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 1-5 of 5 results.

A221967 T(n,k)=Number of -k..k arrays of length n with the sum ahead of each element differing from the sum following that element by k or less.

Original entry on oeis.org

3, 5, 9, 7, 25, 15, 9, 49, 65, 33, 11, 81, 175, 225, 63, 13, 121, 369, 833, 705, 129, 15, 169, 671, 2241, 3647, 2305, 255, 17, 225, 1105, 4961, 12609, 16513, 7425, 513, 19, 289, 1695, 9633, 34111, 73089, 73983, 24065, 1023, 21, 361, 2465, 17025, 78273, 241153
Offset: 1

Views

Author

R. H. Hardin Feb 01 2013

Keywords

Comments

Table starts
....3.......5.........7..........9..........11...........13............15
....9......25........49.........81.........121..........169...........225
...15......65.......175........369.........671.........1105..........1695
...33.....225.......833.......2241........4961.........9633.........17025
...63.....705......3647......12609.......34111........78273........159615
..129....2305.....16513......73089......241153.......653185.......1535745
..255....7425.....73983.....419841.....1690623......5407233......14661375
..513...24065....332801....2419713....11888129.....44890625.....140355585
.1023...77825...1495039...13930497....83512319....372332545....1342437375
.2049..251905...6719489...80230401...586864641...3089205249...12843782145
.4095..815105..30195711..462012417..4123582463..25628045313..122870296575
.8193.2637825.135700481.2660655105.28975366145.212618141697.1175482548225

Examples

			Some solutions for n=6 k=4
..4...-2....4....1...-4...-1...-2....1...-2...-1....1....3....4....1...-1...-1
.-4....4...-4....0....4....4....3....2....3....2...-2...-4...-2...-3....3....3
..1...-3....3...-2...-1...-2...-3...-2...-2....2....0....3....1....2....0...-1
..0....2...-1....3...-2....0....2....2....3...-3....4....1...-3...-2...-2....1
..3...-4...-2...-3....3....3...-2....1...-1....0...-1...-3....0....3...-3...-2
..1....1....2....1...-1...-2...-1....1...-1....1....0....1....2...-4....4....2

Links

Crossrefs

Column 1 is A062510(n+1)
Column 2 is A189318
Row 2 is A016754
Row 3 is A005917(n+1)
Row 4 is A142993

Formula

Empirical for column k:
k=1: a(n) = a(n-1) +2*a(n-2)
k=2: a(n) = 3*a(n-1) +2*a(n-2) -4*a(n-3)
k=3: a(n) = 3*a(n-1) +8*a(n-2) -4*a(n-3) -8*a(n-4)
k=4: a(n) = 5*a(n-1) +8*a(n-2) -20*a(n-3) -8*a(n-4) +16*a(n-5)
k=5: a(n) = 5*a(n-1) +18*a(n-2) -20*a(n-3) -48*a(n-4) +16*a(n-5) +32*a(n-6)
k=6: a(n) = 7*a(n-1) +18*a(n-2) -56*a(n-3) -48*a(n-4) +112*a(n-5) +32*a(n-6) -64*a(n-7)
k=7: a(n) = 7*a(n-1) +32*a(n-2) -56*a(n-3) -160*a(n-4) +112*a(n-5) +256*a(n-6) -64*a(n-7) -128*a(n-8)
Empirical for row n:
n=1: a(n) = 2*n + 1
n=2: a(n) = 4*n^2 + 4*n + 1
n=3: a(n) = 4*n^3 + 6*n^2 + 4*n + 1
n=4: a(n) = (16/3)*n^4 + (32/3)*n^3 + (32/3)*n^2 + (16/3)*n + 1
n=5: a(n) = (20/3)*n^5 + (50/3)*n^4 + 20*n^3 + (40/3)*n^2 + (16/3)*n + 1
n=6: a(n) = (128/15)*n^6 + (128/5)*n^5 + (112/3)*n^4 + 32*n^3 + (272/15)*n^2 + (32/5)*n + 1
n=7: a(n) = (488/45)*n^7 + (1708/45)*n^6 + (2912/45)*n^5 + (602/9)*n^4 + (2072/45)*n^3 + (952/45)*n^2 + (32/5)*n + 1

A189315 Expansion of g.f. 5*(1-3*x+x^2)/(1-5*x+5*x^2).

Original entry on oeis.org

5, 10, 30, 100, 350, 1250, 4500, 16250, 58750, 212500, 768750, 2781250, 10062500, 36406250, 131718750, 476562500, 1724218750, 6238281250, 22570312500, 81660156250, 295449218750, 1068945312500, 3867480468750, 13992675781250, 50625976562500, 183166503906250, 662702636718750
Offset: 0

Views

Author

L. Edson Jeffery, Apr 20 2011

Keywords

Comments

Let A be the unit-primitive matrix (see [Jeffery])
A=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).
Then a(n) = Trace(A^(2*n)).
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
From Tom Copeland, Dec 08 2015: (Start)
These are also the non-vanishing traces for the adjacency matrices of the simple Lie algebras B_5 and C_5. See links for B_4, A265185, and B_3, A025192.
a(n+1) = 10 * A081567(n), and, ignoring a(0), a G.F. is 10 *(1-2*x)/(1-5*x+5*x^2) whose denominator is y^5 * A127672(5,1/y) with y = sqrt(x).
-log(1 - 5x^2 + 5x^4) = 10 x^2/2 + 30 x^4/4 + ... provides a logarithmic series for the traces of both the odd and even powers of the matrix beginning with the first power. (End)

Links

Crossrefs

Cf. A147748, A189316, A189317, A189318.
Cf. A025192, A081567, A127672, A265185.

Programs

  • Magma
    I:=[5,10,30]; [n le 3 select I[n] else 5*Self(n-1)-5*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Dec 09 2015
  • Mathematica
    CoefficientList[Series[5(1-3x+x^2)/(1-5x+5x^2),{x,0,40}],x] (* or *)
Join[{5},LinearRecurrence[{5,-5},{10,30},40]]  (* Harvey P. Dale, Apr 25 2011 *)
  • PARI
    Vec(5*(1-3*x+x^2)/(1-5*x+5*x^2)+O(x^99)) \\ Charles R Greathouse IV, Sep 24 2012

Formula

a(n) = 5*a(n-1)-5*a(n-2), n>2, a(0)=5, a(1)=10, a(2)=30.
a(n) = Sum_{k=1..5} (w_k)^(2*n), w_k=2*cos((2*k-1)*Pi/10).
a(n) = 2^(1-n)*((5-Sqrt(5))^n+(5+Sqrt(5))^n), for n>0, with a(0)=5.
a(n) = 5*A147748(n).
E.g.f.: 1 + 4*exp(5*x/2)*cosh(sqrt(5)*x/2). - 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

Links

Crossrefs

Cf. A000045, A061646, A189315, A189317, A189318.

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

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

Views

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

Links

Crossrefs

A189315, A189316, A189318.

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

Views

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

Links

Crossrefs

Cf. A189315, A189316, A189317, A189318.

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 1-5 of 5 results.

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