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.

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A085292 Product of Lucas (A000204) and a Pell companion series (A001333).

Original entry on oeis.org

1, 9, 28, 119, 451, 1782, 6931, 27119, 105868, 413649, 1615681, 6311522, 24654241, 96306849, 376200748, 1469546399, 5740457491, 22423834422, 87593763331, 342165736199, 1336595027068, 5221113899769, 20395130698081, 79669083012482
Offset: 1

Views

Author

Gary W. Adamson, Jun 24 2003

Keywords

Crossrefs

Programs

  • Mathematica
    L[0] = 2; L[1] = 1; L[n] = L[n - 1] + L[n - 2]; P[0] = P[1] = 1; P[n_] := P[n] = 2P[n - 1] + P[n - 2]; Table[ L[n]P[n], {n, 1, 24}]
    With[{nn=30},Rest[LinearRecurrence[{2,1},{1,1},nn]LucasL[Range[0,nn-1]]]] (* Harvey P. Dale, Apr 20 2012 *)
    LinearRecurrence[{2, 7, 2, -1},{1, 9, 28, 119},24] (* Ray Chandler, Aug 03 2015 *)

Formula

a(n) = A000204(n) * A001333(n).
a(n) = 2*a(n-1)+7*a(n-2)+2*a(n-3)-a(n-4). G.f.: -x*(2*x^3-3*x^2-7*x-1) / (x^4-2*x^3-7*x^2-2*x+1). - Colin Barker, Oct 15 2013
2* A085292(n) = A085293(n).

Extensions

Edited and extended by Robert G. Wilson v, Jun 24 2003

A111108 a(n) = A001333(n) - (-2)^(n-1), n > 0.

Original entry on oeis.org

0, 5, 3, 25, 25, 131, 175, 705, 1137, 3875, 7095, 21649, 43225, 122435, 259423, 698625, 1541985, 4011971, 9107175, 23143825, 53559817, 133933475, 314086735, 776787009, 1838300625, 4512108515, 10745077143, 26237143825, 62749602745
Offset: 1

Views

Author

Creighton Dement, Oct 14 2005

Keywords

Comments

Conjecture: for odd primes p, p divides a(p). Note that (a(n)) and A001333 have different offsets.
The conjecture follows from the formula A001333(n) = ((1-sqrt(2))^n + (1+sqrt(2))^n)/2. - Max Alekseyev, Oct 16 2005

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{0,5,2},{0,5,3},30] (* Harvey P. Dale, May 03 2022 *)
  • PARI
    concat(0, Vec(x^2*(5 + 3*x) / ((1 + 2*x)*(1 - 2*x - x^2)) + O(x^35))) \\ Colin Barker, May 01 2019

Formula

From Colin Barker, Apr 30 2019: (Start)
G.f.: x^2*(5 + 3*x) / ((1 + 2*x)*(1 - 2*x - x^2)).
a(n) = 5*a(n-2) + 2*a(n-3) for n>3.
(End)

A131607 Pell companion numbers A001333 without last digit.

Original entry on oeis.org

1, 4, 9, 23, 57, 139, 336, 811, 1960, 4732, 11424, 27580, 66585, 160752, 388089, 936931, 2261953, 5460839, 13183632, 31828103, 76839840, 185507784, 447855408, 1081218600, 2610292609, 6301803820, 15213900249, 36729604319, 88673108889, 214075822099, 516824753088
Offset: 4

Views

Author

Paul Curtz, Oct 02 2007

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Floor[(((1 - Sqrt[2])^n + (1 + Sqrt[2])^n)/2)/10], {n, 4, 29}] (* Metin Sariyar, Jan 03 2020 *)
  • PARI
    a(n)={polcoef((1 - x) / (1 - 2*x - x^2) + O(x*x^n), n)\10} \\ Andrew Howroyd, Jan 02 2020

Formula

a(n) = floor(A001333(n) / 10). - Andrew Howroyd, Jan 02 2020
Conjectures from Colin Barker, Jan 03 2020: (Start)
G.f.: x^4*(1 + x - 2*x^2 + x^3 + x^4) / ((1 - x)*(1 + x^2)*(1 - 2*x - x^2)*(1 - x^2 + x^4)).
a(n) = 3*a(n-1) - a(n-2) - a(n-3) - a(n-6) + 3*a(n-7) - a(n-8) - a(n-9) for n>12.
(End)

Extensions

Offset changed and terms a(24) and beyond from Andrew Howroyd, Jan 02 2020

A062133 Triangle of coefficients of polynomials (rising powers) useful for convolutions of A001333(n+1), n >= 0 (associated Pell numbers).

Original entry on oeis.org

0, 1, 2, 20, 36, 16, 456, 944, 672, 160, 14304, 33760, 28800, 10880, 1536, 575040, 1466752, 1413120, 666880, 157440, 14848, 27659520, 74774784, 79278080, 43330560, 13153280, 2128896, 143360, 1548126720
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

The row polynomials pPL1(n,x) := Sum_{m=0..n} a(n,m)*x^m, and pPL2(n,x) := Sum_{m=0..n} A062134(n,m)*x^m appear in the k-fold convolution of the associated Pell numbers PL(n) := A001333(n+1), n >= 0, as follows: PL(k; n) := A054458(n+k,k) = (2*pPL1(k,n)*PL(n+1)+pPL2(k,n)*PL(n))/(k!*8^k), k >= 0.

