A283961 - OEIS

%I #19 Mar 19 2017 12:32:44

%S 1,2,4,3,6,10,5,8,13,18,7,11,16,22,29,9,14,20,26,34,43,12,17,24,31,39,

%T 49,59,15,21,28,36,45,55,66,78,19,25,33,41,51,62,73,86,99,23,30,38,47,

%U 57,69,81,94,108,123,27,35,44,53,64,76,89,103,117,133

%N Rank array, R, of (golden ratio)^2, by antidiagonals.

%C Antidiagonals n = 1..60, flattened.

%C Every row intersperses all other rows, and every column intersperses all other columns.  The array is the dispersion of the complement of column 1, where column 1 is given by c(n) = c(n-1) + 1 + U(n), where U = upper Wythoff sequence (A001950).

%H Clark Kimberling, <a href="/A283961/b283961.txt">Table of n, a(n) for n = 1..1829</a>

%H Clark Kimberling and John E. Brown, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL7/Kimberling/kimber67.html">Partial Complements and Transposable Dispersions</a>, J. Integer Seqs., Vol. 7, 2004.

%F R(i,j) = R(i,0) + R(0,j) + i*j - 1, for i>=1, j>=1.

%e Northwest corner of R:

%e 1  2  3  5  7  9  12 15

%e 4  6  8  11 14 17 21 25

%e 10 13 16 20 24 28 33 38

%e 18 22 26 31 36 41 47 53

%e 29 34 39 45 51 57 64 71

%e 43 49 55 62 69 76 84 92

%e Let t = (golden ratio)^2 = (3 + sqrt(5))/2; then R(i,j) = rank of (j,i) when all nonnegative integer pairs (a,b) are ranked by the relation << defined as follows: (a,b) << (c,d) if a + b*t < c + d*t, and also (a,b) << (c,d) if a + b*t = c + d*t and b < d. Thus R(2,0) = 10 is the rank of (0,2) in the list (0,0) << (1,0) << (2,0) << (0,1) << (3,0) << (1,1) << (4,0) << (2,1) << (5,0) << (0,2).

%e From _Indranil Ghosh_, Mar 19 2017: (Start)

%e Triangle formed when the array is read by antidiagonals:

%e    1;

%e    2,  4;

%e    3,  6, 10;

%e    5,  8, 13, 18;

%e    7, 11, 16, 22, 29;

%e    9, 14, 20, 26, 34, 43;

%e   12, 17, 24, 31, 39, 49, 59;

%e   15, 21, 28, 36, 45, 55, 66, 78;

%e   19, 25, 33, 41, 51, 62, 73, 86, 99;

%e   23, 30, 38, 47, 57, 69, 81, 94, 108, 123;

%e   ...

%e (End)

%t r = GoldenRatio^2; z = 100;

%t s[0] = 1; s[n_] := s[n] = s[n - 1] + 1 + Floor[n*r];

%t u = Table[n + 1 + Sum[Floor[(n - k)/r], {k, 0, n}], {n, 0, z}]; (* A283968, row 1 of A283961 *)

%t v = Table[s[n], {n, 0, z}]; (* A283969, col 1 of A283961 *)

%t w[i_, j_] := v[[i]] + u[[j]] + (i - 1)*(j - 1) - 1;

%t Grid[Table[w[i, j], {i, 1, 10}, {j, 1, 10}]] (* A283961 *)

%t v1 = Flatten[Table[w[k, n - k + 1], {n, 1, 60}, {k, 1, n}]] (* A283961,sequence *)

%o (PARI)

%o \\ This code produces the triangle mentioned in the example section

%o r = (3 +sqrt(5))/2;

%o z = 100;

%o s(n) = if(n<1, 1, s(n - 1) + 1 + floor(n*r));

%o p(n) = n + 1 + sum(k=0, n, floor((n - k)/r));

%o u = v = vector(z + 1);

%o for(n=1, 101, (v[n] = s(n - 1)));

%o for(n=1, 101, (u[n] = p(n - 1)));

%o w(i,j) = v[i] + u[j] + (i - 1) * (j - 1) - 1;

%o tabl(nn) = {for(n=1, nn, for(k=1, n, print1(w(k, n - k + 1),", ");); print(););};

%o tabl(10) \\ _Indranil Ghosh_, Mar 19 2017

%Y Cf. A001622, A118276, A283961, A283968, A283969.

%K nonn,tabl,easy

%O 1,2

%A _Clark Kimberling_, Mar 18 2017