Gray code, these terms are equivalent to the array A147995 (then take antidiagonals).
Following the numbers in succession in the array (0 > 1 > 2 >...63) allows for us to have a linear system of Codons with only a 1 letter change from one Codon to the next, as follows: CCC > CCA > CCG > CAU...> through 63 = UCC. The other entries as of this date in the OEIS do not have the 1letter (only) change from one associated decimal term to the next. For example, take entry A163235: If the decimal number system (given) is superimposed upon the 64 Codon array, the term 3 corresponds to CCG, but 4 in the left column corresponds to CAC, having a 2 letter change. Similarly, take A163239: If the decimal array in that entry is superimposed on the 64 Codon array, "3" corresponds in position to CCU, but "4" corresponds to CAC; again a 2 letter change. The system given in A147995 preserves the unique 1 (bit/letter) change from one Codon to any neighbor, going in any direction; along with the corresponding linear system with a 1 letter change from one Codon to the next.
Last, we submit for each Codon the number of hydrogen bonds per codon/anti codon using the following substitution rules: (C,G) = 3; (A,U) = 2, then add.
This gives following array which we superimpose on the Codon array, giving the correct number of Hydrogen bonds for each Codon and antiCodon:
.
9 8 7 8 7 6 7 8
8 9 8 7 6 7 8 7
7 8 9 8 7 8 7 6
8 7 8 9 8 7 6 7
7 6 7 8 9 8 7 8
6 7 8 7 8 9 8 7
6 8 7 6 7 8 9 8
8 7 6 7 8 7 8 9
...(a semimagic square with a binomial distribution of (1, 3, 3, 1) as to (6, 7, 8, 9) in every row and column.
Example: CUG (3rd from left, row next to top) has (C=3, U=2, G=3), total 8.
The antiCodon of CUG = GAC and likewise has 8 hydrogen bonds. (End)
From Gary W. Adamson, Aug 04 2009: (Start)
The final outcome: superimposing the Codon map onto the decimal term map, we obtain a linear sequence of Codons with a 1 letter change between neighbors (which begs the question of how many such permutations are possible with the 1 letter change). The method of A147995 gives:
.
0...CCC;.....16...AUC;.....32...GGC;.....48...UAC
1...CCA;.....17...AUA;.....33...GGA;.....49...UAA
2...CCG;.....18...AUG;.....34...GGG;.....50...UAG
3...CCU;.....19...AUU;.....35...GGU;.....51...UAU
4...CAU;.....20...ACU;.....36...GUU;.....52...UGU
5...CAC;.....21...ACC;.....37...GUC;.....53...UGC
6...CAA;.....22...ACA;.....38...GUA;.....54...UGA
7...CAG;.....23...ACG;.....39...GUG;.....55...UGG
8...CGG;.....24...AAG;.....40...GCG;.....56...UUG
9...CGU;.....25...AAU;.....41...GCU;.....57...UUU
10..CGC;.....26...AAC;.....42...GCC;.....58...UUC
11..CGA;.....27...AAA;.....43...GCA;.....59...UUA
12..CUA;.....28...AGA;.....44...GAA;.....60...UCA
13..CUG;.....29...AGG;.....45...GAG;.....61...UCG
14..CUU;.....30...AGU;.....46...GAU;.....62...UCU
15..CUC;.....31...AGC;.....47...GAC;.....63...UCC
(End)
From Gary W. Adamson, Aug 08 2009: (Start)
The 8 X 8 array of hydrogen bonds can be derived from the 3rd row of A088696 (1, 2, 3, 2, 3, 4, 3, 2) using a simple conversion rule. Given the terms of A088696, each is replaced with its complement to 10: (1>9; 2>8; 3>7; 4>6) Note that the leftmost column going down should read: (9, 8, 7, 8, 7, 6, 7, 8) matching the top row from left to right. (End)
From Gary W. Adamson, Aug 13 2009: (Start)
Gray code > < Binary conversion rules: in either direction for any base; "NAry Gray code" > "Nary" or in the other direction.
