A060587 -id:A060587 - cp's OEIS frontend

A060583 A ternary code related to the Tower of Hanoi.

0, 2, 1, 7, 6, 8, 5, 4, 3, 23, 22, 21, 18, 20, 19, 25, 24, 26, 16, 15, 17, 14, 13, 12, 9, 11, 10, 70, 69, 71, 68, 67, 66, 63, 65, 64, 54, 56, 55, 61, 60, 62, 59, 58, 57, 77, 76, 75, 72, 74, 73, 79, 78, 80, 50, 49, 48, 45, 47, 46, 52, 51, 53, 43, 42, 44, 41, 40, 39, 36, 38, 37

Offset: 0

Henry Bottomley, Apr 04 2001

Write n in base 3, then (working from left to right) if the k-th digit of n is equal to the corresponding digit to the left of the k-th digit of a(n) then this is the k-th digit of a(n), otherwise the k-th digit of a(n) is the element of {0,1,2} which has not just been compared, then read result as a base 3 number.

			a(46) = 76 since 43 = 1201_3; this gives a first digit of 2(=3-1-0), a second digit of 2(=2=2), a third digit of 1(=3-2-0) and a fourth digit of 1(=1=1); 2211_3 = 76.

Cf. A060586, A060587 (inverse).

a(n) = 3*a(floor(n/3)) + ((-a(floor(n/3))-n) mod 3) = 3*a(floor(n/3)) + A060582(n) with a(0)=0.

A253586 The sum of the i-th ternary digits of n, k, and A(n,k) equals 0 (mod 3) for each i>=0 (leading zeros included); square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

0, 2, 2, 1, 1, 1, 6, 0, 0, 6, 8, 8, 2, 8, 8, 7, 7, 7, 7, 7, 7, 3, 6, 6, 3, 6, 6, 3, 5, 5, 8, 5, 5, 8, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 4, 18, 3, 3, 0, 3, 3, 0, 3, 3, 18, 20, 20, 5, 2, 2, 5, 2, 2, 5, 20, 20, 19, 19, 19, 1, 1, 1, 1, 1, 1, 19, 19, 19, 24, 18, 18, 24, 0, 0, 6, 0, 0, 24, 18, 18, 24

Offset: 0

Alois P. Heinz, Jan 04 2015

			Square array A(n,k) begins:
  0, 2, 1, 6, 8, 7, 3, 5, 4, ...
  2, 1, 0, 8, 7, 6, 5, 4, 3, ...
  1, 0, 2, 7, 6, 8, 4, 3, 5, ...
  6, 8, 7, 3, 5, 4, 0, 2, 1, ...
  8, 7, 6, 5, 4, 3, 2, 1, 0, ...
  7, 6, 8, 4, 3, 5, 1, 0, 2, ...
  3, 5, 4, 0, 2, 1, 6, 8, 7, ...
  5, 4, 3, 2, 1, 0, 8, 7, 6, ...
  4, 3, 5, 1, 0, 2, 7, 6, 8, ...

Alois P. Heinz, Antidiagonals n = 0..200, flattened
Rémy Sigrist, Colored representation of the table for 0 <= n, k < 3^7 (where the hue is function of T(n, k))
Wikipedia, Set (game)

Column k=0 and row n=0 gives A004488.
Main diagonal gives A001477.
A(n,floor(n/3)) gives A060587.
Cf. A090245, A253587.

A:= proc(n, k) local i, j; `if`(n=0 and k=0, 0,
      A(iquo(n, 3, 'i'), iquo(k, 3, 'j'))*3 +irem(6-i-j, 3))
    end:
seq(seq(A(n, d-n), n=0..d), d=0..14);

A(n,k) = A(floor(n/3),floor(k/3))*3+(6-(n mod 3)-(k mod 3) mod 3), A(0,0) = 0.

A253587 The sum of the i-th ternary digits of n, k, and T(n,k) equals 0 (mod 3) for each i>=0 (leading zeros included); triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

0, 2, 1, 1, 0, 2, 6, 8, 7, 3, 8, 7, 6, 5, 4, 7, 6, 8, 4, 3, 5, 3, 5, 4, 0, 2, 1, 6, 5, 4, 3, 2, 1, 0, 8, 7, 4, 3, 5, 1, 0, 2, 7, 6, 8, 18, 20, 19, 24, 26, 25, 21, 23, 22, 9, 20, 19, 18, 26, 25, 24, 23, 22, 21, 11, 10, 19, 18, 20, 25, 24, 26, 22, 21, 23, 10, 9, 11

Offset: 0

Alois P. Heinz, Jan 04 2015

			Triangle T(n,k) begins:
  0;
  2, 1;
  1, 0, 2;
  6, 8, 7, 3;
  8, 7, 6, 5, 4;
  7, 6, 8, 4, 3, 5;
  3, 5, 4, 0, 2, 1, 6;
  5, 4, 3, 2, 1, 0, 8, 7;
  4, 3, 5, 1, 0, 2, 7, 6, 8;

Alois P. Heinz, Rows n = 0..200, flattened
Wikipedia, Set (game)

Column k=0 gives A004488.
Main diagonal gives A001477.
T(n,floor(n/3)) gives A060587.
Cf. A090245, A253586.

