keyword:obsc - cp's OEIS frontend

A147995 Array of N X N grid hopping "almost-walk", read by antidiagonals.

0, 1, 3, 6, 2, 14, 5, 7, 13, 15, 26, 4, 8, 12, 58, 27, 25, 9, 11, 59, 57, 22, 24, 30, 10, 54, 56, 62, 21, 23, 29, 31, 53, 55, 61, 63, 106, 20, 18, 28, 32, 52, 50, 60, 234, 107, 105, 19, 17, 33, 35, 51, 49, 235, 233, 108, 104, 100, 16, 38, 34, 46, 48, 236, 232, 228, 111

Offset: 0

Roger L. Bagula and Gary W. Adamson, Nov 18 2008

The original name was: "The sequence is an anti-diagonal of the decimal of a mapped 4-ary Gray code matrix as a triangular sequence."
Gary W. Adamson's explanation of the sequence: Here's the conversion rules for the codons, 4-Ary 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 "H-bond codon-anticodon 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 4-ary 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 4-unit 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 4-ary 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) = 4-ary 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 wrap-arounds.
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 4-ary 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 1-letter (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 anti-Codon:
.
  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 semi-magic 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 anti-Codon 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; "N-Ary Gray code" -> "N-ary" or in the other direction.
.
First, N-Ary Gray code to N-Ary conversion. Write the N-Ary 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 N-Ary 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 (n-th) 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 8-ary. 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 8-Ary
3 6 4 1 0 6 3 8-Ary Gray code
.
Using the 8 X 8 4-Ary chart, convert 133 (bottom row, 4th from the left) to 4-Ary then to decimal. Our setup is:
1
1 3 3
getting (1, 0, 3). Then placing powers of 4 above the 4-Ary, = 1*16 + 3 = 19 as shown in the accompanying chart, 4-Ary Gray code 133 = 19 decimal.
.
Rules for converting an N-Ary number to the corresponding N-Ary Gray code:
As before, we place the N-Ary on the top row with ongoing results on the bottom row = N-Ary Gray code.
In the top row from left to right, through through the entire number looking at pairs (n-th and (n+1)-th terms), if (n+1)-th is > than n-th, take the difference and write it down. If term (n+1) = n-th term, write down a "0".
If term (n+1) < n-th term we ADD N (as N-Ary) 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 4-Ary 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 8-Ary number 3156647, base 8. Using steps (1-2) 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; = 8-ary Gray code given 8-Ary (3 1 5 6 6 4 7).
.
Given the foregoing rules and examples, access the charts accompanying the DNA codons. 3 digit terms = 4-Ary Gray code. Convert 133 (bottom row) to 4-Ary then to decimal. We get:
1
1 0 3 = (16 + 0 + 3) = 19
Convert 39 decimal to 4-Ary then to 4-Ary Gray code. 39 = 213 4-Ary = (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 = 4-Ary 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 (n-1) writing the Gray code on the top row and Gray code for n-th 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 n-th term. Examples: 40 in 4-Ary Gray code = 202, while 41 = 203. Change is in column 0 so 203 can be divided by 4^0. But 44 in 4-Ary Gray code = 211 while 43 = 201. Bit change is in column 1 so 4^1 divides 44. (End)

			Antidiagonals begin:
  { 0},
  { 1,  3},
  { 6,  2, 14},
  { 5,  7, 13, 15},
  {26,  4,  8, 12, 58},
  {27, 25,  9, 11, 59, 57},
  {22, 24, 30, 10, 54, 56, 62},
  {21, 23, 29, 31, 53, 55, 61, 63}

Clifford A. Pickover, The Zen of Magic Squares, Circles, and Stars: An Exhibition of Surprising Structures across Dimensions, Princeton University Press, 2002, pp. 285-289.

A. Karttunen, Table of n, a(n) for n = 0..8255
Jay Kappraff and Gary W. Adamson, Generalized Genomic Matrices, Silver Means, & Pythagorean Triples, FORMA 2009, v24 p.41-48.
Index entries for sequences that are permutations of the natural numbers

a(n) = A163545(A061579(n)), i.e., transpose of A163545. Antidiagonal sums: A163484. Inverse: A163544. See also A163233, A163235, A163237, A163239, A163357, A163359.

