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 1-10 of 18 results. Next

A094648 An accelerator sequence for Catalan's constant.

Original entry on oeis.org

3, -1, 5, -4, 13, -16, 38, -57, 117, -193, 370, -639, 1186, -2094, 3827, -6829, 12389, -22220, 40169, -72220, 130338, -234609, 423065, -761945, 1373466, -2474291, 4459278, -8034394, 14478659, -26088169, 47011093, -84708772, 152642789, -275049240
Offset: 0

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Author

Paul Barry, May 18 2004

Keywords

Comments

The pair A094648 and the alternating sequence A033304 when joined form a two-sided sequence defined by the recurrence formula x(n+3) + x(n+2) - 2x(n+1) - x(n) = 0, n in Z, x(-1)=-2, x(0)=3, x(1)=-1 - for details see Witula's comments to A033304. - Roman Witula, Jul 25 2012
From Roman Witula, Aug 09 2012: (Start)
There exist two interesting subsequences b(n) and c(n) of the given above sequence x(n) defined by the following relations: b(n)=a(2^n) and c(n)=x(-2^n). These subsequences satisfy the following system of recurrence equations:
b(n+1)=b(n)^2-2*c(n), and c(n+1)=c(n)^2-2*b(n),
which easily follow from the general identity: x(n)^2=x(2*n)-2*x(-n), n in Z. We note that b(0)=-1, b(1)=5, b(2)=13, b(3)=117, c(0)=-2, c(1)=6, c(2)=26, c(3)=650. From the above system we deduce that all b(n) are odd, whereas all c(n) are even. Moreover we obtain c(n+1)-b(n+1)=(c(n)-b(n))*(b(n)+c(n)+2), which yields b(n+1)-c(n+1)=product{k=1,..,n}(b(k)+c(k)+2)=13*product{k=2,..,n}(b(k)+c(k)+2)=13^2*41*product{k=3,..,n}(b(k)+c(k)+2). It follows that b(n)-c(n) is divisible by 13^2*41 for every n=3,4,..., and after using the above system again each b(n) and c(n), for n=2,3,..., is divisible by 13. (End)
If we set W(n):=3*A077998(n)-A006054(n+1)-A006054(n), n=0,1,..., then a(n)=(W(n)^2-W(2*n))/2 and W(n) = (-c(1))^(-n) + (-c(2))^(-n) + (-c(4))^(-n) = (-c(1)*c(2))^n + (-c(1)*c(4))^n + (-c(2)*c(4))^n = (-1-c(1))^n + (-1-c(2))^n + (-1-c(4))^n, where c(j):=2*cos(2*Pi*j/7) - for the proof see Witula-Slota-Warzynski's paper. Moreover it follows from the comment at the top and from comments to A033304 that W(n+1)=A033304(n)=(-1)^(n+1)*x(-n-1). - Roman Witula, Aug 11 2012
The following trigonometric type identitities hold true: (1) -a(n-1)-a(n) = c(1)*c(2)^n + c(2)*c(4)^n + c(4)*c(1)^n and (2) a(n)-a(n+2) = c(4)*c(2)^(n+1) + c(1)*c(4)^(n+1) + c(2)*c(1)^(n+1), where a(-1)=-2 and c(j) is defined as above (see also the respective comment to A033304). For the proof see Remark 6 in Witula's paper. - Roman Witula, Aug 14 2012
It can be proved that A033304(n-1)*(-1)^n = (a(n)^2 - a(2*n))/2, n=1,2,... - Roman Witula, Sep 30 2012
With respect to the form of the trigonometric formulas describing a(n), we call this sequence the Berndt-type sequence number 19 for the argument 2*Pi/7. The A-numbers of other Berndt-type sequences numbers are given in below. - Roman Witula, Sep 30 2012

Examples

			We have a(17) = a(19) + 50000, a(4) + a(5) = -3, 2*a(7) + a(8) = 3, and 2*a(9) + a(10) = a(5). - _Roman Witula_, Sep 14 2012
		

Crossrefs

Programs

  • Magma
    I:=[3,-1,5]; [n le 3 select I[n]  else -Self(n-1)+2*Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jul 25 2015
    
