A081342 -id:A081342 - 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

A081341 Expansion of exp(3x)cosh(3*x).

Original entry on oeis.org

1, 3, 18, 108, 648, 3888, 23328, 139968, 839808, 5038848, 30233088, 181398528, 1088391168, 6530347008, 39182082048, 235092492288, 1410554953728, 8463329722368, 50779978334208, 304679870005248, 1828079220031488, 10968475320188928, 65810851921133568

Offset: 0

Paul Barry, Mar 18 2003

Binomial transform of A081340. 3rd binomial transform of (1,0,9,0,81,0,729,0,...).
For m > 1, n > 0, A166469(A002110(m)*a(n)) = (n+1)*A000045(m+1). For n > 0, A166469(a(n)) = 2n. - Matthew Vandermast, Nov 05 2009
Number of compositions of even natural numbers in n parts <= 5. - Adi Dani, May 29 2011

			From _Adi Dani_, May 29 2011: (Start)
a(2)=18: there are 18 compositions of even natural numbers into 2 parts <= 5:
  for 0: (0,0);
  for 2: (0,2),(2,0),(1,1);
  for 4: (0,4),(4,0),(1,3),(3,1),(2,2);
  for 6: (1,5),(5,1),(2,4),(4,2),(3,3);
  for 8: (3,5),(5,3),(4,4);
  for 10: (5,5).  (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..125
Index entries for linear recurrences with constant coefficients, signature (6).

Cf. A000244, A034494, A081340, A081342.

a:= proc(n) option remember; `if`(n=0, 1,
      add(3^j*a(n-j), j=1..n))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 22 2017

Table[Ceiling[1/2(6^n)], {n, 0, 25}]
CoefficientList[Series[-(-1 + 3 x)/(1 - 6 x), {x, 0, 50}], x] (* Vladimir Joseph Stephan Orlovsky, Jun 21 2011 *)
Join[{1},NestList[6#&,3,30]] (* Harvey P. Dale, May 25 2019 *)

x='x+O('x^66); /* that many terms */
Vec((1-3*x)/(1-6*x)) /* show terms */ /* Joerg Arndt, May 29 2011 */

a(0)=1, a(n) = 6^n/2, n > 0.
G.f.: (1-3*x)/(1-6*x).
E.g.f.: exp(3*x)*cosh(3*x).
a(n) = A000244(n)*A011782(n). - Philippe Deléham, Dec 01 2008
a(n) = ((3+sqrt(9))^n + (3-sqrt(9))^n)/2. - Al Hakanson (hawkuu(AT)gmail.com), Dec 08 2008
a(n) = Sum_{k=0..n} A134309(n,k)*3^k = Sum_{k=0..n} A055372(n,k)*2^k. - Philippe Deléham, Feb 04 2012
From Sergei N. Gladkovskii, Jul 19 2012: (Start)
a(n) = ((8*n-4)*a(n-1) - 12*(n-2)*a(n-2))/n, a(0)=1, a(1)=3.
E.g.f. (exp(6*x) + 1)/2 = 1 + 3*x/(G(0) - 6*x) where G(k) = 6*x + 1 + k - 6*x*(k+1)/G(k+1) (continued fraction, Euler's 1st kind, 1-step). (End)
"INVERT" transform of A000244. - Alois P. Heinz, Sep 22 2017

Typo in A-number fixed by Klaus Brockhaus, Apr 04 2010

A034494 a(n) = (7^n+1)/2.

Original entry on oeis.org

1, 4, 25, 172, 1201, 8404, 58825, 411772, 2882401, 20176804, 141237625, 988663372, 6920643601, 48444505204, 339111536425, 2373780754972, 16616465284801, 116315256993604, 814206798955225, 5699447592686572, 39896133148806001, 279272932041642004

