A019446 -id:A019446 - cp's OEIS frontend

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

Clark Kimberling, Oct 31 2009

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

			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

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

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

Cf. A001622, A084531, A084531, A283734.

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 *)

\\ 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

# 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

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

A194968 Fractalization of (1+[n/r]), where [ ]=floor, r=(1+sqrt(5))/2 (the golden ratio), and n>=1.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling, Sep 07 2011

nonn

See A194959 for a discussion of fractalization and the interspersion fractally induced by a sequence. The sequence (1+[n/r]) is A019446.

Cf. A194959, A019446, A194969, A194970.

r = GoldenRatio; p[n_] := 1 + Floor[n/r]
Table[p[n], {n, 1, 90}]  (* A019446 *)
g[1] = {1}; g[n_] := Insert[g[n - 1], n, p[n]]
f[1] = g[1]; f[n_] := Join[f[n - 1], g[n]]
f[20]  (* A194968 *)
row[n_] := Position[f[30], n];
u = TableForm[Table[row[n], {n, 1, 5}]]
v[n_, k_] := Part[row[n], k];
w = Flatten[Table[v[k, n - k + 1], {n, 1, 13},
{k, 1, n}]]  (* A194969 *)
q[n_] := Position[w, n]; Flatten[Table[q[n],
{n, 1, 80}]]  (* A194970 *)

A371381 Main diagonal of A219875.

Original entry on oeis.org

2, 8, 13, 25, 41, 52, 74, 89, 117, 149, 170, 208, 250, 277, 325, 356, 410, 468, 505, 569, 610, 680, 754, 801, 881, 965, 1018, 1108, 1165, 1261, 1361, 1424, 1530, 1640, 1709, 1825, 1898, 2020, 2146, 2225, 2357, 2440, 2578, 2720, 2809, 2957, 3109, 3204, 3362, 3461

Offset: 1

Paolo Xausa, Mar 20 2024

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A219875.

Cf. A001622, A019446, A101332, A101863, A101867, A371382.

Array[#^2 + Ceiling[# / GoldenRatio]^2 &, 100]

from math import isqrt
def A371381(n): return (n<<1)*(n-1)+1+(q:=n+isqrt(5*n**2)>>1)*(q-(n-1<<1)) # Chai Wah Wu, Mar 21 2024

a(n) = n^2 + ceiling(n/(1 + sqrt(5))/2)^2 = n^2 + A019446(n)^2.

A187911 Rank transform of the sequence (1+floor(n*r)), where r=(-1+sqrt(5))/2; complement of A187912.

Original entry on oeis.org

1, 3, 4, 5, 7, 8, 10, 11, 13, 14, 15, 17, 19, 20, 21, 22, 24, 26, 27, 28, 29, 31, 33, 34, 36, 37, 38, 40, 41, 42, 44, 45, 47, 49, 50, 52, 53, 54, 56, 57, 58, 59, 61, 63, 64, 66, 68, 69, 70, 71, 73, 74, 75, 77, 78, 80, 81, 82, 84, 86, 87, 89, 90, 91, 93, 94, 96, 98, 99, 100, 101, 103, 105, 106, 107, 108, 110, 111, 112, 114, 116, 117, 118, 119

Offset: 1

Clark Kimberling, Mar 15 2011

nonn

See A187224.

Cf. A187224, A187912.

r=(-1+5^(1/2))/2;
seqA = Table[1+Floor[r*n], {n, 1, 220}] (* A019446 *)
seqB = Table[n, {n, 1, 220}];  (* A000027 *)
jointRank[{seqA_,
   seqB_}] := {Flatten@Position[#1, {_, 1}],
    Flatten@Position[#1, {_, 2}]} &[
  Sort@Flatten[{{#1, 1} & /@ seqA, {#1, 2} & /@ seqB}, 1]];
limseqU =
 FixedPoint[jointRank[{seqA, #1[[1]]}] &,
   jointRank[{seqA, seqB}]][[1]] (* A187911 *)
Complement[Range[Length[seqA]], limseqU]  (* A187912 *)
(* Peter J. C. Moses, Mar 15 2011 *)

A339765 a(n) = 2floor(nphi) - 3*n, where phi = (1+sqrt(5))/2.

