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 1-4 of 4 results.

A171160 a(n) = a(n-1) + 2*a(n-2) with a(0)=3, a(1)=4.

Original entry on oeis.org

3, 4, 10, 18, 38, 74, 150, 298, 598, 1194, 2390, 4778, 9558, 19114, 38230, 76458, 152918, 305834, 611670, 1223338, 2446678, 4893354, 9786710, 19573418, 39146838, 78293674, 156587350, 313174698, 626349398, 1252698794, 2505397590, 5010795178, 10021590358
Offset: 0

Views

Author

Paul Curtz, Dec 04 2009

Keywords

Crossrefs

Programs

Formula

a(n) = (1/3)*(2*(-1)^n + 7*2^n), with n>=0. - Paolo P. Lava, Dec 14 2009
G.f.: -(x+3) / ((x+1)*(2*x-1)). - Colin Barker, Feb 10 2015
From Paul Curtz, Jun 03 2022: (Start)
a(n) = A078008(n) + A078008(n+1) + A078008(n+2).
a(n) = 2^(n+1) + A078008(n).
a(n) = A001045(n+3) - A001045(n).
(a(n) + a(n+1) = a(n+2) - a(n) = A005009(n).)
a(n) + a(n+3) = A175805(n).
a(n) = A062510(n) + A083582(n-1) with A083582(-1) = 3.
a(n) = A092297(n) + A154879(n). (End)
a(n) = 2*A062092(n-1), for n>0; 2*a(n) = A083595(n+1). - Paul Curtz, Jun 08 2022

Extensions

Edited by N. J. A. Sloane, Dec 05 2009
More terms from J. Mulder (jasper.mulder(AT)planet.nl), Jan 28 2010
More terms from Max Alekseyev, Apr 24 2010

A083594 a(n) = (7 - 4*(-2)^n)/3.

Original entry on oeis.org

1, 5, -3, 13, -19, 45, -83, 173, -339, 685, -1363, 2733, -5459, 10925, -21843, 43693, -87379, 174765, -349523, 699053, -1398099, 2796205, -5592403, 11184813, -22369619, 44739245, -89478483, 178956973, -357913939, 715827885, -1431655763, 2863311533, -5726623059
Offset: 0

Views

Author

Paul Barry, May 02 2003

Keywords

Comments

Also generalized k-bonacci sequence a(n)=2*a(n-2)-a(n-1). - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 30 2007

Crossrefs

Cf. A083595.

Programs

  • Mathematica
    (7-4(-2)^Range[0,40])/3 (* or *) LinearRecurrence[{-1,2},{1,5},40] (* Harvey P. Dale, Feb 25 2012 *)

Formula

G.f.: (1+6*x)/((1-x)*(1+2*x)).
E.g.f.: (7*exp(x)-4*exp(-2*x))/3.

A168673 Binomial transform of A169609.

Original entry on oeis.org

1, 4, 10, 20, 38, 74, 148, 298, 598, 1196, 2390, 4778, 9556, 19114, 38230, 76460, 152918, 305834, 611668, 1223338, 2446678, 4893356, 9786710, 19573418, 39146836, 78293674, 156587350, 313174700, 626349398, 1252698794, 2505397588, 5010795178, 10021590358
Offset: 0

Views

Author

Paul Curtz, Dec 02 2009

Keywords

Comments

Sequence and successive differences are identical to their third differences. See principal sequence A024495. Main diagonal of the array of successive differences is A083595 (1,6,8,20,36,...).

Crossrefs

Programs

  • Magma
    I:=[1,4,10]; [n le 3 select I[n] else 3*Self(n-1)- 3*Self(n-2)+2*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jul 30 2016
    
  • Mathematica
    LinearRecurrence[{3,-3,2},{1,4,10},25] (* G. C. Greubel, Jul 29 2016 *)
    RecurrenceTable[{a[0] == 1, a[1] == 4, a[2] == 10, a[n] == 3 a[n-1] - 3 a[n-2] + 2 a[n-3]}, a, {n, 40}] (* Vincenzo Librandi, Jul 30 2016 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; 2,-3,3]^n*[1;4;10])[1,1] \\ Charles R Greathouse IV, Jul 30 2016

Formula

a(n+1) - 2a(n) = A130772(n).
a(n) = A062092(n) - A130151(n+1).
a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) for n > 2; a(0) = 1, a(1) = 4, a(2) = 10.
G.f.: (1 + x + x^2)/(1 -3*x +3*x^2 -2*x^3). - Philippe Deléham, Dec 03 2009

Extensions

Edited and extended by Klaus Brockhaus, Dec 03 2009

A338198 Triangle read by rows, T(n,k) = ((k+1)*2^(n-k)-(k-2)*(-1)^(n-k))/3 for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 2, 1, 1, 2, 3, 2, 1, 6, 5, 4, 3, 1, 10, 11, 8, 5, 4, 1, 22, 21, 16, 11, 6, 5, 1, 42, 43, 32, 21, 14, 7, 6, 1, 86, 85, 64, 43, 26, 17, 8, 7, 1, 170, 171, 128, 85, 54, 31, 20, 9, 8, 1, 342, 341, 256, 171, 106, 65, 36, 23, 10, 9, 1, 682, 683, 512, 341, 214, 127, 76, 41, 26, 11, 10, 1
Offset: 0

Views

Author

Werner Schulte, Oct 15 2020

Keywords

Comments

This triangle is related to the Jacobsthal numbers (A001045).

Examples

			The triangle T(n,k) for 0 <= k <= n starts:
n\k :    0     1     2    3    4    5    6   7   8   9
======================================================
  0 :    1
  1 :    0     1
  2 :    2     1     1
  3 :    2     3     2    1
  4 :    6     5     4    3    1
  5 :   10    11     8    5    4    1
  6 :   22    21    16   11    6    5    1
  7 :   42    43    32   21   14    7    6   1
  8 :   86    85    64   43   26   17    8   7   1
  9 :  170   171   128   85   54   31   20   9   8   1
etc.
		

Crossrefs

For columns k = 0 to 8 see A078008, A001045, A000079, A001045, A084214, A014551, A083595, A083582, A259713 respectively.

Programs

  • Mathematica
    Table[((k + 1)*2^(n - k) - (k - 2)*(-1)^(n - k))/3, {n, 0, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Oct 15 2020 *)

Formula

T(n,n) = 1 for n >= 0; T(n,n-1) = n-1 for n > 0.
T(n,k) = T(n-1,k) + 2 * T(n-2,k) for 0 <= k <= n-2.
T(n,k) = 2 * T(n-1,k) - (k-2) * (-1)^(n-k) for 0 <= k < n.
T(n,k) = T(n+1-k,1) + (k-1) * T(n-k,1) for 0 <= k < n.
T(n+1,k) * T(n-1,k) - T(n,k+1) * T(n,k-1) = T(n-k,1)^2 for 0 < k < n.
Row sums are A083579(n+1) for n >= 0.
G.f. of column k >= 0: (1+(k-1)*t) * t^k / (1-t-2*t^2).
G.f.: Sum_{n>=0, k=0..n} T(n,k) * x^k * t^n = (1 - (1+x)*t + 2*x*t^2) / ((1 - x*t)^2 * (1 - t - 2*t^2)).
Conjecture: Let M(n,k) be the matrix inverse of T(n,k), seen as a matrix. Then M(i,j) = 0 if j < 0 or j > i, M(n,n) = 1 for n >= 0, M(n,n-1) = 1-n for n > 0, and M(n,k) = (-1)^(n-k) * (k^2-2) * (n-2)! / k! for 0 <= k <= n-2.
Showing 1-4 of 4 results.