Offset: 0

N. J. A. Sloane

Binomial transform of A081341. Inverse binomial transform of A081342. - R. J. Mathar, Oct 23 2008
Number of compositions of even natural numbers into n parts <=6. - Adi Dani, May 28 2011
From Charlie Marion, Jun 24 2011: (Start)
a(n)+(a(n)+1)+...+(a(n+1)-7^n-1)=(a(n+1)-7^n)+...+(a(n+1)-1). Let S(2n) and S(2n+1) be the sets of addends on the left- and right-hand sides, respectively, of the preceding equations. Then, since the intersection of any 2 different S(i) is null and the union of all of them is the positive integers, {S(i)} forms a partition of the positive integers. See also A034659.
In general, for k>0, let b(n)=((4k+3)^n+1)/2. Then b(n)+(b(n)+1)+ ... +(b(n+1)-(4k+3)^n-1)=k*((b(n+1)-(4k+3)^n)+ ... +(b(n+1)-1)). Then, for each k, the set of addends on the two sides of these equations also forms a partition of the positive integers. Also, with b(0)=1, b(n)=(4k+3)*b(n-1)-(2k+1).
For k>0, let c(0)=1 and, for n>0, c(n)=(2*(2k+1))^n/2. Then the sequence b(0),b(1),... is the binomial transform of the sequence c(0),c(1),....
For k>0, let d(2n)=(2k+1)^(2n) and d(2n+1)=0. Then the sequence b(0),b(1),... is the (2k+2)nd binomial transform of the sequence d(0),d(1),.... (End)

			From _Adi Dani_, May 28 2011: (Start)
a(2)=25: there are 25 compositions of even numbers into 2 parts <=6:
  (0,0)
  (0,2),(2,0),(1,1)
  (0,4),(4,0),(1,3),(3,1),(2,2)
  (0,6),(6,0),(1,5),(5,1),(2,4),(4,2),(3,3)
  (2,6),(6,2),(3,5),(5,3),(4,4)
  (4,6),(6,4),(5,5)
  (6,6)
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (8,-7).

[(7^n+1)/2: n in [0..30]]; // Vincenzo Librandi, Jun 16 2011

A034494:=n->(7^n+1)/2: seq(A034494(n), n=0..30); # Wesley Ivan Hurt, Apr 09 2017

(7^Range[0, 25] + 1)/2 (* or *)
LinearRecurrence[{8, -7}, {1, 4}, 26] (* Paolo Xausa, Aug 14 2025 *)

a(n)=(7^n+1)/2 \\ Charles R Greathouse IV, Jul 02 2013

Vec((1-4*x)/((1-x)*(1-7*x)) + O(x^100)) \\ Altug Alkan, Nov 01 2015

E.g.f.: exp(4*x)*cosh(3*x). - Paul Barry, Apr 20 2003
a(n) = 7*a(n-1) - 3, a(0) = 1.
G.f.: (1-4*x)/((1-x)*(1-7*x)). - Philippe Deléham, Jul 11 2005
a(n) = 8*a(n-1)-7*a(n-2), a(0)=1, a(1)=4. - Philippe Deléham, Nov 15 2008
a(n) = ((4+sqrt(9))^n+(4-sqrt(9))^n)/2. - Al Hakanson (hawkuu(AT)gmail.com), Dec 08 2008

A003665 a(n) = 2^(n-1)*( 2^n + (-1)^n ).

Original entry on oeis.org

1, 1, 10, 28, 136, 496, 2080, 8128, 32896, 130816, 524800, 2096128, 8390656, 33550336, 134225920, 536854528, 2147516416, 8589869056, 34359869440, 137438691328, 549756338176, 2199022206976, 8796095119360, 35184367894528, 140737496743936, 562949936644096, 2251799847239680

Offset: 0

N. J. A. Sloane

Binomial transform of expansion of cosh(3*x), the aerated version of A001019, 1,0,9,0,81,0,729,... - Paul Barry, Apr 05 2003
Alternatively: start with the fraction 1/1, take the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 9 times the bottom to get the new top. The limit of the sequence of fractions used to generate this sequence is sqrt(9). - Cino Hilliard, Sep 25 2005
This sequence also gives the number of ordered pairs of subsets (A, B) of {1, 2, ..., n} such that |A u B| is even. (Here "u" stands for the set-theoretic union.) The special case n = 13 can be found as in CRUX Problem 3257. - Walther Janous (walther.janous(AT)tirol.com), Mar 01 2008
For n > 0, a(n) is term (1,1) in the n-th power of the 2 X 2 matrix [1,3; 3,1]. - Gary W. Adamson, Aug 06 2010
a(n) is the number of compositions of n when there are 1 type of 1 and 9 types of other natural numbers. - Milan Janjic, Aug 13 2010
a(n) = ((1+3)^n+(1-3)^n)/2. In general, if b(0),b(1),... is the k-th binomial transform of the sequence ((3^n+(-3)^n)/2), then b(n) = ((k+3)^n+(k-3)^n)/2. More generally, if b(0),b(1),... is the k-th binomial transform of the sequence ((m^n+(-m)^n)/2), then b(n) = ((k+m)^n+(k-m)^n)/2. See A034494, A081340-A081342, A034659. - Charlie Marion, Jun 25 2011
Pisano period lengths: 1, 1, 1, 1, 4, 1, 6, 1, 1, 4, 5, 1, 12, 6, 4, 1, 8, 1, 9, 4, ... - R. J. Mathar, Aug 10 2012
Starting with offset 1 the sequence is the INVERT transform of (1, 9, 9, 9, ...). - Gary W. Adamson, Aug 06 2016

