A081336 -id:A081336 - cp's OEIS frontend

A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

1, 1, 3, 5, 15, 33, 91, 221, 583, 1465, 3795, 9653, 24831, 63441, 162763, 416525, 1067575, 2733673, 7003971, 17938661, 45954543, 117709185, 301527355, 772364093, 1978473511

Offset: 0

Clark Kimberling, Aug 02 2011

nonn

Suppose that p=p(0)*x^n+p(1)*x^(n-1)+...+p(n-1)*x+p(n) is a polynomial of positive degree and that Q is a sequence of polynomials: q(k,x)=t(k,0)*x^k+t(k,1)*x^(k-1)+...+t(k,k-1)*x+t(k,k), for k=0,1,2,... The Q-downstep of p is the polynomial given by D(p)=p(0)*q(n-1,x)+p(1)*q(n-2,x)+...+p(n-1)*q(0,x)+p(n).

Since degree(D(p))

Example: let p(x)=2*x^3+3*x^2+4*x+5 and q(k,x)=(x+1)^k.

D(p)=2(x+1)^2+3(x+1)+4(1)+5=2x^2+7x+14

D(D(p))=2(x+1)+7(1)+14=2x+23

D(D(D(p)))=2(1)+23=25;

the Q-residue of p is 25.

We may regard the sequence Q of polynomials as the triangular array formed by coefficients:

t(0,0)

t(1,0)....t(1,1)

t(2,0)....t(2,1)....t(2,2)

t(3,0)....t(3,1)....t(3,2)....t(3,3)

and regard p as the vector (p(0),p(1),...,p(n)). If P is a sequence of polynomials [or triangular array having (row n)=(p(0),p(1),...,p(n))], then the Q-residues of the polynomials form a numerical sequence.

Following are examples in which Q is the triangle given by t(i,j)=1 for 0<=i<=j:

Q.....P...................Q-residue of P

1.....1...................A000079, 2^n

1....(x+1)^n..............A007051, (1+3^n)/2

1....(x+2)^n..............A034478, (1+5^n)/2

1....(x+3)^n..............A034494, (1+7^n)/2

1....(2x+1)^n.............A007582

1....(3x+1)^n.............A081186

1....(2x+3)^n.............A081342

1....(3x+2)^n.............A081336

1.....A040310.............A193649

1....(x+1)^n+(x-1)^n)/2...A122983

1....(x+2)(x+1)^(n-1).....A057198

1....(1,2,3,4,...,n)......A002064

1....(1,1,2,3,4,...,n)....A048495

1....(n,n+1,...,2n).......A087323

1....(n+1,n+2,...,2n+1)...A099035

1....p(n,k)=(2^(n-k))*3^k.A085350

1....p(n,k)=(3^(n-k))*2^k.A090040

1....A008288 (Delannoy)...A193653

1....A054142..............A101265

1....cyclotomic...........A193650

1....(x+1)(x+2)...(x+n)...A193651

1....A114525..............A193662

More examples:

Q...........P.............Q-residue of P

(x+1)^n...(x+1)^n.........A000110, Bell numbers

(x+1)^n...(x+2)^n.........A126390

(x+2)^n...(x+1)^n.........A028361

(x+2)^n...(x+2)^n.........A126443

(x+1)^n.....1.............A005001

(x+2)^n.....1.............A193660

A094727.....1.............A193657

(k+1).....(k+1)...........A001906 (even-ind. Fib. nos.)

(k+1).....(x+1)^n.........A112091

(x+1)^n...(k+1)...........A029761

(k+1)......A049310........A193663

(In these last four, (k+1) represents the triangle t(n,k)=k+1, 0<=k<=n.)

A051162...(x+1)^n.........A193658

A094727...(x+1)^n.........A193659

A049310...(x+1)^n.........A193664

A075362....A075362........A193665

Changing the notation slightly leads to the Mathematica program below and the following formulation for the Q-downstep of p: first, write t(n,k) as q(n,k). Define r(k)=Sum{q(k-1,i)*r(k-1-i) : i=0,1,...,k-1} Then row n of D(p) is given by v(n)=Sum{p(n,k)*r(n-k) : k=0,1,...,n}.

