A176015 -id:A176015 - cp's OEIS frontend

A244847 Decimal expansion of rho_c = (5-sqrt(5))/10, the asymptotic critical density for the hard hexagon model.

2, 7, 6, 3, 9, 3, 2, 0, 2, 2, 5, 0, 0, 2, 1, 0, 3, 0, 3, 5, 9, 0, 8, 2, 6, 3, 3, 1, 2, 6, 8, 7, 2, 3, 7, 6, 4, 5, 5, 9, 3, 8, 1, 6, 4, 0, 3, 8, 8, 4, 7, 4, 2, 7, 5, 7, 2, 9, 1, 0, 2, 7, 5, 4, 5, 8, 9, 4, 7, 9, 0, 7, 4, 3, 6, 2, 1, 9, 5, 1, 0, 0, 5, 8, 5, 5, 8, 5, 5, 9, 1, 6, 2, 1, 2, 1, 7, 7, 2, 5, 0, 3

Offset: 0

Jean-François Alcover, Nov 12 2014

The vertical distance between the accumulation point and the outermost point of a golden spiral inscribed inside a golden rectangle with dimensions phi and 1 along the x and y axes, respectively (the horizontal distance is A176015). - Amiram Eldar, May 18 2021

The limiting frequency of the digit 1 in the base phi representation of real numbers in the range [0,1], where phi is the golden ratio (A001622) (Rényi, 1957). - Amiram Eldar, Mar 18 2025

			0.2763932022500210303590826331268723764559381640388474275729102754589479...

Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 1.2 The Golden Mean, phi, p. 7.
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 5.12.1 Phase transitions in Lattice Gas Models, p. 347.

Rodney J. Baxter Hard hexagons: exact solution, Journal of Physics A: Mathematical and General, Vol. 13, No. 3 (1980), pp. L61-L70, alternative link.
P. S. Bruckman and I. J. Good, A Generalization of a Series of De Morgan with Applications of Fibonacci Type, The Fibonacci Quarterly, Vol. 14, No. 3 (1976), pp. 193-196.
Alfréd Rényi, Representations for real numbers and their ergodic properties, Acta Math. Acad. Sci. Hungar., Vol.8, No. 3-4 (1957), pp. 477-493.
Eric Weisstein's MathWorld, Hard Hexagon Entropy Constant.
Wikipedia, Hard Hexagon Model.
Index entries for algebraic numbers, degree 2.

Cf. A242671, A244593.
Cf. A000032, A000045.
Essentially the same sequence of digits as A229760 and A187799.

RealDigits[(5 - Sqrt[5])/10, 10, 102] // First

Equals 1/(sqrt(5)*phi), where phi = (1+sqrt(5))/2 = A001622. - Vaclav Kotesovec, Nov 13 2014
Equals lim_{n -> infinity} A000045(n)/A000032(n+1). - Bruno Berselli, Jan 22 2018
Equals Sum_{n>=1} A000045(3^(n-1))/A000032(3^n) = Sum_{n>=1} A045529(n-1)/A006267(n). - Amiram Eldar, Dec 20 2018
Equals 1 - A242671. - Amiram Eldar, Mar 18 2025

A214986 Power ceiling array for the golden ratio, by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 7, 8, 5, 1, 1, 12, 21, 22, 7, 1, 1, 20, 55, 94, 48, 12, 1, 1, 33, 144, 399, 329, 134, 18, 1, 1, 54, 377, 1691, 2255, 1487, 323, 30, 1, 1, 88, 987, 7164, 15456, 16492, 5796, 872, 47, 1, 1, 143, 2584, 30348, 105937, 182900

Offset: 1

Clark Kimberling, Oct 28 2012

row 0: A000012 ... row 6: A049660
row 1: A000071 ... row 8: A049668
row 2: A001906 ... col 0: A000012
row 3: A049652 ... col 1: A169986
row 4: A004187
For x>1, define c(x,0) = 1 and c(x,n) = ceiling(x*c(x,n-1)) for n>0.  Row m of A214986 is the sequence c(r^m,n), where r = golden ratio = (1 + sqrt(5))/2.  The name of the array corresponds to the power ceiling function f(x) = limit of c(x,n)/x^n as n increases without bound; f(x) generalizes the case for x = 3/2 as described under "Power Ceilings" at MathWorld.  For a graph of f(x), see the Mathematica program at A083286.
The term "power ceiling sequence" extends to sequences generated by recurrences P(n) = ceiling(x*P(n-1)) + g(n), and "power ceiling functions" f(x) to the limit of P(n)/x^n in case x>1 and g(n)/x^n -> 0.
Suppose that h is a nonnegative integer and g(n) is a constant.  If x is a positive integer power of the golden ratio r, then f(x), in many cases, lies in the field Q(sqrt(5)).  Examples matching rows of A214986, using g(n) = 0, follow:
...
x ... P ........ f(x)
r ... A000071 .. (5 + 2*sqrt(5))/2 = 1.8944... (A010532)
r^2 . A001906 .. (5 + 3*sqrt(5))/10 = 1.7082...(A176015)
r^3 . A049652 .. (25 + 11*sqrt(5))/40 = 1.2399...
r^4 . A004187 .. (15 + 7*sqrt(5))/10 = 1.0219...
...
If k is odd, then f(r^k) = r^k((b(k) + c(k))/d(k)), where
b(k) = L(j)^2 + L(j-1)^2, where j=[(k+1)/2], L=A000032 (Lucas numbers); c(k) = (L(k)+2)*sqrt(5); d(k) = 10*F(k)*L(k), where F=A000045 (Fibonacci numbers).  If k is even, then f(r^k) = r^k/(F(k)*sqrt(5)).

