A103435 -id:A103435 - cp's OEIS frontend

A305492 a(n) = ((1 + y)^n - (1 - y)^n)/y with y = sqrt(8).

0, 2, 4, 22, 72, 298, 1100, 4286, 16272, 62546, 238996, 915814, 3504600, 13419898, 51371996, 196683278, 752970528, 2882724002, 11036241700, 42251551414, 161756794728, 619274449354, 2370846461804, 9076614069086, 34749153370800

Offset: 0

Peter Luschny, Jun 02 2018

			Array ((1+y)^n - (1-y)^n)/y with y = sqrt(k).
[k\n]
[1]   1, 2, 4,  8, 16, 32,   64,  128,    256,   512,   1024, ...
[2]   0, 2, 4, 10, 24, 58,  140,  338,    816,  1970,   4756, ...
[3]   0, 2, 4, 12, 32, 88,  240,  656,   1792,  4896,  13376, ...
[4]   0, 2, 4, 14, 40, 122, 364,  1094,  3280,  9842,  29524, ...
[5]   0, 2, 4, 16, 48, 160, 512,  1664,  5376, 17408,  56320, ...
[6]   0, 2, 4, 18, 56, 202, 684,  2378,  8176, 28242,  97364, ...
[7]   0, 2, 4, 20, 64, 248, 880,  3248, 11776, 43040, 156736, ...
[8]   0, 2, 4, 22, 72, 298, 1100, 4286, 16272, 62546, 238996, ...
[9]   0, 2, 4, 24, 80, 352, 1344, 5504, 21760, 87552, 349184, ...

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,7).

Let f(n, y) = ((1 + y)^n - (1 - y)^n)/y.
f(n,      1 ) = A000079(n);
f(n, sqrt(2)) = A163271(n+1);
f(n, sqrt(3)) = A028860(n+2);
f(n,      2 ) = A152011(n) for n>0;
f(n, sqrt(5)) = A103435(n);
f(n, sqrt(6)) = A083694(n);
f(n, sqrt(7)) = A274520(n);
f(n, sqrt(8)) = a(n);
f(n,      3 ) = A192382(n+1);
Cf. A305491.
Equals 2 * A015519.

egf :=  (n,x) -> 2*exp(x)*sinh(sqrt(n)*x)/sqrt(n):
ser := series(egf(8,x), x, 26):
seq(n!*coeff(ser,x, n), n=0..24);

Table[Simplify[((1 + Sqrt[8])^n - (1 - Sqrt[8])^n)/ Sqrt[8]], {n, 0, 24}]

concat(0, Vec(2*x / (1 - 2*x - 7*x^2) + O(x^40))) \\ Colin Barker, Jun 05 2018

E.g.f.: 2*exp(x)*sinh(sqrt(n)*x)/sqrt(n) for n = 8.
From Colin Barker, Jun 02 2018: (Start)
G.f.: 2*x / (1 - 2*x - 7*x^2).
a(n) = 2*a(n-1) + 7*a(n-2) for n>1.
(End)

A171648 a(1) = 1, a(n) = 2a(n-1) if n is even; a(n) = a(n-1)Fibonacci((n+1)/2)/Fibonacci((n-1)/2) if n is odd.

Original entry on oeis.org

1, 2, 2, 4, 8, 16, 24, 48, 80, 160, 256, 512, 832, 1664, 2688, 5376, 8704, 17408, 28160, 56320, 91136, 182272, 294912, 589824, 954368, 1908736, 3088384, 6176768, 9994240, 19988480, 32342016, 64684032, 104660992, 209321984, 338690048, 677380096, 1096024064

Offset: 1

Gary W. Adamson, Dec 13 2009

a(n)/a(n-1) apparently tends to phi = A001622 if n=odd; e.g. a(21)/a(20) = 91136/56320 = 1.61818...
a(n)/a(n-2) apparently tends to 1+sqrt(5) = 3.236...= A134945; where a(21)/a(19) = 91136/28160 = 3.23636...
a(1)=1, a(2)=2, a(3)=2, for n>3 a(n)=2*a(n-1) if n is even and a(n)=2*(a(n-1)-a(n-2)+a(n-3)) if n is odd. - Vincenzo Librandi, Dec 06 2010

			a(8) = 48 = 2*a(7) = 2*24. a(9) = 80 = (5/3)*48 since Fibonacci(5) = 5 and Fibonacci(4) = 3.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,4).

