cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A081567 Second binomial transform of F(n+1).

Original entry on oeis.org

1, 3, 10, 35, 125, 450, 1625, 5875, 21250, 76875, 278125, 1006250, 3640625, 13171875, 47656250, 172421875, 623828125, 2257031250, 8166015625, 29544921875, 106894531250, 386748046875, 1399267578125, 5062597656250, 18316650390625, 66270263671875, 239768066406250
Offset: 0

Views

Author

Paul Barry, Mar 22 2003

Keywords

Comments

Binomial transform of F(2*n-1), index shifted by 1, where F is A000045. - corrected by Richard R. Forberg, Aug 12 2013
Case k=2 of family of recurrences a(n) = (2k+1)*a(n-1) - A028387(k-1)*a(n-2), a(0)=1, a(1)=k+1.
Number of (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1, 2, ..., 2*n+1, s(0) = 3, s(2*n+1) = 4.
a(n+1) gives the number of periodic multiplex juggling sequences of length n with base state <2>. - Steve Butler, Jan 21 2008
a(n) is also the number of idempotent order-preserving partial transformations (of an n-element chain) of waist n (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Sep 14 2008
Counts all paths of length (2*n+1), n>=0, starting at the initial node on the path graph P_9, see the Maple program. - Johannes W. Meijer, May 29 2010
Given the 3 X 3 matrix M = [1,1,1; 1,1,0; 1,1,3], a(n) = term (1,1) in M^(n+1). - Gary W. Adamson, Aug 06 2010
Number of nonisomorphic graded posets with 0 and 1 of rank n+2, with exactly 2 elements of each rank level between 0 and 1. Also the number of nonisomorphic graded posets with 0 of rank n+1, with exactly 2 elements of each rank level above 0. (This is by Stanley's definition of graded, that all maximal chains have the same length.) - David Nacin, Feb 26 2012
a(n) = 3^n a(n;1/3) = Sum_{k=0..n} C(n,k) * F(k-1) * (-1)^k * 3^(n-k), which also implies the Deleham formula given below and where a(n;d), n=0,1,...,d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also the papers of Witula et al.). - Roman Witula, Jul 12 2012
The limiting ratio a(n)/a(n-1) is 1 + phi^2. - Bob Selcoe, Mar 17 2014
a(n) counts closed walks on K_2 containing 3 loops on the index vertex and 2 loops on the other. Equivalently the (1,1) entry of A^n where the adjacency matrix of digraph is A=(3,1; 1,2). - David Neil McGrath, Nov 18 2014

Examples

			a(4)=125: 35*(3 + (35 mod 10 - 10 mod 3)/(10-3)) = 35*(3 + 4/7) = 125. - _Bob Selcoe_, Mar 17 2014
		

References

  • R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1997, pages 96-100.

Crossrefs

a(n) = 5*A052936(n-1), n > 1.
Row sums of A114164.
Cf. A000045, A007051 (INVERTi transform), A007598, A028387, A030191, A039717, A049310, A081568 (binomial transform), A086351 (INVERT transform), A090041, A093129, A094441, A111776, A147748, A178381, A189315.

Programs

  • Magma
    I:=[1, 3]; [n le 2 select I[n] else 5*Self(n-1)-5*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 27 2012
    
  • Maple
    with(GraphTheory):G:=PathGraph(9): A:= AdjacencyMatrix(G): nmax:=23; n2:=nmax*2+2: for n from 0 to n2 do B(n):=A^n; a(n):=add(B(n)[1,k],k=1..9); od: seq(a(2*n+1),n=0..nmax); # Johannes W. Meijer, May 29 2010
  • Mathematica
    Table[MatrixPower[{{2,1},{1,3}},n][[2]][[2]],{n,0,44}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
    LinearRecurrence[{5,-5},{1,3},30] (* Vincenzo Librandi, Feb 27 2012 *)
  • PARI
    Vec((1-2*x)/(1-5*x+5*x^2)+O(x^99)) \\ Charles R Greathouse IV, Mar 18 2014
  • Python
    def a(n, adict={0:1, 1:3}):
        if n in adict:
            return adict[n]
        adict[n]=5*a(n-1) - 5*a(n-2)
        return adict[n] # David Nacin, Mar 04 2012
    

