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.

Previous Showing 11-15 of 15 results.

A172319 10th column of A172119.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1023, 2044, 4084, 8160, 16304, 32576, 65088, 130048, 259840, 519168, 1037313, 2072582, 4141080, 8274000, 16531696, 33030816, 65996544, 131863040, 263466240, 526413312, 1051789311
Offset: 0

Views

Author

Richard Choulet, Jan 31 2010

Keywords

Links

Crossrefs

Cf. A172318, A172317, A172316, A172119, A001949, A107066.
Partial sums of A104144.

Programs

  • Maple
    for k from 0 to 20 do for n from 0 to 30 do b(n):=sum((-1)^j*binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j),j=0..floor(n/(k+1))):od:k: seq(b(n),n=0..30):od;
  • Mathematica
    LinearRecurrence[{2,0,0,0,0,0,0,0,0,-1},{1,2,4,8,16,32,64,128,256,512},40] (* Harvey P. Dale, Sep 22 2020 *)

Formula

G.f.: 1/(1-2*z+z^10).
a(n)=sum((-1)^j*binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j),j=0..floor(n/(k+1))). a(n+10)=2*a(n+9)-a(n).

A119407 Number of nonempty subsets of {1,2,...,n} with no gap of length greater than 4 (a set S has a gap of length d if a and b are in S but no x with a < x < b is in S, where b-a=d).

Original entry on oeis.org

1, 3, 7, 15, 31, 62, 122, 238, 462, 894, 1727, 3333, 6429, 12397, 23901, 46076, 88820, 171212, 330028, 636156, 1226237, 2363655, 4556099, 8782171, 16928187, 32630138, 62896622, 121237146, 233692122, 450456058, 868281979, 1673667337, 3226097529, 6218502937, 11986549817, 23104817656
Offset: 1

Views

Author

John W. Layman, Jul 25 2006

Keywords

Comments

The numbers of subsets of {1,2,...,n} with no gap of length greater than d, for d=1,2 and 3, seem to be given in A000217, A001924 and A062544, respectively.

Examples

			G.f. = x + 3*x^2 + 7*x^3 + 15*x^4 + 31*x^5 + 62*x^6 + 122*x^7 + 238*x^8 + 462*x^9 + ...

Links

Crossrefs

Cf. A000217, A001924, A062544, A107066.

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( x/((1-x)*(1-2*x+x^5)) )); // G. C. Greubel, Jun 05 2019
  • Mathematica
    Rest@CoefficientList[Series[x/((1-x)*(1-2*x+x^5)), {x, 0, 40}], x] (* G. C. Greubel, Jun 05 2019 *)
LinearRecurrence[{3,-2,0,0,-1,1},{1,3,7,15,31,62},40] (* Harvey P. Dale, Dec 04 2019 *)
  • PARI
    {a(n) = if( n<0, n = -n; polcoeff( x^5 / ((1 - x)^2 * (1 + x + x^2 + x^3 - x^4)) + x * O(x^n), n), polcoeff( x / ((1 - x)^2 * (1 - x - x^2 - x^3 - x^4)) + x * O(x^n), n))} /* Michael Somos, Dec 28 2012 */
  • PARI
    my(x='x+O('x^40)); Vec(x/((1-x)*(1-2*x+x^5))) \\ G. C. Greubel, Jun 05 2019
  • Sage
    a=(x/((1-x)*(1-2*x+x^5))).series(x, 40).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Jun 05 2019

Formula

G.f. for number of nonempty subsets of {1,2,...,n} with no gap of length greater than d is x/((1-x)*(1-2*x+x^(d+1))). - Vladeta Jovovic, Apr 27 2008
From Michael Somos, Dec 28 2012: (Start)
G.f.: x/((1-x)^2*(1-x-x^2-x^3-x^4)) = x/((1-x)*(1-2*x+x^5)).
First difference is A107066. (End)
a(n-3) = Sum_{k=0..n} (n-k)*A000078(k) for n>3. - Greg Dresden, Jan 01 2021

Extensions

Terms a(25) onward added by G. C. Greubel, Jun 05 2019

A018922 Define the generalized Pisot sequence T(a(0),a(1)) by: a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n). This is T(8,16).