John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, p. 16.
M. Gardner, Riddles of Sphinx, M.A.A., 1987, p. 145.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
R. K. Guy, s-Additive sequences, Preprint, 1994. (Annotated scanned copy)
Bill Sands, Problem 3257, Crux Math. 33 (2007), No.5, p. 298.
Index entries for linear recurrences with constant coefficients, signature (2,8).

Cf. A001019, A001045, A014551, A078008, A098158.

Cf. A034494, A081340-A081342, A034659.

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

[2^(n-1)*( 2^n + (-1)^n ): n in [0..30]]; // Vincenzo Librandi, Aug 19 2011

A003665:=n->2^(n-1)*( 2^n + (-1)^n ): seq(A003665(n), n=0..30); # Wesley Ivan Hurt, Apr 28 2017

CoefficientList[Series[(1+8x)/(1-2x-8x^2), {x,0,30}], x] (* or *)
LinearRecurrence[{2,8}, {1,1}, 30] (* Robert G. Wilson v, Sep 18 2013 *)

PARI
```
a(n)=2^(n-1)*( 2^n + (-1)^n );
```

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

From Paul Barry, Mar 01 2003: (Start)
a(n) = 2*a(n-1) + 8*a(n-2), a(0)=a(1)=1.
a(n) = (4^n + (-2)^n)/2.
G.f.: (1-x)/((1+2*x)*(1-4*x)). (End)
From Paul Barry, Apr 05 2003: (Start)
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*9^k.
E.g.f. exp(x)*cosh(3*x). (End)
a(n) = (A078008(n) + A001045(n+1))*2^(n-1) = A014551(n)*2^(n-1). - Paul Barry, Feb 12 2004
Given a(0)=1, b(0)=1 then for i=1, 2, ..., a(i)/b(i) = (a(i-1) + 9*b(i-1)) / (a(i-1) + b(i-1)). - Cino Hilliard, Sep 25 2005
a(n) = Sum_{k=0..n} A098158(n,k)*9^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = ((1+sqrt(9))^n + (1-sqrt(9))^n)/2. - Al Hakanson (hawkuu(AT)gmail.com), Dec 08 2008
If p[1]=1, and p[i]=9, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Apr 29 2010
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(9*k-1)/(x*(9*k+8) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013

Entry revised by N. J. A. Sloane, Nov 22 2006

A081343 a(n) = (10^n + 4^n)/2.

Original entry on oeis.org

1, 7, 58, 532, 5128, 50512, 502048, 5008192, 50032768, 500131072, 5000524288, 50002097152, 500008388608, 5000033554432, 50000134217728, 500000536870912, 5000002147483648, 50000008589934592, 500000034359738368

Offset: 0

Paul Barry, Mar 18 2003

Binomial transform of A025551. 7th binomial transform of {1, 0, 9, 0, 81, 0, 729, 0, ...}.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (14,-40).

Cf. A081342.

List([0..25], n-> (10^n + 4^n)/2); # G. C. Greubel, Jan 07 2020

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

seq( (10^n + 4^n)/2, n=0..25); # G. C. Greubel, Jan 07 2020

CoefficientList[Series[(1-7x)/((1-4x)(1-10x)), {x, 0, 20}], x] (* Vincenzo Librandi, Aug 08 2013 *)

a(n)=(10^n+4^n)/2 \\ Charles R Greathouse IV, Oct 07 2015

[(10^n + 4^n)/2 for n in (0..25)] # G. C. Greubel, Jan 07 2020

a(n) = 14*a(n-1) -40*a(n-2), a(0)=1, a(1)=7.
G.f.: (1-7*x)/((1-4*x)*(1-10*x)).
E.g.f.: exp(7*x) * cosh(3*x).
a(n) = ((7+sqrt(9))^n + (7-sqrt(9))^n)/2. - Al Hakanson (hawkuu(AT)gmail.com), Dec 08 2008

A162516 Triangle of coefficients of polynomials defined by Binet form: P(n,x) = ((x+d)^n + (x-d)^n)/2, where d=sqrt(x+4).