.
First, NAry Gray code to NAry conversion. Write the NAry on a top row with the Gray code on the bottom row in both conversion variants. Given a Gray code on the bottom row, the NAry may be defined as "running sums MOD N" of the bottom row; then use the following rules: Leftmost term is the same.
Next, use the sum of term (nth) in the top row from the left, and the (n+1)th term in the bottom row, MOD N. By way of example:
Convert Gray code base 8, 3641063 to 8ary. This gives initially,
3..................
3..6..4..1..0..6..3
.
Then (3 + 6) MOD 8 = 1 so we place a "1" above the 6 going to the right.
Then (1 + 4) MOD 8 = 5 so we place a "5" above the 5.
Continuing with this procedure, we obtain:
3 1 5 6 6 4 7 8Ary
3 6 4 1 0 6 3 8Ary Gray code
.
Using the 8x8 4Ary chart, convert 133 (bottom row, 4th from the left) to 4Ary then to decimal. Our setup is:
1
1 3 3
getting (1, 0, 3). Then placing powers of 4 above the 4Ary, = 1*16 + 3 = 19 as shown in the accompanying chart, 4Ary Gray code 133 = 19 decimal.
.
Rules for converting an NAry number to the corresponding NAry Gray code:
As before, we place the NAry on the top row with ongoing results on the bottom row = NAry Gray code.
In the top row from left to right, through through the entire number looking at pairs (nth and (n+1)th terms), if (n+1)th is > than nth, take the difference and write it down. If term (n+1) = nth term, write down a "0".
If term (n+1) < nth term we ADD N (as NAry) to (n+1)th term then take the difference. Examples:
Find the Gray code counterpart to 2 1 base 4 = 9 decimal.
Ans.: next term (1) < (2) so we add 4 to the 1 getting 5, then take (5  2) = 3. So given 4Ary 21, the corresponding Gray code term = 23
.
Find the Gray code counterpart to Binary 10110 = 22 decimal. First, go through the terms writing down the difference if next term > current: (and writing "0" if next term = current term)
1, 0, 1, 1, 0
1.....1..0...
Add "2" to the terms above the vacant places and take the difference from previous term, top row:
1, 1, 1, 0, 1 final result = Gray code for 22 decimal.
.
Given 8Ary number 3156647, base 8. Using steps (12) we get
3, 1, 5, 6, 6, 4, 7
3.....4..1..0.....3; then add 8 to top term for vacant places then take the difference, getting:
3..6..4..1..0..6..3; = 8ary Gray code given 8Ary (3 1 5 6 6 4 7).
.
Given the foregoing rules and examples, access the charts accompanying the DNA codons. 3 digit terms = 4Ary Gray code. Convert 133 (bottom row) to 4Ary then to decimal. We get:
1
1 0 3 = (16 + 0 + 3) = 19
Convert 39 decimal to 4Ary then to 4Ary Gray code. 39 = 213 4Ary = (2*16 + 4 + 3); then
2 1 3
2...2; then add "4" to the 1 and take the difference = (5  2) = 3. = 2 3 2 = 4Ary Gray code for decimal 39 as shown in the dual charts, next to bottom row, third from the right: (232 corresponds to 39) in the accompanying chart.
Properties of Gray code: sum of terms MOD N = decimal MOD N. Example: 232 corresponds to 19, then (2 + 3 + 2) MOD 4 = 3, and 19 == 3 MOD 4.
Another property: Highest exponent of N dividing a decimal term.
Access term (n1) writing the Gray code on the top row and Gray code for nth term on the bottom. Determine column change = (0, 1, 2,...) starting from the right. Let the column = c. then c is the highest exponent of N dividing nth term. Examples: 40 in 4Ary Gray code = 202, while 41 = 203. Change is in column 0 so 203 can be divided by 4^0. But 44 in 4Ary Gray code = 211 while 43 = 201. Bit change is in column 1 so 4^1 divides 44. (End)
