A065189 -id:A065189 - cp's OEIS frontend

A269526 Square array T(n,k) (n>=1, k>=1) read by antidiagonals upwards in which each term is the least positive integer satisfying the condition that no row, column, diagonal, or antidiagonal contains a repeated term.

1, 2, 3, 3, 4, 2, 4, 1, 5, 6, 5, 2, 6, 1, 4, 6, 7, 3, 2, 8, 5, 7, 8, 1, 5, 9, 3, 10, 8, 5, 9, 4, 1, 7, 6, 11, 9, 6, 4, 7, 2, 8, 5, 12, 13, 10, 11, 7, 3, 5, 6, 9, 4, 14, 8, 11, 12, 8, 9, 6, 10, 3, 7, 15, 16, 14, 12, 9, 13, 10, 11, 14, 4, 15, 16, 17, 7, 18, 13, 10, 14, 11, 3, 4, 8, 16, 9, 6, 12, 15, 7

Offset: 1

Alec Jones, Apr 07 2016

An infinite Sudoku-type array.
In the definition, "diagonal" means a diagonal line of slope -1, and "antidiagonal" means a diagonal line of slope +1.
Theorem C (Bob Selcoe, Jul 01 2016): Every column is a permutation of the natural numbers.
Proof: Fix k, and suppose j is the smallest number missing from that column. For this to happen, every entry T(n,k) for sufficiently large n in that column must see a j in the NW diagonal through that cell or in the row to the W of that cell. But there are at most k-1 copies of j in the columns to the left of the k-th column, and if n is very large the entry T(n,k) will be unaffected by those j's, and so T(n,k) would then be set to j, a contradiction. QED
Theorem R (Rob Pratt, Bob Selcoe, N. J. A. Sloane, Jul 02 2016): Every row is a permutation of the natural numbers.
Proof: Fix n, and suppose j is the smallest number missing from that row. For this to happen, every entry T(n,k) for sufficiently large k in that row must see a j in the column to the N, or in the NW diagonal through that cell or in the SW diagonal through that cell.
Rows 1 through n-1 contain at most n-1 copies of j, and their influence on the entries in the n-th row only extend out to the entry T(n,k_0), say. We take k to be much larger than k_0 and consider the entry T(n,k). We will show that for large enough k it can (and therefore must) be equal to j, which is a contradiction.
Consider the triangle bounded by row n, column 1, and the SW antidiagonal through cell (n,k). Replace every copy of j in this triangle by a queen and think of these cells as a triangular chessboard. These are non-attacking queens, by definition of the sequence, and by the result in A274616 there can be at most 2*k/3 + 1 such queens. However, there are k-k_0 cells in row n that have to be attacked, and for large k this is impossible since k-k_0 > 2*k/3+1. If a cell (n,k) is not attacked by a queen, then T(n,k) can take the value j. QED
Presumably every diagonal is also a permutation of the natural numbers, but the proof does not seem so straightforward. Of course the antidiagonals are not permutations of the natural numbers, since they are finite in length. - N. J. A. Sloane, Jul 02 2016
For an interpretation of this array in terms of Sprague-Grundy values, see A274528.
From Don Reble, Jun 30 2016: (Start)
Let b(n) be the position in column n where 1 appears, i.e., such that T(b(n),n) = 1. Then b(n) is A065188, which is Antti Karttunen's "Greedy Queens" permutation.
Let b'(n) be the position in row n where 1 appears, i.e., such that T(n,b'(n)) = 1. Then b'(n) is A065189, the inverse "Greedy Queens" permutation. (End)
The same sequence arises if we construct a triangle, by reading from left to right in each row, always choosing the smallest positive number which does not produce a duplicate number in any row or diagonal. - N. J. A. Sloane, Jul 02 2016
It appears that the numbers generally appear for the first time in or near the first few rows. - Omar E. Pol, Jul 03 2016
The last comment in the FORMULA section seems wrong: It seems that columns 4, 5, 6, 7, 8, 9, ...(?) all have first differences which become 16-periodic from, respectively, term 8, 17, 52, 91, 92, 131, ... on, rather than having period 4^(k-1) from term k on. - M. F. Hasler, Sep 26 2022