T:= proc(n, k) local i, j; `if`(n=0 and k=0, 0,
      T(iquo(n, 3, 'i'), iquo(k, 3, 'j'))*3 +irem(6-i-j, 3))
    end:
seq(seq(T(n, k), k=0..n), n=0..14);

T(n,k) = T(floor(n/3),floor(k/3))*3+(6-(n mod 3)-(k mod 3) mod 3), T(0,0) = 0.

A361818 For any number k >= 0, let T_k be the triangle whose base corresponds to the ternary expansion of k (without leading zeros) and other values, say t above u and v, satisfy t = (-u-v) mod 3; this sequence lists the numbers k such that T_k has 3-fold rotational symmetry.

Original entry on oeis.org

0, 1, 2, 4, 8, 13, 26, 34, 40, 46, 59, 65, 80, 112, 121, 130, 224, 233, 242, 304, 364, 424, 518, 578, 728, 772, 862, 925, 1003, 1093, 1183, 1261, 1324, 1414, 1535, 1598, 1688, 1766, 1856, 1919, 2006, 2096, 2186, 2257, 2509, 2734, 3028, 3280, 3532, 3826, 4051

Offset: 1

Rémy Sigrist, Mar 25 2023

We can devise a similar sequence for any fixed base b >= 2; the present sequence corresponds to b = 3, and A334556 corresponds to b = 2.
This sequence is infinite as it contains A048328.
If k belongs to the sequence, then A004488(k) and A030102(k) belong to the sequence.
Empirically, there are 2*3^floor((w-1)/3) positive terms with w ternary digits.
For any k, if t appears above u and v in T_k, then t + u + v = 0 (mod 3) and #{t, u, v} = 1 or 3 (the three values are either equal or all distinct); each value is uniquely determined by the two others in the same way: t = (-u-v) mod 3, u = (-t-v) mod 3, v = (-t-u) mod 3; this means that we can reconstruct T_k from any of its three sides.
If some row of T_k, say r, has w values and corresponds to the ternary expansion of m, then the row above r corresponds to the w-1 rightmost digits of the ternary expansion of A060587(m).
All positive terms belong to A297250 (their most significant digit equals their least significant digit in base 3).

			The ternary expansion of 304 is "102021", and the corresponding triangle is:
             1
            0 2
           2 1 0
          0 1 1 2
         2 1 1 1 0
        1 0 2 0 2 1
As this triangle has 3-fold rotational symmetry, 304 belongs to the sequence.

Rémy Sigrist, Triangles illustrating initial terms
Rémy Sigrist, PARI program
Index entries for sequences related to XOR-triangles

Cf. A004488, A048328, A060587, A297250, A334556.

PARI
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See Links section.
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A060588 If the final two digits of n written in base 3 are the same then this digit, otherwise mod 3-sum of these two digits.

Original entry on oeis.org

0, 2, 1, 2, 1, 0, 1, 0, 2, 0, 2, 1, 2, 1, 0, 1, 0, 2, 0, 2, 1, 2, 1, 0, 1, 0, 2, 0, 2, 1, 2, 1, 0, 1, 0, 2, 0, 2, 1, 2, 1, 0, 1, 0, 2, 0, 2, 1, 2, 1, 0, 1, 0, 2, 0, 2, 1, 2, 1, 0, 1, 0, 2, 0, 2, 1, 2, 1, 0, 1, 0, 2, 0, 2, 1, 2, 1, 0, 1, 0, 2, 0, 2, 1, 2, 1, 0, 1, 0, 2, 0, 2, 1, 2, 1, 0, 1, 0, 2, 0, 2, 1, 2, 1, 0

Offset: 0

Henry Bottomley, Apr 04 2001

From William Walkington, Sep 14 2016: (Start)
With offset 1, the y-coordinates of position vectors from the origin (point 1) to the points numbered 1 to N^2 of the magic tori that display the Agrippa odd-order-N magic squares can be expressed as follows: a(n) = (-(n-1)-floor((n-1)/N)) mod N.
This generates the y-coordinates of the magic tori that display the Agrippa order-3 "Saturn," order-5 "Mars," order-7 "Venus," order-9 "Luna," and higher-odd-order-N magic squares.
Therefore, if the odd-order-N of the torus is 3, then the resulting sequence 0,2,1,2,1,0,1,0,2 represents the y-coordinates of position vectors from the origin (point number 1) to the point numbered 1 to 9 of the magic torus that displays the Agrippa order-3 "Saturn" magic square. (End)

			a(22)=1 since 22 is written in base 3 as 211 and the final two digits are 1; a(23)=0 since 23 is written in base 3 as 212 and the final two digits are 1 and 2 and 3-(1+2)=0.