Formula

a(n) = 5*a(n-1) - 5*a(n-2) for n >= 2, with a(0) = 1 and a(1) = 3.
a(n) = (1/2 - sqrt(5)/10) * (5/2 - sqrt(5)/2)^n + (sqrt(5)/10 + 1/2) * (sqrt(5)/2 + 5/2)^n.
G.f.: (1 - 2*x)/(1 - 5*x + 5*x^2).
a(n-1) = Sum_{k=1..n} binomial(n, k)*F(k)^2. - Benoit Cloitre, Oct 26 2003
a(n) = A090041(n)/2^n. - Paul Barry, Mar 23 2004
The sequence 0, 1, 3, 10, ... with a(n) = (5/2 - sqrt(5)/2)^n/5 + (5/2 + sqrt(5)/2)^n/5 - 2(0)^n/5 is the binomial transform of F(n)^2 (A007598). - Paul Barry, Apr 27 2004
From Paul Barry, Nov 15 2005: (Start)
a(n) = Sum_{k=0..n} Sum_{j=0..n} binomial(n, j)*binomial(j+k, 2k);
a(n) = Sum_{k=0..n} Sum_{j=0..n} binomial(n, k+j)*binomial(k, k-j)2^(n-k-j);
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} binomial(n+k-j, n-k-j)*binomial(k, j)(-1)^j*2^(n-k-j). (End)
a(n) = A111776(n, n). - Abdullahi Umar, Sep 14 2008
a(n) = Sum_{k=0..n} A094441(n,k)*2^k. - Philippe Deléham, Dec 14 2009
a(n+1) = Sum_{k=-floor(n/5)..floor(n/5)} ((-1)^k*binomial(2*n, n+5*k)/2). -Mircea Merca, Jan 28 2012
a(n) = A030191(n) - 2*A030191(n-1). - R. J. Mathar, Jul 19 2012
G.f.: Q(0,u)/x - 1/x, where u=x/(1-2*x), Q(k,u) = 1 + u^2 + (k+2)*u - u*(k+1 + u)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 07 2013
For n>=3: a(n) = a(n-1)*(3+(a(n-1) mod a(n-2) - a(n-2) mod a(n-3))/(a(n-2) - a(n-3))). - Bob Selcoe, Mar 17 2014
a(n) = sqrt(5)^(n-1)*(3*S(n-1, sqrt(5)) - sqrt(5)*S(n-2, sqrt(5))) with Chebyshev's S-polynomials (see A049310), where S(-1, x) = 0 and S(-2, x) = -1. This is the (1,1) entry of A^n with the matrix A=(3,1;1,2). See the comment by David Neil McGrath, Nov 18 2014. - Wolfdieter Lang, Dec 04 2014
Conjecture: a(n) = 2*a(n-1) + A039717(n). - Benito van der Zander, Nov 20 2015
a(n) = A189315(n+1) / 10. - Tom Copeland, Dec 08 2015
a(n) = A093129(n) + A030191(n-1). - Gary W. Adamson, Apr 24 2023
E.g.f.: exp(5*x/2)*(5*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Jun 03 2024

A147748 Row sums of Riordan array ((1-3x+x^2)/(1-4x+3x^2), x(1-2x)/(1-4x+3x^2)).

