A143095 -id:A143095 - cp's OEIS frontend

A048655 Generalized Pellian with second term equal to 5.

1, 5, 11, 27, 65, 157, 379, 915, 2209, 5333, 12875, 31083, 75041, 181165, 437371, 1055907, 2549185, 6154277, 14857739, 35869755, 86597249, 209064253, 504725755, 1218515763, 2941757281, 7102030325, 17145817931, 41393666187, 99933150305, 241259966797, 582453083899

Offset: 0

Barry E. Williams

Equals binomial transform of A143095: (1, 4, 2, 8, 4, 16, 8, 32, ...). - Gary W. Adamson, Jul 23 2008

T. D. Noe, Table of n, a(n) for n = 0..300
M. Bicknell, A primer on the Pell sequence and related sequences, Fibonacci Quarterly, Vol. 13, No. 4, 1975, pp. 345-349.
A. F. Horadam, Basic properties of a certain generalized sequence of numbers, Fibonacci Quarterly, Vol. 3, No. 3, 1965, pp. 161-176.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434.
A. F. Horadam, Pell identities, Fib. Quart., Vol. 9, No. 3, 1971, pp. 245-252.
Tanya Khovanova, Recursive sequences
Index entries for linear recurrences with constant coefficients, signature (2,1)

Cf. A001333, A000129, A048654, A143095.

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+3*x)/(1-2*x-x^2))); // G. C. Greubel, Jul 26 2018

with(combinat): a:=n->3*fibonacci(n, 2)+fibonacci(n+1, 2): seq(a(n), n=0..26); # Zerinvary Lajos, Apr 04 2008

a[n_]:=(MatrixPower[{{1,2},{1,1}},n].{{4},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
LinearRecurrence[{2,1},{1,5},30] (* Harvey P. Dale, Nov 05 2011 *)

a[0]:1$
a[1]:5$
a[n]:=2*a[n-1]+a[n-2]$
A048655(n):=a[n]$
makelist(A048655(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

a(n)=([0,1; 1,2]^n*[1;5])[1,1] \\ Charles R Greathouse IV, Feb 09 2017

a(n) = 2*a(n-1) + a(n-2); a(0)=1, a(1)=5.
a(n) = ((4+sqrt(2))(1+sqrt(2))^n - (4-sqrt(2))(1-sqrt(2))^n)/2*sqrt(2).
a(n) = P(n) - 3*P(n+1) + 2*P(n+2). - Creighton Dement, Jan 18 2005
G.f.: (1+3*x)/(1 - 2*x - x^2). - Philippe Deléham, Nov 03 2008
E.g.f.: exp(x)*(cosh(sqrt(2)*x) + 2*sqrt(2)*sinh(sqrt(2)*x)). - Vaclav Kotesovec, Feb 16 2015
a(n) = 3*Pell(n) + Pell(n+1), where Pell = A000129. - Vladimir Reshetnikov, Sep 27 2016

A081180 4th binomial transform of (0,1,0,2,0,4,0,8,0,16,...).

Original entry on oeis.org

0, 1, 8, 50, 288, 1604, 8800, 47944, 260352, 1411600, 7647872, 41420576, 224294400, 1214467136, 6575615488, 35602384000, 192760455168, 1043650265344, 5650555750400, 30593342288384, 165638957801472, 896804870374400

Offset: 0

Paul Barry, Mar 11 2003

Vincenzo Librandi, Table of n, a(n) for n = 0..300
S. Falcon, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.
Index entries for linear recurrences with constant coefficients, signature (8,-14).

Binomial transform of A081179.

Cf. A081182.

I:=[0, 1]; [n le 2 select I[n] else 8*Self(n-1)-14*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 06 2013

Join[{a=0,b=1},Table[c=8*b-14*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 19 2011 *)
CoefficientList[Series[x / (1 - 8 x + 14 x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 06 2013 *)
LinearRecurrence[{8,-14},{0,1},30] (* Harvey P. Dale, Aug 17 2019 *)

[lucas_number1(n,8,14) for n in range(0, 22)] # Zerinvary Lajos, Apr 23 2009

a(n) = 8a(n-1) - 14a(n-2), a(0)=0, a(1)=1.
G.f.: x/(1 - 8x + 14x^2).
a(n) = ((4 + sqrt(2))^n - (4 - sqrt(2))^n)/(2*sqrt(2)).
a(n) = Sum_{k=0..n} C(n,2k+1) 2^k*4^(n-2k-1).
If shifted once left, fourth binomial transform of A143095. - Al Hakanson (hawkuu(AT)gmail.com), Jul 25 2009, R. J. Mathar, Oct 15 2009
E.g.f.: exp(4*x)*sinh(sqrt(2)*x)/sqrt(2). - Ilya Gutkovskiy, Aug 12 2017

Modified the completing comment on the fourth binomial transform - R. J. Mathar, Oct 15 2009

A163349 a(n) = 10a(n-1) - 23a(n-2) for n > 1; a(0) = 1, a(1) = 9.

