A236428 -id:A236428 - cp's OEIS frontend

A265762 Coefficient of x in minimal polynomial of the continued fraction [1^n,2,1,1,1,...], where 1^n means n ones.

-3, -5, -15, -37, -99, -257, -675, -1765, -4623, -12101, -31683, -82945, -217155, -568517, -1488399, -3896677, -10201635, -26708225, -69923043, -183060901, -479259663, -1254718085, -3284894595, -8599965697, -22515002499, -58945041797, -154320122895

Offset: 0

Clark Kimberling, Jan 04 2016

In the following guide to related sequences, d(n), e(n), f(n) represent the coefficients in the minimal polynomial written as d(n)*x^2 + e(n)*x + f(n), except, in some cases, for initial terms. All of these sequences (eventually) satisfy the linear recurrence relation a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
continued fractions       d(n)     e(n)     f(n)
[1^n,2,1,1,1,...]       A236428  A265762  A236428
[1^n,3,1,1,1,...]       A236428  A265762  A236428
[1^n,4,1,1,1,...]       A265802  A265803  A265802
[1^n,5,1,1,1,...]       A265804  A265805  A265804
[1^n,1/2,1,1,1,...]     A266699  A266700  A266699
[1^n,1/3,1,1,1,...]     A266701  A266702  A266701
[1^n,2/3,1,1,1,...]     A266703  A266704  A266703
[1^n,sqrt(5),1,1,1,...] A266705  A266706  A266705
[1^n,tau,1,1,1,...]     A266707  A266708  A266707
[2,1^n,2,1,1,1,...]     A236428  A266709  A236428
The following forms of continued fractions have minimal polynomials of degree 4 and, after initial terms, satisfy the following linear recurrence relation:
a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5).
[1^n,sqrt(2),1,1,1,...]:  A266710, A266711, A266712, A266713, A266710
[1^n,sqrt(3),1,1,1,...]:  A266799, A266800, A266801, A266802, A266799
[1^n,sqrt(6),1,1,1,...]:  A266804, A266805, A266806, A266807, A277804
Continued fractions [1^n,2^(1/3),1,1,1,...] have minimal polynomials of degree 6.  The coefficient sequences are linearly recurrenct with signature {13, 104, -260, -260, 104, 13, -1, 0, 0}; see A267078-A267083.
Continued fractions [1^n,sqrt(2)+sqrt(3),1,1,1,...] have minimal polynomials of degree 8.  The coefficient sequences are linearly recurrenct with signature {13, 104, -260, -260, 104, 13, -1}; see A266803, A266808, A267061-A267066.

			Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[2,1,1,1,1,...] = (3 + sqrt(5))/2 has p(0,x) = x^2 - 3x + 1, so a(0) = -3;
[1,2,1,1,1,...] = (5 - sqrt(5))/2 has p(1,x) = x^2 - 5x + 5, so a(1) = -5;
[1,1,2,1,1,...] = (15 + sqrt(5))/10 has p(2,x) = 5x^2 - 15x + 11, so a(2) = -15.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A236428, A265802.

I:=[-3,-5,-15]; [n le 3 select I[n] else 2*Self(n-1)+2*Self(n-2)-Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jan 05 2016

Program 1:
u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[u[n], {2}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0] (* A236428 *)
Coefficient[t, x, 1] (* A265762 *)
Coefficient[t, x, 2] (* A236428 *)
Program 2:
LinearRecurrence[{2, 2, -1}, {-3, -5, -15}, 50] (* Vincenzo Librandi, Jan 05 2016 *)

Vec((-3+x+x^2)/(1-2*x-2*x^2+x^3) + O(x^100)) \\ Altug Alkan, Jan 04 2016

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.:  (-3 + x + x^2)/(1 - 2 x - 2 x^2 + x^3).
a(n) = (-1)*(2^(-n)*(3*(-2)^n+2*((3-sqrt(5))^(1+n)+(3+sqrt(5))^(1+n))))/5. - Colin Barker, Sep 27 2016

A175395 a(n) = 2*Fibonacci(n)^2.