Original entry on oeis.org

1, 2, 6, 20, 70, 250, 900, 3250, 11750, 42500, 153750, 556250, 2012500, 7281250, 26343750, 95312500, 344843750, 1247656250, 4514062500, 16332031250, 59089843750, 213789062500, 773496093750, 2798535156250, 10125195312500
Offset: 0

Views

Author

Paul Barry, Nov 11 2008

Keywords

Comments

Row sums of A147747. Binomial transform of A061646.
Counts all paths of length (2*n), n>=0, starting at the initial node on the path graph P_9, see the Maple program. - Johannes W. Meijer, May 29 2010
From L. Edson Jeffery, Apr 19 2011: (Start)
For the 5 X 5 unit-primitive matrix (see [Jeffery])
A_(10,1) = [0,1,0,0,0; 1,0,1,0,0; 0,1,0,1,0; 0,0,1,0,1; 0,0,0,2,0],
a(n) = (Trace([A_(10,1)]^(2*n)))/5. (See also A189315.) (End)

Crossrefs

Programs

  • Maple
    with(GraphTheory): G:=PathGraph(9): A:= AdjacencyMatrix(G): nmax:=24; n2:=nmax*2: for n from 0 to n2 do B(n):=A^n; a(n):= add(B(n)[1,k], k=1..9); od: seq(a(2*n), n=0..nmax); # Johannes W. Meijer, May 29 2010
  • Mathematica
    (1 - 3x + x^2)/(1 - 5x + 5x^2) + O[x]^25 // CoefficientList[#, x]& (* Jean-François Alcover, Oct 05 2016 *)

Formula

G.f.: (1-3*x+x^2)/(1-5*x+5*x^2).
a(n) = 5*a(n-1) - 5*a(n-2) for n > 2, a(0)=1, a(1)=2, a(2)=6. - Philippe Deléham, Nov 13 2008
For n >= 1: a(n) = (2/5)*((5-sqrt(5))/2)^n + (2/5)*((5+sqrt(5))/2)^n. - Richard Choulet, Nov 14 2008
G.f.: 1/(1-2x/(1-x/(1-x/(1-x)))) (hence sequence approximates A000984 in first few terms). - Paul Barry, Aug 05 2009
a(n) = (1/5)*Sum_{k=1..5} (x_k)^(2*n), x_k=2*cos((2*k-1)*Pi/10). - L. Edson Jeffery, Apr 19 2011
From R. J. Mathar, Apr 20 2011: (Start)
a(n) = A030191(n) - 3*A030191(n-1) + A030191(n-2).
a(n) = 2*A081567(n-1), n > 0. (End)
a(n) = Sum_{k=0..n} A147746(n,k)*2^k. - Philippe Deléham, Oct 30 2011
E.g.f.: (1 + 4*exp(5*x/2)*cosh(sqrt(5)*x/2))/5. - Stefano Spezia, Jul 09 2024

A189316 Expansion of g.f. 5*(1-x-x^2)/((1+x)*(1-3*x+x^2)).

Original entry on oeis.org

5, 5, 15, 35, 95, 245, 645, 1685, 4415, 11555, 30255, 79205, 207365, 542885, 1421295, 3720995, 9741695, 25504085, 66770565, 174807605, 457652255, 1198149155, 3136795215, 8212236485, 21499914245, 56287506245, 147362604495, 385800307235, 1010038317215
Offset: 0

Views

Author

L. Edson Jeffery, Apr 20 2011

Keywords

Comments

(Start) Let A be the unit-primitive matrix (see [Jeffery])
A=A_(10,2)=
(0 0 1 0 0)
(0 1 0 1 0)
(1 0 1 0 1)
(0 1 0 2 0)
(0 0 2 0 1).
Then a(n)=Trace(A^n). For m=1,2,..., A^(m) can also be written
A^(m)=
[ F(m-1)^2 0 F(m)^2 0 F(m-1)*F(m) ]
[ 0 F(2*m-1) 0 F(2*m) 0 ]
[ F(m)^2 0 F(m+1)^2 0 F(m)*F(m+1) ]
[ 0 F(2*m) 0 F(2*m+1) 0 ]
[ 2*F(m-1)*F(m) 0 2*F(m)*F(m+1) 0 F(2*m+1)-F(m)*F(m+1) ],
where F(m-1)=A000045(n) are the Fibonacci numbers and m=n+1. Hence also a(n+1)=Trace(A^(n+1))=F(m-1)^2+F(2*m-1)+F(m+1)^2+2*F(2*m+1)-F(m)*F(m+1). (End)
Evidently one of a class of accelerator sequences for Catalan's constant based on traces of successive powers of a unit-primitive matrix A_(N,r), 0