Original entry on oeis.org

1, 1, 0, 1, 1, 4, 1, 3, 12, 0, 1, 6, 25, 8, 16, 1, 10, 45, 40, 80, 0, 1, 15, 75, 121, 252, 48, 64, 1, 21, 119, 287, 644, 336, 448, 0, 1, 28, 182, 588, 1457, 1360, 1888, 256, 256, 1, 36, 270, 1092, 3033, 4176, 6240, 2304, 2304, 0, 1, 45, 390, 1890, 5925, 10801, 17780, 11680, 12160, 1280, 1024

Offset: 0

Clark Kimberling, Jul 05 2009

			First six rows:
  1;
  1,  0;
  1,  1,  4;
  1,  3, 12,  0;
  1,  6, 25,  8, 16;
  1, 10, 48, 40, 80, 0;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A162514, A162515, A162517.
Cf. A000217, A026150, A084057, A125818, A199572.
For fixed k, the sequences P(n,k), for n=1,2,3,4,5, are A084057, A084059, A146963, A081342, A081343, respectively.

m:=12;
p:= func< n,x | ((x+Sqrt(x+4))^n + (x-Sqrt(x+4))^n)/2 >;
R:=PowerSeriesRing(Rationals(), m+1);
T:= func< n,k | Coefficient(R!( p(n,x) ), n-k) >;
[T(n,k): k in [0..n], n in [0..m]]; // G. C. Greubel, Jul 09 2023

P[n_, x_]:= P[n, x]= ((x+Sqrt[x+4])^n + (x-Sqrt[x+4])^n)/2;
T[n_, k_]:= Coefficient[Series[P[n, x], {x,0,n-k+1}], x, n-k];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 08 2020; Jul 09 2023 *)

def p(n,x): return ((x+sqrt(x+4))^n + (x-sqrt(x+4))^n)/2
def T(n,k):
    P. = PowerSeriesRing(QQ)
    return P( p(n,x) ).list()[n-k]
flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 09 2023

P(n,x) = 2*x*P(n-1,x) - (x^2 -x -4)*P(n-2,x).
From G. C. Greubel, Jul 09 2023: (Start)
T(n, k) = [x^(n-k)] ( ((x+sqrt(x+4))^n + (x-sqrt(x+4))^n)/2 ).
T(n, 1) = A000217(n-1), n >= 1.
T(n, n) = A199572(n).
Sum_{k=0..n} T(n, k) = A084057(n).
Sum_{k=0..n} 2^k*T(n, k) = A125818(n).
Sum_{k=0..n} (-1)^k*T(n, k) = A026150(n).
Sum_{k=0..n} (-2)^k*T(n, k) = A133343(n). (End)

A025551 a(n) = 3^n*(3^n + 1)/2.

Original entry on oeis.org

1, 6, 45, 378, 3321, 29646, 266085, 2392578, 21526641, 193720086, 1743421725, 15690618378, 141215033961, 1270933711326, 11438398618965, 102945573221778, 926510115949281, 8338590914403366, 75047317842209805, 675425859417626778

Offset: 0

N. J. A. Sloane

nonn

G. C. Greubel, Table of n, a(n) for n = 0..750
Index entries for linear recurrences with constant coefficients, signature (12, -27).

Cf. A081342, A081343.

List([0..20], n-> Binomial(3^n+1,2) ); # G. C. Greubel, Jan 08 2020

[Binomial(3^n+1,2): n in [0..20]]; // G. C. Greubel, Jan 08 2020

seq( binomial(3^n +1,2), n=0..20); # G. C. Greubel, Jan 08 2020

LinearRecurrence[{12,-27}, {1,6}, 20] (* G. C. Greubel, Jan 08 2020 *)
Table[3^n(3^n+1)/2,{n,0,20}] (* Harvey P. Dale, Mar 13 2022 *)

Vec( (1-6*x)/((1-3*x)*(1-9*x)) + O(x^66) ) \\ Joerg Arndt, Sep 01 2013

[binomial(3^n+1,2) for n in (0..20)] # G. C. Greubel, Jan 08 2020

From Philippe Deléham, Jul 11 2005: (Start)
Binomial transform of A081342.
6th binomial transform of (1, 0, 9, 0, 81, 0, 729, 0, . . ).
Inverse binomial transform of A081343.
a(n) = 12*a(n-1) - 27*a(n-2), a(0) = 1, a(1) = 6.
G.f.: (1-6*x)/((1-3*x)*(1-9*x)).
E.g.f.: exp(7*x)*cosh(3*x). (End)
a(n) = ((6+sqrt(9))^n + (6-sqrt(9))^n)/2. - Al Hakanson (hawkuu(AT)gmail.com), Dec 08 2008
a(n) = Sum_{k=1..3^n} k. - Joerg Arndt, Sep 01 2013

A120689 a(n) = 10a(n-1) - 16a(n-2), with a(0) = 0 and a(1) = 3.