Original entry on oeis.org

1, 9, 67, 463, 3089, 20241, 131363, 848087, 5459521, 35089209, 225323107, 1446179263, 9279361169, 59531488641, 381889579523, 2449671556487, 15713255235841, 100790106559209, 646496195167747, 4146789500815663

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Jul 25 2009

nonn

Binomial transform of A081180 without initial 0. Fifth binomial transform of A143095.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10,-23).

Cf. A081180, A143095.

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); S:=[ ((1+2*r)*(5+r)^n+(1-2*r)*(5-r)^n)/2: n in [0..19] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Jul 26 2009

LinearRecurrence[{10,-23}, {1,9}, 50] (* G. C. Greubel, Dec 19 2016 *)

Vec((1-x)/(1-10*x+23*x^2) + O(x^50)) \\ G. C. Greubel, Dec 19 2016

a(n) = 10*a(n-1) - 23*a(n-2) for n > 1; a(0) = 1, a(1) = 9.
a(n) = ((1+2*sqrt(2))*(5+sqrt(2))^n + (1-2*sqrt(2))*(5-sqrt(2))^n)/2.
G.f.: (1-x)/(1-10*x+23*x^2).
E.g.f.: exp(5*x)*( cosh(sqrt(2)*x) + 2*sqrt(2)*sinh(sqrt(2)*x) ). - G. C. Greubel, Dec 19 2016

Edited and extended beyond a(5) by Klaus Brockhaus, Jul 26 2009

A163416 a(n) = 20a(n-1) - 98a(n-2) for n>1, a(0)=1, a(1)=14.

Original entry on oeis.org

1, 14, 182, 2268, 27524, 328216, 3866968, 45174192, 524520976, 6063348704, 69863918432, 803070195648, 9214739906624, 105593918958976, 1208833868330368, 13828473308627712, 158103747076178176, 1806884557278047744, 20643523932095493632, 235795792028661193728, 2692850495227865498624

Offset: 0

Klaus Brockhaus, Jul 27 2009

Binomial transform of A163415. Tenth binomial transform of A143095.

G. C. Greubel, Table of n, a(n) for n = 0..940
Index entries for linear recurrences with constant coefficients, signature (20, -98).

Cf. A163415, A143095.

[n le 2 select 13*n-12 else 20*Self(n-1)-98*Self(n-2): n in [1..17]];

LinearRecurrence[{20,-98},{1,14},30] (* Harvey P. Dale, Aug 14 2012 *)

Vec((1-6*x)/(1-20*x+98*x^2) + O(x^50)) \\ G. C. Greubel, Dec 21 2016

a(n) = ((1 + 2*sqrt(2))*(10 + sqrt(2))^n + (1 - 2*sqrt(2))*(10 - sqrt(2))^n)/2.
O.g.f.: (1 - 6*x)/(1 - 20*x + 98*x^2).
E.g.f.: exp(10*x)*( cosh(sqrt(2)*x) + 2*sqrt(2)*sinh(sqrt(2)*x) ). - G. C. Greubel, Dec 21 2016

A238549 a(n) is one fourth of the total number of free ends of 4 line segments expansion at n iterations (see Comments lines for definition).

Original entry on oeis.org

1, 2, 6, 8, 16, 20, 36, 44, 76, 92, 156, 188, 316, 380, 636, 764, 1276, 1532, 2556, 3068, 5116, 6140, 10236, 12284, 20476, 24572, 40956, 49148, 81916, 98300, 163836, 196604, 327676, 393212, 655356, 786428, 1310716, 1572860, 2621436, 3145724, 5242876, 6291452, 10485756

Offset: 1

Kival Ngaokrajang, May 01 2015

nonn

The initial pattern consists of 4 straight line segments which are the radii of a square. The next generations are scaled down by a factor of 1/sqrt(2) and rotated by an angle of Pi/4. Their free ends are the ends of elements that do not contact or cross the other ones. Overlaps among different generations are prohibited. See illustration in the links.
We take the official definition to be that provided by the PARI program. From this the assertions in the Formula section follow (they were formerly stated as conjectures). - N. J. A. Sloane, Feb 24 2019
From Georg Fischer, Feb 20 2019: (Start)
The following pattern can be seen for a(n) in base 2:
   n         a(n)
  ==  ==================
   1    1 =          1_2
   2    2 =         10_2
   3    6 =        110_2
   4    8 =       1000_2
   5   16 =      10000_2
   6   20 =      10100_2
   7   36 =     100100_2
   8   44 =     101100_2
   9   76 =    1001100_2
  10   92 =    1011100_2
  11  156 =   10011100_2
  12  188 =   10111100_2
  13  316 =  100111100_2
  14  380 =  101111100_2
  15  636 = 1001111100_2
  16  764 = 1011111100_2
(End)

			The first numbers of free ends (4*a(n)) are 4, 8, 24, 32, 64, 80, 144, 176, 304, 368, 624, ...