Original entry on oeis.org

0, 2, 2, 8, 18, 50, 128, 338, 882, 2312, 6050, 15842, 41472, 108578, 284258, 744200, 1948338, 5100818, 13354112, 34961522, 91530450, 239629832, 627359042, 1642447298, 4299982848, 11257501250, 29472520898, 77160061448, 202007663442, 528862928882, 1384581123200, 3624880440722, 9490060198962, 24845300156168, 65045840269538, 170292220652450, 445830821687808, 1167200244410978, 3055769911545122, 8000109490224392, 20944558559128050

Offset: 0

N. J. A. Sloane, Dec 03 2010

a(n) (n=1..) is half the number of nX2 binary arrays with no element equal to a strict majority of its diagonal and antidiagonal neighbors. - R. H. Hardin, Dec 02 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A007598, A236428.

[2*Fibonacci(n)^2: n in [0..50]]; // Vincenzo Librandi, Apr 24 2011

Table[2 Fibonacci[n]^2, {n, 0, 40}] (* Bruno Berselli, Nov 03 2015 *)
LinearRecurrence[{2,2,-1},{0,2,2},50] (* Harvey P. Dale, May 24 2023 *)

a(n) = round(2*(-2*(-1)^n+(1/2*(3-sqrt(5)))^n+(1/2*(3+sqrt(5)))^n)/5) \\ Colin Barker, Sep 28 2016

Vec(2*x*(1-x)/(1+x)/(1-3*x+x^2) + O(x^30)) \\ Colin Barker, Sep 28 2016

a(n) = 2*A007598(n).
G.f.: 2*x*(1-x)/(1+x)/(1-3*x+x^2). - Colin Barker, Feb 23 2012
a(n) = F(n-1)*F(n+1) + F(n-2)*F(n+2), where F = A000045, -F(-2) = F(-1) = 1. - Bruno Berselli, Nov 03 2015
a(n) = 2*(-2*(-1)^n+(1/2*(3-sqrt(5)))^n+(1/2*(3+sqrt(5)))^n)/5. - Colin Barker, Sep 28 2016
For n>1 a(n) is the denominator of the continued fraction [1, 1, ... 1, 2, 1, 1, ... 1, 2] with n-2 1's before each 2. See A236428 for the numerator. - Greg Dresden and Kevin Zhanming Zheng, Aug 16 2020

A264080 a(n) = 6F(n)F(n+1) + (-1)^n, where F = A000045.

Original entry on oeis.org

1, 5, 13, 35, 91, 239, 625, 1637, 4285, 11219, 29371, 76895, 201313, 527045, 1379821, 3612419, 9457435, 24759887, 64822225, 169706789, 444298141, 1163187635, 3045264763, 7972606655, 20872555201, 54645058949, 143062621645, 374542805987, 980565796315

Offset: 0

Bruno Berselli, Nov 03 2015

a(n) is prime for n = 1, 2, 5, 7, 14, 15, 29, 40, 49, 57, 70, 87, 105, 127, 175, 279, 362, 647, 727, ...

Bruno Berselli, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000032, A000045; A001654, A002878.

Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n: A226205 (k=1); A236428 (k=2); A014742 (k=3); A061647 (k=4); A002878 (k=5).

[6*Fibonacci(n)*Fibonacci(n+1)+(-1)^n: n in [0..30]];

a:= n-> (<<0|1|0>, <0|0|1>, <-1|2|2>>^n. <<1,5,13>>)[1, 1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 28 2016

Table[6 Fibonacci[n] Fibonacci[n + 1] + (-1)^n, {n, 0, 30}]
LinearRecurrence[{2,2,-1},{1,5,13},30] (* Harvey P. Dale, Jul 12 2019 *)

makelist(6*fib(n)*fib(n+1)+(-1)^n, n, 0, 30);

for(n=0, 30, print1(6*fibonacci(n)*fibonacci(n+1)+(-1)^n", "));

a(n) = round((2^(-n)*(-(-2)^n-3*(3-sqrt(5))^n*(-1+sqrt(5))+3*(1+sqrt(5))*(3+sqrt(5))^n))/5) \\ Colin Barker, Sep 28 2016