Cf. A063727 (bisection), A103435 (bisection).

Vec(x*(1+2*x)/(1-2*x^2-4*x^4) + O(x^50)) \\ Colin Barker, Aug 02 2016

a(1) = 1, a(n) = 2*a(n-1) if n is even; a(n) = a(n-1)*A000045((n+1)/2)/A000045((n-1)/2) if n is odd.
From Colin Barker, Aug 02 2016: (Start)
a(n) = 2*a(n-2) + 4*a(n-4) for n>4.
G.f.: x*(1+2*x) / (1-2*x^2-4*x^4).
(End)

Defined "F", removed abundant parentheses, added punctuation to examples, added a factor to the definition, corrected a(13) and added more terms - R. J. Mathar, Dec 15 2009

A261397 a(n) = 3^n*Fibonacci(n).

Original entry on oeis.org

0, 3, 9, 54, 243, 1215, 5832, 28431, 137781, 669222, 3247695, 15766083, 76527504, 371477259, 1803179313, 8752833270, 42487113627, 206236840311, 1001094543576, 4859415193527, 23588096472765, 114499026160038, 555789946734999, 2697861075645339, 13095692747551008, 63567827923461075

Offset: 0

N. J. A. Sloane, Aug 18 2015

Colin Barker, Table of n, a(n) for n = 0..1000
Tom Edgar, Extending Some Fibonacci-Lucas Relations, Fib. Quarterly, 54 (2016), 79.
D. Marques, A new Fibonacci-Lucas relation, Amer. Math. Monthly, 122 (2015), 683.
Ivica Martinjak, Complementary Families of the Fibonacci-Lucas Relations, arXiv:1508.04949 [math.CO], 2015.
Index entries for linear recurrences with constant coefficients, signature (3,9).

Cf. A000045, A103435.

[3^n*Fibonacci(n): n in [0..30]]; // Vincenzo Librandi, Aug 21 2015

RecurrenceTable[{a[0]== 0, a[1]== 3, a[n]== 3*a[n-1]  + 9*a[n-2]}, a, {n,50}] (* G. C. Greubel, Aug 21 2015 *)
LinearRecurrence[{3, 9}, {0, 3}, 30] (* Vincenzo Librandi, Aug 21 2015 *)

vector(30, n, n--; 3^n*fibonacci(n)) \\ Michel Marcus, Aug 21 2015

concat(0, Vec(-3*x/(9*x^2+3*x-1) + O(x^30))) \\ Colin Barker, Sep 01 2015

a(n) = 3*a(n-1) + 9*a(n-2), a(0)=0, a(1)=3. - G. C. Greubel, Aug 21 2015
G.f.: 3*x / (1 - 3*x - 9*x^2). - G. C. Greubel, Aug 21 2015
E.g.f.: (1/(phi - 1/phi))*(e^(3*phi*x) - e^(3*x/phi)), where 2*phi = 1 + sqrt(5). - G. C. Greubel, Aug 21 2015

A277091 a(n) = ((1 + sqrt(15))^n - (1 - sqrt(15))^n)/sqrt(15).

Original entry on oeis.org

0, 2, 4, 36, 128, 760, 3312, 17264, 80896, 403488, 1939520, 9527872, 46209024, 225808256, 1098542848, 5358401280, 26096402432, 127210422784, 619770479616, 3020486878208, 14717760471040, 71722337236992, 349493321068544, 1703099363454976, 8299105221869568, 40441601532108800

Offset: 0

Ilya Gutkovskiy, Sep 29 2016

Number of zeros in substitution system {0 -> 1111111, 1 -> 1001} at step n from initial string "1" (see example).

