A072265 -id:A072265 - cp's OEIS frontend

A072270 A partial product representation of A006131 and A072265.

1, 1, 13, 9, 101, 5, 701, 49, 361, 29, 31021, 33, 204101, 181, 1021, 1889, 8799541, 233, 57746701, 1361, 41581, 7589, 2486401661, 1633, 161532401, 49661, 22810681, 58241, 702418373381, 2245, 4608956945501, 3437249, 74991181, 2135149, 2802699901, 75921, 1302034904649701, 14007941, 3219888061, 3019201

Offset: 1

Miklos Kristof, Jul 09 2002

nonn

Define f(n) = A006131(n-1) and L(n) = 4*f(n-1)+f(n+1), which implies L(n) = A072265(n), n>1.
For even n, f(n) = product_{d|n} a(d) and for odd n, f(n) = product_{d|n} a(2d).
Taking logarithms defines the sequence a(.) via a Mobius transformation (see A072183).
Writing f(n) and L(n) in terms of Binet formulas leads to a representation as cyclotomic polynomials.

			f(12)=a(1)*a(2)*a(3)*a(4)*a(6)*a(12) = 1*1*13*9*5*33 = 19305 for even n=12.
f(9)=a(2)*a(6)*a(18)= 1*5*233 = 1165 for odd n=9.
L(6)=a(4)*a(12) = 9*33 = 297 = 4*f(5)+f(7) = 4*29+181 for even n=6.
L(15)=a(1)*a(3)*a(5)*a(15) = 1*13*101*1021 = 1340573 for odd n=15.

Cf. A006131, A072183.

A072270 := proc(n) if n <=2 then 1; else h := (1+sqrt(17))/2 ; cy := numtheory[cyclotomic](n,x) ; g := degree(cy) ; (h-1)^g*subs(x=h^2/4,cy) ; expand(%) ; end if; end proc: # R. J. Mathar, Nov 17 2010

Let h=(1+sqrt(17))/2, Phi(n, x) = n-th cyclotomic polynomial, so x^n-1= product_{d|n} Phi(d, x), and let g(d) be the order of Phi(d, x). Then a(n)=(h-1)^g(n)*Phi(n, h^2/4), n>2.
a(p) = L(p) for odd prime p.
a(2p) = f(p) for odd prime p.
a(2^k+1) = L(2^k).
a(3*2^k = L(2^k)-4^k.
L(n) = product_{d|n} a(d) for odd n.
L(n*2^k) = product_{d|n} a(d*2^(k+1)) for k>0 and odd n.

Divided argument of Phi by 4; moved comments to formula section - R. J. Mathar, Nov 17 2010

A085488 Duplicate of A072265.

Original entry on oeis.org

2, 1, 9, 13, 49, 101, 297, 701, 1889, 4693, 12249, 31021, 80017, 204101, 524169

Offset: 0

dead

A027286 a(n) = Sum_{k=0..2n} (k+1) * A026584(n, k).

Original entry on oeis.org

1, 4, 18, 56, 190, 564, 1722, 4976, 14454, 40940, 115698, 322728, 896558, 2471588, 6786090, 18537184, 50459366, 136844892, 370030434, 997705240, 2683514526, 7201203988, 19284880794, 51546789456, 137541880150, 366412976332

Offset: 0

Clark Kimberling

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,7,-8,-16).

Cf. A000032, A000045, A006131, A026584, A072265.

I:=[1,4,18,56]; [n le 4 select I[n] else 2*Self(n-1) +7*Self(n-2) -8*Self(n-3) -16*Self(n-4): n in [1..31]]; // G. C. Greubel, Dec 12 2021

LinearRecurrence[{2,7,-8,-16},{1,4,18,56}, 30] (* G. C. Greubel, Dec 12 2021 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -16,-8,7,2]^n*[1;4;18;56])[1,1] \\ Charles R Greathouse IV, Oct 21 2022

[2^(n-1)*(n+1)*(14*lucas_number2(n+2, 1/2, -1) + 5*lucas_number2(n+1, 1/2, -1))/17 for n in (0..30)] # G. C. Greubel, Dec 12 2021