			The array is constructed along its antidiagonals, in the following way:
  a(1)  a(3)  a(6)  a(10)
  a(2)  a(5)  a(9)
  a(4)  a(8)
  a(7)
See the link from Peter Kagey for an animated example.
The beginning of the square array is:
   1,  3,  2,  6,  4,  5, 10, 11, 13,  8, 14, 18,  7, 20, 19,  9, 12, ...
   2,  4,  5,  1,  8,  3,  6, 12, 14, 16,  7, 15, 17,  9, 22, 21, 11, ...
   3,  1,  6,  2,  9,  7,  5,  4, 15, 17, 12, 19, 18, 21,  8, 10, 23, ...
   4,  2,  3,  5,  1,  8,  9,  7, 16,  6, 18, 17, 11, 10, 23, 22, 14, ...
   5,  7,  1,  4,  2,  6,  3, 15,  9, 10, 13,  8, 20, 14, 12, 11, 17, ...
   6,  8,  9,  7,  5, 10,  4, 16,  2,  1,  3, 11, 22, 15, 24, 13, 27, ...
   7,  5,  4,  3,  6, 14,  8,  9, 11, 18,  2, 21,  1, 16, 10, 12, 20, ...
   8,  6,  7,  9, 11,  4, 13,  3, 12, 15,  1, 10,  2,  5, 26, 14, 18, ...
   9, 11,  8, 10,  3,  1, 14,  6,  7, 13,  4, 12, 24, 18,  2,  5, 19, ...
  10, 12, 13, 11, 16,  2, 17,  5, 20,  9,  8, 14,  4,  6,  1,  7,  3, ...
  11,  9, 14, 12, 10, 15,  1,  8, 21,  7, 16, 20,  5,  3, 18, 17, 32, ...
  12, 10, 11,  8,  7,  9,  2, 13,  5, 23, 25, 26, 14, 17, 16, 15, 33, ...
...
  - _N. J. A. Sloane_, Jun 29 2016

Peter Kagey, Table of n, a(n) for n = 1..10000
F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
Peter Kagey, Animated example illustrating the first fifteen terms.

First 4 rows are A274315, A274316, A274317, A274791.
Main diagonal is A274318.
Column 1 is A000027, column 2 is A256008(n) = A004443(n-1)+1 = 1 + (nimsum of n-1 and 2), column 3 is A274614 (or equally, A274615 + 1), and column 4 is A274617 (or equally, A274619 + 1).
Antidiagonal sums give A274530. Other properties of antidiagonals: A274529, A275883.
Cf. A274080 (used in Haskell program), A274616.
A065188 and A065189 say where the 1's appear in successive columns and rows.
If all terms are reduced by 1 and the offset is changed to 0 we get A274528.
A274650 and A274651 are triangles in the shape of a right triangle and with a similar definition.
See A274630 for the case where both queens' and knights' moves must avoid duplicates.

import Data.List ((\\))
a269526 n = head $ [1..] \\ map a269526 (a274080_row n)
-- Peter Kagey, Jun 10 2016

# The following Maple program was provided at my request by Alois P. Heinz, who said that he had not posted it himself because it stores the data in an inefficient way. - N. J. A. Sloane, Jul 01 2016
A:= proc(n, k) option remember; local m, s;
         if n=1 and k=1 then 1
       else s:= {seq(A(i,k), i=1..n-1),
                 seq(A(n,j), j=1..k-1),
                 seq(A(n-t,k-t), t=1..min(n,k)-1),
                 seq(A(n+j,k-j), j=1..k-1)};
            for m while m in s do od; m
         fi
     end:
[seq(seq(A(1+d-k, k), k=1..d), d=1..15)];

A[n_, k_] := A[n, k] = If[n == 1 && k == 1, 1, s = {Table[A[i, k], {i, 1, n-1}], Table[A[n, j], {j, 1, k-1}], Table[A[n-t, k-t], {t, 1, Min[n, k] - 1}], Table[A[n+j, k-j], {j, 1, k-1}]} // Flatten; For[m = 1, True, m++, If[FreeQ[s, m], Return[m]]]];
Table[Table[A[1+d-k, k], {k, 1, d}], {d, 1, 15}] // Flatten (* Jean-François Alcover, Jul 21 2016, translated from Maple *)