  • Mathematica
    CoefficientList[ Series[(3 + 2x - 2x^2)/(1 + x - 2x^2 - x^3), {x, 0, 33}], x] (* Robert G. Wilson v, May 24 2004 *)
    a[n_] := Round[(2Sin[3Pi/14])^n + (-2Sin[Pi/14])^n + (-2Cos[Pi/7])^n]; Table[ a[n], {n, 0, 33}] (* Robert G. Wilson v, May 24 2004 *)
    LinearRecurrence[{-1,2,1}, {3,-1,5}, 50] (* Roman Witula, Aug 09 2012 *)
  • PARI
    x='x+O('x^30); Vec((3+2*x-2*x^2)/(1+x-2*x^2-x^3)) \\ G. C. Greubel, May 09 2018

Formula

G.f.: (3+2*x-2*x^2)/(1+x-2*x^2-x^3);
a(n) = (2*sin(3*Pi/14))^n+(-2*sin(Pi/14))^n+(-2*cos(Pi/7))^n.
a(p) == -1 mod(p), p prime. - Philippe Deléham, Oct 03 2009
a(n) = (2*cos(2*Pi/7))^n + (2*cos(4*Pi/7))^n + (2*cos(8*Pi/7))^n, which is equivalent to the formula given above (for analogous sums with sines see A215493 and A215494). Moreover we have a(n+3) + a(n+2) - 2a(n+1) - a(n) = 0 - for the proof see Witula-Slota's paper. - Roman Witula, Jul 24 2012
a(n) = 3*(-1)^n*A006053(n+2) +2*A078038(n-1). - R. J. Mathar, Nov 03 2020

A274975 Sum of n-th powers of the three roots of x^3-2*x^2-x+1.

Original entry on oeis.org

3, 2, 6, 11, 26, 57, 129, 289, 650, 1460, 3281, 7372, 16565, 37221, 83635, 187926, 422266, 948823, 2131986, 4790529, 10764221, 24186985, 54347662, 122118088, 274396853, 616564132, 1385407029, 3112981337, 6994805571, 15717185450, 35316195134, 79354770147, 178308549978
Offset: 0

Views

Author

Kai Wang, Jul 14 2016

Keywords

Comments

a(n) is x1^n + x2^n + x3^n, where x1, x2, x3 are the roots of the polynomial x^3-2*x^2-x+1.
x1 = 1/(2*cos(Pi/7)),
x2 = 1/(-2*cos(2*Pi/7)),
x3 = 1/(-2*cos(4*Pi/7)).

Crossrefs

Cf. A096975.
3 followed by terms of A033304.

Programs

  • Mathematica
    CoefficientList[Series[-(x^2 + 4 x - 3)/(x^3 - x^2 - 2 x + 1), {x, 0, 32}], x] (* Michael De Vlieger, Jul 14 2016 *)
  • PARI
    Vec(-(x^2+4*x-3)/(x^3-x^2-2*x+1) + O(x^50)) \\ Colin Barker, Aug 02 2016

Formula

G.f.: -(x^2+4*x-3)/(x^3-x^2-2*x+1). - Alois P. Heinz, Jul 14 2016
a(0)=3, a(1)=2, a(2)=6; thereafter a(n)=2*a(n-1)+a(n-2)-a(n-3).
a(n) = (2*cos(Pi/7))^(-n) + (-2*cos(2*Pi/7))^(-n) + (-2*cos(4*Pi/7))^(-n).
a(n) = A033304(n-1) for n>0.

A217274 a(n) = 7*a(n-1) - 14*a(n-2) + 7*a(n-3) with a(0)=0, a(1)=1, a(2)=7.

Original entry on oeis.org

0, 1, 7, 35, 154, 637, 2548, 9996, 38759, 149205, 571781, 2184910, 8333871, 31750824, 120875944, 459957169, 1749692735, 6654580387, 25306064602, 96226175941, 365880389868, 1391138718116, 5289228800247, 20109822277181, 76457523763621, 290689756066542
Offset: 0

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Author

Roman Witula, Sep 29 2012

Keywords

Comments

This is the Berndt-type sequence number 18 for the argument 2*Pi/7 defined by the relation
a(n)*sqrt(7) = c(4)*s(1)^(2n+1) + c(2)*s(4)^(2n+1) + c(1)*s(2)^(2n+1) = (1/s(4))*s(1)^(2n+2) + (1/s(2))*s(4)^(2n+2) + (1/s(1))*s(2)^(2n+2), where c(j) := 2*cos(2*Pi*j/7) and s(j) := 2*sin(2*Pi*j/7) (for the sums of the respective even powers see A094429). For the proof of this formula see the Witula/Slota and Witula references.
The definitions of the other Berndt-type sequences for the argument 2*Pi/7 (with numbers from 1 to 17) are in the cross references.
We note that all numbers of the form a(n)*7^(-floor((n+1)/3)) = A217444(n) are integers.
It can be proved that Sum_{k=2..n}a(k) = 7*(a(n-1) - a(n-2)).