G.f.: (1+2*x+3*x^2)/(1-x-4*x^2)^2.
From G. C. Greubel, Dec 12 2021: (Start)
a(n) = 2^(n-3)*( -6*F(n+1, 1/2) + Sum_{j=0..n} F(n-j+1, 1/2)*( 14*F(j+1, 1/2) + 5*F(j, 1/2) ), where F(n, x) are the Fibonacci polynomials.
a(n) = (2^(n-1)/17)*(n+1)*( 14*L(n+2, 1/2) + 5*L(n+1, 1/2) ), where L(n, x) are the Lucas polynomials.
a(n) = 2*a(n-1) + 7*a(n-2) - 8*a(n-3) - 16*a(n-4). (End)

A206776 a(n) = 3a(n-1) + 2a(n-2) for n>1, a(0)=2, a(1)=3.

Original entry on oeis.org

2, 3, 13, 45, 161, 573, 2041, 7269, 25889, 92205, 328393, 1169589, 4165553, 14835837, 52838617, 188187525, 670239809, 2387094477, 8501763049, 30279478101, 107841960401, 384084837405, 1367938433017, 4871984973861, 17351831787617, 61799465310573

Offset: 0

Bruno Berselli, Jan 10 2013

This is the Lucas sequence V(3,-2).
Inverse binomial transform of this sequence is A072265.
a(n) = A124805(n) - 1 for n>0.

			G.f. = 2 + 3*x + 13*x^2 + 45*x^3 + 161*x^4 + 573*x^5 + 2041*x^6 + 7269*x^7 + ...

Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics, 2nd ed., Addison-Wesley, 1994. Exercise 7.49(c), pages 379, 573.

Bruno Berselli, Table of n, a(n) for n = 0..200
Wikipedia, Lucas sequence: Specific names.
Index entries for linear recurrences with constant coefficients, signature (3,2).

Cf. A189736 (same recurrence but with initial values reversed).

Cf. A007482, A072265, A124805.

[n le 1 select n+2 else 3*Self(n)+2*Self(n-1): n in [0..25]];

A206776 := proc(n)
    option remember ;
    if n <= 1 then
        n+2 ;
    else
        3*procname(n-1)+2*procname(n-2) ;
    end if;
end proc:
seq(A206776(n),n=0..30) ; # R. J. Mathar, Feb 18 2024

RecurrenceTable[{a[n] == 3 a[n - 1] + 2 a[n - 2], a[0] == 2, a[1] == 3}, a[n], {n, 25}]
LinearRecurrence[{3,2},{2,3},30] (* Harvey P. Dale, Apr 29 2014 *)
a[ n_] := If[ n < 0, (-2)^n a[ -n], ((3 + Sqrt[17])/2)^n + ((3 - Sqrt[17])/2)^n // Expand]; (* Michael Somos, Oct 13 2016 *)
a[ n_] := If[ n < 0, (-2)^n a[ -n], Boole[n == 0] + SeriesCoefficient[ ((1 + 3*x + Sqrt[1 + 6*x + 17*x^2])/2)^n, {x, 0, n}]]; (* Michael Somos, Oct 13 2016 *)

a[0]:2$ a[1]:3$ a[n]:=3*a[n-1]+2*a[n-2]$ makelist(a[n], n, 0, 25);

Vec((2-3*x)/(1-3*x-2*x^2) + O(x^30)) \\ Michel Marcus, Jun 26 2015

{a(n) = 2 * real(( (3 + quadgen(68)) / 2 )^n)}; /* Michael Somos, Oct 13 2016 */

{a(n) = my(w = quadgen(-8)); simplify(w^n * subst(2 * polchebyshev(n), x, -3/4*w))}; /* Michael Somos, Oct 13 2016 */

for(n=0,25,print1(round(((3+sqrt(17))/2)^n+((3-sqrt(17))/2)^n),", ")) \\ Hugo Pfoertner, Nov 19 2018

G.f.: (2-3*x)/(1-3*x-2*x^2).
a(n) = ((3-sqrt(17))^n+(3+sqrt(17))^n)/2^n.
a(n) = [x^n] ( (1 + 3*x + sqrt(1 + 6*x + 17*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
a(n) = (-2)^n * a(-n) for all n in Z. - Michael Somos, Oct 13 2016
If c = (3 + sqrt(17))/2, then c^n = (a(n) + sqrt(17)*A007482(n-1)) / 2. - Michael Somos, Oct 13 2016
E.g.f.: 2*exp(3*x/2)*cosh(sqrt(17)*x/2). - Stefano Spezia, Oct 21 2022
a(n) = 2*A007482(n)-3*A007482(n-1). - R. J. Mathar, Feb 18 2024

A075117 Table by antidiagonals of generalized Lucas numbers: T(n,k) = T(n,k-1) + n*T(n,k-2) with T(n,0)=2 and T(n,1)=1.