			First five rows of Q, coefficients of Fibonacci polynomials (A049310):
1
1...0
1...0...1
1...0...2...0
1...0...3...0...1
To obtain a(4)=15, downstep four times:
D(x^4+3*x^2+1)=(x^3+x^2+x+1)+3(x+1)+1: (1,1,4,5) [coefficients]
DD(x^4+3*x^2+1)=D(1,1,4,5)=(1,2,11)
DDD(x^4+3*x^2+1)=D(1,2,11)=(1,14)
DDDD(x^4+3*x^2+1)=D(1,14)=15.

Cf. A192872 (polynomial reduction), A193091 (polynomial augmentation), A193722 (the upstep operation and fusion of polynomial sequences or triangular arrays).

q[n_, k_] := 1;
r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}];
f[n_, x_] := Fibonacci[n + 1, x];
p[n_, k_] := Coefficient[f[n, x], x, k]; (* A049310 *)
v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
Table[v[n], {n, 0, 24}]    (* A193649 *)
TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
Table[r[k], {k, 0, 8}]  (* 2^k *)
TableForm[Table[p[n, k], {n, 0, 6}, {k, 0, n}]]

Conjecture: G.f.: -(1+x)*(2*x-1) / ( (x-1)*(4*x^2+x-1) ). - R. J. Mathar, Feb 19 2015

A074608 a(n) = 3^n + 7^n.

Original entry on oeis.org

2, 10, 58, 370, 2482, 17050, 118378, 825730, 5771362, 40373290, 282534298, 1977503890, 13841818642, 96890604730, 678227855818, 4747575858850, 33232973616322, 232630643127370, 1628413985330938, 11398896347634610

Offset: 0

Robert G. Wilson v, Aug 25 2002

G. C. Greubel and Jon E. Schoenfield, Table of n, a(n) for n = 0..1000 (terms 0..160 from G. C. Greubel).
Index entries for linear recurrences with constant coefficients, signature (10,-21).

Cf. A000051, A034472, A052539, A034474, A062394, A034491, A062395, A062396, A007689, A063376, A063481, A074600 - A074624.

Cf. A081336.

Table[3^n + 7^n, {n, 0, 25}]
RecurrenceTable[{a[0]== 2, a[1]== 10, a[n]== 10*a[n-1]  - 21*a[n-2]}, a, {n,30}] (* G. C. Greubel, Aug 20 2015 *)

first(m)=vector(m,i,i--;3^i + 7^i) \\ Anders Hellström, Aug 20 2015

Vec(1/(1-3*x) + 1/(1-7*x) + O(x^50)) \\ Altug Alkan, Oct 12 2015

From Mohammad K. Azarian, Jan 11 2009: (Start)
G.f.: 1/(1-3*x) + 1/(1-7*x).
E.g.f.: exp(3*x) + exp(7*x). (End)
a(n) = 10*a(n-1)-21*a(n-2) with a(0)=2, a(1)=10. - Vincenzo Librandi, Jul 21 2010
a(n) = 2 * A081336(n). - Michel Marcus, Oct 07 2015

A081335 a(n) = (6^n + 2^n)/2.

Original entry on oeis.org

1, 4, 20, 112, 656, 3904, 23360, 140032, 839936, 5039104, 30233600, 181399552, 1088393216, 6530351104, 39182090240, 235092508672, 1410554986496, 8463329787904, 50779978465280, 304679870267392, 1828079220555776

Offset: 0

Paul Barry, Mar 18 2003

Binomial transform of A034478. 4th binomial transform of (1, 0, 4, 0, 16, 0, 64, ...).

Case k=4 of the family of recurrences a(n) = 2*k*a(n-1) - (k^2-4)*a(n-2), a(0)=1, a(1)=k.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (8,-12).

Cf. A081336.