Cf. A088696. - Gary W. Adamson, Aug 08 2009

M = {{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}}; Table[Table[M[[n - m + 1, m]], {m, 1, n}], {n, 1, Length[M]}]; Flatten[%]

M = {{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}}; t(n,m) = antidiagonal(M).

a(n) = A163485(A057300(A054238(n))). - Antti Karttunen, Aug 01 2009

Edited, extended, keywords tabl and obsc added and offset changed from 1 to 0 by Antti Karttunen, Aug 01 2009

A245921 Index sequence for limit-reversing the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.

Original entry on oeis.org

0, 2, 5, 7, 15, 20, 28, 36, 41, 54, 75, 96, 109, 130, 143, 164, 185, 198, 219, 240, 253, 274, 308, 329, 363, 397, 418, 452, 473, 507, 541, 562, 596, 617, 651, 685, 706, 740, 774, 795, 829, 850, 884, 918, 973, 1007, 1062, 1117, 1151, 1206, 1261, 1295, 1350

Offset: 0

Clark Kimberling and Peter J. C. Moses, Aug 07 2014

Suppose S = (s(0), s(1), s(2), ...) is an infinite sequence such that every finite block of consecutive terms occurs infinitely many times in S. (It is assumed that A014675 is such a sequence.) Let B = B(m,k) = (s(m-k), s(m-k+1),...,s(m)) be such a block, where m >= 0 and k >= 0. Let m(1) be the least i > m such that (s(i-k), s(i-k+1),...,s(i)) = B(m,k), and put B(m(1),k+1) = (s(m(1)-k-1), s(m(1)-k),...,s(m(1))). Let m(2) be the least i > m(1) such that (s(i-k-1), s(i-k),...,s(i)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)-k-2), s(m(2)-k-1),...,s(m(2))). Continuing in this manner gives a sequence of blocks B(m(n),k+n). Let B'(n) = reverse(B(m(n),k+n)), so that for n >= 1, B'(n) comes from B'(n-1) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limit-reverse of S with initial block B(m,k)", denoted by S*(m,k), or simply S*. The sequence (m(i)), where m(0) = 0, is the "index sequence for limit-reversing S with initial block B(m,k)" or simply the index sequence for S*, as in A245921.

			S = infinite Fibonacci word A014675, B = (s(0)); that is, (m,k) = (0,0);
S = (2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...)
B'(0) = (2)
B'(1) = (2,1)
B'(2) = (2,1,2)
B'(3) = (2,1,2,1)
B'(4) = (2,1,2,1,2)
B'(5) = (2,1,2,1,2,2)
S* = (2,1,2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...),
with index sequence (0,2,5,7,15,...)

Cf. A245920, A245922.

z = 100; seqPosition2[list_, seqtofind_] := Last[Last[Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 2]]] &[seqtofind] (*finds the position of the SECOND appearance of seqtofind. Example: seqPosition2[{1,2,3,4,2,3},{2}] = 5*)
A014675 = Nest[Flatten[# /. {1 -> 2, 2 -> {2, 1}}] &, {1}, 25]; ans = Join[{A014675[[p[0] = pos = seqPosition2[A014675, #] - 1]]}, #] &[{A014675[[1]]}]; cfs = Table[A014675 = Drop[A014675, pos - 1]; ans = Join[{A014675[[p[n] = pos = seqPosition2[A014675, #] - 1]]}, #] &[ans], {n, z}]; q = -1+Accumulate[Join[{1}, Table[p[n], {n, 0, z}]]] (* A245921 *)
q1 = Differences[q] (* A245922 *)

A056863 Related to triangle A056858 of number of rises in set partitions of n at a given index i.

Original entry on oeis.org

1, -1, 1, -2, 1, 1, -3, 4, 2, 1, -4, 9, 10, 4, 1, -5, 16, 28, 24, 8, 1, -6, 25, 60, 80, 56, 16

Offset: 1

Winston C. Yang (winston(AT)cs.wisc.edu), Aug 31 2000

			1,-1;
1,-2,1;
1,-3,4,2;
1,-4,9,10,4;
...