Vec((1+3*x+x^2)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 28 2016

[6*fibonacci(n)*fibonacci(n+1)+(-1)^n for n in (0..30)]

G.f.: (1+3*x+x^2) / ((1+x)*(1-3*x+x^2)). - Corrected by Colin Barker, Sep 28 2016
a(n) = -a(-n-1) = 2*a(n-1) + 2*a(n-2) - a(n-3) for all n in Z.
a(n) = L(2*n+1) + F(n)*F(n+1) = A002878(n) + A001654(n). See similar identity for A061647.
a(n) = A001654(n+1) + 3*A001654(n) + A001654(n-1).
a(n) - a(n-1) = 2*A099016(n) with a(-1)=-1.
a(n) + a(n-1) = 2*A097134(n) for n>0.
Sum_{i>=0} 1/a(i) = 1.3232560865206157372628688449331...
a(n) = (2^(-n)*(-(-2)^n-3*(3-sqrt(5))^n*(-1+sqrt(5))+3*(1+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Sep 28 2016
E.g.f.: (1/5)*exp(-x)*(-1 + 6*exp(5*x/2)*(cosh((sqrt(5)*x)/2) + sqrt(5)*sinh((sqrt(5)*x)/2))). - Stefano Spezia, Dec 09 2019

A192873 Coefficient of x in the reduction by (x^2->x+1) of the polynomial p(n,x) given in Comments.

Original entry on oeis.org

0, 1, 2, 7, 18, 49, 128, 337, 882, 2311, 6050, 15841, 41472, 108577, 284258, 744199, 1948338, 5100817, 13354112, 34961521, 91530450, 239629831, 627359042, 1642447297, 4299982848, 11257501249, 29472520898, 77160061447, 202007663442, 528862928881, 1384581123200

Offset: 0

Clark Kimberling, Jul 11 2011

The polynomial p(n,x) is defined by p(0,x) = 1, p(1,x) = x, and p(n,x) = x*p(n-1,x) + (x^2)*p(n-1,x) + 1. See A192872.

First differences give A236428. - Richard R. Forberg, Feb 23 2014

			The coefficients of all the polynomials p(n,x) are Fibonacci numbers (A000045).  The first 6 and their reductions:
p(0,x) = 1 -> 1
p(1,x) = x -> x
p(2,x) = 1 +2*x^2 -> 3 +2*x
p(3,x) = 1 +x +3*x^3 -> 4 +7*x
p(4,x) = 1 +x +2*x^2 +5*x^4 -> 13 +18*x
p(5,x) = 1 +x +2*x^2 +3*x^3 +8*x^5 -> 30 +49*x
G.f. = x + 2*x^2 + 7*x^3 + 18*x^4 + 49*x^5 + 128*x^6 + 337*x^7 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A192872, A192232, A192744.

a:=[0,1,2,7];; for n in [5..40] do a[n]:=3*a[n-1]-3*a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jan 07 2019

I:=[0,1,2,7]; [n le 4 select I[n] else 3*Self(n-1) - 3*Self(n-3) +Self(n-4): n in [1..40]]; // G. C. Greubel, Jan 07 2019

seq(coeff(series(x*(x^2-x+1)/((1-x)*(1+x)*(x^2-3*x+1)),x,n+1), x, n), n = 0 .. 35); # Muniru A Asiru, Jan 08 2019

(See A192872.)
a[ n_] := SeriesCoefficient[ x * (1 - x + x^2) / ((1 - x^2) * (1 - 3*x + x^2)), {x, 0, Abs @ n}]; (* Michael Somos, Apr 08 2014 *)
LinearRecurrence[{3,0,-3,1}, {0,1,2,7}, 40] (* G. C. Greubel, Jan 07 2019 *)

concat(0, Vec(-x*(x^2-x+1)/((x-1)*(x+1)*(x^2-3*x+1)) + O(x^40))) \\ Colin Barker, Apr 01 2014