Examples

			Writing c(j) as cj and s(k) as sk,
we have 7*sqrt(7) = c4*s1^5 + c2*s4^5 + c1*s2^5
and c4*s1^13 + c2*s4^13 + c1*s2^13 = 4(c4*s1^11 + c2*s4^11 + c1*s2^11).
We note that a(9) = 87*a(3)*a(2)^2 and a(11) = 2*a(3)*a(5)*a(2)^2.
		

Crossrefs

Programs

  • Magma
    I:=[0,1,7]; [n le 3 select I[n] else 7*Self(n-1)-14*Self(n-2)+7*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jul 26 2015
  • Mathematica
    LinearRecurrence[{7,-14,7}, {0,1,7}, 30]
    CoefficientList[Series[x/(1 - 7*x + 14*x^2 - 7*x^3), {x,0,50}], x] (* G. C. Greubel, Apr 16 2017 *)
  • Maxima
    a[0]:0$
    a[1]:1$
    a[2]:7$
    a[n]:=7*a[n-1] - 14*a[n-2] + 7*a[n-3];
    makelist(a[n], n, 0, 25); /* Martin Ettl, Oct 11 2012 */
    
  • PARI
    concat(0, Vec(x/(1-7*x+14*x^2-7*x^3) + O(x^40))) \\ Michel Marcus, Jul 25 2015
    

Formula

G.f.: x/(1-7*x+14*x^2-7*x^3).

A274032 Sum of n-th powers of the roots of x^3 + 9*x^2 - x - 1.

Original entry on oeis.org

3, -9, 83, -753, 6851, -62329, 567059, -5159009, 46935811, -427014249, 3884905043, -35344223825, 321555905219, -2925462465753, 26615373873171, -242142271419073, 2202970354179075, -20042260085157577, 182341168849178195, -1658909809373582257
Offset: 0

Views

Author

Kai Wang, Jun 07 2016

Keywords

Comments

A Berndt-type sequence for tan(2*Pi/7).
a(n) is always an integer.
a(n) is x1^n + x2^n + x3^n, where x1, x2, x3 are the roots of the polynomial
x^3 + 9*x^2 - x - 1.
x1 = tan(Pi/7)/tan(2*Pi/7),
x2 = tan(2*Pi/7)/tan(4*Pi/7),
x3 = tan(4*Pi/7)/tan(Pi/7).
This is a two sided sequence. The other half is A274075. - Kai Wang, Aug 02 2016

Crossrefs

Programs

Formula

a(n) = (tan(Pi/7)/tan(2*Pi/7))^n + (-tan(2*Pi/7)/tan(3*Pi/7))^n + (-tan(3*Pi/7)/tan(Pi/7))^n.
From Colin Barker, Jun 07 2016: (Start)
a(n) = -9*a(n-1)+a(n-2)+a(n-3) for n>2.
G.f.: (3+18*x-x^2) / (1+9*x-x^2-x^3).
(End)

Extensions

Edited by N. J. A. Sloane, Jun 07 2016

A274075 Sum of n-th powers of the roots of x^3 + x^2 - 9*x - 1.

Original entry on oeis.org

3, -1, 19, -25, 195, -401, 2131, -5545, 24323, -72097, 285459, -910009, 3407043, -11311665, 41065043, -139462985, 497736707, -1711838529, 6052005907, -20960815961, 73717030595, -256312368337, 898804827731, -3131899112169, 10964830193411, -38253117375201
Offset: 0

Views

Author

Kai Wang, Jun 09 2016

Keywords

Comments

a(n) is always an integer.
This is the other half of A274032.
a(n) is x1^n + x2^n + x3^n, where x1, x2, x3 are the roots of the polynomial
x^3 + x^2 - 9*x - 1.
x1 = tan(Pi/7)/tan(4*Pi/7),
x2 = tan(4*Pi/7)/tan(2*Pi/7),
x3 = tan(2*Pi/7)/tan(Pi/7).

