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 1-6 of 6 results.

A238379 Expansion of (1 - x)/(1 - 36*x + x^2).

Original entry on oeis.org

1, 35, 1259, 45289, 1629145, 58603931, 2108112371, 75833441425, 2727895778929, 98128414600019, 3529895029821755, 126978092658983161, 4567681440693572041, 164309553772309610315, 5910576254362452399299, 212616435603275976764449
Offset: 0

Views

Author

Bruno Berselli, Feb 25 2014

Keywords

Comments

First bisection of A041611.

Crossrefs

Cf. similar sequences with g.f. (1-x)/(1-k*x+x^2): A122367 (k=3), A079935 (k=4), A004253 (k=5), A001653 (k=6), A049685 (k=7), A070997 (k=8), A070998 (k=9), A138288 (k=10), A078922 (k=11), A077417 (k=12), A085260 (k=13), A001570 (k=14), A160682 (k=15), A157456 (k=16), A161595 (k=17). From 18 to 38, even k only, except k=27 and k=31: A007805 (k=18), A075839 (k=20), A157014 (k=22), A159664 (k=24), A153111 (k=26), A097835 (k=27), A159668 (k=28), A157877 (k=30), A111216 (k=31), A159674 (k=32), A077420 (k=34), this sequence (k=36), A097315 (k=38).

Programs

  • Magma
    [n le 2 select 35^(n-1) else 36*Self(n-1)-Self(n-2): n in [1..20]];
    
  • Magma
    R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1 - x)/(1 - 36*x + x^2))); // Marius A. Burtea, Jan 14 2020
    
  • Mathematica
    CoefficientList[Series[(1 - x)/(1 - 36 x + x^2), {x, 0, 20}], x] (* or *) LinearRecurrence[{36, -1}, {1, 35}, 20]
  • PARI
    a(n)=([0,1; -1,36]^n*[1;35])[1,1] \\ Charles R Greathouse IV, May 10 2016
  • Sage
    m = 20; L. = PowerSeriesRing(ZZ, m); f = (1-x)/(1-36*x+x^2)
    print(f.coefficients())
    

Formula

G.f.: (1 - x)/(1 - 36*x + x^2).
a(n) = a(-n-1) = 36*a(n-1) - a(n-2).
a(n) = ((19-sqrt(323))/38)*(1+(18+sqrt(323))^(2*n+1))/(18+sqrt(323))^n.
a(n+1) - a(n) = 34*A144128(n+1).
323*a(n+1)^2 - ((a(n+2)-a(n))/2)^2 = 34.
Sum_{n>0} 1/(a(n) - 1/a(n)) = 1/34.
See also Tanya Khovanova in Links field:
a(n) = 35*a(n-1) + 34*Sum_{i=0..n-2} a(i).
a(n+2)*a(n) - a(n+1)^2 = 36-2 = 34 = 34*1,
a(n+3)*a(n) - a(n+1)*a(n+2) = 36*(36-2) = 1224 = 34*36.
Generalizing:
a(n+4)*a(n) - a(n+1)*a(n+3) = 44030 = 34*1295,
a(n+5)*a(n) - a(n+1)*a(n+4) = 1583856 = 34*46584,
a(n+6)*a(n) - a(n+1)*a(n+5) = 56974786 = 34*1675729, etc.,
where 1, 36, 1295, 46584, 1675729, ... is the sequence A144128, which is the second bisection of A041611.
a(n)^2 - 36*a(n)*a(n+1) + a(n+1)^2 + 34 = 0 (see comments by Colin Barker in similar sequences).

A157014 Expansion of x*(1-x)/(1 - 22*x + x^2).

Original entry on oeis.org

1, 21, 461, 10121, 222201, 4878301, 107100421, 2351330961, 51622180721, 1133336644901, 24881784007101, 546265911511321, 11992968269241961, 263299036011811821, 5780585823990618101, 126909589091781786401, 2786230374195208682721, 61170158643202809233461
Offset: 1