Original entry on oeis.org

2, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 4, 5, 1, 2, 1, 7, 7, 7, 1, 2, 1, 11, 17, 10, 9, 1, 2, 1, 18, 31, 31, 13, 11, 1, 2, 1, 29, 65, 61, 49, 16, 13, 1, 2, 1, 47, 127, 154, 101, 71, 19, 15, 1, 2, 1, 76, 257, 337, 297, 151, 97, 22, 17, 1, 2, 1, 123, 511, 799, 701, 506, 211, 127, 25, 19, 1, 2

Offset: 0

Henry Bottomley, Sep 02 2002

			Array starts as:
  2, 1,  1,  1,  1,   1, ...;
  2, 1,  3,  4,  7,  11, ...;
  2, 1,  5,  7, 17,  31, ...;
  2, 1,  7, 10, 31,  61, ...;
  2, 1,  9, 13, 49, 101, ...;
  2, 1, 11, 16, 71, 151, ...; etc.

G. C. Greubel, Antidiagonals n = 0..100, flattened

Cf. A060959.
Rows include: A054977, A000032, A014551, A075118, A072265.
Columns include: A007395, A000012, A005408, A016777, A056220, A062786.

[2^(1+k-n)*(&+[Binomial(n-k,2*j)*(1+4*k)^j: j in [0..Floor((n-k)/2)]]): k in [0..n], n in [0..13]]; // G. C. Greubel, Jan 27 2020

seq(seq( 2^(1+k-n)*add( binomial(n-k, 2*j)*(1+4*k)^j, j=0..floor((n-k)/2)), k=0..n), n=0..13); # G. C. Greubel, Jan 27 2020

T[n_, k_]:= ((1 + Sqrt[1+4n])/2)^k + ((1 - Sqrt[1+4n])/2)^k; Table[If[n==0 && k==0, 2, T[k, n-k]]//Simplify, {n,0,13}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 27 2020 *)

def T(n, k): return 2^(1-k)*sum( binomial(k, 2*j)*(1+4*n)^j for j in (0..floor(k/2)) )
[[T(k,n-k) for k in (0..n)] for n in (0..13)] # G. C. Greubel, Jan 27 2020

T(n, k) = ((1+sqrt(4*n+1))/2)^k + ((1-sqrt(4*n+1))/2)^k = 2*A060959(n, k+1) - A060959(n, k).

T(n, k) = 2^(1-k)*Sum_{j=0..floor(k/2)} binomial(k, 2*j)*(1+4*n)^j. - G. C. Greubel, Jan 27 2020

A075118 Variant on Lucas numbers: a(n) = a(n-1) + 3*a(n-2) with a(0)=2 and a(1)=1.

Original entry on oeis.org

2, 1, 7, 10, 31, 61, 154, 337, 799, 1810, 4207, 9637, 22258, 51169, 117943, 271450, 625279, 1439629, 3315466, 7634353, 17580751, 40483810, 93226063, 214677493, 494355682, 1138388161, 2621455207, 6036619690, 13900985311, 32010844381, 73713800314, 169746333457

Offset: 0

Henry Bottomley, Sep 02 2002

The sequence 4,1,7,.. = 2*0^n+A075118(n) is given by trace(A^n) where A=[1,1,1,1;1,0,0,0;1,0,0,0;1,0,0,0]. - Paul Barry, Oct 01 2004

For n>2, a(n) is the numerator of the value of the continued fraction 1+3/(1+3/(1+...+3/7)) where there are n-2 1's. - Alexander Mark, Aug 16 2020

			a(4) = a(3)+3*a(2) = 10+3*7 = 31.

Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", Wiley, 2001, p. 471.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Lucas sequence
Index entries for linear recurrences with constant coefficients, signature (1,3).

Cf. A000032, A006130, A014551, A072265, A075117, A274977.

a:=[2,1];; for n in [3..40] do a[n]:=a[n-1]+3*a[n-2]; od; a; # G. C. Greubel, Jan 15 2020

I:=[2,1]; [n le 2 select I[n] else Self(n-1)+3*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Jul 20 2013

R:=PowerSeriesRing(Integers(), 33); Coefficients(R!((2-x)/(1-x-3*x^2))); // Marius A. Burtea, Jan 15 2020

a:= n-> (Matrix([[1,2]]). Matrix([[1,1], [3,0]])^n)[1,2]:
seq(a(n), n=0..35);  # Alois P. Heinz, Aug 15 2008

a[0]=2; a[1]=1; a[n_]:= a[n]= a[n-1] +3a[n-2]; Table[a[n], {n, 0, 30}]
CoefficientList[Series[(2-x)/(1-x-3x^2), {x,0,40}], x] (* Vincenzo Librandi, Jul 20 2013 *)
LinearRecurrence[{1,3},{2,1},40] (* Harvey P. Dale, Jun 18 2017 *)
Table[Round[Sqrt[3]^n*LucasL[n, 1/Sqrt[3]]], {n,0,40}] (* G. C. Greubel, Jan 15 2020 *)

my(x='x+O('x^30)); Vec((2-x)/(1-x-3*x^2)) \\ G. C. Greubel, Dec 21 2017

polsym(x^2-x-3, 44) \\ Joerg Arndt, Jan 22 2023

[lucas_number2(n,1,-3) for n in range(0, 30)] # Zerinvary Lajos, Apr 30 2009

a(n) = ((1+sqrt(13))/2)^n + ((1-sqrt(13))/2)^n.
a(n) = 2*A006130(n) - A006130(n-1) = A075117(3, n).
G.f.: (2-x)/(1-x-3*x^2). - Philippe Deléham, Nov 15 2008
a(n) = [x^n] ( (1 + x + sqrt(1 + 2*x + 13*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
a(n) = 3^(n/2) * Lucas(n, 1/sqrt(3)). - G. C. Greubel, Jan 15 2020

A081708 a(n) = a(n-1) + 64*a(n-2) starting with a(0) = 2 and a(1) = 1.

Original entry on oeis.org

2, 1, 129, 193, 8449, 20801, 561537, 1892801, 37831169, 158970433, 2580165249, 12754272961, 177884848897, 994158318401, 12378788647809, 76004921025473, 868247394485249, 5732562340115521, 61300395587171457, 428184385354564801, 4351409702933538049

Offset: 0

Helmut Schmiedel (a.positiv(AT)web.de), Apr 03 2003

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,64).

Cf. A072265.

a(n) = my(w = quadgen(257)); w^n + (1 - w)^n;

Vec((2-x)/(1-x-64*x^2) + O(x^30)) \\ Colin Barker, Feb 22 2016

a(n) = ((1 + sqrt(257))/2)^n + ((1 - sqrt(257))/2)^n.

G.f.: (2-x) / (1-x-64*x^2). - Colin Barker, Feb 22 2016

More terms from Michel Marcus, Aug 24 2013

cp's OEIS Frontend

A072270 A partial product representation of A006131 and A072265.

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A085488 Duplicate of A072265.

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A027286 a(n) = Sum_{k=0..2n} (k+1) * A026584(n, k).

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A206776 a(n) = 3*a(n-1) + 2*a(n-2) for n>1, a(0)=2, a(1)=3.

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A075117 Table by antidiagonals of generalized Lucas numbers: T(n,k) = T(n,k-1) + n*T(n,k-2) with T(n,0)=2 and T(n,1)=1.

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A075118 Variant on Lucas numbers: a(n) = a(n-1) + 3*a(n-2) with a(0)=2 and a(1)=1.

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A081708 a(n) = a(n-1) + 64*a(n-2) starting with a(0) = 2 and a(1) = 1.

A206776 a(n) = 3a(n-1) + 2a(n-2) for n>1, a(0)=2, a(1)=3.