Original entry on oeis.org

8, 16, 31, 60, 116, 224, 432, 833, 1606, 3096, 5968, 11504, 22175, 42744, 82392, 158816, 306128, 590081, 1137418, 2192444, 4226072, 8146016, 15701951, 30266484, 58340524, 112454976, 216763936, 417825921, 805385358, 1552430192, 2992405408, 5768046880
Offset: 0

Views

Author

R. K. Guy

Keywords

Comments

Not to be confused with the Pisot T(8,16), which is essentially A000079. - R. J. Mathar, Feb 13 2016

Links

Crossrefs

Cf. A107066.

Programs

  • Magma
    Tiv:=[8,16]; [n le 2 select Tiv[n] else Ceiling(Self(n-1)^2/Self(n-2))-1: n in [1..40]]; // Bruno Berselli, Feb 17 2016
  • Mathematica
    Drop[CoefficientList[Series[1/(1 - 2*z + z^5), {z, 0, 100}], z], 3] (* Vladimir Joseph Stephan Orlovsky, Jul 08 2011 *)
RecurrenceTable[{a[1] == 8, a[2] == 16, a[n] == Ceiling[a[n-1]^2/a[n-2]] - 1}, a, {n, 40}] (* Bruno Berselli, Feb 17 2016 *)
LinearRecurrence[{2,0,0,0,-1},{8,16,31,60,116},40] (* Harvey P. Dale, Sep 21 2024 *)
  • PARI
    T(a0, a1, maxn) = a=vector(maxn); a[1]=a0; a[2]=a1; for(n=3, maxn, a[n]=ceil(a[n-1]^2/a[n-2])-1); a
T(8, 16, 40) \\ Colin Barker, Feb 14 2016

Formula

a(n) = 2*a(n-1) - a(n-5).
a(n) = A107066(n+3). - Vladimir Joseph Stephan Orlovsky, Jul 08 2011
O.g.f: -(-8+x^2+2*x^3+4*x^4)/((x-1)*(x^4+x^3+x^2+x-1)) = (1/3)/(x-1)+(1/3)*(-13*x^3-20*x^2-24*x-25)/(x^4+x^3+x^2+x-1) . - R. J. Mathar, Dec 02 2007

A107065 Riordan array (1/(1-x),x(1+x+x^2+x^3)).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 4, 10, 10, 5, 1, 1, 4, 13, 20, 15, 6, 1, 1, 4, 15, 32, 35, 21, 7, 1, 1, 4, 16, 44, 66, 56, 28, 8, 1, 1, 4, 16, 54, 106, 121, 84, 36, 9, 1, 1, 4, 16, 60, 150, 222, 204, 120, 45, 10, 1, 1, 4, 16, 63, 190, 357, 420, 323, 165, 55, 11, 1, 1, 4, 16
Offset: 0

Views

Author

Paul Barry, May 10 2005

Keywords

Comments

Row sums are A107066. Diagonal sums are A023436(n+1).

Examples

			Triangle begins
  1;
  1, 1;
  1, 2, 1;
  1, 3, 3, 1;
  1, 4, 6, 4, 1;
  1, 4, 10, 10, 5, 1;
  1, 4, 13, 20, 15, 6, 1;
  ...

Crossrefs

Cf. A107066, A023436.

A172320 11th column of A172119.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2047, 4092, 8180, 16352, 32688, 65344, 130624, 261120, 521984, 1043456, 2085888, 4169729, 8335366, 16662552, 33308752, 66584816, 133104288, 266077952, 531894784, 1063267584
Offset: 0

Views

Author

Richard Choulet, Jan 31 2010

Keywords

Examples

			a(12)=C(12,12)*2^12-C(2,1)*2^1=4092.

Links

Crossrefs

Cf. A172319, A172318, A172317, A172316, A172119, A001949, A107066.
Partial sums of A122265.

Programs

  • Maple
    k:=10:taylor(1/(1-2*z+z^(k+1)),z=0,30); for k from 0 to 20 do for n from 0 to 30 do b(n):=sum((-1)^j*binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j),j=0..floor(n/(k+1))):od:k: seq(b(n),n=0..30):od;

Formula

a(n)=sum((-1)^j*binomial(n-k*j,n-(k+1)*j)*2^(n-(k+1)*j),j=0..floor(n/(k+1))) with k=10.
G.f: f(z)=1/(1-2*z+z^(11)).
a(n+11)=2*a(n+10)-a(n).
Previous Showing 11-15 of 15 results.

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