Views

Author

Paul Weisenhorn, Feb 21 2009

Keywords

Comments

This sequence is part of a solution of a general problem involving 2 equations, three sequences a(n), b(n), c(n) and a constant A:
A * c(n)+1 = a(n)^2,
(A+1) * c(n)+1 = b(n)^2, where solutions are given by the recurrences:
a(1) = 1, a(2) = 4*A+1, a(n) = (4*A+2)*a(n-1)-a(n-2) for n>2, resulting in a(n) terms 1, 4*A+1, 16*A^2+12*A+1, 64*A^3+80*A^2+24*A+1, ...;
b(1) = 1, b(2) = 4*A+3, b(n) = (4*A+2)*b(n-1)-b(n-2) for n>2, resulting in b(n) terms 1, 4*A+3, 16*A^2+20*A+5, 64*A^3+112*A^2+56*A+7, ...;
c(1) = 0, c(2) = 16*A+8, c(3) = (16*A^2+16*A+3)*c(2), c(n) = (16*A^2+16*A+3) * (c(n-1)-c(n-2)) + c(n-3) for n>3, resulting in c(n) terms 0, 16*A+8, 256*A^3+384*A^2+176*A+24, 4096*A^5 + 10240*A^4 + 9472*A^3 + 3968*A^2 + 736*A + 48, ... .
A157014 is the a(n) sequence for A=5.
For other A values the a(n), b(n) and c(n) sequences are in the OEIS:
A a-sequence b-sequence c-sequence
2 A072256 A054320(n-1) A045502(n-1)
9 A097315(n-1) A097314(n-1) A157881
Positive values of x (or y) satisfying x^2 - 22xy + y^2 + 20 = 0. - Colin Barker, Feb 19 2014
From Klaus Purath, Apr 22 2025: (Start)
Nonnegative solutions to the Diophantine equation 5*b(n)^2 - 6*a(n)^2 = -1. The corresponding b(n) are A133283(n). Note that (b(n+1)^2 - b(n)*b(n+2))/4 = 6 and (a(n)*a(n+2) - a(n+1)^2)/4 = 5.
(a(n) + b(n))/2 = (b(n+1) - a(n+1))/2 = A077421(n-1) = Lucas U(22,1). Also b(n)*a(n+1) - b(n+1)*a(n) = -2.
a(n)=(t(i+2*n-1) + t(i))/(t(i+n) + t(i+n-1)) as long as t(i+n) + t(i+n-1) != 0 for any integer i and n >= 1 where (t) is a sequence satisfying t(i+3) = 21*t(i+2) - 21*t(i+1) + t(i) or t(i+2) = 22*t(i+1) - t(i) without regard to initial values and including this sequence itself. (End)

Crossrefs

Cf. similar sequences listed in A238379.

Programs

  • GAP
    a:=[1,21];; for n in [3..20] do a[n]:=22*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 14 2020
  • Magma
    I:=[1,21]; [n le 2 select I[n] else 22*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 21 2014
    
  • Maple
    seq( simplify(ChebyshevU(n-1,11) - ChebyshevU(n-2,11)), n=1..20); # G. C. Greubel, Jan 14 2020
  • Mathematica
    CoefficientList[Series[(1-x)/(1-22x+x^2), {x,0,20}], x] (* Vincenzo Librandi, Feb 21 2014 *)
    a[c_, n_] := Module[{},
       p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
       d := Denominator[Convergents[Sqrt[c], n p]];
       t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
       Return[t];
    ] (* Complement of A041049 *)
    a[30, 20] (* Gerry Martens, Jun 07 2015 *)
    Table[ChebyshevU[n-1, 11] - ChebyshevU[n-2, 11], {n,20}] (* G. C. Greubel, Jan 14 2020 *)
  • PARI
    Vec((1-x)/(1-22*x+x^2)+O(x^20)) \\ Charles R Greathouse IV, Sep 23 2012
    
  • Sage
    [chebyshev_U(n-1,11) - chebyshev_U(n-2,11) for n in (1..20)] # G. C. Greubel, Jan 14 2020
    

Formula

G.f.: x*(1-x)/(1-22*x+x^2).
a(1) = 1, a(2) = 21, a(n) = 22*a(n-1) - a(n-2) for n>2.
5*A157460(n)+1 = a(n)^2 for n>=1.
6*A157460(n)+1 = A133283(n)^2 for n>=1.
a(n) = (6+sqrt(30)-(-6+sqrt(30))*(11+2*sqrt(30))^(2*n))/(12*(11+2*sqrt(30))^n). - Gerry Martens, Jun 07 2015
a(n) = ChebyshevU(n-1, 11) - ChebyshevU(n-2, 11). - G. C. Greubel, Jan 14 2020

Extensions

Edited by Alois P. Heinz, Sep 09 2011

A157879 Expansion of 120*x^2 / (-x^3+899*x^2-899*x+1).

