A167267 Interspersion of the signature sequence of (1+sqrt(5))/2.

1, 3, 2, 7, 5, 4, 12, 10, 8, 6, 19, 16, 14, 11, 9, 28, 24, 21, 18, 15, 13, 38, 34, 30, 26, 23, 20, 17, 50, 45, 41, 36, 32, 29, 25, 22, 63, 58, 53, 48, 43, 39, 35, 31, 27, 78, 72, 67, 61, 56, 51, 46, 42, 37, 33

Offset: 1

Author

Clark Kimberling, Oct 31 2009

Keywords

• nonn
• tabl

Comments

Row n is the ordered sequence of numbers k such that A084531(k)=n. Is the difference sequence of column 1 equal to A019446? Is the difference sequence of row 1 essentially equal to A026351?

As a sequence, A167267 is a permutation of the positive integers. As an array, A167267 is the joint-rank array (defined at A182801) of the numbers {i+j*r}, for i>=1, j>=1, where r = golden ratio = (1+sqrt(5))/2. - Clark Kimberling, Nov 10 2012

This is a transposable interspersion; i.e., its transpose, A283734, is also an interspersion. - Clark Kimberling, Mar 16 2017

Examples

			Northwest corner:
1....3....7....12...19...28...38
2....5....10...16...24...34...45
4....8....14...21...30...41...53
6....11...18...26...36...48...61
9....15...23...32...43...56...70
13...20...29...39...51...65...80
		

References

• Clark Kimberling, "Fractal Sequences and Interspersions," Ars Combinatoria 45 (1997) 157-168.

Links

• Clark Kimberling, Antidiagonals n = 1..60, flattened
• N. Carey, Lambda Words: A Class of Rich Words Defined Over an Infinite Alphabet, J. Int. Seq. 16 (2013) #13.3.4

Crossrefs

Cf. A001622, A084531, A084531, A283734.

Programs

Mathematica

v = GoldenRatio;
x = Table[Sum[Ceiling[i*v], {i, q}], {q, 0, end = 35}];
y = Table[Sum[Ceiling[i*1/v], {i, q}], {q, 0, end}];
tot[p_, q_] := x[[p + 1]] + p q + 1 + y[[q + 1]]
row[r_] := Table[tot[n, r], {n, 0, (end - 1)/v}]
Grid[Table[row[n], {n, 0, (end - 1)}]]
(* Norman Carey, Jul 03 2012 *)

PARI

\\ Produces the triangle when the array is read by antidiagonals
r = (1+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 26 2017

Python

# Produces the triangle when the array is read by antidiagonals
import math
from sympy import sqrt
r=(1 + sqrt(5))/2
def s(n): return 1 if n<1 else s(n - 1) + 1 + int(math.floor(n*r))
def p(n): return n + 1 + sum(int(math.floor((n - k)/r)) for k in range(n+1))
v=[s(n) for n in range(101)]
u=[p(n) for n in range(101)]
def w(i, j): return u[i - 1] + v[j - 1] + (i - 1) * (j - 1) - 1
for n in range(1, 11):
    print([w(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 26 2017

Formula

R(m,n) = sum{[(m-i+n+r)/r], i=1,2,...z(m,n)}, where r = (1+sqrt(5))/2 and z(m,n) = m + [(n-1)*r]. - Clark Kimberling, Nov 10 2012