Crossrefs

Programs

  • Mathematica
    FullSimplify[Table[(Tan[Pi/7]/Tan[4*Pi/7])^n + (Tan[4*Pi/7]/Tan[2*Pi/7])^n + (Tan[2*Pi/7]/Tan[Pi/7])^n, {n, 0, 12}]] (* Wesley Ivan Hurt, Jun 11 2016 *)
  • PARI
    Vec((3+2*x-9*x^2)/(1+x-9*x^2-x^3) + O(x^30)) \\ Colin Barker, Jun 11 2016
    
  • PARI
    polsym(x^3 + x^2 - 9*x - 1, 30) \\ Charles R Greathouse IV, Jul 20 2016

Formula

a(n) = (tan(Pi/7)/tan(4*Pi/7))^n + (tan(4*Pi/7)/tan(2*Pi/7))^n + (tan(2*Pi/7)/tan(Pi/7))^n.
a(n) = -a(n-1) + 9*a(n-2) + a(n-3) for n>2.
G.f.: (3+2*x-9*x^2) / (1+x-9*x^2-x^3). - Colin Barker, Jun 11 2016

A096975 Trace sequence of a path graph plus loop.

Original entry on oeis.org

3, 1, 5, 4, 13, 16, 38, 57, 117, 193, 370, 639, 1186, 2094, 3827, 6829, 12389, 22220, 40169, 72220, 130338, 234609, 423065, 761945, 1373466, 2474291, 4459278, 8034394, 14478659, 26088169, 47011093, 84708772, 152642789, 275049240
Offset: 0

Views

Author

Paul Barry, Jul 16 2004

Keywords

Comments

Let A be the adjacency matrix of the graph P_3 with a loop added at the end. Then a(n) = trace(A^n). A is a 'reverse Jordan matrix' [0,0,1;0,1,1;1,1,0]. a(n) = abs(A094648(n)).
From L. Edson Jeffery, Mar 22 2011: (Start)
Let A be the unit-primitive matrix (see [Jeffery])
A = A_(7,1) =
(0 1 0)
(1 0 1)
(0 1 1).
Then a(n) = Trace(A^n). (End)

Crossrefs

A033304(n) = a(-1-n). - Michael Somos, Aug 03 2006

Programs

  • Mathematica
    CoefficientList[Series[(3 - 2 x - 2 x^2)/(1 - x - 2 x^2 + x^3), {x, 0, 33}], x] (* Michael De Vlieger, Aug 21 2019 *)
  • PARI
    {a(n)=if(n>=0, n+=1; polsym(x^3-x^2-2*x+1,n-1)[n], n=1-n; polsym(1-x-2*x^2+x^3,n-1)[n])} /* Michael Somos, Aug 03 2006 */
    
  • PARI
    a(n)=trace([0,1,0;1,0,1;0,1,1]^n); /* Joerg Arndt, Apr 30 2011 */

Formula

G.f.: (3-2*x-2*x^2)/(1-x-2*x^2+x^3);
a(n) = a(n-1) + 2*a(n-2) - a(n-3);
a(n) = (2*sqrt(7)*sin(atan(sqrt(3)/9)/3)/3+1/3)^n + (1/3-2*sqrt(7)*sin(atan(sqrt(3)/9)/3+Pi/3)/3)^n + (2*sqrt(7)*cos(acot(-sqrt(3)/9)/3)/3+1/3)^n.
a(n) = 2^n*((cos(Pi/7))^n+(cos(3*Pi/7))^n+(cos(5*Pi/7))^n). - Vladimir Shevelev, Aug 25 2010
a(n) = (-1)^n*A094648(n). - R. J. Mathar, Nov 05 2024

A274220 a(n) = (-cos(Pi/7)/cos(2*Pi/7))^n + (-cos(2*Pi/7)/cos(3*Pi/7))^n + (cos(3*Pi/7)/cos(Pi/7))^n.

Original entry on oeis.org

3, -4, 10, -25, 66, -179, 493, -1369, 3818, -10672, 29865, -83626, 234237, -656205, 1838483, -5151080, 14432666, -40438941, 113306686, -317477255, 889550021, -2492461633, 6983719214, -19567941936, 54828148469, -153625048854, 430447808073, -1206087937261, 3379383275971, -9468821484028
Offset: 0

Views

Author

Kai Wang, Jun 14 2016

Keywords

Comments

a(n) is an integer.
This is other half of A215076.
a(n) is the sum of n-th powers of the roots of x^3 + 4*x^2 + 3*x - 1. - Greg Dresden, Mar 11 2020

Examples

			a(0) = 3, a(1) = -4, a(2) = 10, a(3) = -25.
		