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[5 (1-x-x^2)/((1+x)(1-3x+x^2)),{x,0,40}],x] (* or *) LinearRecurrence[{2,2,-1},{5,5,15},40] (* Harvey P. Dale, Nov 26 2016 *)

Formula

G.f.: 5*(1-x-x^2)/((1+x)*(1-3*x+x^2)).
a(n) = 2*a(n-1)+2*a(n-2)-a(n-3), n>2, a(0)=5, a(1)=5, a(2)=15.
a(n) = Sum_{k=1..5} ((w_k)^2-1)^n, w_k = 2*cos((2*k-1)*Pi/10).
a(n) = (-1)^n+2*(1/tau^(2*n)+tau^(2*n)), tau = (1+sqrt(5))/2=1.618033....
a(n) = 5*A061646(n), n>=0 (offset for A061646 is -1).
E.g.f.: cosh(x) + 4*exp(3*x/2)*cosh(sqrt(5)*x/2) - sinh(x). - Stefano Spezia, Jul 09 2024

A189318 Expansion of 5*(1-2*x)/(1-3*x-2*x^2+4*x^3).

Original entry on oeis.org

5, 5, 25, 65, 225, 705, 2305, 7425, 24065, 77825, 251905, 815105, 2637825, 8536065, 27623425, 89391105, 289275905, 936116225, 3029336065, 9803137025, 31723618305, 102659784705, 332214042625, 1075067224065, 3478990618625, 11258250133505
Offset: 0

Author

L. Edson Jeffery, Apr 20 2011

Keywords

Comments

(Start) Let A be the unit-primitive matrix (see [Jeffery])
A=A_(10,4)=
(0 0 0 0 1)
(0 0 0 2 0)
(0 0 2 0 1)
(0 2 0 2 0)
(2 0 2 0 1).
Then a(n)=Trace(A^n). (End)
Evidently one of a class of accelerator sequences for Catalan's constant based on traces of successive powers of a unit-primitive matrix A_(N,r) (0

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[5(1-2x)/(1-3x-2x^2+4x^3),{x,0,30}],x] (* or *) LinearRecurrence[{3,2,-4},{5,5,25},30] (* Harvey P. Dale, Jun 02 2014 *)
  • PARI
    Vec(5*(1-2*x)/(1-3*x-2*x^2+4*x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 25 2012

Formula

G.f.: 5*(1-2*x)/(1-3*x-2*x^2+4*x^3).
a(n)=3*a(n-1)+2*a(n-2)-4*a(n-3), n>3, a(0)=5, a(1)=5, a(2)=25, a(3)=65.
a(n)=Sum_{k=1..5} ((w_k)^4-3*(w_k)^2+1)^n, w_k=2*cos((2*k-1)*Pi/10).
a(n)=1+2*(1-Sqrt(5))^n+2*(1+Sqrt(5))^n.
a(n)=5*A052899(n).

A265185 Non-vanishing traces of the powers of the adjacency matrix for the simple Lie algebra B_4: 2 * ((2 + sqrt(2))^n + (2 - sqrt(2))^n).

Original entry on oeis.org

4, 8, 24, 80, 272, 928, 3168, 10816, 36928, 126080, 430464, 1469696, 5017856, 17132032, 58492416, 199705600, 681837568, 2327939072, 7948081152, 27136446464, 92649623552, 316325601280, 1080003158016, 3687361429504, 12589439401984, 42983034748928
Offset: 0

