A283961
Rank array, R, of (golden ratio)^2, by antidiagonals.
Original entry on oeis.org
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, 49, 59, 15, 21, 28, 36, 45, 55, 66, 78, 19, 25, 33, 41, 51, 62, 73, 86, 99, 23, 30, 38, 47, 57, 69, 81, 94, 108, 123, 27, 35, 44, 53, 64, 76, 89, 103, 117, 133
Offset: 1
Northwest corner of R:
1 2 3 5 7 9 12 15
4 6 8 11 14 17 21 25
10 13 16 20 24 28 33 38
18 22 26 31 36 41 47 53
29 34 39 45 51 57 64 71
43 49 55 62 69 76 84 92
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).
From _Indranil Ghosh_, Mar 19 2017: (Start)
Triangle formed when the array is read by antidiagonals:
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, 49, 59;
15, 21, 28, 36, 45, 55, 66, 78;
19, 25, 33, 41, 51, 62, 73, 86, 99;
23, 30, 38, 47, 57, 69, 81, 94, 108, 123;
...
(End)
-
r = GoldenRatio^2; z = 100;
s[0] = 1; s[n_] := s[n] = s[n - 1] + 1 + Floor[n*r];
u = Table[n + 1 + Sum[Floor[(n - k)/r], {k, 0, n}], {n, 0, z}]; (* A283968, row 1 of A283961 *)
v = Table[s[n], {n, 0, z}]; (* A283969, col 1 of A283961 *)
w[i_, j_] := v[[i]] + u[[j]] + (i - 1)*(j - 1) - 1;
Grid[Table[w[i, j], {i, 1, 10}, {j, 1, 10}]] (* A283961 *)
v1 = Flatten[Table[w[k, n - k + 1], {n, 1, 60}, {k, 1, n}]] (* A283961,sequence *)
-
\\ This code produces the triangle mentioned in the example section
r = (3 +sqrt(5))/2;
z = 100;
s(n) = if(n<1, 1, s(n - 1) + 1 + floor(n*r));
p(n) = n + 1 + sum(k=0, n, floor((n - k)/r));
u = v = vector(z + 1);
for(n=1, 101, (v[n] = s(n - 1)));
for(n=1, 101, (u[n] = p(n - 1)));
w(i,j) = v[i] + u[j] + (i - 1) * (j - 1) - 1;
tabl(nn) = {for(n=1, nn, for(k=1, n, print1(w(k, n - k + 1),", ");); print(););};
tabl(10) \\ Indranil Ghosh, Mar 19 2017
A283968
a(n) = a(n-1) + 1 + floor(n*(3 + sqrt(5))/2), a(0) = 1.
Original entry on oeis.org
1, 2, 3, 5, 7, 9, 12, 15, 19, 23, 27, 32, 37, 42, 48, 54, 61, 68, 75, 83, 91, 100, 109, 118, 128, 138, 148, 159, 170, 182, 194, 206, 219, 232, 245, 259, 273, 288, 303, 318, 334, 350, 367, 384, 401, 419, 437, 455, 474, 493, 513, 533, 553, 574, 595, 617, 639
Offset: 0
-
r = GoldenRatio^2; z = 120;
s[0] = 1; s[n_] := s[n] = s[n - 1] + 1 + Floor[n*r];
Table[n + 1 + Sum[Floor[(n - k)/r], {k, 0, n}], {n, 0, z}] (* A283968 *)
Table[s[n], {n, 0, z}] (* A283969 *)
-
r = (3 + sqrt(5))/2;
a(n) = n + 1 + sum(k=0, n, floor((n - k)/r));
for(n=0, 30, print1(a(n),", ")) \\ Indranil Ghosh, Mar 19 2017
-
from sympy import sqrt
import math
def a(n):
return n + 1 + sum([int(math.floor((n - k)/r)) for k in range(n + 1)])
print([a(n) for n in range(61)]) # Indranil Ghosh, Mar 19 2017
A283938
Interspersion of the signature sequence of tau^2, where tau = (1 + sqrt(5))/2 = golden ratio.
Original entry on oeis.org
1, 4, 2, 10, 6, 3, 18, 13, 8, 5, 29, 22, 16, 11, 7, 43, 34, 26, 20, 14, 9, 59, 49, 39, 31, 24, 17, 12, 78, 66, 55, 45, 36, 28, 21, 15, 99, 86, 73, 62, 51, 41, 33, 25, 19, 123, 108, 94, 81, 69, 57, 47, 38, 30, 23, 150, 133, 117, 103, 89, 76, 64, 53, 44, 35
Offset: 1
Northwest corner:
1 4 10 18 29 43 59 78 99 123
2 6 13 22 34 49 66 86 108 133
3 8 16 26 39 55 73 94 117 143
5 11 20 31 45 62 81 103 127 154
7 14 24 36 51 69 89 112 137 165
9 17 28 41 57 76 97 121 147 176
From _Indranil Ghosh_, Mar 19 2017: (Start)
Triangle formed when the array is read by antidiagonals:
1;
4, 2;
10, 6, 3;
18, 13, 8, 5;
29, 22, 16, 11, 7;
43, 34, 26, 20, 14, 9;
59, 49, 39, 31, 24, 17, 12;
78, 66, 55, 45, 36, 28, 21, 15;
99, 86, 73, 62, 51, 41, 33, 25, 19;
123, 108, 94, 81, 69, 57, 47, 38, 30, 23;
...
(End)
-
r = GoldenRatio^2; z = 100;
s[0] = 1; s[n_] := s[n] = s[n - 1] + 1 + Floor[n*r];
u = Table[n + 1 + Sum[Floor[(n - k)/r], {k, 0, n}], {n, 0, z}] (* A283968, row 1 of A283938 *)
v = Table[s[n], {n, 0, z}] (* A283969, col 1 of A283938 *)
w[i_, j_] := v[[i]] + u[[j]] + (i - 1)*(j - 1) - 1;
Grid[Table[w[i, j], {i, 1, 10}, {j, 1, 10}]] (* A283938, array *)
Flatten[Table[w[k, n - k + 1], {n, 1, 20}, {k, 1, n}]] (* A283938, sequence *)
-
\\ This code produces the triangle mentioned in the example section
r = (3 +sqrt(5))/2;
z = 100;
s(n) = if(n<1, 1, s(n - 1) + 1 + floor(n*r));
p(n) = n + 1 + sum(k=0, n, floor((n - k)/r));
u = v = vector(z + 1);
for(n=1, 101, (v[n] = s(n - 1)));
for(n=1, 101, (u[n] = p(n - 1)));
w(i,j) = v[i] + u[j] + (i - 1) * (j - 1) - 1;
tabl(nn) = {for(n=1, nn, for(k=1, n, print1(w(n - k + 1, k),", ");); print(););};
tabl(10) \\ Indranil Ghosh, Mar 19 2017
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