References

  • R. Witula, E. Hetmaniok, D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the Fifteenth International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(3 + 8 x + 3 x^2)/(1 + 4 x + 3 x^2 - x^3), {x, 0, 29}], x] (* Michael De Vlieger, Jun 14 2016 *)
  • PARI
    Vec((3+8*x+3*x^2)/(1+4*x+3*x^2-x^3) + O(x^30)) \\ Colin Barker, Jun 14 2016
    
  • PARI
    polsym(x^3 + 4*x^2 + 3*x - 1,33) \\ Joerg Arndt, Mar 12 2020

Formula

a(n) = -4*a(n-1)-3*a(n-2)+a(n-3).
G.f.: (3+8*x+3*x^2) / (1+4*x+3*x^2-x^3). - Colin Barker, Jun 14 2016

Extensions

Many terms corrected by Colin Barker, Jun 14 2016

A062883 (1-2*cos(1/11*Pi))^n+(1+2*cos(2/11*Pi))^n+(1-2*cos(3/11*Pi))^n+(1+2*cos(4/11*Pi))^n+(1-2*cos(5/11*Pi))^n.

Original entry on oeis.org

4, 12, 25, 64, 159, 411, 1068, 2808, 7423, 19717, 52529, 140251, 375015, 1003770, 2688570, 7204696, 19313075, 51782613, 138861732, 372414289, 998851473, 2679146955, 7186319506, 19276417059, 51707411684, 138702360471
Offset: 1

Views

Author

Vladeta Jovovic, Jun 27 2001

Keywords

Comments

From L. Edson Jeffery, Apr 20 2011: (Start)
Let U be the unit-primitive matrix (see [Jeffery])
U = U_(11,2) =
(0 0 1 0 0)
(0 1 0 1 0)
(1 0 1 0 1)
(0 1 0 1 1)
(0 0 1 1 1).
Then a(n) = Trace(U^(n+1)). Evidently this is one of a class of accelerator sequences for Catalan's constant based on traces of successive powers of a unit-primitive matrix U_(N,r) (0 < r < floor(N/2)) and for which the closed-form expression for a(n) is derived from the eigenvalues of U_(N,r). (End)
a(n) = A(n;1), where A(n;d), d in C, is the sequence of polynomials defined in Witula's comments to A189235 (see also Witula-Slota's paper for compatible sequences). - Roman Witula, Jul 26 2012

References

  • R. Witula, D. Slota, Quasi-Fibonacci Numbers of Order 11, 10 (2007), Article 07.8.5.

Crossrefs

Programs

  • Maple
    Digits := 1000:q := seq(floor(evalf((1-2*cos(1/11*Pi))^n+(1+2*cos(2/11*Pi))^n+(1-2*cos(3/11*Pi))^n+(1+2*cos(4/11*Pi))^n+(1-2*cos(5/11*Pi))^n)),n=1..50);
  • Mathematica
    a[n_] := (1 - 2*Cos[Pi/11])^n + (2*Cos[(2*Pi)/11] + 1)^n + (1 - 2*Sin[Pi/22])^n + (2*Sin[(3*Pi)/22] + 1)^n + (1 - 2*Sin[(5*Pi)/22])^n; Table[a[n] // FullSimplify, {n, 1, 26}] (* Jean-François Alcover, Mar 26 2013 *)
    u = {{0, 0, 1, 0, 0}, {0, 1, 0, 1, 0}, {1, 0, 1, 0, 1}, {0, 1, 0, 1, 1}, {0, 0, 1, 1, 1}}; a[n_] := Tr[MatrixPower[u, n]]; Table[a[n], {n, 1, 26}] (* Jean-François Alcover, Oct 16 2013, after L. Edson Jeffery *)
    LinearRecurrence[{4,-2,-5,2,1},{4,12,25,64,159},30] (* Harvey P. Dale, Dec 30 2024 *)
  • PARI
    { default(realprecision, 200); for (n=1, 200, a=(1 - 2*cos(1/11*Pi))^n + (1 + 2*cos(2/11*Pi))^n + (1 - 2*cos(3/11*Pi))^n + (1 + 2*cos(4/11*Pi))^n + (1 - 2*cos(5/11*Pi))^n; write("b062883.txt", n, " ", round(a)) ) } \\ Harry J. Smith, Aug 12 2009