Original entry on oeis.org

0, 120, 107880, 96876240, 86994755760, 78121193796360, 70152745034375640, 62997086919675528480, 56571313901123590199520, 50800976886122064323640600, 45619220672423712639039059400, 40966009362859607827792751700720, 36787430788627255405645251988187280
Offset: 1

Views

Author

Paul Weisenhorn, Mar 08 2009

Keywords

Comments

This sequence is part of a solution of a more general problem involving 2 equations, three sequences a(n), b(n), c(n) and a constant A:
A * c(n)+1 = a(n)^2,
(A+1) * c(n)+1 = b(n)^2, for details see comment in A157014.
A157879 is the c(n) sequence for A=7.

Crossrefs

7*A157879(n)+1 = A157877(n)^2.
8*A157879(n)+1 = A157878(n)^2.
Cf. A245031.

Programs

  • Mathematica
    CoefficientList[Series[120x^2/(-x^3+899x^2-899x+1),{x,0,30}],x] (* or *) LinearRecurrence[{899,-899,1},{0,0,120},30] (* Harvey P. Dale, Jan 14 2014 *)
  • PARI
    concat(0, Vec(120*x^2/(-x^3+899*x^2-899*x+1)+O(x^20))) \\ Charles R Greathouse IV, Sep 25 2012
    
  • PARI
    a(n) = round(-((449+120*sqrt(14))^(-n)*(-1+(449+120*sqrt(14))^n)*(15+4*sqrt(14)+(-15+4*sqrt(14))*(449+120*sqrt(14))^n))/224) \\ Colin Barker, Jul 25 2016

Formula

G.f.: 120*x^2/(-x^3+899*x^2-899*x+1).
c(1) = 0, c(2) = 120, c(3) = 899*c(2), c(n) = 899 * (c(n-1)-c(n-2)) + c(n-3) for n>3.
a(n) = -((449+120*sqrt(14))^(-n)*(-1+(449+120*sqrt(14))^n)*(15+4*sqrt(14)+(-15+4*sqrt(14))*(449+120*sqrt(14))^n))/224. - Colin Barker, Jul 25 2016

Extensions

Edited by Alois P. Heinz, Sep 09 2011

A157878 Expansion of x*(1+x)/(x^2-30*x+1).

Original entry on oeis.org

1, 31, 929, 27839, 834241, 24999391, 749147489, 22449425279, 672733610881, 20159558901151, 604114033423649, 18103261443808319, 542493729280825921, 16256708616980969311, 487158764780148253409, 14598506234787466632959, 437468028278843850735361
Offset: 1

Views

Author

Paul Weisenhorn, Mar 08 2009

Keywords

Comments

This sequence is part of a solution of a more general problem involving 2 equations, three sequences a(n), b(n), c(n) and a constant A:
A * c(n)+1 = a(n)^2,
(A+1) * c(n)+1 = b(n)^2, for details see comment in A157014.
this is the b(n) sequence for A=7.

Crossrefs

7*A157879(n)+1 = A157877(n)^2.
8*A157879(n)+1 = A157878(n)^2.

Programs

  • Maple
    f:= gfun:-rectoproc({a(1) = 1, a(2) = 31, a(n) = 30*a(n-1)-a(n-2)}, a(n), remember):
    map(f, [$1..30]); # Robert Israel, Jul 09 2015
  • Mathematica
    CoefficientList[Series[x*(1 + x)/(x^2 - 30 x + 1), {x, 0, 17}], x] (* Michael De Vlieger, Jul 09 2015 *)
    LinearRecurrence[{30,-1},{1,31},20] (* Harvey P. Dale, Sep 05 2021 *)
  • PARI
    Vec((1+x)/(x^2-30*x+1)+O(x^99)) \\ Charles R Greathouse IV, Sep 25 2012

Formula

G.f.: x*(1+x)/(x^2-30*x+1).
a(1) = 1, a(2) = 31, a(n) = 30*a(n-1)-a(n-2) for n>2.
a(n) = ((15-4*sqrt(14))^(n-1)*(7-2*sqrt(14))+(7+2*sqrt(14))*(15+4*sqrt(14))^(n-1))/14. - Gerry Martens, Jul 09 2015

Extensions

Edited by Alois P. Heinz, Sep 09 2011

A253514 Centered heptagonal numbers (A069099) which are also centered octagonal numbers (A016754).