Original entry on oeis.org

0, 3, 30, 252, 2040, 16368, 131040, 1048512, 8388480, 67108608, 536870400, 4294966272, 34359736320, 274877902848, 2199023247360, 17592186028032, 140737488322560, 1125899906777088, 9007199254609920, 72057594037665792

Offset: 0

Gary W. Adamson, Jun 25 2006

a(n) is a leg in a Pythagorean triangle along with A081342(n) (the hypotenuse) and 4^n. Example: a(4) = 2040, A081342(4) = 2056; then sqrt(2056^2 - 2040^2) = 256 = 4^4. Characteristic polynomial of M = x^2 -10*x + 16.

Order of modular group of degree 2^(n-1)+1. - Artur Jasinski, Aug 04 2007

G. C. Greubel, Table of n, a(n) for n = 0..1000
E. Mathieu, Mémoire sur le nombre de valeurs que peut acquérir une fonction quand on y permute ses variables de toutes les manières possibles, Journ. de math. pures et appliquées (2) 5 (1860), 9-42 (see p. 39).
Index entries for linear recurrences with constant coefficients, signature (10,-16).

Cf. A016131, A081342.

[2^(n-1)*(4^n-1): n in [0..30]]; // G. C. Greubel, Dec 27 2022

a[0]:=0: a[1]:=3; for n from 2 to 20 do a[n]:=10*a[n-1]-16*a[n-2] end do: seq(a[n], n = 0 .. 20); # Emeric Deutsch, Aug 16 2007
seq(binomial(2^n,2)*(2^n + 1),n=0..19); # Zerinvary Lajos, Jan 07 2008

Table[2^(n-1) (4^n-1), {n,0,20}] (* Artur Jasinski, Aug 04 2007 *)

A120689=BinaryRecurrenceSequence(10,-16,0,3)
[A120689(n) for n in range(31)] # G. C. Greubel, Dec 27 2022

a(n) = 8^n - A081342(n).
Given M = 2 X 2 matrix [5,3; 3,5]; M^n * [1,0] = [A081342(a), a(n)]. E.g. a(4) = 2040, right term in = M^4 * [1,0] = [2056, 2040] = [A081342(4), a(4)].
a(n) = 2^(n-1)*(4^n - 1). - Artur Jasinski, Aug 04 2007
From R. J. Mathar, Feb 16 2011: (Start)
a(n) = 3*A016131(n-1).
G.f.: 3*x / ( (1-2*x)*(1-8*x) ). (End)
E.g.f.: (1/2)*(exp(8*x) - exp(2*x)). - G. C. Greubel, Dec 27 2022

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jul 13 2007

More terms from Emeric Deutsch, Aug 16 2007

A083332 a(n) = 10a(n-2) - 16a(n-4) for n > 3, a(0) = 1, a(1) = 5, a(2) = 14, a(3) = 34.

Original entry on oeis.org

1, 5, 14, 34, 124, 260, 1016, 2056, 8176, 16400, 65504, 131104, 524224, 1048640, 4194176, 8388736, 33554176, 67109120, 268434944, 536871424, 2147482624, 4294968320, 17179867136, 34359740416, 137438949376, 274877911040

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Apr 24 2003

a(n)/A083333(n) converges to 3.

Index entries for linear recurrences with constant coefficients, signature (0, 10, 0, -16).

