A034478 -id:A034478 - cp's OEIS frontend

A113405 Expansion of x^3/(1 - 2x + x^3 - 2x^4) = x^3/( (1-2x)(1+x)*(1-x+x^2) ).

0, 0, 0, 1, 2, 4, 7, 14, 28, 57, 114, 228, 455, 910, 1820, 3641, 7282, 14564, 29127, 58254, 116508, 233017, 466034, 932068, 1864135, 3728270, 7456540, 14913081, 29826162, 59652324, 119304647, 238609294, 477218588, 954437177, 1908874354, 3817748708

Offset: 0

Paul Barry, Oct 28 2005

A transform of the Jacobsthal numbers. A059633 is the equivalent transform of the Fibonacci numbers.
Paul Curtz, Aug 05 2007, observes that the inverse binomial transform of 0,0,0,1,2,4,7,14,28,57,114,228,455,910,1820,... gives the same sequence up to signs. That is, the extended sequence is an eigensequence for the inverse binomial transform (an autosequence).
The round() function enables the closed (non-recurrence) formula to take a very simple form: see Formula section. This can be generalized without loss of simplicity to a(n) = round(b^n/c), where b and c are very small, incommensurate integers (c may also be an integer fraction). Particular choices of small integers for b and c produce a number of well-known sequences which are usually defined by a recurrence - see Cross Reference. - Ross Drewe, Sep 03 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, arXiv:math/0205301 [math.CO], 2002. [Link to arXiv version]
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to Lin. Alg. Applic. version together with omitted figures]
N. J. A. Sloane, Transforms
Index entries for linear recurrences with constant coefficients, signature (2,0,-1,2).

From Ross Drewe, Sep 03 2009: (Start)
Other sequences a(n) = round(b^n / c), where b and c are very small integers:
A001045 b = 2; c = 3
A007910 b = 2; c = 5
A016029 b = 2; c = 5/3
A077947 b = 2; c = 7
abs(A078043) b = 2; c = 7/3
A007051 b = 3; c = 2
A015518 b = 3; c = 4
A034478 b = 5; c = 2
A003463 b = 5; c = 4
A015531 b = 5; c = 6
(End)

[Round(2^n/9): n in [0..40]]; // Vincenzo Librandi, Aug 11 2011

A010892 := proc(n) op((n mod 6)+1,[1,1,0,-1,-1,0]) ; end proc:
A113405 := proc(n) (2^n-(-1)^n)/9 -A010892(n-1)/3; end proc: # R. J. Mathar, Dec 17 2010

CoefficientList[Series[x^3/(1-2x+x^3-2x^4),{x,0,40}],x] (* or *) LinearRecurrence[{2,0,-1,2},{0,0,0,1},40] (* Harvey P. Dale, Apr 30 2011 *)

a(n)=2^n\/9 \\ Charles R Greathouse IV, Jun 05 2011

def A113405(n): return ((1<Chai Wah Wu, Apr 17 2025

a(n) = 2a(n-1) - a(n-3) + 2a(n-4).
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k,k)*A001045(k).
a(n) = Sum_{k=0..n} binomial((n+k)/2,k)*A001045((n-k)/2)*(1+(-1)^(n-k))/2.
a(3n) = A015565(n), a(3n+1) = 2*A015565(n), a(3n+2) = 4*A015565(n). - Paul Curtz, Nov 30 2007
From Paul Curtz, Dec 16 2007: (Start)
a(n+1) - 2a(n) = A131531(n).
a(n) + a(n+3) = 2^n. (End)
a(n) = round(2^n/9). - Ross Drewe, Sep 03 2009
9*a(n) = 2^n + (-1)^n - 3*A010892(n). - R. J. Mathar, Mar 24 2018

Edited by N. J. A. Sloane, Dec 13 2007

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.)

Original entry on oeis.org

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

A020699 Expansion of (1-3x)/(1-5x).

Original entry on oeis.org

1, 2, 10, 50, 250, 1250, 6250, 31250, 156250, 781250, 3906250, 19531250, 97656250, 488281250, 2441406250, 12207031250, 61035156250, 305175781250, 1525878906250, 7629394531250, 38146972656250, 190734863281250, 953674316406250

Offset: 0

David W. Wilson

Partial sums are A034478.
Except for the first two terms 1 and 2, these are the integers that satisfy phi(n) = 2*n/5. - Michel Marcus, Jul 14 2015
For n>=1, period of powers of 4 mod 10^n. See A000302. - Martin Renner, Jun 12 2020

Nathaniel Johnston, Table of n, a(n) for n = 0..250
D Bevan, D Levin, P Nugent, J Pantone, L Pudwell, Pattern avoidance in forests of binary shrubs, arXiv preprint arXiv:1510:08036 [math.CO], 2015-2016.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1037
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1037 (archived version of page)
M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Index entries for linear recurrences with constant coefficients, signature (5).