{M269526=Map(); A269526=T(r,c)=c>1 && !mapisdefined(M269526, [r,c], &r) && mapput(M269526, [r,c], r=sum(k=1, #c=Set(concat([[T(r+k,c+k)|k<-[1-min(r, c)..-1]], [T(r,k)|k<-[1..c-1]], [T(k,c)|k<-[1..r-1]], [T(r+c-k,k)|k<-[1..c-1]]])), c[k]==k)+1); r} \\ M. F. Hasler, Sep 26 2022

Theorem 1: T(n,1) = n.
Proof by induction. T(1,1)=1 by definition. When calculating T(n,1), the only constraint is that it be different from all earlier entries in the first column, which are 1,2,3,...,n-1. So T(n,1)=n. QED
Theorem 2 (Based on a message from Bob Selcoe, Jun 29 2016): Write n = 4t+i with t >= 0, i=1,2,3, or 4. Then T(n,2) = 4t+3 if i=1, 4t+4 if i=2, 4t+1 if i=3, 4t+2 if i=4. This implies that the second column is the permutation A256008.
Proof: We check that the first 4 entries in column 2 are 2,5,6,3. From then on, to calculate the entry T(n,2), we need only look to the N, NW, W, and SW (we need never look to the East). After we have found the first 4t entries in the column, the column contains all the numbers from 1 to 4t. The four smallest free numbers are 4t+1, 4t+2, 4t+3, 4t+4. Entry T(4t+1,2) cannot be 4t+1 or 4t+2, but it can (and therefore must) be 4t+3. Similarly T(4t+2,2)=4t+4, T(4t+3,2)=4t+1, and T(4t+4,2)=4t+2. The column now contains all the numbers from 1 to 4t+4. Repeating this argument established the theorem. QED
Comments from Bob Selcoe, Jun 29 2016: (Start)
From Theorem 2, column 2 (i.e., terms a((j^2+j+4)/2), j>=1) is a permutation. After a(3)=3, the differences of successive terms follow the pattern a(n) = 3 [+1, -3, +1, +5], so a(5)=4, a(8)=1, a(12)=2, a(17)=7, a(23)=8, a(30)=5...
Similarly, column 3 (i.e., terms a((j^2+j+6)/2), j>=2) appears to be a permutation, but with the pattern after a(6)=2 and a(9)=5 being 5 [+1, -3, -2, +8, -5, +3, +1, +5, +1, -3, +1, -2, +8, -3, +1, +5]. (See A274614 and A274615.)
I conjecture that other similar cyclical difference patterns should hold for any column k (i.e., terms a((j^2+j+2*k)/2), j>=k-1), so that each column is a permutation.
Also, the differences in column 1 are a 1-cycle ([+1]), in column 2 a 4-cycle after the first term, and in column 3 a 16-cycle after the second term. Perhaps the cycle lengths are 4^(k-1) starting after j=k-1. (End)  WARNING: These comments may be wrong - see COMMENTS section. - N. J. A. Sloane, Sep 26 2022

Definition clarified by Omar E. Pol, Jun 29 2016

A065188 "Greedy Queens" permutation of the positive integers.

Original entry on oeis.org

1, 3, 5, 2, 4, 9, 11, 13, 15, 6, 8, 19, 7, 22, 10, 25, 27, 29, 31, 12, 14, 35, 37, 39, 41, 16, 18, 45, 17, 48, 20, 51, 53, 21, 56, 58, 60, 23, 63, 24, 66, 28, 26, 70, 72, 74, 76, 78, 30, 32, 82, 84, 86, 33, 89, 34, 92, 38, 36, 96, 98, 100, 102, 40, 105, 107, 42, 110, 43, 113