Cf. A147590, A081342 (bisections). [R. J. Mathar, Jul 13 2009]

Cf. A199710. [Bruno Berselli, Nov 11 2011]

CoefficientList[Series[(1+5x+4x^2-16x^3)/(1-10x^2+16x^4), {x, 0, 30}], x]

(a[0] : 1, a[1] : 5, a[2] : 14, a[3] : 34, a[n] := 10*a[n - 2] - 16*a[n - 4], makelist(a[n], n, 0, 50));/* Franck Maminirina Ramaharo, Nov 12 2018 */

G.f.: (1 + 5*x + 4*x^2 - 16*x^3)/(1 - 10*x^2 + 16*x^4).
a(n) = A016116(n)*A014551(n+1). - R. J. Mathar, Jul 08 2009
From Franck Maminirina Ramaharo, Nov 12 2018: (Start)
a(n) = sqrt(2)^(3*n - 1)*(1 + sqrt(2) + (-1)^n*(sqrt(2) - 1)) + sqrt(2)^(n - 3)*(1 - sqrt(2) - (-1)^n*(sqrt(2) + 1)).
E.g.f.: (sinh(sqrt(2)*x) + 2*sinh(2*sqrt(2)*x))/sqrt(2) - cosh(sqrt(2)*x) + 2*cosh(2*sqrt(2)*x). (End)

A146988 Triangle, read by rows, T(n, k) = binomial(n, k) for n < 2 and binomial(n, k) + 4^(n-1) * binomial(n-2, k-1) otherwise.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 19, 19, 1, 1, 68, 134, 68, 1, 1, 261, 778, 778, 261, 1, 1, 1030, 4111, 6164, 4111, 1030, 1, 1, 4103, 20501, 40995, 40995, 20501, 4103, 1, 1, 16392, 98332, 245816, 327750, 245816, 98332, 16392, 1, 1, 65545, 458788, 1376340, 2293886, 2293886, 1376340, 458788, 65545, 1

Offset: 0

Roger L. Bagula, Nov 04 2008

Row sums are {1, 2, 8, 40, 272, 2080, 16448, 131200, 1048832, 8389120, 67109888, ...} = {1, 2, 8*A081342(n)}. (modified by G. C. Greubel, Jan 09 2020)

			Triangle begins as:
  1;
  1,    1;
  1,    6,    1;
  1,   19,   19,    1;
  1,   68,  134,   68,    1;
  1,  261,  778,  778,  261,    1;
  1, 1030, 4111, 6164, 4111, 1030, 1;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A028262, A081342.

T:= function(n,k,q)
    if n<2 then return Binomial(n,k);
    else return Binomial(n,k) + q^(n-1)*Binomial(n-2,k-1);
    fi; end;
Flat(List([0..10], n-> List([0..n], k-> T(n,k,4) ))); # G. C. Greubel, Jan 09 2020

T:= func< n,k,q | n lt 2 select Binomial(n,k) else Binomial(n,k) + q^(n-1)*Binomial(n-2,k-1) >;
[T(n,k,4): k in [0..n], n in [0..10]]; // G. C. Greubel, Jan 09 2020

q:=4; seq(seq( `if`(n<2, binomial(n,k), binomial(n,k) + q^(n-1)*binomial(n-2,k-1)), k=0..n), n=0..10); # G. C. Greubel, Jan 09 2020

Table[If[n<2, Binomial[n, m], Binomial[n, m] + 4^(n-1)*Binomial[n-2, m-1]], {n, 0, 10}, {m, 0, n}]//Flatten

T(n,k) = if(n<2, binomial(n,k), binomial(n,k) + 4^(n-1)*binomial(n-2,k-1) ); \\ G. C. Greubel, Jan 09 2020

@CachedFunction
def T(n, k, q):
    if (n<2): return binomial(n,k)
    else: return binomial(n,k) + q^(n-1)*binomial(n-2,k-1)
[[T(n, k, 4) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Jan 09 2020

T(n, k) = binomial(n, k) for n < 2 and binomial(n, k) + 2^(n-1) * binomial(n-2, k-1) otherwise.

Sum_{k=0..n} T(n,k) = n+1 for n < 2 and 4*(2^n + 8^n) otherwise. - G. C. Greubel, Jan 09 2020

Edited by G. C. Greubel, Jan 09 2020

Showing 1-10 of 10 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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A081341 Expansion of exp(3*x)*cosh(3*x).

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A034494 a(n) = (7^n+1)/2.

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A003665 a(n) = 2^(n-1)*( 2^n + (-1)^n ).

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A081343 a(n) = (10^n + 4^n)/2.

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A162516 Triangle of coefficients of polynomials defined by Binet form: P(n,x) = ((x+d)^n + (x-d)^n)/2, where d=sqrt(x+4).

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A081341 Expansion of exp(3x)cosh(3*x).

A120689 a(n) = 10a(n-1) - 16a(n-2), with a(0) = 0 and a(1) = 3.

A083332 a(n) = 10a(n-2) - 16a(n-4) for n > 3, a(0) = 1, a(1) = 5, a(2) = 14, a(3) = 34.