Author

Tom Copeland, Dec 04 2015

Keywords

Comments

a(n) is the trace of the 2*n-th power of the adjacency matrix M for the simple Lie algebra B_4, given in the Damianou link. M = Matrix[row 1; row 2; row 3; row 4] = Matrix[0,1,0,0; 1,0,1,0; 0,1,0,2; 0,0,1,0]. Equivalently, the trace tr(M^(2*k)) is the sum of the 2*n-th powers of the eigenvalues of M. The eigenvalues are the zeros of the characteristic polynomial of M, which is det(x*I - M) = x^4 - 4*x^2 + 2 = A127672(4,x), and are (+-) sqrt(2 + sqrt(2)) and (+-) sqrt(2 - sqrt(2)), or the four unique values generated by 2*cos((2*n+1)*Pi/8). Compare with A025192 for B_3. The odd power traces vanish.
-log(1 - 4*x^2 + 2*x^4) = 8*x^2/2 + 24*x^4/4 + 80*x^6/6 + ... = Sum_{n>0} tr(M^k) x^k / k = Sum_{n>0} a(n) x^(2k) / 2k gives an aerated version of the sequence a(n), excluding a(0), and exp(-log(1 - 4*x + 2*x^2)) = 1 / (1 - 4*x + 2*x^2) is the e.g.f. for A007070.
As in A025192, the cycle index partition polynomials P_k(x[1],...,x[k]) of A036039 evaluated with the negated power sums, the aerated a(n), are P_2(0,-a(1)) = P_2(0,-8) = -8, P_4(0,-a(1),0,-a(2)) = P_4(0,-8,0,-24) = 48, and all other P_k(0,-a(1),0,-a(2),0,...) = 0 since 1 - 4*x^2 + 2*x^4 = 1 - 8*x^2/2! + 48*x^4/4! = det(I - x M) = exp(-Sum_{k>0} tr(M^k) x^k / k) = exp[P.(-tr(M),-tr(M^2),...)x] = exp[P.(0,-a(1),0,-a(2),...)x].
Because of the inverse relation between the Faber polynomials F_n(b1,b2,...,bn) of A263916 and the cycle index polynomials, F_n(0,-4,0,2,0,0,0,...) = tr(M^n) gives aerated a(n), excluding a(0). E.g., F_2(0,-4) = -2 * -4 = 8, F_4(0,-4,0,2) = -4 * 2 + 2 * (-4)^2 = 24, and F_6(0,-4,0,2,0,0) = -2*(-4)^3 + 6*(-4)*2 = 80.

Programs

  • Magma
    [Floor(2 * ((2 + Sqrt(2))^n + (2 - Sqrt(2))^n)): n in [0..30]]; // Vincenzo Librandi, Dec 06 2015
    
  • Mathematica
    4 LinearRecurrence[{4, -2}, {1, 2}, 30] (* Vincenzo Librandi, Dec 06 2015 and slightly modified by Robert G. Wilson v, Feb 13 2018 *)
  • PARI
    my(x='x+O('x^30)); Vec((4-8*x)/(1-4*x+2*x^2)) \\ G. C. Greubel, Feb 12 2018

Formula

a(n) = 2 * ((2 + sqrt(2))^n + (2 - sqrt(2))^n) = Sum_{k=0..3} 2^(2n) (cos((2k+1)*Pi/8))^(2n) = 2*2^(2n) (cos(Pi/8)^(2n) + cos(3*Pi/8)^(2n)) = 2 Sum_{k=0..1} (exp(i(2k+1)*Pi/8) + exp(-i*(2k+1)*Pi/8))^(2n).
E.g.f.: 2 * e^(2*x) * (e^(sqrt(2)*x) + e^(-sqrt(2)*x)) = 4*e^(2*x)*cosh(sqrt(2)*x) = 2*(exp(4*x*cos(Pi/8)^2) + exp(4*x cos(3*Pi/8)^2) ).
a(n) = 4*A006012(n) = 8*A007052(n-1) = 2*A056236(n).
G.f.: (4-8*x)/(1-4*x+2*x^2). - Robert Israel, Dec 07 2015
Note the preceding o.g.f. is four times that of A006012 and the denominator is y^4 * A127672(4,1/y) with y = sqrt(x). Compare this with those of A025192 and A189315. - Tom Copeland, Dec 08 2015

Extensions

More terms from Vincenzo Librandi, Dec 06 2015

A189317 Expansion of 5*(1-6*x+x^2)/(1-10*x+5*x^2).