Offset: 1

Antti Karttunen, Oct 19 2001

nonn

This permutation is produced by a simple greedy algorithm: starting from the top left corner, walk along each successive antidiagonal of an infinite chessboard and place a queen in the first available position where it is not threatened by any of the existing queens. In other words, this permutation satisfies the condition that p(i+d) <> p(i)+-d for all i and d >= 1.
p(n) = k means that a queen appears in column n in row k. - N. J. A. Sloane, Aug 18 2016
That this is a permutation follows from the proof in A269526 that every row and every column in that array is a permutation of the positive integers. In particular, every row and every column contains a 1 (which translates to a queen in the present sequence). - N. J. A. Sloane, Dec 10 2017
The graph of this sequence shows two straight lines of respective slope equal to the Golden Ratio A001622, Phi = 1+phi = (sqrt(5)+1)/2 and phi = 1/Phi = (sqrt(5)-1)/2. - M. F. Hasler, Jan 13 2018
One has a(42) = 28 and a(43) = 26. Such irregularities make it difficult to get an explicit formula. They would not occur if the squares on the antidiagonals had been checked for possible positions starting from the opposite end, so as to ensure that the subsequences corresponding to the points on either line would both be increasing. Then one would have that a(n-1) is either round(n*phi)+1 or round(n/phi)+1. (The +-1's could all be avoided if the origin were taken as a(0) = 0 instead of a(1) = 1.) Presently most values are such that either round(n*phi) or round(n/phi) does not differ by more than 1 from a(n-1)-1, except for very few exceptions of the above form (a(42) being the first of these). - M. F. Hasler, Jan 15 2018
Equivalently, a(n) is the least positive integer not occurring earlier and so that |a(n)-a(k)| <> |n-k| for all k < n; i.e., fill the first quadrant column by column with lowest possible peaceful queens. - M. F. Hasler, Jan 11 2022

			The top left corner of the board is:
  +------------------------
  | Q x x x x x x x x x ...
  | x x x Q x x x x x x ...
  | x Q x x x x x x x x ...
  | x x x x Q x x x x x ...
  | x x Q x x x x x x x ...
  | x x x x x x x x x Q ...
  | x x x x x x x x x x ...
  | x x x x x x x x x x ...
  | x x x x x Q x x x x ...
  | ...
which illustrates p(1)=1, p(2)=3, p(3)=5, p(4)=2, etc. - _N. J. A. Sloane_, Aug 18 2016, corrected Aug 21 2016

Alois P. Heinz, Table of n, a(n) for n = 1..10000
F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
Matteo Fischetti and Domenico Salvagnin, Chasing First Queens by Integer Programming, 2018.
Matteo Fischetti and Domenico Salvagnin, Finding First and Most-Beautiful Queens by Integer Programming, arXiv:1907.08246 [cs.DS], 2019.
N. J. A. Sloane, Scatterplot of first 100 terms
N. J. A. Sloane, Table of n, a(n) for n = 1..50000 [Obtained using the Maple program of _Alois P. Heinz_]
Index entries for sequences that are permutations of the natural numbers

A065185 gives the associated p(i)-i delta sequence. A065186 gives the corresponding permutation for "promoted rooks" used in Shogi, A065257 gives "Quintal Queens" permutation.
A065189 gives inverse permutation.
See A199134, A275884, A275890, A275891, A275892 for information about the split of points below and above the diagonal.
Cf. A269526.
If we subtract 1 and change the offset to 0 we get A275895, A275896, A275893, A275894.
Tracking at which squares along the successive antidiagonals the queens appear gives A275897 and A275898.
Antidiagonal and diagonal indices give A276324 and A276325.

