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.

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

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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

A245031 Numbers m such that 3*m+1 and 8*m+1 are both squares.

Original entry on oeis.org

0, 1, 21, 120, 2080, 11781, 203841, 1154440, 19974360, 113123361, 1957283461, 11084934960, 191793804840, 1086210502741, 18793835590881, 106437544333680, 1841604094101520, 10429793134197921, 180458407386358101, 1022013289607062600
Offset: 1

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Author

Bruno Berselli, Jul 15 2014

Keywords

Comments

Naturally, all terms are triangular numbers.
Numbers m such that k*m+1 and 8*m+1 are both squares:
k=1: A006454;
k=3: this sequence;
k=4: A029549;
k=5: 0, 3, 231, 4560, 333336, 6575751, ...
k=6: A200999;
k=7: A157879.
Numbers m such that 3*m+1 and k*m+1 are both squares:
k=1: A045899;
k=2: A045502;
k=4: A059989;
k=5: A159683;
k=6: 8*A029546;
k=7: A160695;
k=8: this sequence.

Crossrefs

Programs

  • Magma
    I:=[0,1,21,120,2080]; [n le 5 select I[n] else Self(n-1)+98*Self(n-2)-98*Self(n-3)-Self(n-4)+Self(n-5): n in [1..20]];
  • Mathematica
    LinearRecurrence[{1, 98, -98, -1, 1}, {0, 1, 21, 120, 2080}, 20] (* or *) CoefficientList[Series[x (1 + 20 x + x^2)/((1 - x) (1 - 10 x + x^2) (1 + 10 x + x^2)), {x, 0, 20}], x]
  • Maxima
    a[1]:0$ a[2]:1$ a[3]:21$ a[4]:120$ a[5]:2080$ a[n]:=a[n-1]+98*a[n-2]-98*a[n-3]-a[n-4]+a[n-5]$ makelist(a[n], n, 1, 20);
    
  • PARI
    a=vector(20); a[1]=0; a[2]=1; a[3]=21; a[4]=120; a[5]=2080; for(i=6, #a, a[i]=a[i-1]+98*a[i-2]-98*a[i-3]-a[i-4]+a[i-5]); a
    

Formula

G.f.: x^2*(1 + 20*x + x^2)/((1 - x)*(1 - 10*x + x^2)*(1 + 10*x + x^2)).
a(n) = a(n-1) + 98*a(n-2) - 98*a(n-3) - a(n-4) + a(n-5).
G.f. of the quadrisections:
a(4k+1): 40*x*(52 + 3*x)/((1 - x)*(1 - 9602*x + x^2));
a(4k+2): (1 + 2178*x + 21*x^2)/((1 - x)*(1 - 9602*x + x^2));
a(4k+3): (21 + 2178*x + x^2)/((1 - x)*(1 - 9602*x + x^2));
a(4k+4): 40*(3 + 52*x)/((1 - x)*(1 - 9602*x + x^2)).

Extensions

Changed offset from 0 to 1 and adapted formulas by Bruno Berselli, Mar 03 2016

A157877 Expansion of (1-x)*x/(x^2-30*x+1).

Original entry on oeis.org

1, 29, 869, 26041, 780361, 23384789, 700763309, 20999514481, 629284671121, 18857540619149, 565096933903349, 16934050476481321, 507456417360536281, 15206758470339607109, 455695297692827676989, 13655652172314490702561, 409213869871741893399841
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.
A157877 is the a(n) sequence for A=7.
Positive values of x (or y) satisfying x^2 - 30xy + y^2 + 28 = 0. - Colin Barker, Feb 23 2014

Crossrefs

Cf. similar sequences listed in A238379.

Programs

  • Magma
    I:=[1,29,869]; [n le 3 select I[n] else 30*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 25 2014
  • Mathematica
    LinearRecurrence[{30,-1},{1,29},30] (* Harvey P. Dale, Dec 14 2011 *)
    CoefficientList[Series[(1 - x)/(x^2 - 30 x + 1), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 25 2014 *)
  • PARI
    Vec((1-x)*x/(x^2-30*x+1)+O(x^99)) \\ Charles R Greathouse IV, Sep 25 2012
    

Formula

G.f.: (1-x)*x/(x^2-30*x+1).
a(1)=1, a(2)=29; for n>2, a(n) = 30*a(n-1)-a(n-2).
7*A157879(n)+1 = a(n)^2.
8*A157879(n)+1 = A157878(n)^2.
a(n) = (1/8)*(4-sqrt(14))*(1+(15+4*sqrt(14))^(2*n-1))/(15+4*sqrt(14))^(n-1). - Bruno Berselli, Feb 25 2014
From Andrea Pinos, Oct 05 2022: (Start)
a(n) = ceiling((C^n)/(C+1)), where C = 15 + 4*sqrt(14) = sqrt(225) + sqrt(224).
Limit_{n->oo} a(n+1)/a(n) = C. (End)

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)
Showing 1-5 of 5 results.