W. C. Yang, Conjectures on some sequences involving set partitions and Bell numbers, preprint, 2000. [Apparently unpublished. - R. J. Mathar, Oct 06 2011]

Cf. Bell numbers A000110.

Cf. A056857-A056862.

A122172 Triangle read by rows relating A074139, A074141, A078436 and A079025.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 2, 3, 2, 1, 1, 3, 4, 3, 1, 1, 4, 6, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 3, 3, 2, 1, 1, 3, 4, 4, 3, 1, 1, 3, 5, 5, 3, 1, 1, 4, 7, 7, 4, 1, 1, 5, 10, 10, 5, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Alford Arnold, Aug 23 2006

A proper definition is needed for this sequence.
Are the row sums A074139(n) and the row lengths A000041(n)? - R. J. Mathar, May 08 2019 [Not exactly: see below. - M. F. Hasler, Jan 07 2024]
From M. F. Hasler, Jan 06 2024: (Start)
I get this triangle as T(n,k) = # { v in S(p_n), |v| = k }, where p_n is the n-th partition as listed in A036036 or A036037 (which has a nice table of the p's), and S(p) = {0, ..., p[1]} x ... x {0, ..., p[#p]}, the set of vectors v with 0 <= v[i] <= p[i] for all indices i from 1 to #p = number of parts in p.
Then the row sums are indeed the total number of elements in S(p_n) which is equal to the product (p[1]+1)*...*(p[#p]+1) which is also the number of divisors of the Heinz number of p (cf. A185974).
The row lengths are 1 + |p| = 1 + sum of all parts of p (corresponding to the possible values of |v| ranging from 0 to |p|), repeated A000041(|p|) times: A000041(0) = 1 row of length 0+1 for the partition () of 0, A000041(1) = 1 row of length 1+1 for partition (1) of 1; A000041(2) = 2 rows of length 2+1 for the two partitions (2) and (1,1) of 2;  A000041(3) = 3 rows of length 3+1 for the 3 partitions {(3), (2,1), (1,1,1)} of 3; etc. (End)

			The triangle begins:
  1
  1 1
  1 1 1
  1 2 1
  1 1 1 1
  1 2 2 1
  1 3 3 1
  1 1 1 1 1
  1 2 2 2 1
  1 2 3 2 1
  1 3 4 3 1
  1 4 6 4 1
  1 1 1 1 1 1
  1 2 2 2 2 1
  1 2 3 3 2 1
  1 3 4 4 3 1
  1 3 5 5 3 1
  1 4 7 7 4 1
  1 5 10 10 5 1

M. F. Hasler, Table of n, a(n) for n = 0..9172 (up to partitions of 15), Jan 07 2024

Cf. A036036 (partitions in A-S order), A036037 (the same, parts reversed), A185974 (corresponding Heinz numbers).

A122172_row(n, p=part(n))={my(c=Vec(0, vecsum(p)+1)); forvec(v=[[0, k]| k<-p], c[vecsum(v)+1]++); c} \\ instead of n one can directly give p as 2nd arg
/* helper function: n-th partition as listed in A036036, A036037 or A185974 */
part(n)={my(c, r=0); while(n >= c = numbpart(r), n -= c; r++); partitions(r)[n+1]}
for(n=0,5, forpart(p=n, print(A122172_row(, Vec(p))) )) \\ Illustration. \\ M. F. Hasler, Jan 06 2024

More terms from M. F. Hasler, Jan 07 2024

A260747 Consolidated Dragon Curve triple points. If D:[0,1] is a Dragon curve, then besides n, there are two other integers p and q with D(A(n)/(152^k)) = D(A(p)/(152^k)) = D(A(q)/(15*2^k)), where k is any integer > log_2(max(A(n),A(p),A(q))/15).