			Evolution from initial string "1": 1 -> 1001 -> 1001111111111111111001 -> ...
Therefore, number of zeros at step n:
a(0) = 0;
a(1) = 2;
a(2) = 4, etc.

Ilya Gutkovskiy, Illustration (substitution system {0 -> 1111111, 1 -> 1001}) and similar sequences
Eric Weisstein's World of Mathematics, Substitution System
Index entries for linear recurrences with constant coefficients, signature (2,14).

Cf. A010472, A103435, A274520, A274526.

Mathematica
```
LinearRecurrence[{2, 14}, {0, 2}, 26]
```

concat(0, Vec(2*x/(1-2*x-14*x^2) + O(x^99))) \\ Altug Alkan, Oct 01 2016

O.g.f.: 2*x/(1 - 2*x - 14*x^2).
E.g.f.: 2*sinh(sqrt(15)*x)*exp(x)/sqrt(15).
a(n) = 2*a(n-1) + 14*a(n-2).
Lim_{n->infinity} a(n+1)/a(n) = 1 + sqrt(15) = 1 + A010472.

A152036 Triangular product sequence based 2^n times the Fibonacci version and 4 replaced with m: t(m,n)=2^nProduct[(1 + mCos[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}].

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 2, 4, 14, 1, 2, 4, 16, 48, 1, 2, 4, 18, 56, 202, 1, 2, 4, 20, 64, 248, 880, 1, 2, 4, 22, 72, 298, 1100, 4286, 1, 2, 4, 24, 80, 352, 1344, 5504, 21760, 1, 2, 4, 26, 88, 410, 1612, 6914, 28336, 118898, 1, 2, 4, 28, 96, 472, 1904, 8528, 36096, 157472

Offset: 0

Roger L. Bagula and Gary W. Adamson, Nov 20 2008

The row sums are: {1, 3, 7, 21, 71, 283, 1219, 5785, 29071, 156291, 880507,...}. A sequence of sequences with the row numbers m instead of n: and the ratio increases with each row: at (1+Sqrt[5]) for m=4.

			1;
1, 2;
1, 2, 4;
1, 2, 4, 14;
1, 2, 4, 16, 48;
1, 2, 4, 18, 56, 202;
1, 2, 4, 20, 64, 248, 880;
1, 2, 4, 22, 72, 298, 1100, 4286;
1, 2, 4, 24, 80, 352, 1344, 5504, 21760;
1, 2, 4, 26, 88, 410, 1612, 6914, 28336, 118898;
1, 2, 4, 28, 96, 472, 1904, 8528, 36096, 157472, 675904;

Cf. A103435 (row 4), A083694 (row 5)

f[n_, m_] = 2^n*Product[(1 + m*Cos[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}]; Table[Table[FullSimplify[ExpandAll[f[n, m]]], {n, 0, m}], {m, 0, 10}]

t(m,n)=2^n*Product[(1 + m*Cos[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}].

cp's OEIS Frontend

A305492 a(n) = ((1 + y)^n - (1 - y)^n)/y with y = sqrt(8).

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A171648 a(1) = 1, a(n) = 2*a(n-1) if n is even; a(n) = a(n-1)*Fibonacci((n+1)/2)/Fibonacci((n-1)/2) if n is odd.

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A261397 a(n) = 3^n*Fibonacci(n).

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A277091 a(n) = ((1 + sqrt(15))^n - (1 - sqrt(15))^n)/sqrt(15).

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A152036 Triangular product sequence based 2^n times the Fibonacci version and 4 replaced with m: t(m,n)=2^n*Product[(1 + m*Cos[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}].

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Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A171648 a(1) = 1, a(n) = 2a(n-1) if n is even; a(n) = a(n-1)Fibonacci((n+1)/2)/Fibonacci((n-1)/2) if n is odd.

A152036 Triangular product sequence based 2^n times the Fibonacci version and 4 replaced with m: t(m,n)=2^nProduct[(1 + mCos[k*Pi/n]^2), {k, 1, Floor[(n - 1)/2]}].