Examples

			Triangle begins:
  {0};
  {1,2};
  {20,36,16};
  {456,944,672,160};
  ...
pPL1(2,n) = 4*(5+9*n+4*n^2) = 4*(1+n)*(5+4*n).
pPL2(2,n) = 8*(1+3*n+2*n^2) = 8*(1+n)*(1+2*n).
PL(2; n) = A054460(n) = (1+n)*((5+4*n)*PL(n+1)+(1+2*n)*PL(n))/16.
		

Crossrefs

Cf. A062134(n, m) (companion triangle), A054458(n, m) (convolution triangle).

A131721 Overlay of Pell companion numbers: a(n) = A001333(n) + A001333(n-6).

Original entry on oeis.org

1, 1, 3, 7, 17, 41, 100, 240, 580, 1400, 3380, 8160, 19700, 47560, 114820, 277200, 669220, 1615640, 3900500, 9416640, 22733780, 54884200, 132502180, 319888560, 772279300, 1864447160, 4501173620, 10866794400, 26234762420, 63336319240
Offset: 0

Views

Author

Paul Curtz, Sep 15 2007

Keywords

Crossrefs

Cf. A131710.

Programs

  • Mathematica
    LinearRecurrence[{2, 1}, {1, 1, 3, 7, 17, 41, 100, 240}, 50] (* Paolo Xausa, Mar 01 2024 *)

Formula

a(n) = A001333(n), n<6. a(n) = A001333(n)+A001333(n-6) = 20*A000129(n-3), n>=6.
O.g.f.: (x-1)(x^2+1)(x^4-x^2+1)/(-1+2x+x^2). - R. J. Mathar, Jul 16 2008
For n>5, a(n) = 100*A000129(n+1) -240*A000129(n). - R. J. Mathar, Feb 05 2024

Extensions

Edited and extended by R. J. Mathar, Jul 16 2008

A138683 a(1)=1 and, for n>1, a(n) is the smallest integer greater than a(n-1) such that a(n) and a(k) do not sum to a term of A001333 (Numerators of continued fraction convergents to sqrt(2)).

Original entry on oeis.org

2, 3, 6, 7, 8, 12, 13, 16, 17, 18, 19, 20, 26, 27, 30, 31, 32, 36, 37, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 60, 61, 64, 65, 66, 70, 71, 74, 75, 76, 77, 78, 84, 85, 88, 89, 90, 94, 95, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115
Offset: 1

Views

Author

John W. Layman, Mar 26 2008

Keywords

Comments

The graph of the first differences (A138684) of this sequence is fractal-like.

Crossrefs

A143966 Eigentriangle with row sums = A001333 starting (1, 3, 7, 17, 41, 99, ...).

Original entry on oeis.org

1, 2, 1, 2, 2, 3, 2, 2, 6, 7, 2, 2, 6, 14, 17, 2, 2, 6, 14, 34, 41, 2, 2, 6, 14, 34, 82, 99, 2, 2, 6, 14, 34, 82, 198, 239, 2, 2, 6, 14, 34, 82, 198, 478, 577
Offset: 0

Views

Author

Gary W. Adamson, Sep 06 2008

Keywords

Comments

Right border = A001333.
Row sums = shifted A001333: (1, 3, 7, 17, 41, 99, ...) = INVERT transform of (1, 2, 2, 2, ...). Sum of n-th row terms = rightmost term of next row.

Examples

			First few rows of the triangle:
  1;
  2, 1;
  2, 2, 3;
  2, 2, 6,  7;
  2, 2, 6, 14, 17;
  2, 2, 6, 14, 34, 41;
  2, 2, 6, 14, 34, 82, 99;
  2, 2, 6, 14, 34, 82, 198, 239;
  ...
		

Crossrefs

Cf. A001333.

Formula

n-th row of the triangle = n terms of twice A001333: (1, 1, 3, 7, 17, 41, 99, ...) followed by A001333(n).

A158843 G.f.: A(x) = exp( Sum_{n>=1} A001333(n)^n * 2^n*x^n/n ).

Original entry on oeis.org

1, 2, 20, 952, 336112, 742166496, 10043945021760, 814531629739559808, 393150002983518264270592, 1123538097532735360702239462912, 18948231465474675384343860006353603584, 1881331085022567366434813565917484763975526400
Offset: 0

Views

Author

Paul D. Hanna, Apr 09 2009

Keywords

Comments

Compare to g.f.: exp( Sum_{n>=1} 2*A001333(n)*x^n/n ) = 1/(1-2*x-x^2), which is the g.f. of the Pell numbers A000129 (with offset), where A001333(n) = A000129(n+1) - A000129(n).