Original entry on oeis.org

13, 21, 23, 26, 37, 39, 42, 46, 47, 52, 73, 74, 78, 81, 83, 84, 92, 94, 97, 99, 103, 104, 107, 111, 113, 133, 141, 143, 146, 148, 156, 157, 159, 162, 163, 166, 167, 168, 171, 173, 184, 188, 193, 194, 198, 199, 201, 203, 206, 207, 208, 209, 211, 213, 214, 217, 219, 221, 222, 223, 226, 227, 231, 233, 253, 261, 263, 266, 277, 279, 282, 283, 286, 287

Offset: 1

Bill Gosper, Jul 30 2015

It appears that every Dragon triple point is an image of A(n)/(15*2^k) for three different n and some k.
For the triples grouped, use
Dragon(A260748(n)) = Dragon(A260749(n)) = Dragon(A260750(n)).  (I.e., they're "conformal".)
The first differences of this sequence appear to comprise only 1, 2, 3, 4, 5, 8, 11, 20, and 21.  21 occurs only twice for A(n) < 30720.
See dragun in the MATHEMATICA section for an exact evaluator of a continuous, spacefilling Dragon function, and undrag, its multivalued inverse.
Even excluding multiples of 5, it is NOT the case that A260747 contains 7*A260747, e.g., 7*13=91 is missing.

			For definiteness, we choose the Dragon in the complex plane with Dragon(0) = 0, Dragon(1) = 1, Dragon(1/3) = 1/5+2i/5
Then using A(1) = 13, for k=0,1,2, {dragun[13/15], dragun[13/30], dragun[13/60]}
-> {{2/3 - I/3}, {1/2 + I/6}, {1/6 + I/3}} (where I^2:=-1)
These have inverse images undrag/@First/@%
{{13/15}, {13/30, 7/10, 23/30}, {13/60, 7/20, 23/60}}
k=0 is too small--7/5 and 23/15 are off the end of the curve!
dragun[13/15/2^k] = dragun[7/5/2^k] = dragun[23/15/2^k], which empirically = (2/3 - I/3) (1/2 + I/2)^k

Brady Haran and Don Knuth, Wrong turn on the Dragon, Numberphile video (2014)
Wikipedia, Dragon curve

A260747 = A260748 U A260749 U A260750 = Superset of 3*A260482.

(* by Julian Ziegler Hunts *)
piecewiserecursivefractal[x_, f_, which_, iters_, fns_] := piecewiserecursivefractal[x, g_, which, iters, fns] = ((piecewiserecursivefractal[x, h_, which, iters, fns] := Block[{y}, y /. Solve[f[y] == h[y], y]]); Union @@ ((fns[[#]] /@ piecewiserecursivefractal[iters[[#]][x], Composition[f, fns[[#]]], which, iters, fns]) & /@ which[x]));
dragun[t_] := piecewiserecursivefractal[t, Identity, Piecewise[{{{1}, 0 <= # <= 1/2}, {{2}, 1/2 <= # <= 1}}, {}] &, {2*# &, 2*(1 - #) &}, {(1 + I)*#/2 &, (I - 1)*#/2 + 1 &}]
undrag[z_] := piecewiserecursivefractal[z, Identity, If[-(1/3) <= Re[#] <= 7/6 && -(1/3) <= Im[#] <= 2/3, {1, 2}, {}] &, {#*(1 - I) &, (1 - #)*(1 + I) &}, {#/2 &, 1 - #/2 &}]
Reap[Do[If[Length[undrag[dragun[k/15/32][[1]]]] > 2, Sow[k]],{k,0,288}]][[2, 1]]

Corrected subtle bug in NAME section, plus three tweaks to EXAMPLE. Tweaked comment. - Bill Gosper, Jul 31 2015

A035403 Related to Rogers-Ramanujan Identities.

Original entry on oeis.org

4, 8, 12, 32, 37, 45, 68, 74, 80, 85, 97, 197, 202, 215, 218, 225, 239, 243, 253, 272, 374, 380, 387, 392, 420, 424, 428, 438, 463, 470, 484

Offset: 1

Olivier Gérard

G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, p. 109.
Contact author at address below [ Note: this form of description is not acceptable! - N. J. A. Sloane ]

Cf. A003114, A035401, A035402, A035404, A035405.