			Northwest corner:
1...1....1.....1......1.......1
1...2....4.....7......12......20
1...3....8.....21.....55......144
1...5....22....94.....399.....1691
1...7....48....329....2255....15456
1...19...134...1487...16492...182900

Clark Kimberling, Antidiagonals n = 1..35, flattened
Eric Weisstein's World of Mathematics, Power Ceilings

Cf. A000045, A214978, A214984, A214987.

r = GoldenRatio;
s[x_, 0] := 1; s[x_, n_] := Ceiling[x*s[x, n - 1]];
t = TableForm[Table[s[r^m, n], {m, 0, 10}, {n, 0, 10}]  ]
u = Flatten[Table[s[r^m, n - m], {n, 0, 10}, {m, 0, n}]]

The odd-numbered rows of A214986 are even-numbered rows of A213978; the even-numbered rows of A214986 are odd-numbered rows of A214984.

A115339 a(2n-1)=F(n+1), a(2n)=L(n), where F(n) and L(n) are the Fibonacci and the Lucas sequences.

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 5, 7, 8, 11, 13, 18, 21, 29, 34, 47, 55, 76, 89, 123, 144, 199, 233, 322, 377, 521, 610, 843, 987, 1364, 1597, 2207, 2584, 3571, 4181, 5778, 6765, 9349, 10946, 15127, 17711, 24476, 28657, 39603, 46368, 64079, 75025, 103682, 121393, 167761

Offset: 1

Giuseppe Coppoletta, Mar 06 2006

Alternate Fibonacci and Lucas sequence respecting their natural order.
See A116470 for an essentially identical sequence.
The ratio a(n+1)/a(n) increasingly approximates two constants: (5-sqrt(5))/2 (A094874) and (5+3*sqrt(5))/10 (A176015) according to whether n is odd or even. - Davide Rotondo, Oct 27 2024

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Fibonacci Number
Eric Weisstein's World of Mathematics, Lucas Number.
Index entries for linear recurrences with constant coefficients, signature (0,1,0,1).

Cf. A000045, A000032.
Cf. A000930.
Cf. A094874, A176015.

a115339 n = a115339_list !! (n-1)
a115339_list = [1, 1, 2, 3] ++
               zipWith (+) a115339_list (drop 2 a115339_list)
-- Reinhard Zumkeller, Aug 03 2013

f[n_] := If[OddQ@n, Fibonacci[(n + 3)/2], Fibonacci[n/2 - 1] + Fibonacci[n/2 + 1]]; Array[f, 50] (* Robert G. Wilson v, Apr 29 2006 *)

x='x+O('x^50); Vec(x*(-1-x-x^2-2*x^3)/(-1+x^2+x^4)) \\ G. C. Greubel, Apr 27 2017

a(n+2) = a(n) + a(n-2).

G.f.: x*( -1-x-x^2-2*x^3 ) / ( -1+x^2+x^4 ). - R. J. Mathar, Mar 08 2011

More terms from Robert G. Wilson v, Apr 29 2006

A214987 Power round array for the golden ratio, by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 5, 8, 4, 1, 1, 8, 21, 17, 7, 1, 1, 13, 55, 72, 48, 11, 1, 1, 21, 144, 305, 329, 122, 18, 1, 1, 34, 377, 1292, 2255, 1353, 323, 29, 1, 1, 55, 987, 5473, 15456, 15005, 5796, 842, 47, 1, 1, 89, 2584, 23184, 105937, 166408, 104005

