A085449 -id:A085449 - cp's OEIS frontend

A209084 a(n) = 2a(n-1) + 4a(n-2) with n>1, a(0)=0, a(1)=4.

0, 4, 8, 32, 96, 320, 1024, 3328, 10752, 34816, 112640, 364544, 1179648, 3817472, 12353536, 39976960, 129368064, 418643968, 1354760192, 4384096256, 14187233280, 45910851584, 148570636288, 480784678912, 1555851902976, 5034842521600, 16293092655104

Offset: 0

Seiichi Kirikami, Mar 06 2012

a(n)/A063727(n) are convergents for A134972.

Abs(Sum_{i=0..n} C(n,n-i)*a(i)-(sqrt(5)-1)* A033887(n))->0. - Seiichi Kirikami, Jan 20 2016

E. W. Cheney, Introduction to Approximation Theory, McGraw-Hill, Inc., 1966.

Bruno Berselli, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (2,4).

Cf. A063727, A033887, A085449, A103435, A134972, A269991.

Cf. A086344 (this sequence with signs).

I:=[0,4]; [n le 2 select I[n] else 2*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jan 16 2016

RecurrenceTable[{a[n]==2*a[n-1]+4*a[n-2], a[0]==0, a[1]==4}, a, {n, 30}]
LinearRecurrence[{2, 4}, {0, 4}, 40] (* Vincenzo Librandi, Jan 16 2016 *)

concat(0, Vec(4*x/(1-2*x-4*x^2) + O(x^40))) \\ Michel Marcus, Jan 16 2016

a(n) = (2/sqrt(5))*((1+sqrt(5))^n-(1-sqrt(5))^n).
G.f.: 4*x/(1-2*x-4*x^2). - Bruno Berselli, Mar 08 2012
a(n) = 4*A085449(n) = 2*A103435(n). - Bruno Berselli, Mar 08 2012
Sum_{n>=1} 1/a(n) = (1/4) * A269991. - Amiram Eldar, Feb 01 2021

A099133 4^(n-1)*Fibonacci(n).

Original entry on oeis.org

0, 1, 4, 32, 192, 1280, 8192, 53248, 344064, 2228224, 14417920, 93323264, 603979776, 3909091328, 25300041728, 163745628160, 1059783180288, 6859062771712, 44392781971456, 287316132233216, 1859549040476160, 12035254277636096, 77893801758162944

Offset: 0

Paul Barry, Sep 29 2004

Binomial transform of A099134.
Second binomial transform of x/(1-20x^2), or (0,1,0,20,0,400,0,8000,....).
In general k^(n-1)*Fibonacci(n) has g.f. x/(1-kx-k^2x^2).
The ratio a(n+1)/a(n) converges to 4 times the golden ratio as n approaches infinity. In general, the ratio a(n+1)/a(n) of the sequence which is the solution to the linear recurrence relation a(n) = m*a(n-1)+m^2*a(n-2) with a(0)=0 and a(1) = 1 converges to m times the golden ratio as n approaches infinity where m is a positive integer. - Felix P. Muga II, Mar 10 2014

			G.f. = x + 4*x^2 + 32*x^3 + 192*x^4 + 1280*x^5 + 8192*x^6 + 53248*x^7 + ...

F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivré Formula, March 2014; Preprint on ResearchGate.

Index entries for linear recurrences with constant coefficients, signature (4,16).

Cf. A000045, A099012, A085449. Fourth row of A234357.

Join[{a=0,b=1},Table[c=4*b+16*a;a=b;b=c,{n,40}]] (* Vladimir Joseph Stephan Orlovsky, Mar 29 2011*)
Table[4^(n-1) Fibonacci[n],{n,0,20}] (* Harvey P. Dale, Aug 22 2012 *)
LinearRecurrence[{4,16},{0,1},30] (* Harvey P. Dale, Aug 22 2012 *)

a(n) = 4^(n-1)*fibonacci(n); \\ Michel Marcus, Jan 10 2014

G.f.: x/(1-4*x-16*x^2).
a(n) = 4*a(n-1) + 16*a(n-2).
a(n) = (2+2*sqrt(5))^n/(4*sqrt(5))-(2-sqrt(5))^n/(4*sqrt(5)).
a(-n) = -(-1)^n * a(n) / 16^n for all n in Z. - Michael Somos, Mar 18 2014

A109447 Binomial coefficients C(n,k) with n-k odd, read by rows.