A116672 Triangle read by rows in which the binomial transform of the n-th row gives the Euler transform of the n-th diagonal of Pascal's triangle (A007318).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 6, 11, 7, 1, 1, 10, 27, 29, 12, 1, 1, 14, 57, 96, 72, 21, 1, 1, 21, 117, 277, 319, 176, 38, 1

Offset: 1

Alford Arnold, Feb 22 2006

For example, the Euler transform of 1,3,6,... is 1,1,4,10,26,59,141,... (A000294) differing slightly from A000293 which counts the solid partitions.
The NAME does not reproduce the DATA, COMMENTS, or EXAMPLES.  - R. J. Mathar, Jul 19 2017
The binomial transforms of the rows form the rows of A289656. - N. J. A. Sloane, Jul 19 2017

			Row 6 is 1 10 27 29 12 1 generating 1 11 48 141 ... (A008780) the seventh term in the Euler transforms of 1,1,1,...; 1,2,3,...; 1,3,6,... 1,4,10,... etc.
Triangle begins:
1;
1, 1;
1, 2, 1;
1, 4, 4, 1;
1, 6, 11, 7, 1;
1, 10, 27, 29, 12, 1;
1, 14, 57, 96, 72, 21, 1;
1, 21, 117, 277, 319, 176, 38, 1;
...

N. J. A. Sloane, Transforms

Cf. A000293, A116673 (row sums), A008778 - A008780, A289656.

A260748 Dragon Curve triple point lower inverses. If D:[0,1] is a Dragon curve, then besides n, there are two larger integers p, q (with p < q) with D(A(n)/(152^k)) = D(A(p)/(152^k)) = D(A(q)/(15*2^k)), where k is any integer > log_2(A(q)/15).

Original entry on oeis.org

13, 26, 37, 52, 73, 74, 97, 103, 104, 111, 133, 146, 148, 157, 193, 194, 199, 206, 207, 208, 209, 217, 221, 222, 223, 231, 253, 266, 277, 292, 296, 307, 313, 314, 317, 337, 373, 386, 388, 397, 398, 409, 412, 414, 416, 417, 418, 419, 431, 433, 434, 439, 442, 444, 446, 447, 449, 457, 461, 462, 463, 471, 493, 506, 517, 532, 553, 554, 577, 583, 584, 591, 592, 613, 614, 619, 626, 627, 628, 629, 631, 634, 637, 667, 673, 674, 677, 697, 733, 746, 757, 772, 776, 787, 793, 794, 797, 817, 853

Offset: 1

Bill Gosper, Jul 30 2015

For the triples grouped, use Dragon(A260748(n)) = Dragon(A260749(n)) = Dragon(A260750(n)).  (I.e., they're "conformal".)
Unlike A260747, A260749, A260750, and A260482, the first differences show a fractal pattern of repetitions.
See dragun in the MATHEMATICA section for an exact evaluator of a continuous, spacefilling Dragon function, and undrag, its multivalued inverse.

			For definiteness, we choose the Dragon in the complex plane with Dragon(0) = 0, Dragon(1) = 1, Dragon(1/3) = 1/5+2i/5
Then using A(1) = 13, for k=0,1,2, {dragun[13/15], dragun[13/30], dragun[13/60]}
-> {{2/3 - I/3}, {1/2 + I/6}, {1/6 + I/3}}
These have inverse images undrag/@First/@%
{{13/15}, {13/30, 7/10, 23/30}, {13/60, 7/20, 23/60}}
k=0 is too small--7/5 and 23/15 are off the end of the curve!
dragun[13/15/2^k] = dragun[21/15/2^k] = dragun[23/15/2^k], which empirically = (2/3 - I/3) (1/2 + I/2)^k

Brady Haran and Don Knuth, Wrong turn on the Dragon, Numberphile video (2014)
Wikipedia, Dragon curve

A260747 = A260748 U A260749 U A260750 = Superset of 3*A260482.