Cf. A034478, A001045, A020729, A000302.

seq(`if`(n=0,1,2*5^(n-1)), n=0..22); # Nathaniel Johnston, Jun 26 2011

CoefficientList[Series[(1 - 3 x)/(1 - 5 x), {x, 0, 22}], x] (* Michael De Vlieger, Jul 14 2015 *)

Vec((1-3*x)/(1-5*x) + O(x^30)) \\ Michel Marcus, Jul 14 2015

a(n) = 2*5^(n-1) for n>0.
E.g.f.: (2*exp(5*x)+3)/5; a(n)=(2*5^n+3*0^n)/5. - Paul Barry, Sep 03 2003
a(n) = sum{k=0..n, C(n-1, k)*(Jac(2n-2k)+Jac(2n-2k-1))}+0^n/2, where Jac(n)=A001045(n). - Paul Barry, Jun 07 2005
a(0)=1, a(1)=2, a(n) = 5*a(n-1) for n>=2. [Vincenzo Librandi, Jan 01 2011]
a(n) = A020729(n-1), n>0. - R. J. Mathar, Sep 16 2016

A083075 Square array read by antidiagonals: T(n,k) = (k(2k+3)^n + 1)/(k+1).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 13, 1, 1, 7, 33, 63, 1, 1, 9, 61, 229, 313, 1, 1, 11, 97, 547, 1601, 1563, 1, 1, 13, 141, 1065, 4921, 11205, 7813, 1, 1, 15, 193, 1831, 11713, 44287, 78433, 39063, 1, 1, 17, 253, 2893, 23801, 128841, 398581, 549029, 195313, 1, 1, 19, 321

Offset: 0

Paul Barry, Apr 23 2003

			Array begins:
  1     1     1     1     1 ...
  1     3    13    63   313 ...
  1     5    33   229  1601 ...
  1     7    61   547  4921 ...
  1     9    97  1065 11713 ...
  ...

Nathaniel Johnston, Antidiagonals 0..100, flattened

Rows include A034478, A083076, A066443, A083077, A083078.
Columns include odds, A082109, A083079.
Diagonals include A083079, A083080, A083081, A083082.

T := proc(n,k) return (k*(2*k+3)^n+1)/(k+1): end: seq(seq(T(k,n-k),k=0..n),n=0..10); # Nathaniel Johnston, Jun 26 2011

A083076 Third row of number array A083075.

Original entry on oeis.org

1, 5, 33, 229, 1601, 11205, 78433, 549029, 3843201, 26902405, 188316833, 1318217829, 9227524801, 64592673605, 452148715233, 3165041006629, 22155287046401, 155087009324805, 1085609065273633, 7599263456915429, 53194844198408001

Offset: 0

Paul Barry, Apr 23 2003

Binomial transform of A067411. Inverse binomial transform of A082412.

Trinomial transform of Jacobsthal numbers A001045. - Paul Barry, Sep 10 2007

Nathaniel Johnston, Table of n, a(n) for n = 0..250
Index entries for linear recurrences with constant coefficients, signature (8,-7).

Cf. A034478, A066443, A082761, A083077.

[(2*7^n+1)/3: n in [0..30]]; // Vincenzo Librandi, Nov 06 2011

seq((2*7^n+1)/3,n=0..20); # Nathaniel Johnston, Jun 26 2011

a(n) = (2*7^n + 1)/3.
G.f.: (1-3*x)/((1-x)*(1-7*x)).
E.g.f.: (2*exp(7*x) + exp(x))/3.
a(n) = Sum_{k=0..2*n} trinomial(n,k)*Fibonacci(k+1), where trinomial(n,k) are the trinomial coefficients (A027907). - Paul Barry, Sep 10 2007
a(n) = 7*a(n-1) - 2, a(n) = 8*a(n-1) - 7*a(n-2). - Vincenzo Librandi, Nov 06 2011

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

A097165 Expansion of (1-3x)/((1-x)(1-4x)(1-5x)).

Original entry on oeis.org

1, 7, 41, 227, 1221, 6447, 33601, 173467, 889181, 4533287, 23015961, 116477907, 587981941, 2962279327, 14900875121, 74862289547, 375743103501, 1884442140567, 9445117195081, 47317211944387, 236952563597861

Offset: 0

Paul Barry, Jul 30 2004

Partial sums of A085351. Convolution of A034478 and 4^n. Convolution of A047849 and 5^n. a(n)=A097162(2n+1)/3. Third binomial transform of A097164.

Index entries for linear recurrences with constant coefficients, signature (10,-29,20).

CoefficientList[Series[(1-3x)/((1-x)(1-4x)(1-5x)),{x,0,30}],x] (* or *) LinearRecurrence[{10,-29,20},{1,7,41},30] (* Harvey P. Dale, Jan 24 2012 *)

a(n)=5*5^n/2-4*4^n/3-1/6; a(n)=sum{k=0..n, (5^k+1)4^(n-k)/2}; a(n)=sum{k=0..n, (4^k+2)5^(n-k)/3}; a(n)=10a(n-1)-29a(n-2)+20a(n-3).

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cp's OEIS Frontend

A034474 a(n) = 5^n + 1.

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A050621 Smallest n-digit number divisible by 2^n.

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A113405 Expansion of x^3/(1 - 2*x + x^3 - 2*x^4) = x^3/( (1-2*x)*(1+x)*(1-x+x^2) ).

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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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A020699 Expansion of (1-3*x)/(1-5*x).

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A083075 Square array read by antidiagonals: T(n,k) = (k*(2*k+3)^n + 1)/(k+1).

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A083076 Third row of number array A083075.

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A113405 Expansion of x^3/(1 - 2x + x^3 - 2x^4) = x^3/( (1-2x)(1+x)*(1-x+x^2) ).

A020699 Expansion of (1-3x)/(1-5x).

A083075 Square array read by antidiagonals: T(n,k) = (k(2k+3)^n + 1)/(k+1).

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