Examples

			G.f.: A(x) = 1 + 2*x + 20*x^2 + 952*x^3 + 336112*x^4 + 742166496*x^5 +...
log(A(x)) = 2*x + 6^2*x^2/2 + 14^3*x^3/3 + 34^4*x^4/4 + 82^5*x^5/5 +...
log(G(x)) = 2*x + 6*x^2/2 + 14*x^3/3 + 34*x^4/4 + 82*x^5/5 +...
G(x) = 1 + 2*x + 5*x^2 + 12*x^3 + 29*x^4 + 70*x^5 + 169*x^6 +... (A000129).
		

Crossrefs

Programs

  • PARI
    {a(n)=local(LD=Vec(2*(1+x)/(1-2*x-x^2 +x*O(x^n)))); polcoeff(exp(sum(m=1,n,LD[m]^m*x^m/m)+x*O(x^n)),n)}

A160756 Triangle read by rows, infinite lower triangular Toeplitz matrix with A078008 in every column convolved with A001333.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 2, 2, 0, 3, 6, 2, 2, 0, 7, 10, 6, 2, 6, 0, 17, 22, 10, 6, 6, 14, 0, 41, 42, 22, 10, 18, 14, 34, 0, 99, 86, 42, 22, 30, 42, 34, 82, 0, 239, 170, 86, 42, 66, 70, 102, 82, 198, 0, 577
Offset: 0

Views

Author

Gary W. Adamson, May 25 2009

Keywords

Comments

Row sums = A001333: (1, 1, 3, 7, 17, 41,...). Sum of n-th row terms = rightmost term of next row.

Examples

			First few rows of the triangle =
1;
0, 1;
2, 0, 1;
2, 2, 0, 3;
6, 2, 2, 0, 7;
10, 6, 2, 6, 0, 17;
22, 10, 6, 6, 14, 0, 41;
42, 22, 10, 18, 14, 34, 0, 99;
86, 42, 22, 30, 42, 34, 82, 0, 239;
170, 86, 42, 66, 70, 102, 82, 198, 0, 577;
...
Example: row 4 = (6, 2, 2, 0, 7) = (6, 2, 2, 0, 1) * (1, 1, 1, 3, 7).
		

Crossrefs

Formula

Let M = an infinite lower triangular Toeplitz matrix with A078008 (1, 0, 2, 2, 6, 10, 22, 42, 86, 170,...). Let Q = the eigensequence of that triangle prefaced with a 1: (1, 1, 1, 3, 7, 17,...) where A001333 = (1, 1, 3, 7, 17,...). The triangle = M * Q.

A176481 Triangle, read by rows, defined by T(n, k) = b(n) - b(k) - b(n-k) + 2, where b(n) = A001333(n).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 11, 13, 11, 1, 1, 25, 33, 33, 25, 1, 1, 59, 81, 87, 81, 59, 1, 1, 141, 197, 217, 217, 197, 141, 1, 1, 339, 477, 531, 545, 531, 477, 339, 1, 1, 817, 1153, 1289, 1337, 1337, 1289, 1153, 817, 1, 1, 1971, 2785, 3119, 3249, 3283, 3249, 3119, 2785, 1971, 1
Offset: 0

Views

Author

Roger L. Bagula, Apr 18 2010

Keywords

Comments

Row sums are: {1, 2, 5, 12, 37, 118, 369, 1112, 3241, 9194, 25533, ...}.

Examples

			Triangle begins as:
  1;
  1,    1;
  1,    3,    1;
  1,    5,    5,    1;
  1,   11,   13,   11,    1;
  1,   25,   33,   33,   25,    1;
  1,   59,   81,   87,   81,   59,    1;
  1,  141,  197,  217,  217,  197,  141,    1;
  1,  339,  477,  531,  545,  531,  477,  339,    1;
  1,  817, 1153, 1289, 1337, 1337, 1289, 1153,  817,    1;
  1, 1971, 2785, 3119, 3249, 3283, 3249, 3119, 2785, 1971, 1;
		

Crossrefs

Cf. A001333.

Programs

  • Magma
    b:= func< n| Round(((1+Sqrt(2))^n + (1-Sqrt(2))^n)/2) >; [[b(n)-b(k)-b(n-k)+2: k in [0..n]]: n in [0..10]]; // G. C. Greubel, May 06 2019
    
  • Mathematica
    b[n_]:= LucasL[n, 2]/2; T[n_, k_]:= b[n] -b[k] -b[n-k] +2;
    Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, May 06 2019 *)
  • PARI
    {b(n) = round(((1+sqrt(2))^n + (1-sqrt(2))^n)/2)};
    {T(n,k) = b(n) -b(k) -b(n-k) +2};
    for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, May 06 2019
    
  • Sage
    def b(m): return lucas_number2(m,2,-1)/2
    def T(n, k): return b(n) - b(k) - b(n-k) + 2
    [[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, May 06 2019

Formula

Let b(n) = ((1+sqrt(2))^n + (1-sqrt(2))^n)/2 = A001333(n), then T(n, k) = b(n) - b(k) - b(n-k) + 2.

Extensions

Edited by G. C. Greubel, May 06 2019
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