(* by Julian Ziegler Hunts *)
piecewiserecursivefractal[x_, f_, which_, iters_, fns_] := piecewiserecursivefractal[x, g_, which, iters, fns] = ((piecewiserecursivefractal[x, h_, which, iters, fns] := Block[{y}, y /. Solve[f[y] == h[y], y]]); Union @@ ((fns[[#]] /@ piecewiserecursivefractal[iters[[#]][x], Composition[f, fns[[#]]], which, iters, fns]) & /@ which[x]));
dragun[t_] := piecewiserecursivefractal[t, Identity, Piecewise[{{{1}, 0 <= # <= 1/2}, {{2}, 1/2 <= # <= 1}}, {}] &, {2*# &, 2*(1 - #) &}, {(1 + I)*#/2 &, (I - 1)*#/2 + 1 &}]
undrag[z_] := piecewiserecursivefractal[z, Identity, If[-(1/3) <= Re[#] <= 7/6 && -(1/3) <= Im[#] <= 2/3, {1, 2}, {}] &, {#*(1 - I) &, (1 - #)*(1 + I) &}, {#/2 &, 1 - #/2 &}]
DeleteDuplicates[Reap[Do[If[Length[#] > 2, Sow[15*64*#[[1]]]] &@
     undrag[dragun[k/15/64][[1]]], {k, 0, 288*3}]][[2, 1]]]
(* or 128 or 256 or ...*)

A129783 Conjectured numbers n with the property that there exist two consecutive primes p and q such that pq + n is a square.

Original entry on oeis.org

1, 3, 4, 9, 10, 14, 16, 19, 21, 23, 25, 26, 29, 30, 34, 35, 36, 38, 43, 44, 46, 47, 49, 53, 58, 62, 64, 65, 66, 67, 68, 75, 77, 78, 81, 82, 83, 85, 86, 92, 94, 95, 100, 103, 106, 109, 110, 113, 115, 117, 118, 119, 121, 122, 125, 129, 134, 138, 139, 140, 143, 144, 146, 148

Offset: 1

Cino Hilliard, May 18 2007

Complement of A129816. [From Omar E. Pol, Dec 26 2008]

David W. Wilson, Table of n, a(n) for n = 1..10000

Cf. A029707, A074927.

Cf. A129816. [From Omar E. Pol, Dec 26 2008]

primesq(n,m) = \square of the form prime(x)*prime(x+1) + k {local(c,k,x,p1,p2,j); c=0; for(k=1,m, for(x=1,n, p1=prime(x); p2=(prime(x+1)); y=p1*p2+k; if(issquare(y), c++; print1(k","); break; ) ) ); c; }

I don't know how many of the missing terms have been proved to be missing. Has it been proved that 2 is missing? - N. J. A. Sloane, May 20 2007

A152396 Let f(M,k) denote the decimal concatenation of k numbers starting with M: M | M-1 | M-2 | ... | M-k+1, k > 1. Then a(n) is the smallest M such that for all m in {1,..,n} an m-th prime occurs as f(M,k) for the smallest possible k, order prioritized m = 1 through n.

