A077444 -id:A077444 - cp's OEIS frontend

A259131 Numbers n such that 13*n^2 + 52 is a square.

3, 36, 393, 4287, 46764, 510117, 5564523, 60699636, 662131473, 7222746567, 78788080764, 859446141837, 9375119479443, 102266868132036, 1115560429972953, 12168897861570447, 132742316047301964, 1447996578658751157, 15795220049198960763, 172299423962529817236, 1879498443538629028833, 20502183454962389499927

Offset: 1

Derek Orr, Jun 18 2015

The limit of a(n)/a(n-1) approaches (11+sqrt(117))/2 as n -> infinity.
The continued fraction [a(n); a(n), a(n), ...] = ((3+sqrt(13))/2)^(2*n-1).
Equivalently, numbers n such that (n^2+4)/13 is a square.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(13)). - Greg Dresden, Jul 22 2019
As 13*n^2 + 52 = 13 * (n^2 + 4), n == 3 (mod 13) or n == 10 (mod 13) alternately. - Bernard Schott, Jul 23 2019

Index entries for linear recurrences with constant coefficients, signature (11,-1).

Cf. A002878, A077444.

I:=[3,36]; [n le 2 select I[n] else 11*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 23 2019

Table[Floor[((3 + Sqrt[13])/2)^(2*n + 1) + ((3 + Sqrt[13])/2)^(1 - 2 n)], {n, 21}] (* Michael De Vlieger, Jun 20 2015 *)
LinearRecurrence[{11, -1}, {3, 36}, 25] (* Vincenzo Librandi, Jul 23 2019 *)

for(n=1,20,q=((3+sqrt(13))/2)^(2*n-1);print1(contfrac(q)[1],", "))

G.f.: 3*x*(1+x)/(1-11*x+x^2).
a(n) = 11*a(n-1) - a(n-2); a(0) = 3, a(1) = 36.
a(n) = floor(((3+sqrt(13))/2)^(2*n+1)+((3+sqrt(13))/2)^(1-2*n)).
a(n) = 3*A097783(n-1). - R. J. Mathar, Jun 07 2016

A270444 Expansion of 2(1+2x) / (1-8x+4x^2).

Original entry on oeis.org

2, 20, 152, 1136, 8480, 63296, 472448, 3526400, 26321408, 196465664, 1466439680, 10945654784, 81699479552, 609813217280, 4551707820032, 33974409691136, 253588446248960, 1892809931227136, 14128125664821248, 105453765593661440

Offset: 1

Altug Alkan, Mar 17 2016

nonn

If p is an odd prime, a((p+1)/2) == 2 mod p. In other words, a((p+1)/2) - 2^p is divisible by p where p is an odd prime.

			a(2) = 20 because (1 + sqrt(3))^3 + (1 - sqrt(3))^3 = 20.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (8, -4).

Cf. A077444, A080040, A080041, A107903.

CoefficientList[Series[2(1+2x)/(1-8x+4x^2),{x,0,30}],x] (* or *) LinearRecurrence[{8,-4},{2,20},30] (* Harvey P. Dale, Jun 09 2020 *)

PARI
```
Vec(2*(1+2*x)/(1-8*x+4*x^2) + O(x^100))
```

G.f.: 2*(1+2*x)/(1-8*x+4*x^2).
a(n) = (1+sqrt(3))^(2*n-1) + (1-sqrt(3))^(2*n-1).
a(n) = 2 * A107903(n-1).

A270445 Expansion of 2x(1+4x) / (1-12x+16*x^2).

Original entry on oeis.org

2, 32, 352, 3712, 38912, 407552, 4268032, 44695552, 468058112, 4901568512, 51329892352, 537533612032, 5629125066752, 58948963008512, 617321555034112, 6464675252273152, 67698958146732032, 708952693724413952, 7424248994345254912

Offset: 1

Altug Alkan, Mar 17 2016

If p is an odd prime, a((p+1)/2) == 2 mod p. In other words, a((p+1)/2) - 2^p is divisible by p where p is an odd prime.

			a(2) = 32 because (1 + sqrt(5))^3 + (1 - sqrt(5))^3 = 32.