Original entry on oeis.org

1, 2, 1, 3, 4, 4, 1, 10, 5, 6, 20, 6, 1, 21, 35, 7, 8, 56, 56, 8, 1, 36, 126, 84, 9, 10, 120, 252, 120, 10, 1, 55, 330, 462, 165, 11, 12, 220, 792, 792, 220, 12, 1, 78, 715, 1716, 1287, 286, 13, 14, 364, 2002, 3432, 2002, 364, 14, 1, 105, 1365, 5005, 6435, 3003, 455, 15

Offset: 1

Philippe Deléham, Aug 27 2005

The same as A119900 without 0's. A reflected version of A034867 or A202064. - Alois P. Heinz, Feb 07 2014
From Vladimir Shevelev, Feb 07 2014: (Start)
Also table of coefficients of polynomials  P_1(x)=1, P_2(x)=2, for n>=2, P_(n+1)(x) = 2*P_n(x)+(x-1)* P_(n-1)(x).  The polynomials P_n(x)/2^(n-1) are connected with sequences A000045 (x=5), A001045 (x=9), A006130 (x=13), A006131 (x=17), A015440 (x=21), A015441 (x=25), A015442 (x=29), A015443 (x=33), A015445 (x=37), A015446 (x=41), A015447 (x=45), A053404 (x=49); also the polynomials P_n(x) are connected with sequences A000129, A002605, A015518, A063727, A085449, A002532, A083099, A015519, A003683, A002534, A083102, A015520. (End)

			Starred terms in Pascal's triangle (A007318), read by rows:
1;
1*, 1;
1, 2*, 1;
1*, 3, 3*, 1;
1, 4*, 6, 4*, 1;
1*, 5, 10*, 10, 5*, 1;
1, 6*, 15, 20*, 15, 6*, 1;
1*, 7, 21*, 35, 35*, 21, 7*, 1;
1, 8*, 28, 56*, 70, 56*, 28, 8*, 1;
1*, 9, 36*, 84, 126*, 126, 84*, 36, 9*, 1;
Triangle T(n,k) begins:
1;
2;
1,    3;
4,    4;
1,   10,  5;
6,   20,  6;
1,   21,  35,   7;
8,   56,  56,   8;
1,   36, 126,  84,  9;
10, 120, 252, 120, 10;

Alois P. Heinz, Rows n = 1..200, flattened

Cf. A109446.

T:= (n, k)-> binomial(n, 2*k+1-irem(n, 2)):
seq(seq(T(n, k), k=0..ceil((n-2)/2)), n=1..20);  # Alois P. Heinz, Feb 07 2014

Flatten[ Table[ If[ OddQ[n - k], Binomial[n, k], {}], {n, 0, 15}, {k, 0, n}]] (* Robert G. Wilson v *)

More terms from Robert G. Wilson v, Aug 30 2005

Corrected offset by Alois P. Heinz, Feb 07 2014

A272263 a(n) = numerator of A000032(n) - 1/2^n.

Original entry on oeis.org

1, 1, 11, 31, 111, 351, 1151, 3711, 12031, 38911, 125951, 407551, 1318911, 4268031, 13811711, 44695551, 144637951, 468058111, 1514668031, 4901568511, 15861809151, 51329892351, 166107021311, 537533612031, 1739495309311, 5629125066751, 18216231370751

Offset: 0

Paul Curtz, Apr 24 2016

A000032(n), Lucas numbers, and 1/2^n are autosequences of the second kind.
Then a(n)/2^n is also an autosequence of the second kind.
Their corresponding autosequences of the first kind are A000045(n) and n/2^n, the Oresme numbers.
Difference table of A000032(n) - 1/2^n:
1, 1/2,  11/4,   31/8,  111/16,    351/32,  1151/64, ...
   9/4,   9/8,  49/16,  129/32,    449/64, 1409/128, ...
        31/16,  31/32,  191/64,   511/128, 1791/256, ...
               129/64, 129/128,   769/256, ...
                       511/256,   511/256, ...
                                2049/1024, ... .
The first upper diagonal is A140323(n)/A004171(n). The main diagonal is the double, i.e. A140323(n)/A000302(n). The inverse binomial transform is the signed sequence.
Quintisections from a(2):
     11,      31,      111,      351,      1151,
   3711,   12031,    38911,   125951,    407551,
1318911, 4268031, 13811711, 44695551, 144637951,
etc.

			Numerators of a(0) =2-1=1, a(1)=1-1/2=1/2, a(2)=3-1/4=11/4, a(3)=4-1/8=31/8, ... .

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,2,-4).

Cf. A000012, A000032, A000045, A000079, A000302, A004171, A085449, A087131, A140323.

CoefficientList[Series[(1 - 2*x + 6*x^2)/((1 - x)*(1 - 2*x - 4*x^2)), {x, 0, 30}], x] (* Robert Price, Apr 24 2016 *)
Table[Numerator[LucasL@ n - 1/2^n], {n, 0, 26}] (* Michael De Vlieger, Apr 24 2016 *)

Vec((1-2*x+6*x^2)/((1-x)*(1-2*x-4*x^2)) + O(x^50)) \\ Colin Barker, Apr 24 2016

a(n) = a(n-1) + 10*A085449(n), for n>0, a(0)=1.
a(n) = A087131(n) - 1.
From Colin Barker, Apr 24 2016: (Start)
a(n) = (-1+(1-sqrt(5))^n+(1+sqrt(5))^n).
a(n) = 3*a(n-1)+2*a(n-2)-4*a(n-3) for n>2.
G.f.: (1-2*x+6*x^2) / ((1-x)*(1-2*x-4*x^2)).
(End)

A319053 a(n) is the exponent of the largest power of 2 that appears in the factorization of the entries in the matrix {{3,1},{1,-1}}^n.