SquareThreatened := proc(a,i,j,upto_n,senw,nesw) local k; for k from 1 to i do if a[k,j] > 0 then RETURN(1); fi; od; for k from 1 to j do if a[i,k] > 0 then RETURN(1); fi; od; if 1 = i and 1 = j then RETURN(0); fi; for k from 1 to `if`((-1 = senw),min(i,j)-1,senw) do if a[i-k,j-k] > 0 then RETURN(1); fi; od; for k from 1 to `if`((-1 = nesw),i-1,nesw) do if a[i-k,j+k] > 0 then RETURN(1); fi; od; for k from 1 to `if`((-1 = nesw),j-1,nesw) do if a[i+k,j-k] > 0 then RETURN(1); fi; od; RETURN(0); end;
GreedyNonThreateningPermutation := proc(upto_n,senw,nesw) local a,i,j; a := array(1..upto_n,1..upto_n); for i from 1 to upto_n do for j from 1 to upto_n do a[i,j] := 0; od; od; for j from 1 to upto_n do for i from 1 to j do if 0 = SquareThreatened(a,i,(j-i+1),upto_n,senw,nesw) then a[i,j-i+1] := 1; fi; od; od; RETURN(eval(a)); end;
PM2PL := proc(a,upto_n) local b,i,j; b := []; for i from 1 to upto_n do for j from 1 to upto_n do if a[i,j] > 0 then break; fi; od; b := [op(b),`if`((j > upto_n),0,j)]; od; RETURN(b); end;
GreedyQueens := upto_n -> PM2PL(GreedyNonThreateningPermutation(upto_n,-1,-1),upto_n);GreedyQueens(256);
# From Alois P. Heinz, Aug 19 2016: (Start)
max_diagonal:= 3 * 100: # make this about 3*max number of terms
h:= proc() true end:   # horizontal line free?
v:= proc() true end:   # vertical   line free?
u:= proc() true end:   # up     diagonal free?
d:= proc() true end:   # down   diagonal free?
a:= proc() 0 end:      # for A065188
b:= proc() 0 end:      # for A065189
for t from 2 to max_diagonal do
   if u(t) then
      for j to t-1 do
        i:= t-j;
        if v(j) and h(i) and d(i-j) then
          v(j),h(i),d(i-j),u(i+j):= false$4;
          a(j):= i;
          b(i):= j;
          break
        fi
      od
   fi
od:
seq(a(n), n=1..100); # this is A065188
seq(b(n), n=1..100); # this is A065189 # (End)

Fold[Function[{a, n}, Append[a, 2 + LengthWhile[Differences@ Union@ Apply[Join, MapIndexed[Select[#2 + #1 {-1, 0, 1}, # > 0 &] & @@ {n - First@ #2, #1} &, a]], # == 1 &]]], {1}, Range[2, 70]] (* Michael De Vlieger, Jan 14 2018 *)

A065188_first(N, a=List(), u=[0])={for(n=1,N, for(x=u[1]+1,oo, setsearch(u,x) && next; for(i=1,n-1, abs(x-a[i])==n-i && next(2)); u=setunion(u,[x]); while(#u>1 && u[2]==u[1]+1, u=u[^1]); listput(a,x); break));a} \\ M. F. Hasler, Jan 11 2022

It would be nice to have a formula! - N. J. A. Sloane, Jun 30 2016

a(n) = A275895(n-1)-1. - M. F. Hasler, Jan 11 2022

A065256 Quintal Queens permutation of N: halve or multiply by 3 (mod 5) each digit (0->0, 1->3, 2->1, 3->4, 4->2) of the base 5 representation of n.

Original entry on oeis.org

0, 3, 1, 4, 2, 15, 18, 16, 19, 17, 5, 8, 6, 9, 7, 20, 23, 21, 24, 22, 10, 13, 11, 14, 12, 75, 78, 76, 79, 77, 90, 93, 91, 94, 92, 80, 83, 81, 84, 82, 95, 98, 96, 99, 97, 85, 88, 86, 89, 87, 25, 28, 26, 29, 27, 40, 43, 41, 44, 42, 30, 33, 31, 34, 32, 45, 48, 46, 49, 47, 35, 38

Offset: 0

Antti Karttunen, Oct 26 2001

All the permutations A004515 and A065256-A065258 consist of the first fixed term ("Queen on the corner") plus infinitely many 4-cycles and they satisfy the "nonattacking queen condition" that p(i+d) <> p(i)+-d for all i and d >= 1.

The corresponding infinite permutation matrix is a scale-invariant fractal (cf. A048647) and any subarray (5^i) X (5^i) (i >= 1) cut from its corner gives a solution to the case n=5^i of the n nonattacking queens on n X n chessboard (A000170). Is there any permutation of N which would give solutions to the queen problem with more frequent intervals than A000351?

Index entries for sequences that are permutations of the natural numbers

Inverse permutation: A004515. A065256[n] = A065258[n+1]-1. Cf. also A065187, A065189.