Formula

G.f.: x*(4-4*x-15*x^2+8*x^3+5*x^4)/(1-4*x+2*x^2+5*x^3-2*x^4-x^5). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009
-A062883 = series expansion of (5-8*x-15*x^2+4*x^3+4*x^4)/(1-2*x-5*x^2+2*x^3+4*x^4+x^5) at x=infinity. (See also A189236.) - L. Edson Jeffery, Apr 20 2011
Also, a(n) = Sum_{k = 1..5} ((w_k)^2-1)^(n+1), w_k = 2*(-1)^(k-1)*cos(k*Pi/11), in which the polynomials {(w_k)^2-1} give the spectrum of the matrix U_(11,2) above. - L. Edson Jeffery, Apr 20 2011

Extensions

G.f. proposed by Maksym Voznyy checked and corrected by R. J. Mathar, Sep 16 2009
More terms from Sascha Kurz, Mar 24 2002

A120747 Sequence relating to the 11-gon (or hendecagon).

Original entry on oeis.org

0, 1, 4, 14, 50, 175, 616, 2163, 7601, 26703, 93819, 329615, 1158052, 4068623, 14294449, 50221212, 176444054, 619907431, 2177943781, 7651850657, 26883530748, 94450905714, 331837870408, 1165858298498, 4096053203771, 14390815650209, 50559786403254
Offset: 1

Views

Author

Gary W. Adamson, Jul 01 2006

Keywords

Comments

The hendecagon is an 11-sided polygon. The preferred word in the OEIS is 11-gon.
The lengths of the diagonals of the regular 11-gon are r[k] = sin(k*Pi/11)/sin(Pi/11), 1 <= k <= 5, where r[1] = 1 is the length of the edge.
The value of limit(a(n)/a(n-1),n=infinity) equals the longest diagonal r[5].
The a(n) equal the matrix elements M^n[1,2], where M = Matrix([[1,1,1,1,1], [1,1,1,1,0], [1,1,1,0,0], [1,1,0,0,0], [1,0,0,0,0]]). The characteristic polynomial of M is (x^5 - 3x^4 - 3x^3 + 4x^2 + x - 1) with roots x1 = -r[4]/r[3], x2 = -r[2]/r[4], x3 = r[1]/r[2], x4 = r[3]/r[5] and x5 = r[5]/r[1].
Note that M^4*[1,0,0,0,0] = [55, 50, 41, 29, 15] which are all terms of the 5-wave sequence A038201. This is also the case for the terms of M^n*[1,0,0,0,0], n>=1.

Examples

			From _Johannes W. Meijer_, Aug 03 2011: (Start)
The lengths of the regular hendecagon edge and diagonals are:
  r[1] = 1.000000000, r[2] = 1.918985948, r[3] = 2.682507066,
  r[4] = 3.228707416, r[5] = 3.513337092.
The first few rows of the T(n,k) array are, n>=1, 1 <= k <=5:
    0,   0,   0,   0,   1, ...
    1,   1,   1,   1,   1, ...
    1,   2,   3,   4,   5, ...
    5,   9,  12,  14,  15, ...
   15,  29,  41,  50,  55, ...
   55, 105, 146, 175, 190, ...
  190, 365, 511, 616, 671, ... (End)
		

Crossrefs

From Johannes W. Meijer, Aug 03 2011: (Start)
Cf. A006358 (T(n+2,1) and T(n+1,5)), A069006 (T(n+1,2)), A038342 (T(n+1,3)), this sequence (T(n,4)) (m=5: hendecagon or 11-gon).
Cf. A000045 (m=2; pentagon or 5-gon); A006356, A006054 and A038196 (m=3: heptagon or 7-gon); A006357, A076264, A091024 and A038197 (m=4: enneagon or 9-gon); A006359, A069007, A069008, A069009, A070778 (m=6; tridecagon or 13-gon); A025030 (m=7: pentadecagon or 15-gon); A030112 (m=8: heptadecagon or 17-gon). (End)

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x^2*(1+x-x^2)/(1-3*x-3*x^2+4*x^3+x^4-x^5) )); // G. C. Greubel, Nov 13 2022
    