Original entry on oeis.org

0, 1, 5, 3, 4, 8, 6, 7, 12, 9, 10, 15, 12, 13, 18, 15, 16, 20, 18, 19, 25, 21, 22, 28, 24, 25, 31, 27, 28, 32, 30, 31, 36, 33, 34, 39, 36, 37, 42, 39, 40, 44, 42, 43, 50, 45, 46, 53, 48, 49, 56, 51, 52, 56, 54, 55, 60, 57, 58, 63, 60, 61, 66, 63, 64, 68, 66, 67, 73, 69

Offset: 1

Greg Dresden, Sep 09 2018

nonn

a(n) appears to equal n-1 for n not a multiple of 3.
The matrix entries of M^n, with n >= 0, are M^n(1, 1) = 2^(n-1)*F(n+3) = A063782(n), M^n(2, 2) = 2^(n-1)*F(n-3) = A319196(n), M^n(1, 2) = M^n(2, 1) = 2^(n-1)*F(n) = A085449(n), where i = sqrt(-1), F = A000045, and F(-1) = 1, F(-2) = -1, F(-3) = 2. Proof by Cayley-Hamilton, with S(n, -i) = (-i)^n*F(n+1), where S(n, x) is given in A049310. - Wolfdieter Lang, Oct 08 2018
The above conjecture is true. From the preceding formulas for the elements of M^n this claims that the Fibonacci numbers F(n-3), F(n) and F(n+3) are always odd for n == 1 or 2 (mod 3). This is true because F(n) is even iff n == 0 (mod 3) (see e.g. Vajda, p.73), and each of the three indices is == 1 or 2 (mod 3) for n == 1 or 2 (mod 3), respectively. - Wolfdieter Lang, Oct 09 2018

			For n = 3, the matrix {{3,1},{1,-1}}^3 = {{32,8},{8,0}} and the largest power of 2 appearing in the factorization of any entry is 2^5 = 32. Hence, a(3) = 5.

S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989, p. 73.

Cf. A000045, A049310, A063782, A085449, A130481, A319196.

Join[{0, 1, 5}, Table[Max[ IntegerExponent[Flatten[MatrixPower[{{3, 1}, {1, -1}}, n]], 2]], {n, 4, 40}]]

a(n) = vecmax(apply(x->if (x, valuation(x, 2), 0), [3,1;1,-1]^n)); \\ Michel Marcus, Sep 09 2018

A319196 a(n) = 2^(n-1)*Fibonacci(n-3), n >= 0.

Original entry on oeis.org

1, -1, 2, 0, 8, 16, 64, 192, 640, 2048, 6656, 21504, 69632, 225280, 729088, 2359296, 7634944, 24707072, 79953920, 258736128, 837287936, 2709520384, 8768192512, 28374466560, 91821703168, 297141272576, 961569357824, 3111703805952, 10069685043200, 32586185310208, 105451110793216

Offset: 0

Wolfdieter Lang, Oct 09 2018

This sequence gives the elements M^n(2, 2) of the matrix M = [[3, 1], [1, -1]].
Motivation to look into these matrix powers came from A319053. M^n[1, 1] = A063782 and M^n(1, 2) =  M^n(2, 1) = A085449(n). Proof by Cayley-Hamilton, using S(n, -I) = (-I)^n*F(n+1), and S(n, x) from A049310 and F = A000045.
For a similar signed sequence see A087205.

Index entries for linear recurrences with constant coefficients, signature (2,4).

Cf. A000045, A001622, A049310, A085449, A063782, A087205, A319053.

[2^(n-1)*Fibonacci(n-3): n in [0..30]]; // Vincenzo Librandi, Oct 09 2018

Table[2^(n-1) Fibonacci[n-3], {n, 0, 40}] (* Vincenzo Librandi, Oct 09 2018 *)
LinearRecurrence[{2,4},{1,-1},40] (* Harvey P. Dale, Mar 29 2020 *)

a(n) = 2^(n-1)*F(n-3), n >= 0, with F = A000045 with F(-1) = 1, F(-2) = -1 and F(-3) = 1.
G.f.: (1-3*x)/(1-2*x-(2*x)^2).
a(n) = 2*(a(n-1) + 2*a(n-2)), n >= 2, with a(0) = 1 and a(1) = -1.
a(n) = 2^(n-1)*(phi^(n-3) - (1 - phi)^(n-3))/(2*phi - 1) with the golden section phi =  A001622.

cp's OEIS Frontend

A209084 a(n) = 2*a(n-1) + 4*a(n-2) with n>1, a(0)=0, a(1)=4.

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A099133 4^(n-1)*Fibonacci(n).

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A109447 Binomial coefficients C(n,k) with n-k odd, read by rows.

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A272263 a(n) = numerator of A000032(n) - 1/2^n.

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A319053 a(n) is the exponent of the largest power of 2 that appears in the factorization of the entries in the matrix {{3,1},{1,-1}}^n.

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A319196 a(n) = 2^(n-1)*Fibonacci(n-3), n >= 0.

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A209084 a(n) = 2a(n-1) + 4a(n-2) with n>1, a(0)=0, a(1)=4.