Kival Ngaokrajang, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (1, 2, -2).

Cf. A143095, A256641.

{print1(1,", "); for (n=1,100,s=1; for (i=0,n-1,s=s+(5-3*(-1)^i)*2^(1/4*(2*i-1+(-1)^i))/2); print1(s,", "))}

def a():
    s, n = 2, 1
    yield 1
    while True:
        yield s
        s += (5-3*(-1)^n)*2^((2*n-1+(-1)^n)//4)//2
        n += 1
A238549 = a(); [next(A238549) for  in range(43)] # _Peter Luschny, Feb 24 2019

a(n) =  1 + Sum_{i=1..n-1} A143095(i).
G.f.: x*(2*x^2+x+1) / ((x-1)*(2*x^2-1)). - Colin Barker, May 02 2015
From Georg Fischer, Feb 20 2019: (Start)
With p = floor((n + 2) / 2) for n >= 4: if n even then a(n) = 2^p + 4 * (2^(p - 4) - 1); if n odd then a(n) = 2^p + 4 * (2^(p - 3) - 1).
a(n) = a(n - 1) + 2 * a(n - 2) - 2 * a(n - 3).
(End)

A163348 a(n) = 6a(n-1) - 7a(n-2) for n > 1; a(0) = 1, a(1) = 7.

Original entry on oeis.org

1, 7, 35, 161, 721, 3199, 14147, 62489, 275905, 1218007, 5376707, 23734193, 104768209, 462469903, 2041441955, 9011362409, 39778080769, 175588947751, 775087121123, 3421400092481, 15102790707025, 66666943594783

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Jul 25 2009

Binomial transform of A111566. Third binomial transform of A143095. Inverse binomial transform of A081180 without initial 0.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-7).

Cf. A111566, A143095 (1,4,2,8,4,16,...), A081180.

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); S:=[ ((1+2*r)*(3+r)^n+(1-2*r)*(3-r)^n)/2: n in [0..21] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Jul 26 2009

LinearRecurrence[{6, -7}, {1, 7}, 50] (* G. C. Greubel, Dec 19 2016 *)

Vec((1+x)/(1-6*x+7*x^2) + O(x^50)) \\ G. C. Greubel, Dec 19 2016

a(n) = 6*a(n-1) - 7*a(n-2) for n > 1; a(0) = 1, a(1) = 7.
a(n) = ((1+2*sqrt(2))*(3+sqrt(2))^n + (1-2*sqrt(2))*(3-sqrt(2))^n)/2.
G.f.: (1+x)/(1-6*x+7*x^2).
E.g.f.: exp(3*x)*( cosh(sqrt(2)*x) + 2*sqrt(2)*sinh(sqrt(2)*x) ). - G. C. Greubel, Dec 19 2016
a(n) = A081179(n)+A081179(n+1). - R. J. Mathar, Feb 04 2021

Edited and extended beyond a(5) by Klaus Brockhaus, Jul 26 2009

cp's OEIS Frontend

A048655 Generalized Pellian with second term equal to 5.

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A081180 4th binomial transform of (0,1,0,2,0,4,0,8,0,16,...).

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A163349 a(n) = 10*a(n-1) - 23*a(n-2) for n > 1; a(0) = 1, a(1) = 9.

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Magma

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A163416 a(n) = 20*a(n-1) - 98*a(n-2) for n>1, a(0)=1, a(1)=14.

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Programs

Magma

Mathematica

PARI

Formula

A238549 a(n) is one fourth of the total number of free ends of 4 line segments expansion at n iterations (see Comments lines for definition).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Sage

Formula

A163348 a(n) = 6*a(n-1) - 7*a(n-2) for n > 1; a(0) = 1, a(1) = 7.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A163349 a(n) = 10a(n-1) - 23a(n-2) for n > 1; a(0) = 1, a(1) = 9.

A163416 a(n) = 20a(n-1) - 98a(n-2) for n>1, a(0)=1, a(1)=14.

A163348 a(n) = 6a(n-1) - 7a(n-2) for n > 1; a(0) = 1, a(1) = 7.