Original entry on oeis.org

-1, 0, -1, 0, 1, 0, 1, 0, 1, 2, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 6, 5, 6, 5, 6, 7, 6, 7, 8, 7, 8, 7, 8, 9, 8, 9, 8, 9, 10, 9, 10, 11, 10, 11, 10, 11, 12, 11, 12, 11, 12, 13, 12, 13, 14, 13, 14, 13, 14, 15, 14, 15, 16, 15, 16, 15, 16, 17, 16, 17, 16

Offset: 1

Primoz Pirnat, Dec 16 2020

sign

a(n) are coefficients in the formulas for multiplication of fractional parts of multiples of the golden mean:
(I)    frac(b*phi)*frac(c*phi) = 1-frac(d*phi); d = 2*b*c+a(b)*c/2+a(c)*b/2;
(IIa)  frac(b*phi)*(1-frac(c*phi)) = frac(e*phi); e = d+b;
(IIb)  (1-frac(b*phi))*frac(c*phi) = frac(f*phi); f = d+c;
(III)  (1-frac(b*phi))*(1-frac(c*phi)) = 1-frac(g*phi); g = d+b+c;
where frac() = FractionalPart(), phi = (1+sqrt(5))/2 and b,c are positive integers.
The parameters d,e,f,g are also positive integers.

			For b=3, c=10, a(3)=-1, a(10)=2 are solutions of upper formulas:
(I) frac(3*phi)*frac(10*phi) = 1-frac(58*phi); d = 2*3*10+a(3)*10/2+a(10)*3/2 = 58;
(IIa)  frac(3*phi)*(1-frac(10*phi)) = frac(61*phi); e = d+3 = 61;
(IIb)  (1-frac(3*phi))*frac(10*phi) = frac(68*phi); f = d+10 = 68;
(III)  (1-frac(3*phi))*(1-frac(10*phi)) = 1-frac(71*phi); g = d+3+10 = 71.

Cf. A000201, A001622, A005206, A050140, A189663, A019446, A060144.

Table[2Floor[n*GoldenRatio]-3n,{n,76}] (* Stefano Spezia, Dec 18 2020 *)

a(n) = 2*floor(n*quadgen(5)) - 3*n; \\ Michel Marcus, Jan 05 2021

from math import isqrt
def A339765(n): return ((n+isqrt(5*n**2))&~1)-3*n # Chai Wah Wu, Aug 09 2022

a(n) = 2*A000201(n) - 3*n.
a(n) = A005206(n-1) - A189663(n+1).
a(n) = A019446(n) - A060144(n) - sign(abs(n)) - 1.
From Primoz Pirnat, May 15 2024: (Start)
a(n) = A050140(n) - 2*n.
a(n) = 2*A005206(n-1) - n.
a(n) = n - 2*A189663(n+1). (End)

A134299 Maximal length of a sequence such that v(0)=n, v(k+2) = v(k)-v(k+1), v(k) >= 0.

Original entry on oeis.org

4, 5, 6, 5, 7, 6, 5, 8, 6, 7, 6, 6, 9, 6, 7, 8, 6, 7, 6, 7, 10, 6, 7, 8, 7, 9, 6, 7, 8, 7, 7, 8, 7, 11, 7, 7, 8, 7, 9, 8, 7, 10, 7, 7, 8, 7, 9, 8, 7, 8, 7, 9, 8, 7, 12, 8, 7, 8, 7, 9, 8, 7, 10, 8, 9, 8, 7, 11, 8, 7, 8, 8, 9, 8, 7, 10, 8, 9, 8, 8, 9, 8, 7, 10, 8, 9, 8, 8, 13, 8, 9, 8, 8, 9, 8, 8, 10, 8, 9, 8

Offset: 1

M. F. Hasler, Oct 18 2007

nonn

This is also the maximal index at which n can occur in a Fibonacci-like sequence u(k+2) = u(k)+u(k+1) of nonnegative numbers.

A sequence of this length is obtained for v(0) = n, v(1) = A019446(n) = ceiling(n/tau) or A060143(n) = floor(n/tau).

			a(2007)=11 since there is no such sequence longer than v = (2007, 1240, 767, 473, 294, 179, 115, 64, 51, 13, 38).

Eric Angelini, Max Alekseyev and M. F. Hasler, Longest Lucas sequence from start to n, postings to Sequence Fans mailing list, Oct 18, 2007.

Cf. A101803, A019446, A060143.

A134299( goal, mi=0, mx=0, new=0 ) = { for( j=mi,goal, a=[goal,new=j]; while( mi<=new=a[ #a-1]-new, a=concat(a,new)); if( #a>mx, mx=#a)); mx }

A300763 a(n) = ceiling(n/g^3), where g = (1+sqrt(5))/2 is the golden ratio.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 17, 18, 18

Offset: 0

Jeffrey Shallit, Jul 02 2018

nonn

Altug Alkan, Table of n, a(n) for n = 0..1000

Cf. A001622, A019446, which is ceiling(n/g), A189663, which is ceiling(n/g^2) (but shifted by one).

Cf. A098317.