Original entry on oeis.org

4, 10, 1000, 21910420, 1113475000, 67483920430

Offset: 1

James G. Merickel, Oct 20 2009

The sequence of k values is 2, 4, 8, 10, 14, 20, 28, 32, 34, 40, 50, etc. a(n) is necessarily congruent to 10 modulo 30 starting with n=4 (and is coincidentally so for n=2 and 3). Each successive unknown term's existence is only conjectural, but is supported by standard heuristics. The values in the lead sentence come through considerations modulo the smallest primes.
Concatenation of 22 numbers is ruled out by consideration modulo 11 in concert with the prior need for the stem to be 10 mod 30: A prime concatenation of 22 values necessarily involves passage from one length number to another, with a power of 10 not leading. Beginning with a(11), however, with each time a multiple of 22 for the number of concatenated values is passed there is a measure of uncertainty. Theoretically, it seems a(11) could include for largest prime a concatenation of 44 (rather than 50) values, but with these primes very sparse it is a near certainty heuristically that this is not so. Mathematically, a conjectured a(11) would have a higher bar for strict proof than a(n) for n < 11, and the same holds repeatedly for even less accessible terms.
The alternative sequence where only the rapidity of arrival of the n-th prime determines a(n) (k minimal for the largest prime, with no constraint on k for the smaller prime concatenations) necessarily shares its first 5 terms in common with this one. It also shares its 6th by virtue of the fact that this sequence's a(6) is the only value less than 10^12 producing its 6th prime with the attachment of the 20th value, whether alternative length possibilities for primes are allowed or not (i.e., the first cases giving 5 other smaller primes -- in addition to one of 20 concatenated values -- where there is a prime concatenation of 16 values, in place of one of either 2 or 8 values, are both at least this large). However, it does necessarily differ at a(7) and a(8) (but then not necessarily at a(9)), as the resolution of the theoretical problem for the twin sequence is given for a(7) by the possibility of 5, 7, 11, 13, 17, 23 and 25 numbers being concatenated to give primes, and for a(8) by the replacement of 2 concatenated values with concatenations of both 16 and 26 of them (with result that a(8) for this alternative sequence appears already with concatenation of 28 values, while here that corresponds to a(7)).
This necessity to distinguish between whether or not only the last of the primes comes as quickly as possible does not arise as an issue in the situation where the number itself is required to be prime (A172257). [Comments re-edited from Feb 2014] - James G. Merickel, Aug 07 2015
The definition is not clear to me. - N. J. A. Sloane, Aug 11 2015

			43 is prime while 32 and 21 are not, so a(1)=4; 109 and 10987 are both prime, and like concatenations for values 4 through 9 do not produce 2 primes, so a(2)=10; 1000999, 1000999998997 and 1000999998997996995994993 are all prime and no smaller value produces 3 primes so quickly, so a(3)=1000.

Cf. A152397, A172257, A232657.

Two more terms from James G. Merickel, Dec 09 2009
6th term added by James G. Merickel, Jan 29 2010
Title changed by James G. Merickel, Feb 18 2014
Title changed by James G. Merickel, Aug 06 2015

cp's OEIS Frontend

A147995 Array of N X N grid hopping "almost-walk", read by antidiagonals.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A245921 Index sequence for limit-reversing the (2,1)-version of the infinite Fibonacci word A014675 with first term as initial block.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A056863 Related to triangle A056858 of number of rises in set partitions of n at a given index i.

Views

Author

Keywords

Examples

References

Crossrefs

A122172 Triangle read by rows relating A074139, A074141, A078436 and A079025.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Extensions

A260747 Consolidated Dragon Curve triple points. If D:[0,1] is a Dragon curve, then besides n, there are two other integers p and q with D(A(n)/(15*2^k)) = D(A(p)/(15*2^k)) = D(A(q)/(15*2^k)), where k is any integer > log_2(max(A(n),A(p),A(q))/15).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Extensions

A035403 Related to Rogers-Ramanujan Identities.

Views

Author

Keywords

References

Crossrefs

A116672 Triangle read by rows in which the binomial transform of the n-th row gives the Euler transform of the n-th diagonal of Pascal's triangle (A007318).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A260748 Dragon Curve triple point lower inverses. If D:[0,1] is a Dragon curve, then besides n, there are two larger integers p, q (with p < q) with D(A(n)/(15*2^k)) = D(A(p)/(15*2^k)) = D(A(q)/(15*2^k)), where k is any integer > log_2(A(q)/15).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A129783 Conjectured numbers n with the property that there exist two consecutive primes p and q such that pq + n is a square.

Views

Author

Keywords

A260747 Consolidated Dragon Curve triple points. If D:[0,1] is a Dragon curve, then besides n, there are two other integers p and q with D(A(n)/(152^k)) = D(A(p)/(152^k)) = D(A(q)/(15*2^k)), where k is any integer > log_2(max(A(n),A(p),A(q))/15).

A260748 Dragon Curve triple point lower inverses. If D:[0,1] is a Dragon curve, then besides n, there are two larger integers p, q (with p < q) with D(A(n)/(152^k)) = D(A(p)/(152^k)) = D(A(q)/(15*2^k)), where k is any integer > log_2(A(q)/15).