(x*(x^2-x+1)/((1-x^2)*(x^2-3*x+1))).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Jan 07 2019

a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4).
G.f.: x*(x^2-x+1) / ((1-x)*(1+x)*(x^2-3*x+1)). - Colin Barker, Apr 01 2014
a(n) = (1/10) * (4L(2*n) - 3*(-1)^n - 5), with L(n) the Lucas numbers (A000032). - Ralf Stephan, Apr 06 2014
a(-n) = a(n) for all n in Z. - Michael Somos, Apr 08 2014

More terms from Colin Barker, Apr 01 2014

A266709 Coefficient of x in minimal polynomial of the continued fraction [2,1^n,2,1,1,...], where 1^n means n ones.

Original entry on oeis.org

-7, -25, -59, -161, -415, -1093, -2855, -7481, -19579, -51265, -134207, -351365, -919879, -2408281, -6304955, -16506593, -43214815, -113137861, -296198759, -775458425, -2030176507, -5315071105, -13915036799, -36430039301, -95375081095, -249695203993

Offset: 0

Clark Kimberling, Jan 09 2016

See A265762 for a guide to related sequences.

			Let p(n,x) be the minimal polynomial of the number given by the n-th continued fraction:
[2,2,1,1,1,...] = (7-sqrt(5))/2 has p(0,x) = 11 - 7 x + x^2, so a(0) = -7;
[2,1,2,1,1,1,...] = (25+sqrt(5))/10 has p(1,x) = 31 - 25 x + 5 x^2, so a(1) = -25;
[2,1,1,2,1,...] = (59-sqrt(5))/22 has p(2,x) = 79 - 59 x + 11 x^2, so a(2) = -59.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A265762, A236428.

u[n_] := Table[1, {k, 1, n}]; t[n_] := Join[{2}, u[n], {2}, {{1}}];
f[n_] := FromContinuedFraction[t[n]];
t = Table[MinimalPolynomial[f[n], x], {n, 0, 20}]
Coefficient[t, x, 0]  (* A236428 *)
Coefficient[t, x, 1]  (* A266709 *)
Coefficient[t, x, 2]  (* A236428 *)

a(n) = round((2^(-n)*(9*(-2)^n+2*(3-sqrt(5))^n*(-11+5*sqrt(5))-2*(3+sqrt(5))^n*(11+5*sqrt(5))))/5) \\ Colin Barker, Oct 01 2016

Vec(-(7+11*x-5*x^2)/((1+x)*(1-3*x+x^2)) + O(x^40)) \\ Colin Barker, Oct 01 2016

a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3).
G.f.:  (1 + 3 x - x^2)/(1 - 2 x - 2 x^2 + x^3).
a(n) = (2^(-n)*(9*(-2)^n+2*(3-sqrt(5))^n*(-11+5*sqrt(5))-2*(3+sqrt(5))^n*(11+5*sqrt(5))))/5. - Colin Barker, Oct 01 2016

A279890 Expansion of x(1 - x + 2x^3 - x^4)/((1 - x)(1 + x)(1 - x + x^2)*(1 - x - x^2)).

Original entry on oeis.org

0, 1, 1, 2, 4, 7, 12, 19, 31, 50, 82, 133, 216, 349, 565, 914, 1480, 2395, 3876, 6271, 10147, 16418, 26566, 42985, 69552, 112537, 182089, 294626, 476716, 771343, 1248060, 2019403, 3267463, 5286866, 8554330, 13841197, 22395528, 36236725, 58632253, 94868978, 153501232, 248370211, 401871444, 650241655, 1052113099, 1702354754