List([0..30], n-> 2^(n-1)*(3^n + 1)); # G. C. Greubel, Aug 02 2019

[(6^n+2^n)/2: n in [0..30]]; // Vincenzo Librandi, Aug 08 2013

LinearRecurrence[{8, -12}, {1, 4}, 30] (* Harvey P. Dale, May 03 2013 *)
CoefficientList[Series[(1-4x)/((1-2x)(1-6x)), {x,0,30}], x] (* Vincenzo Librandi, Aug 08 2013 *)

a(n)=(6^n+2^n)/2 \\ Charles R Greathouse IV, Oct 07 2015

[2^(n-1)*(3^n + 1) for n in (0..30)] # G. C. Greubel, Aug 02 2019

a(n) = 8*a(n-1) - 12*a(n-2), a(0)=1, a(1)=4.
G.f.: (1-4*x)/((1-2*x)*(1-6*x)).
E.g.f.: exp(4*x)*cosh(2*x).
a(n) = Sum_{k=0..floor(n/2)} binomial(n,2*k) * 4^(n-k) = Sum_{k=0..n} binomial(n,k) * 4^(n-k/2) * (1+(-1)^k)/2. - Paul Barry, Nov 22 2003
a(n) = Sum_{k=0..n} 4^k*A098158(n,k). - Philippe Deléham, Dec 04 2006

A141041 a(n) = ((3 + 2sqrt(3))^n + (3 - 2sqrt(3))^n)/2.

Original entry on oeis.org

1, 3, 21, 135, 873, 5643, 36477, 235791, 1524177, 9852435, 63687141, 411680151, 2661142329, 17201894427, 111194793549, 718774444575, 4646231048097, 30033709622307, 194140950878133, 1254946834135719, 8112103857448713

Offset: 0

Roger L. Bagula, Aug 18 2008

nonn

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,3).

Cf. A011943, A034478, A081336, A099842, A201701.

[n le 2 select 3^(n-1) else 6*Self(n-1) +3*Self(n-2): n in [1..31]]; // G. C. Greubel, Oct 10 2022

a[n_]= ((3+2*Sqrt[3])^n + (3-2*Sqrt[3])^n)/2; Table[FullSimplify[a[n]], {n,0,30}]
LinearRecurrence[{6,3},{1,3},30] (* Harvey P. Dale, Aug 25 2014 *)

A141041 = BinaryRecurrenceSequence(6,3,1,3)
[A141041(n) for n in range(31)] # G. C. Greubel, Oct 10 2022

a(n) = 3*abs(A099842(n-1)), for n > 0.
G.f.: (1-3*x)/(1-6*x-3*x^2). - Philippe Deléham, Mar 03 2012
a(n) = 6*a(n-1) + 3*a(n-2), a(0) = 1, a(1) = 3. - Philippe Deléham, Mar 03 2012
a(n) = Sum_{k=0..n} A201701(n,k)*3^(n-k). - Philippe Deléham, Mar 03 2012
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(4*k-3)/(x*(4*k+1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 27 2013
a(n) = (-i*sqrt(3))^n * ChebyshevT(n, i*sqrt(3)). - G. C. Greubel, Oct 10 2022

Edited by N. J. A. Sloane, Aug 24 2008

A081337 a(n) = (8^n + 4^n)/2.

Original entry on oeis.org

1, 6, 40, 288, 2176, 16896, 133120, 1056768, 8421376, 67239936, 537395200, 4297064448, 34368126976, 274911461376, 2199157473280, 17592722915328, 140739635838976, 1125908496777216, 9007233614479360, 72057731476881408

Offset: 0

Paul Barry, Mar 18 2003

Binomial transform of A081336. 6th binomial transform of (1,0,4,0,16,0,64,....).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (12,-32).

Cf. A081338.

[(8^n+4^n)/2: n in [0..25]]; // Vincenzo Librandi, Aug 08 2013

CoefficientList[Series[(1 - 6 x) / ((1 - 4 x) (1 - 8 x)),{x, 0, 20}], x] (* Vincenzo Librandi, Aug 08 2013 *)

a(n) = 12*a(n-1) -32*a(n-2), a(0)=1, a(1)=6.
G.f.: (1-6*x)/((1-4*x)*(1-8*x)).
E.g.f.: exp(6*x) * cosh(2*x).