H.C. Agrippa, "De occulta philosophia Libri tres," (1533) translated by "J.F." (John French?) and printed by Moule, London, in 1651, Book II, chapter XXII entitled "Of the tables of the Planets, their vertues,forms, and what Divine names, Intelligencies, and Spirits are set over them."

H. C. Agrippa, De Occulta Philosophia libri tres (Three books of occult philosophy), Book II, chapter XXII, Digital edition Peterson J.F.
W. Walkington, Magic torus coordinate and vector symmetries.
William Walkington, Agrippa odd-order magic squares
Wikipedia, Agrippa's magic squares.
Index entries for sequences related to final digits of numbers

Cf. A060582, A270740.

b3d[n_]:=Module[{d3=Take[IntegerDigits[n,3],-2]},If[MatchQ[d3,{x_, x_}], d3[[1]],3-Total[d3]]]; Join[{0,2,1},Array[b3d,110,3]] (* Harvey P. Dale, Feb 29 2016 *)
Table[If[MatchQ @@ #, First@ #, Mod[3 - Total@ #, 3]] &@ Take[PadLeft[#, 2], -2] &@ IntegerDigits[n, 3], {n, 0, 120}] (* or *)
Table[Mod[-n - Floor[n/3], 3], {n, 0, 120}] (* Michael De Vlieger, Sep 14 2016 *)

a(n) = a(n-9) = (-[n/3]-n) mod 3 = A060587(n) mod 3.

a(n) = (-n - floor(n/3)) mod 3. - William Walkington, Sep 14 2016

A060585 Write n in base 3, then (working from left to right) if the k-th digit of n is not equal to the digit to its left then the k-th digit of a(n) is 1, otherwise it is 0, and finally read the result as a base-2 number.

Original entry on oeis.org

0, 1, 1, 3, 2, 3, 3, 3, 2, 6, 7, 7, 5, 4, 5, 7, 7, 6, 6, 7, 7, 7, 6, 7, 5, 5, 4, 12, 13, 13, 15, 14, 15, 15, 15, 14, 10, 11, 11, 9, 8, 9, 11, 11, 10, 14, 15, 15, 15, 14, 15, 13, 13, 12, 12, 13, 13, 15, 14, 15, 15, 15, 14, 14, 15, 15, 13, 12, 13, 15, 15, 14, 10, 11, 11, 11, 10, 11, 9, 9

Offset: 0

Henry Bottomley, Apr 04 2001

A ternary-to-binary map.

Each k appears A001316(k) times in the sequence.

			a(76) = 10 since 76 = 2211_3 and looking for changing digits produces 1010 which, read as if in binary, is 10.

Cf. A001316, A060584, A060586, A060587.

a(n) = my(v=digits(n,3)); if(#v, forstep(k=#v,2,-1, v[k]=(v[k]!=v[k-1])); v[1]=1); fromdigits(v,2); \\ Kevin Ryde, Apr 15 2021

a(n) = 2*a(floor(n/3)) + A060584(n) = A060586(A060587(n)).

Definition rewritten by Editors of OEIS, Apr 15 2021.

cp's OEIS Frontend

A060583 A ternary code related to the Tower of Hanoi.

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A253586 The sum of the i-th ternary digits of n, k, and A(n,k) equals 0 (mod 3) for each i>=0 (leading zeros included); square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A253587 The sum of the i-th ternary digits of n, k, and T(n,k) equals 0 (mod 3) for each i>=0 (leading zeros included); triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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A361818 For any number k >= 0, let T_k be the triangle whose base corresponds to the ternary expansion of k (without leading zeros) and other values, say t above u and v, satisfy t = (-u-v) mod 3; this sequence lists the numbers k such that T_k has 3-fold rotational symmetry.

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PARI

A060588 If the final two digits of n written in base 3 are the same then this digit, otherwise mod 3-sum of these two digits.

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A060585 Write n in base 3, then (working from left to right) if the k-th digit of n is not equal to the digit to its left then the k-th digit of a(n) is 1, otherwise it is 0, and finally read the result as a base-2 number.

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