Offset: 0

Ilya Gutkovskiy, Dec 22 2016

The integer part of the harmonic mean of Fibonacci(n), Fibonacci(n+1) and Fibonacci(n+2).
The o.g.f. for the numerators of the fractional part of the harmonic mean of Fibonacci(n), Fibonacci(n+1) and Fibonacci(n+2) is 6*x/((1 + x - x^2)*(1 - 4*x - x^2)).
The o.g.f. for the denominators of the fractional part of the harmonic mean of Fibonacci(n), Fibonacci(n+1) and Fibonacci(n+2) is (1 + 3*x - x^2)/((1 + x)*(1 - 3*x + x^2)).
Convolution of Fibonacci numbers and periodic sequence [1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, ...].

			a(1) = floor(3/(1/F(1)+1/F(2)+1/F(3))) = floor(3/(1/1+1/1+1/2)) = 1;
a(2) = floor(3/(1/F(2)+1/F(3)+1/F(4))) = floor(3/(1/1+1/2+1/3)) = 1;
a(3) = floor(3/(1/F(3)+1/F(4)+1/F(5))) = floor(3/(1/2+1/3+1/5)) = 2, etc.

Eric Weisstein's World of Mathematics, Fibonacci Number
Eric Weisstein's World of Mathematics, Harmonic Mean
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,2,0,-1).

Cf. A000045, A001622, A004695, A065563, A236428.
Cf. A062114 (the integer part of the harmonic mean of Fibonacci(n+1) and Fibonacci(n+2) for n>0).
Cf. A074331 (the integer part of the geometric mean of Fibonacci(n), Fibonacci(n+1) and Fibonacci(n+2)).

LinearRecurrence[{2, 0, -2, 2, 0, -1}, {0, 1, 1, 2, 4, 7}, 46]
Table[Floor[3 Fibonacci[n] Fibonacci[n + 1] Fibonacci[n + 2]/(2 Fibonacci[n + 1] Fibonacci[n + 2] - (-1)^n)], {n, 0, 45}]

concat(0, Vec((x*(1-x+2*x^3-x^4)/((1-x)*(1+x)*(1-x+x^2))) + O(x^40))) \\ Felix Fröhlich, Dec 22 2016

G.f.: x*(1 - x + 2*x^3 - x^4)/((1 - x)*(1 + x)*(1 - x + x^2)*(1 - x - x^2)).
a(n) = 2*a(n-1) - 2*a(n-3) + 2*a(n-4) - a(n-6).
a(n) = (9*sqrt(5)*(((1 + sqrt(5))/2)^n - ((1 - sqrt(5))/2)^n) + 5*((-1)^n + 2*cos(Pi*n/3) - 3))/30.
a(n) = floor(3*F(n)*F(n+1)*F(n+2)/(2*F(n+1)*F(n+2)-(-1)^n)), where F(n) is the n-th Fibonacci number (A000045).
a(n) = floor(3*A065563(n)/A236428(n+1)).
a(n) = 3*A000045(n)/2 + ((-1)^n + 2*cos(Pi*n/3) - 3)/6.
a(n) ~ 3*phi^n/(2*sqrt(5)), where phi is the golden ratio (A001622).
Lim_{n->infinity} a(n+1)/a(n) = phi.

cp's OEIS Frontend

A265762 Coefficient of x in minimal polynomial of the continued fraction [1^n,2,1,1,1,...], where 1^n means n ones.

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A175395 a(n) = 2*Fibonacci(n)^2.

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A264080 a(n) = 6*F(n)*F(n+1) + (-1)^n, where F = A000045.

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A192873 Coefficient of x in the reduction by (x^2->x+1) of the polynomial p(n,x) given in Comments.

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A266709 Coefficient of x in minimal polynomial of the continued fraction [2,1^n,2,1,1,...], where 1^n means n ones.

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A279890 Expansion of x*(1 - x + 2*x^3 - x^4)/((1 - x)*(1 + x)*(1 - x + x^2)*(1 - x - x^2)).

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A264080 a(n) = 6F(n)F(n+1) + (-1)^n, where F = A000045.

A279890 Expansion of x(1 - x + 2x^3 - x^4)/((1 - x)(1 + x)(1 - x + x^2)*(1 - x - x^2)).