[seq(QuintalQueens0Inv(j),j=0..124)];
HalveDigit := (d,b) -> op(2,op(1,msolve(2*x=d,b))); # b should be an odd integer >= 3 and d should be in range [0,b-1].
HalveDigits := proc(n,b) local i; add((b^i)*HalveDigit((floor(n/(b^i)) mod b),b),i=0..floor(evalf(log[b](n+1)))+1); end;
QuintalQueens0Inv := n -> HalveDigits(n,5);

HalveDigit[d_, b_ /; OddQ[b] && b >= 3] /; 0 <= d <= b - 1 := Module[{x}, x /. Solve[2*x == d, x, Modulus -> b][[1]]];
HalveDigits[n_, b_] := Sum[b^i*HalveDigit[Mod[Floor[n/b^i] , b], b], {i, 0, Floor[Log[b, n + 1]]}];
QuintalQueens0Inv[n_] := HalveDigits[n, 5];
Table[QuintalQueens0Inv[n], {n, 0, 80}] (* Jean-François Alcover, Mar 05 2016, adapted from Maple *)

Edited by Charles R Greathouse IV, Nov 01 2009

A275895 "Greedy Queens" permutation of the nonnegative integers.

Original entry on oeis.org

0, 2, 4, 1, 3, 8, 10, 12, 14, 5, 7, 18, 6, 21, 9, 24, 26, 28, 30, 11, 13, 34, 36, 38, 40, 15, 17, 44, 16, 47, 19, 50, 52, 20, 55, 57, 59, 22, 62, 23, 65, 27, 25, 69, 71, 73, 75, 77, 29, 31, 81, 83, 85, 32, 88, 33, 91, 37, 35, 95, 97, 99, 101, 39, 104, 106, 41, 109, 42, 112, 43, 115, 117, 119, 45, 122

Offset: 0

N. J. A. Sloane, Aug 23 2016

nonn

This permutation is produced by a simple greedy algorithm: starting from the top left corner of an infinite chessboard placed in the fourth quadrant of the plane, walk along successive antidiagonals and place a queen in the first available position where it is not threatened by any of the existing queens. In other words, this permutation satisfies the condition that p(i+d) <> p(i)+-d for all i and d >= 1.
The rows and columns are indexed starting at 0. p(n) = k means that a queen appears in column n in row k. - N. J. A. Sloane, Aug 18 2016
All of A065188 (same for positive integers), A065189, A199134, A275884 should really have started at 0 rather than 1. Then the graph of A065188, for example, would be comparable with the graph of A002251.
That this is a permutation of the nonnegative integers follows from the proof in A269526 that every row and every column in that array is a permutation of the positive integers. In particular, every row and every column contains a 0 (which translates to a queen in the present sequence). - N. J. A. Sloane, Dec 10 2017

N. J. A. Sloane, Table of n, a(n) for n = 0..9999
F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
N. J. A. Sloane, Table of n, a(n) for n = 0..49999

Cf. A065188 (same for positive integers), A065189 (it's inverse), A199134 (indices of a(n) < n), A275884 (complement), A275894 (same for "nonnegative", i.e., this sequence), A275896 (same for A065189), A002251 (Wythoff pairs).

a(n) = A065188(n+1)-1.

cp's OEIS Frontend

A275896 a(n) = A065189(n+1)-1.

Views

Author

Keywords

Comments

Links

Crossrefs

A269526 Square array T(n,k) (n>=1, k>=1) read by antidiagonals upwards in which each term is the least positive integer satisfying the condition that no row, column, diagonal, or antidiagonal contains a repeated term.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Formula

Extensions

A065188 "Greedy Queens" permutation of the positive integers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A065256 Quintal Queens permutation of N: halve or multiply by 3 (mod 5) each digit (0->0, 1->3, 2->1, 3->4, 4->2) of the base 5 representation of n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Extensions

A275895 "Greedy Queens" permutation of the nonnegative integers.

Views

Author

Keywords

Comments

Links

Crossrefs

Formula

A275894 a(n) = A275884(n+1) - 1.

Views

Author

Keywords

Comments

Links

Crossrefs

A275893 a(n) = A199134(n+1)-1.

Views

Author

Keywords

Comments

Links

Crossrefs

A065258 Quintal Queens permutation of N: halve or multiply by 3 (mod 5) each digit (0->0, 1->3, 2->1, 3->4, 4->2) of the base 5 representation of n-1, add one.

Views

Author

Keywords

Comments

Links

Crossrefs