Original entry on oeis.org

1, 841, 755161, 678133681, 608963290321, 546848356574521, 491069215240629481, 440979608437728699361, 395999197307865131396641, 355606838202854450265484201, 319334544706965988473273415801, 286762065540017254794549261905041
Offset: 1

Views

Author

Colin Barker, Jan 03 2015

Keywords

Examples

			841 is in the sequence because it is the 16th centered heptagonal number and the 15th centered octagonal number.
		

Crossrefs

Programs

  • PARI
    Vec(-x*(x^2-58*x+1)/((x-1)*(x^2-898*x+1)) + O(x^100))

Formula

a(n) = 899*a(n-1)-899*a(n-2)+a(n-3).
G.f.: -x*(x^2-58*x+1) / ((x-1)*(x^2-898*x+1)).
From Peter Bala, Apr 15 2025; (Start)
a(n) = (1/64)*(-4 + sqrt(14))^2*(15 + 4*sqrt(14) + (449 + 120*sqrt(14))^n)^2 *(449 + 120*sqrt(14))^(-n).
a(-n) = a(n+1).
a(n) = (1/16) * (1 - T(2*n+1, -15)), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind. Cf. A001110.
a(n) = A157877(n)^2 = 1 + 7*A157879(n).
a(2) divides a(3*n+2); a(3) divides a(5*n+3); a(4) divides a(7*n+4); a(5) divides a(9*n+5). In general, a(k) divides a((2*k-1)*n + k). (End)

A269028 a(n) = 40*a(n - 1) - a(n - 2) for n>1, a(0) = 1, a(1) = 1.

Original entry on oeis.org

1, 1, 39, 1559, 62321, 2491281, 99588919, 3981065479, 159143030241, 6361740144161, 254310462736199, 10166056769303799, 406387960309415761, 16245352355607326641, 649407706263983649879, 25960062898203738668519, 1037753108221885563090881
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 18 2016

Keywords

Comments

In general, the ordinary generating function for the recurrence relation b(n) = k*b(n - 1) - b(n - 2) with n>1 and b(0)=1, b(1)=1, is (1 - (k - 1)*x)/(1 - k*x +x^2). This recurrence gives the closed form b(n) = (2^( -n - 1)*((k - 2)*(k - sqrt(k^2 - 4))^n + sqrt(k^2 - 4)*(k - sqrt(k^2 - 4))^n - (k - 2)*(sqrt(k^2 - 4) + k)^n + sqrt(k^2 - 4)*(sqrt(k^2 - 4) + k)^n))/sqrt(k^2 - 4).

Crossrefs

Programs

  • Magma
    [n le 2 select 1 else 40*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 19 2016
  • Mathematica
    Table[Cosh[n Log[20 + Sqrt[399]]] - Sqrt[19/21] Sinh[n Log[20 + Sqrt[399]]], {n, 0, 17}]
    Table[(2^(-n - 2) (38 (40 - 2 Sqrt[399])^n + 2 Sqrt[399] (40 - 2 Sqrt[399])^n - 38 (40 + 2 Sqrt[399])^n + 2 Sqrt[399] (40 + 2 Sqrt[399])^n))/Sqrt[399], {n, 0, 17}]
    LinearRecurrence[{40, -1}, {1, 1}, 17]

Formula

G.f.: (1 - 39*x)/(1 - 40*x + x^2).
a(n) = cosh(n*log(20 + sqrt(399))) - sqrt(19/21)*sinh(n*log(20 + sqrt(399))).
a(n) = (2^(-n - 2)*(38*(40 - 2*sqrt(399))^n + 2*sqrt(399)*(40 - 2*sqrt(399))^n - 38*(40 + 2*sqrt(399))^n + 2*sqrt(399)*(40 + 2*sqrt(399))^n))/sqrt(399).
Sum_{n>=0} 1/a(n) = 2.0262989201139499769986...
Showing 1-6 of 6 results.