Offset: 1

Clark Kimberling, Oct 28 2012

The term "power round sequence" (after "power ceiling sequence" at A214986) extends to sequences generated by recurrences P(n) = round(x*P(n-1)) + g(n), and "power round functions" f(x) to the limit of P(n)/x^n in case x>1 and g(n)/x^n -> 0.  Suppose that h is a nonnegative integer and g(n) is a constant.  If x is a positive integer power of the golden ratio r, then f(x), in many cases, lies in the field Q(sqrt(5)).  Examples matching rows of A214987, using g(n) = 0, follow:
...
x ... P . .. . . f(x)
r ... A000045 .. 1/2 + 3*sqrt(5)/10 = 1.1708... (A176015)
r^2 . A001906 .. 1/2 + 3*sqrt(5)/10 = 1.1708... (A176015)
r^3 . A001076 .. 1/2 + sqrt(5)/5 = 0.9472...
r^4 . A004187 .. 1/2 + 7*sqrt(5)/30 = 1.0217...
In general, f(r^k) = 1/2 + sqrt(5)*L(k)/(10*F(k)) for k>1, where L = A000032 (Lucas numbers) and F = A000045 (Fibonacci numbers).
(row 2 of A214987) = (row 1 of A213978 except for its initial 1)
(row n of A214987) = (row n-1 of A213978 for n>2).

			1...1...1....1.....1......1
1...2...3....5.....8......13
1...3...8....21....5......144
1...4...17...72....305....1292
1...7...48...329...2255...15456

Clark Kimberling, Antidiagonals n = 1..35, flattened

Cf. A000045, A214978, A214984, A214986.

r = GoldenRatio;
s[x_, 0] := 1; s[x_, n_] := Round[x*s[x, n - 1]];
t = TableForm[Table[s[r^m, n], {m, 0, 10}, {n, 0, 10}]  ]
u = Flatten[Table[s[r^m, n - m], {n, 0, 10}, {m, 0, n}]]

A380981 Decimal expansion of the medium/short edge length ratio of a disdyakis triacontahedron.

Original entry on oeis.org

1, 5, 7, 0, 8, 2, 0, 3, 9, 3, 2, 4, 9, 9, 3, 6, 9, 0, 8, 9, 2, 2, 7, 5, 2, 1, 0, 0, 6, 1, 9, 3, 8, 2, 8, 7, 0, 6, 3, 2, 1, 8, 5, 5, 0, 7, 8, 8, 3, 4, 5, 7, 7, 1, 7, 2, 8, 1, 2, 6, 9, 1, 7, 3, 6, 2, 3, 1, 5, 6, 2, 7, 7, 6, 9, 1, 3, 4, 1, 4, 6, 9, 8, 2, 4, 3, 2, 4, 3, 2

Offset: 1

Paolo Xausa, Feb 10 2025

			1.57082039324993690892275210061938287063218550788...

Paolo Xausa, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Disdyakis Triacontahedron.
Wikipedia, Disdyakis triacontahedron.

Cf. A380982 (long/short edge length ratio).
Cf. A002163, A379388, A379708, A379709, A379710, A379711, A380940, A380941, A380942.
Apart from leading digits the same as A176015, A134976 and A010499.

First[RealDigits[3/10*(3 + Sqrt[5]), 10, 100]]

Equals (3/10)*(3 + sqrt(5)) = (3/10)*(3 + A002163).

Equals A176015 + 2/5.

A342089 Numbers that have two representations as the sum of distinct non-consecutive Lucas numbers (A000032).

Original entry on oeis.org

5, 12, 16, 23, 30, 34, 41, 45, 52, 59, 63, 70, 77, 81, 88, 92, 99, 106, 110, 117, 121, 128, 135, 139, 146, 153, 157, 164, 168, 175, 182, 186, 193, 200, 204, 211, 215, 222, 229, 233, 240, 244, 251, 258, 262, 269, 276, 280, 287, 291, 298, 305, 309, 316, 320, 327

Offset: 1

Amiram Eldar, Feb 27 2021

nonn

Brown (1969) proved that every positive number has a unique representation as a sum of non-consecutive Lucas numbers, if L(0) = 2 and L(2) = 3 do not appear simultaneously in the representation.
Chu et al. (2020) proved that if L(0) and L(2) are allowed to appear simultaneously, then each positive number can have at most two representations. The terms with two representations are listed in this sequence. They found that the number of terms that do not exceed 10^k, for k = 1, 2, ..., are 1, 17, 171, 1708, 17082, 170820, ..., and proved that the asymptotic density of this sequence is 1/(3*phi+1) = 0.1708203932... (A176015 - 1), where phi is the golden ratio (A001622).
A number n appears in the sequence if and only if the coefficient of phi^{-1} in the base-phi expansion of n is 1.  Alternatively, the last bit of the n-th term of A341722 is 1. - Jeffrey Shallit, May 03 2023

			5 is a term since it has two representations: L(0) + L(2) = 2 + 3 and L(1) + L(3) = 1 + 4.
12 is a term since it has two representations: L(1) + L(5) = 1 + 11 and L(0) + L(2) + L(4) = 2 + 3 + 7.