A141575 A gap prime-type triangular sequence of coefficients: gap(n)=Prime[n+1]-Prime[n]; t(n,m)=If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^ n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]Sqrt[Prime[m]])^n + (Prime[n] - gap[n]Sqrt[Prime[m]])^n)/2]].

Original entry on oeis.org

1, 2, 2, 13, 17, 21, 185, 245, 305, 425, 7361, 12833, 18817, 32321, 47873, 215171, 271051, 328691, 449251, 576851, 853171, 12334505, 21164697, 31341961, 55836009, 86013257, 164203785, 212610281, 532365557, 659940697, 793109789, 1076412613

Offset: 1

Roger L. Bagula, Aug 18 2008

General Lucas-like Binet sequences
where Prime[m]starts at 1:
a(n)=((Prime[n]+gap[n]*Sqrt[Prime[m])^n+(Prime[n]-gap[n]*Sqrt[Prime[m])^n)/2.
Row sums are:
{1, 4, 51, 1160, 119205, 2694186, 583504495, 12222749556, 4868938911913,
3621654266405174, 21636046625243691}

			{1},
{2, 2},
{13, 17, 21},
{185, 245, 305, 425},
{7361, 12833, 18817, 32321, 47873},
{215171, 271051, 328691, 449251, 576851, 853171},
{12334505, 21164697, 31341961, 55836009, 86013257, 164203785, 212610281},
{532365557, 659940697, 793109789, 1076412613, 1382639597, 2065328317, 2442521189, 3270431797},
{40436937953, 68810349217, 102354570337, 185966400481, 293310073697, 587469359713, 778486092257, 1259085279457, 1553019848801},
{7312866926183, 15217609281335, 25813998655559, 56317915837223,
101380456546055, 246072307427783, 351480840333479, 643872497781095,
837435900955463, 1336749872660999}, {512759709537725, 608866569299409,
709085196658213, 922088454409101, 1152233212894709, 1665820807145925,
1950209769575213, 2576571400365309, 2919512658836837, 3667365684348213,
4951533162173037}

Cf. A011943, A081336, A034478.

gap[n_] := Prime[n + 1] - Prime[n]; t[n_, m_] := If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]*Sqrt[Prime[m]])^n + (Prime[n] - gap[n]*Sqrt[Prime[m]])^n)/2]]; Table[Table[FullSimplify[t[n, m]], {m, 0, n}], {n, 0, 10}]; Flatten[%]

gap(n)=Prime[n+1]-Prime[n]; t(n,m)=If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^ n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]*Sqrt[Prime[m]])^n + (Prime[n] - gap[n]*Sqrt[Prime[m]])^n)/2]].

Showing 1-6 of 6 results.

cp's OEIS Frontend

A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

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A074608 a(n) = 3^n + 7^n.

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Mathematica

PARI

PARI

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A081335 a(n) = (6^n + 2^n)/2.

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GAP

Magma

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PARI

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A141041 a(n) = ((3 + 2*sqrt(3))^n + (3 - 2*sqrt(3))^n)/2.

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A081337 a(n) = (8^n + 4^n)/2.

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Magma

Mathematica

Formula

A141575 A gap prime-type triangular sequence of coefficients: gap(n)=Prime[n+1]-Prime[n]; t(n,m)=If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^ n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]*Sqrt[Prime[m]])^n + (Prime[n] - gap[n]*Sqrt[Prime[m]])^n)/2]].

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Author

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Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A141041 a(n) = ((3 + 2sqrt(3))^n + (3 - 2sqrt(3))^n)/2.

A141575 A gap prime-type triangular sequence of coefficients: gap(n)=Prime[n+1]-Prime[n]; t(n,m)=If[n == m == 0, 1, If[m == 0, ((Prime[n] + gap[n])^ n + (Prime[n] - gap[n])^n)/2, ((Prime[n] + gap[n]Sqrt[Prime[m]])^n + (Prime[n] - gap[n]Sqrt[Prime[m]])^n)/2]].