  • Maple
    nmax:=27: m:=5: for k from 1 to m-1 do T(1,k):=0 od: T(1,m):=1: for n from 2 to nmax do for k from 1 to m do T(n,k):= add(T(n-1,k1), k1=m-k+1..m) od: od: for n from 1 to nmax/3 do seq(T(n,k), k=1..m) od; for n from 1 to nmax do a(n):=T(n,4) od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Aug 03 2011
  • Mathematica
    LinearRecurrence[{3, 3, -4, -1, 1}, {0, 1, 4, 14, 50}, 41] (* G. C. Greubel, Nov 13 2022 *)
  • SageMath
    def A120747_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( x*(1+x-x^2)/(1-3*x-3*x^2+4*x^3+x^4-x^5) ).list()
    A120747_list(40) # G. C. Greubel, Nov 13 2022

Formula

a(n) = 3*a(n-1) + 3*a(n-2) - 4*a(n-3) - a(n-4) + a(n-5).
G.f.: x^2*(1+x-x^2)/(1-3*x-3*x^2+4*x^3+x^4-x^5). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 12 2009
From Johannes W. Meijer, Aug 03 2011: (Start)
a(n) = T(n,4) with T(n,k) = Sum_{k1 = 6-k..6} T(n-1, k1), T(1,1) = T(1,2) = T(1,3) = T(1,4) = 0 and T(1,5) = 1, n>=1 and 1 <= k <= 5. [Steinbach]
Sum_{k=1..5} T(n,k)*r[k] = r[5]^n, n>=1. [Steinbach]
r[k] = sin(k*Pi/11)/sin(Pi/11), 1 <= k <= 5. [Kappraff]
Sum_{k=1..5} T(n,k) = A006358(n-1).
Limit_{n -> 00} T(n,k)/T(n-1,k) = r[5], 1 <= k <= 5.
sequence(sequence( T(n,k), k=2..5), n>=1) = A038201(n-4).
G.f.: (x^2*(x - x1)*(x - x2))/((x - x3)*(x - x4)*(x - x5)*(x - x6)*(x - x7)) with x1 = phi, x2 = (1-phi), x3 = r[1] - r[3], x4 = r[3] - r[5], x5 = r[5] - r[4], x6 = r[4] - r[2], x7 = r[2], where phi = (1 + sqrt(5))/2 is the golden ratio A001622. (End)

Extensions

Edited and information added by Johannes W. Meijer, Aug 03 2011

A215492 a(n) = 21*a(n-2) + 7*a(n-3), with a(0)=0, a(1)=3, and a(2)=6.

Original entry on oeis.org

0, 3, 6, 63, 147, 1365, 3528, 29694, 83643, 648270, 1964361, 14199171, 45789471, 311933118, 1060973088, 6871121775, 24463966674, 151720368891, 561841152579, 3357375513429, 12860706786396, 74437773850062, 293576471108319, 1653218198356074, 6686170310225133
Offset: 0

Views

Author

Roman Witula, Aug 13 2012

Keywords

Comments

We have a(n)=B(n;3), where B(n;d), n=1,2,..., d \in C, denote one of the quasi-Fibonacci numbers defined in the comments to A121449 and in the Witula-Slota-Warzynski paper. Its conjugate sequences A(n;3) and C(n;3) are discussed in A121458 and A215484 respectively. Similarly as in A121458 we deduce that each of the following elements a(3*n), a(3*n+1), a(3*n+2) is divided by 3*7^n for every n=0,1,... .

Crossrefs

Programs

  • Magma
    I:=[0,3,6]; [n le 3 select I[n] else 21*Self(n-2)+7*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 18 2015
  • Mathematica
    LinearRecurrence[{0,21,7}, {0,3,6}, 50]
    CoefficientList[Series[(3 x + 6 x^2)/(1 - 21 x^2 - 7 x^3), {x, 0, 33}], x] (* Vincenzo Librandi, Sep 18 2015 *)
  • PARI
    concat(0,Vec((3+6*x)/(1-21*x^2-7*x^3)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012
    

Formula

a(n) = (1/7)*((c(1)-c(4))*(1+3*c(1))^n + (c(2)-c(1))*(1+3*c(2))^n + (c(4)-c(2))*(1+3*c(4))^n), where c(j):=2*cos(2*Pi*j/7) (for the proof see Witula-Slota-Warzynski paper).
G.f.: (3*x+6*x^2)/(1-21*x^2-7*x^3).
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