Robert Israel, Table of n, a(n) for n = 1..10000
John L. Brown, Jr., Unique representation of integers as sums of distinct Lucas numbers, The Fibonacci Quarterly, Vol. 7, No. 3 (1969), pp. 243-252.
Hung V. Chu, David C. Luo and Steven J. Miller, On Zeckendorf Related Partitions Using the Lucas Sequence, arXiv:2004.08316 [math.NT], 2020-2021.
David C. Luo, Java code for calculating non-consecutive partitions of natural numbers in any infinite integer sequence given by a second-order linear recurrence, GitHub.
Jeffrey Shallit, Proving Properties of phi-Representations with the Walnut Theorem-Prover, arXiv:2305.02672 [math.NT], 2023.

Cf. A000032, A001622, A130310, A116543, A176015, A341722.

Java
```
See David C. Luo's GitHub link.
```

L:= [seq(combinat:-fibonacci(n+1)+combinat:-fibonacci(n-1), n=0..40)]:
f1:= proc(n, m) option remember;
      if n = 0 then return 1 fi;
      if m <= 0 then 0
      elif L[m] <= n then procname(n - L[m],m-2) + procname(n, m-1)
      else procname(n,m-1)
      fi
end proc:
filter:= n -> f1(n,ListTools:-BinaryPlace(L,n+1))=2:
select(filter, [$1..1000]); # Robert Israel, Mar 10 2021

L = Table[Fibonacci[n+1] + Fibonacci[n-1], {n, 0, 40}];
f1[n_, m_] := f1[n, m] = If[n == 0, Return[1], Which[m <= 0, 0, L[[m]] <= n, f1[n-L[[m]], m-2] + f1[n, m-1], True, f1[n, m-1]]];
filterQ[n_] := f1[n, FirstPosition[L, b_ /; b > n+1][[1]]-1] == 2;
Select[Range[1000], filterQ] (* Jean-François Alcover, Aug 27 2022, after Robert Israel *)

A344362 Decimal expansion of (5^(1/4) + 5^(-1/4))/2.

Original entry on oeis.org

1, 0, 8, 2, 0, 4, 4, 5, 4, 3, 0, 9, 8, 8, 2, 1, 2, 8, 2, 9, 5, 7, 5, 6, 6, 0, 3, 3, 6, 9, 9, 7, 8, 0, 6, 6, 5, 8, 7, 5, 7, 4, 7, 4, 7, 4, 6, 3, 3, 5, 9, 1, 9, 5, 5, 1, 4, 3, 2, 8, 8, 4, 7, 6, 5, 9, 8, 3, 4, 5, 3, 9, 5, 2, 9, 7, 1, 7, 8, 7, 4, 2, 2, 8, 6, 6, 1, 0, 0, 8, 0, 4, 4, 7, 8, 3, 2, 3, 2, 5, 9, 3, 0, 3, 4

Offset: 1

Daniel Carter, May 15 2021

Solution for x in the system {x = 1/y + 1/z, y = x + 1/z, z = y + 1/x}. The corresponding values of y and z are 5^(1/4) and (5^(1/4) + 5^(3/4))/2.

The smallest aspect ratio of a set of three rectangles which have the property that any two of them can be scaled, rotated, and joined at an edge to obtain a rectangle with the third aspect ratio. The other two aspect ratios are given in the comment above.

			1.082044543098821282957566033699780665875...

Cf. A011003, A344363, A293409, A176015.

RealDigits[Cosh[Log[5]/4], 10, 120][[1]] (* Amiram Eldar, Jun 29 2023 *)

solve(x=1, 2, 5*x^4 - 5*x^2 - 1) \\ Hugo Pfoertner, May 16 2021

my(c=50+30*quadgen(20)); a_vector(len) = digits(sqrtint(floor(c*100^(len-2)))); \\ Kevin Ryde, May 28 2021

Equals 5*A293409.
Equals sqrt(A176015).
Equals cosh(log(5)/4). - Vaclav Kotesovec, May 28 2021

cp's OEIS Frontend

A244847 Decimal expansion of rho_c = (5-sqrt(5))/10, the asymptotic critical density for the hard hexagon model.

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A214986 Power ceiling array for the golden ratio, by antidiagonals.

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A115339 a(2n-1)=F(n+1), a(2n)=L(n), where F(n) and L(n) are the Fibonacci and the Lucas sequences.

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A214987 Power round array for the golden ratio, by antidiagonals.

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A380981 Decimal expansion of the medium/short edge length ratio of a disdyakis triacontahedron.

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A342089 Numbers that have two representations as the sum of distinct non-consecutive Lucas numbers (A000032).

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A344362 Decimal expansion of (5^(1/4) + 5^(-1/4))/2.

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