Colin Barker, Table of n, a(n) for n = 1..950
Index entries for linear recurrences with constant coefficients, signature (12,-16).

Cf. A077444, A087131.

Vec(2*x*(1+4*x)/(1-12*x+16*x^2) + O(x^50)) \\ Colin Barker, Mar 17 2016

a(n) = 12*a(n-1) - 16*a(n-2) for n>2. G.f.: 2*x*(1+4*x) / (1-12*x+16*x^2). - Colin Barker, Mar 17 2016

a(n) = (1+sqrt(5))^(2*n-1) + (1-sqrt(5))^(2*n-1).

A175033 Numbers n such that (ceiling(sqrt(nn/2)))^2 - nn/2 = 17/2.

Original entry on oeis.org

9, 15, 55, 89, 321, 519, 1871, 3025, 10905, 17631, 63559, 102761, 370449, 598935, 2159135, 3490849, 12584361, 20346159, 73347031, 118586105, 427497825, 691170471

Offset: 1

Ctibor O. Zizka, Nov 09 2009

Let (ceiling(sqrt(n*n/2)))^2 - n*n/2 = i. Then for i=1/2 we have A002315, for i=1 we have A005319, for i=2 we have A077444, for i=7/2 we have A077446, for i=4 we have A081554.

Conjecture: a(n) = 6*a(n-2) - a(n-4). - Charles R Greathouse IV, Apr 30 2016

Cf. A005319, A077444, A081554, A002315, A077446,

lista(nn)=for (n=1, nn, if ((ceil(sqrt(n*n/2)))^2 - n*n/2 == 17/2, print1(n, ", "));); \\ Michel Marcus, Jun 02 2013

forstep(n=9,1e9,2, if((sqrtint(n^2\2)+1)^2==(n^2+17)/2, print1(n", "))) \\ Charles R Greathouse IV, Apr 30 2016

More terms from Michel Marcus, Jun 02 2013

a(17)-a(22) from Charles R Greathouse IV, Apr 30 2016

A351898 Decimal expansion of metallic ratio for N = 14.

Original entry on oeis.org

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

Offset: 2

A.H.M. Smeets, Feb 24 2022

Decimal expansion of continued fraction [14; 14, 14, 14, ...].
Also largest solution of x^2 - 14 x - 1 = 0.
Essentially the same digit sequence as A010503, A157214, A174968 and A268683.
The metallic ratio's for N = A077444(n) are equal to powers of the silver ratio, i.e., A014166^(2n-1); this constant represents the special case for N = A077444(2).

			14.0710678118654752440084436210484903928483593...

Michael Penn, 7+5√2 is a perfect cube??, YouTube video, 2022.
Wikipedia, Metallic mean.

Cf. A010503, A077444, A157214, A174968, A268683.

Metallic ratios: A001622 (N=1), A014176 (N=2), A098316 (N=3), A098317 (N=4), A098318 (N=5), A176398 (N=6), A176439 (N=7), A176458 (N=8), A176522 (N=9), A176537 (N=10), A244593 (N=11).

RealDigits[7 + 5*Sqrt[2], 10, 100][[1]] (* Amiram Eldar, Feb 24 2022 *)

PARI
```
(1+sqrt(2))^3
```

Equals 2 + 5*A014176.
Equals A014176^3.
Equals exp(arcsinh(7)). - Amiram Eldar, Jul 04 2023

A352403 Indices of metallic means that are powers of other metallic means.

Original entry on oeis.org

4, 11, 14, 29, 36, 76, 82, 140, 199, 234, 364, 393, 478, 521, 536, 756, 1030, 1364, 1764, 2236, 2786, 3420, 3571, 3775, 4144, 4287, 4964, 5886, 6916, 8060, 8886, 9324, 9349, 10714, 12236, 13896, 15700, 16238, 17654, 18557, 19764, 22036, 24476, 27090, 29884