Array[Ceiling[#/GoldenRatio^3] &, 90, 0] (* Robert G. Wilson v, Jul 02 2018 *)

a(n) = my(t=2+sqrt(5)); ceil(n/t); \\ Altug Alkan, Jul 02 2018

a(n) = ceiling(n/t) where t = 2*g + 1 = 2 + sqrt(5). - Altug Alkan, Jul 02 2018

A332502 Rectangular array read by antidiagonals: T(n,k) = floor(n + k*r), where r = golden ratio = (1+sqrt(5))/2.

Original entry on oeis.org

0, 1, 1, 3, 2, 2, 4, 4, 3, 3, 6, 5, 5, 4, 4, 8, 7, 6, 6, 5, 5, 9, 9, 8, 7, 7, 6, 6, 11, 10, 10, 9, 8, 8, 7, 7, 12, 12, 11, 11, 10, 9, 9, 8, 8, 14, 13, 13, 12, 12, 11, 10, 10, 9, 9, 16, 15, 14, 14, 13, 13, 12, 11, 11, 10, 10, 17, 17, 16, 15, 15, 14, 14, 13

Offset: 0

Clark Kimberling, May 08 2020

Every nonnegative integer occurs exactly once in the union of row 0 and the main diagonal.
Column 0: Nonnegative integers, A001477.
Row 0: Lower Wythoff sequence, A000201.
Row 1: A026351.
Row 2: A026355 (and essentially A099267).
Main Diagonal: Upper Wythoff sequence, A001950.
Diagonal (1,4,6,9,...) = A003622;
Diagonal (3,5,8,11,...) = A026274;
Diagonal (1,3,6,8,...) = A026352;
Diagonal (2,4,7,9,...) = A026356.

			Northwest corner:
  0   1   3   4   6   8    9    11   12   14   16
  1   2   4   5   7   9    10   12   13   15   17
  2   3   5   6   8   10   11   13   14   16   18
  3   4   6   7   9   11   12   14   15   17   19
  4   5   7   8   10  12   13   15   16   18   20
  5   6   8   9   11  13   14   16   17   19   21
As a triangle (antidiagonals):
  0
  1   1
  2   2   3
  3   3   4   4
  4   4   5   5   6
  5   5   6   6   7   8
  6   6   7   7   8   9   9

Cf. A000210, A001950, A001622, A084531, A167267, A019446.

t[n_, k_] := Floor[n + k*GoldenRatio];
Grid[Table[t[n, k], {n, 0, 10}, {k, 0, 10}]] (* array *)
u = Table[t[n - k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten  (* sequence *)

T(n,k) = floor(n + k*r), where r = golden ratio = (1+sqrt(5))/2.

A385177 a(n) = Sum_{k=1..n} ceiling(k/phi), where phi is the golden ratio (A001622).

Original entry on oeis.org

1, 3, 5, 8, 12, 16, 21, 26, 32, 39, 46, 54, 63, 72, 82, 92, 103, 115, 127, 140, 153, 167, 182, 197, 213, 230, 247, 265, 283, 302, 322, 342, 363, 385, 407, 430, 453, 477, 502, 527, 553, 579, 606, 634, 662, 691, 721, 751, 782, 813, 845, 878, 911, 945, 979, 1014, 1050, 1086, 1123

Offset: 1

Paolo Xausa, Jun 20 2025

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A001622, A183136.

Partial sums of A019446.

Accumulate[Ceiling[Range[100]/GoldenRatio]]

from math import isqrt
def A385177(n): return n+sum(isqrt(5*i**2)-i>>1 for i in range(1,n+1)) # Chai Wah Wu, Jun 20 2025

a(n) = A183136(n) + n.

cp's OEIS Frontend

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

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A194968 Fractalization of (1+[n/r]), where [ ]=floor, r=(1+sqrt(5))/2 (the golden ratio), and n>=1.

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A371381 Main diagonal of A219875.

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A187911 Rank transform of the sequence (1+floor(n*r)), where r=(-1+sqrt(5))/2; complement of A187912.

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A339765 a(n) = 2*floor(n*phi) - 3*n, where phi = (1+sqrt(5))/2.

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A134299 Maximal length of a sequence such that v(0)=n, v(k+2) = v(k)-v(k+1), v(k) >= 0.

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A300763 a(n) = ceiling(n/g^3), where g = (1+sqrt(5))/2 is the golden ratio.

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A332502 Rectangular array read by antidiagonals: T(n,k) = floor(n + k*r), where r = golden ratio = (1+sqrt(5))/2.

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A339765 a(n) = 2floor(nphi) - 3*n, where phi = (1+sqrt(5))/2.