Original entry on oeis.org

5, 20, 180, 1700, 16100, 152500, 1444500, 13682500, 129602500, 1227612500, 11628112500, 110143062500, 1043290062500, 9882185312500, 93605402812500, 886643101562500, 8398404001562500, 79550824507812500, 753516225070312500, 7137408128164062500
Offset: 0

Author

L. Edson Jeffery, Apr 20 2011

Keywords

Comments

(Start) Let A be the unit-primitive matrix (see [Jeffery])
A=A_(10,3)=
(0 0 0 1 0)
(0 0 1 0 1)
(0 1 0 2 0)
(1 0 2 0 1)
(0 2 0 2 0).
Then a(n)=Trace(A^(2*n)). (End)
Evidently one of a class of accelerator sequences for Catalan's constant based on traces of successive powers (here they are A^(2*n)) of a unit-primitive matrix A_(N,r) (0

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[5*(1-6x+x^2)/(1-10x+5x^2),{x,0,30}],x] (* or *) Join[ {5},LinearRecurrence[{10,-5},{20,180},30]] (* Harvey P. Dale, Apr 02 2013 *)
  • PARI
    Vec(5*(1-6*x+x^2)/(1-10*x+5*x^2)+O(x^99)) \\ Charles R Greathouse IV, Sep 25 2012

Formula

G.f.: 5*(1-6*x+x^2)/(1-10*x+5*x^2).
a(n)=10*a(n-1)-5*a(n-2), n>2, a(0)=5, a(1)=20, a(2)=180.
a(n)=Sum_{k=1..5} ((w_k)^3-2*w_k)^(2*n), w_k=2*cos((2*k-1)*Pi/10).
a(n)=2*((5-2*Sqrt(5))^n+(5+2*Sqrt(5))^n), for n>0, with a(0)=5.

A189334 Expansion of g.f. (1-6*x+x^2)/(1-10*x+5*x^2).

Original entry on oeis.org

1, 4, 36, 340, 3220, 30500, 288900, 2736500, 25920500, 245522500, 2325622500, 22028612500, 208658012500, 1976437062500, 18721080562500, 177328620312500, 1679680800312500, 15910164901562500, 150703245014062500, 1427481625632812500, 13521300031257812500, 128075592184414062500
Offset: 0

Author

L. Edson Jeffery, Apr 20 2011

Keywords

Comments

(Start) Let A be the unit-primitive matrix (see [Jeffery])
A=A_(10,3)=
(0 0 0 1 0)
(0 0 1 0 1)
(0 1 0 2 0)
(1 0 2 0 1)
(0 2 0 2 0).
Then a(n)=(1/5)*Trace(A^(2*n)). (See also A189317.) (End)
Evidently one of a class of accelerator sequences for Catalan's constant based on traces of successive powers (here they are A^(2*n)) of a unit-primitive matrix A_(N,r) (0

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{10,-5},{1,4,36},22] (* Stefano Spezia, Jul 09 2024 *)

Formula

a(n) = 10*a(n-1) - 5*a(n-2), n>2, a(0)=1, a(1)=4, a(2)=36.
a(n) = (1/5)*Sum_{k=1..5} ((w_k)^3-2*w_k)^(2*n), w_k = 2*cos((2*k-1)*Pi/10).
From Stefano Spezia, Jul 09 2024: (Start)
a(n) = 2*((5 - 2*sqrt(5))^n + (5 + 2*sqrt(5))^n)/5 for n > 0.
E.g.f.: (1 + 4*exp(5*x)*cosh(2*sqrt(5)*x))/5. (End)

Extensions

a(20)-a(21) from Stefano Spezia, Jul 09 2024
Showing 1-7 of 7 results.