Offset: 1

Douglas Blumeyer, Mar 14 2022

nonn

Metallic mean k is mm(k) = (k + sqrt(k^2 + 4))/2.
The 4th metallic mean (mm), sometimes called the "copper" mean, is mm(4) = (4 + sqrt(16 + 4))/2 = 4.236... This value is also = 1.618...^3, where 1.618... is the 1st mm, the "golden" one, or (1 + sqrt(1 + 4))/2. This can be shown algebraically.
Any odd power of an mm will give another mm. The odd powers of the 1st mm are:
phi^1 =  1.618..., the  1st mm,
phi^3 =  4.236..., the  4th mm,
phi^5 = 11.090..., the 11th mm,
phi^7 = 29.034..., the 29th mm,
etc.
The indices of these mm's are 1, 4, 11, 29, 76, ... (A002878).
In parallel, the powers of the 2nd mm are:
slv^1 =  2.414..., the  2nd mm,
slv^3 = 14.071..., the 14th mm,
slv^5 = 82.012..., the 82nd mm,
etc.
The indices of these mm's are 2, 14, 82, 478, 2786, ... (A077444).
The indices of the mm's for the 3rd mm are A259131. The 4th mm's are A267797.
Every mm produces such a sequence.
The union of all such sequences (excluding their first terms) is this sequence.

			76 is a term since mm(76) = mm(1)^9 is a power of an earlier mean (the golden ratio in this case, and 76 is a Lucas number).
From _Peter Luschny_, Mar 16 2022: (Start)
A representation of the values claimed in the definition is given for the first 8 terms by:
( 4  +  2*sqrt(5)) / 2 = ((1 + sqrt(5)) / 2)^3.
(11  +  5*sqrt(5)) / 2 = ((1 + sqrt(5)) / 2)^5.
(14  +  5*sqrt(8)) / 2 = ((2 + sqrt(8)) / 2)^3.
(29  + 13*sqrt(5)) / 2 = ((1 + sqrt(5)) / 2)^7.
(36  + 10*sqrt(13))/ 2 = ((3 + sqrt(13))/ 2)^3.
(76  + 34*sqrt(5)) / 2 = ((1 + sqrt(5)) / 2)^9.
(82  + 29*sqrt(8)) / 2 = ((2 + sqrt(8)) / 2)^5.
(140 + 26*sqrt(29))/ 2 = ((5 + sqrt(29))/ 2)^3.
(End)

Union of A002878, A077444, A259131, A267797, etc., minus each sequence's first entry.

getMetallicMean[n_] := (n + Power[Power[n, 2] + 4, 1 / 2]) / 2;
getMetallicCompositesUpTo[maxCandidateIndex_] := Module[
  {sequence, metallicMeanIndex, metallicMean, oddPower, candidateIndex},
  sequence = {};
  metallicMeanIndex = 1;
  While[
    True,
    (* skip metallic means already shown to be a power of another *)
    If[MemberQ[sequence, metallicMeanIndex], metallicMeanIndex++];
    metallicMean = getMetallicMean[metallicMeanIndex];
    oddPower = 3;
    While[
      True,
      candidateIndex = Floor[Power[metallicMean, oddPower]];
      If[
        candidateIndex <= maxCandidateIndex,
        AppendTo[sequence, candidateIndex];
        oddPower += 2,
        Break[]
      ]
    ];
    If[
      oddPower == 3,
      (* no chance of finding further results below the max, if even the first candidate at this index exceeded it *)
      Break[],
      metallicMeanIndex++
    ];
  ];
  Sort[sequence]
];
getMetallicCompositesUpTo[50000]

cp's OEIS Frontend

A259131 Numbers n such that 13*n^2 + 52 is a square.

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A270444 Expansion of 2*(1+2*x) / (1-8*x+4*x^2).

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A270445 Expansion of 2*x*(1+4*x) / (1-12*x+16*x^2).

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A175033 Numbers n such that (ceiling(sqrt(n*n/2)))^2 - n*n/2 = 17/2.

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PARI

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A351898 Decimal expansion of metallic ratio for N = 14.

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Mathematica

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A352403 Indices of metallic means that are powers of other metallic means.

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Mathematica

A270444 Expansion of 2(1+2x) / (1-8x+4x^2).

A270445 Expansion of 2x(1+4x) / (1-12x+16*x^2).

A175033 Numbers n such that (ceiling(sqrt(nn/2)))^2 - nn/2 = 17/2.