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 1-3 of 3 results.

A115113 a(n) = 3*a(n-1) + 4*a(n-2), with a(0) = 2, a(1) = 6, a(2) = 10.

Original entry on oeis.org

2, 6, 10, 54, 202, 822, 3274, 13110, 52426, 209718, 838858, 3355446, 13421770, 53687094, 214748362, 858993462, 3435973834, 13743895350, 54975581386, 219902325558, 879609302218, 3518437208886, 14073748835530, 56294995342134, 225179981368522, 900719925474102
Offset: 0

Views

Author

Roger L. Bagula, Mar 06 2006

Keywords

Links

Crossrefs

Cf. A115164, A115335.

Programs

  • Magma
    I:=[6,10]; [2] cat [n le 2 select I[n] else 3*Self(n-1) + 4*Self(n-2): n in [1..49]]; // G. C. Greubel, Nov 23 2018
  • Mathematica
    Join[{2}, LinearRecurrence[{3, 4}, {6, 10}, 50]]
  • Maxima
    (a[0] : 2, a[1] : 6, a[2] : 10, a[n] := 3*a[n-1] + 4*a[n-2], makelist(a[n], n, 0, 50)); /* Franck Maminirina Ramaharo, Nov 23 2018 */
  • PARI
    x='x+O('x^50); Vec(2*(8*x^2-1)/((x+1)*(4*x-1))) \\ G. C. Greubel, Nov 23 2018
  • Sage
    s=(2*(8*x^2-1)/((x+1)*(4*x-1))).series(x,50); s.coefficients(x, sparse=False) # G. C. Greubel, Nov 23 2018

Formula

From Colin Barker, Nov 13 2012: (Start)
a(n) = (-2*(7*(-1)^n - 2^(1 + 2*n)))/5 for n > 0.
a(n) = 3*a(n-1) + 4*a(n-2) for n > 2.
G.f.: 2*(8*x^2 - 1)/((x + 1)*(4*x - 1)). (End)
E.g.f.: (20 - 14*exp(-x) + 4*exp(4*x))/5. - Franck Maminirina Ramaharo, Nov 23 2018

Extensions

Edited, and new name from Franck Maminirina Ramaharo, Nov 23 2018, after Colin Barker's formula

A115164 a(n) = 3*a(n-1) + 4*a(n-2), with a(0) = 3, a(1) = 7, a(3) = 9, for n > 2.

Original entry on oeis.org

3, 7, 9, 55, 201, 823, 3273, 13111, 52425, 209719, 838857, 3355447, 13421769, 53687095, 214748361, 858993463, 3435973833, 13743895351, 54975581385, 219902325559, 879609302217, 3518437208887, 14073748835529, 56294995342135
Offset: 0

Views

Author

Roger L. Bagula, Mar 06 2006

Keywords

Links

Crossrefs

Cf. A115113, A115335.

Programs

  • Magma
    [3] cat [(4^(1+n) -19*(-1)^n)/5: n in [1..50]]; // G. C. Greubel, Nov 23 2018
  • Mathematica
    Join[{3}, LinearRecurrence[{3, 4}, {7, 9}, 50]]
  • Maxima
    (a[0] : 3, a[1] : 7, a[2] : 9, a[n] := 3*a[n-1] + 4*a[n-2], makelist(a[n], n, 0, 50)); /* Franck Maminirina Ramaharo, Nov 23 2018 */
  • PARI
    vector(50, n, n--; if(n==0, 3, (4^(1+n) -19*(-1)^n)/5)) \\ G. C. Greubel, Nov 23 2018
  • Sage
    [3] + [(4^(1+n) -19*(-1)^n)/5 for n in (1..50)] # G. C. Greubel, Nov 23 2018

Formula

From Colin Barker, Oct 31 2012: (Start)
a(n) = (4^(1 + n) - 19*(-1)^n)/5 for n > 0.
a(n) = 3*a(n-1) + 4*a(n-2) for n > 2.
G.f.: (24*x^2 + 2*x - 3)/((x + 1)*(4*x - 1)). (End)
From Franck Maminirina Ramaharo, Nov 23 2018: (Start)
a(n) = A115113(n) + A165326(n).
E.g.f.: (30 - 19*exp(-x) + 4*exp(4*x))/5. (End)

Extensions

Edited, and new name from Franck Maminirina Ramaharo, Nov 23 2018, after Colin Barker

A352692 a(n) + a(n+1) = 2^n for n >= 0 with a(0) = 4.

Original entry on oeis.org

4, -3, 5, -1, 9, 7, 25, 39, 89, 167, 345, 679, 1369, 2727, 5465, 10919, 21849, 43687, 87385, 174759, 349529, 699047, 1398105, 2796199, 5592409, 11184807, 22369625, 44739239, 89478489, 178956967, 357913945, 715827879, 1431655769, 2863311527, 5726623065, 11453246119, 22906492249
Offset: 0

Views

Author

Paul Curtz, Mar 29 2022

Keywords

Comments

Difference table D(n,k) = D(n-1,k+1) - D(n-1,k), D(0,k) = a(k):
4, -3, 5, -1, 9, 7, 25, ...
-7, 8, -6, 10, -2, 18, 14, 50, ...
15, -14, 16, -12, 20, -4, 36, 28, 100, ...
-29, 30, -28, 32, -24, 40, -8, 72, 56, 200, ...
59, -58, 60, -56, 64, -48, 80, -16, 144, 112, 400, ...
...
The diagonals are given by D(n,n+k) = a(k)*2^n.
D(n,1) = -(-1)^n* A340627(n).
a(n) - a(n) = 0, 0, 0, 0, 0, ... (trivially)
a(n+1) + a(n) = 1, 2, 4, 8, 16, ... = 2^n (by definition)
a(n+2) - a(n) = 1, 2, 4, 8, 16, ... = 2^n
a(n+3) + a(n) = 3, 6, 12, 24, 48, ... = 2^n*3
a(n+4) - a(n) = 5, 10, 20, 40, 80, ... = 2^n*5
a(n+5) + a(n) = 11, 22, 44, 88, 176, ... = 2^n*11
(...)
This table is given by T(r,n) = A001045(r)*2^n with r, n >= 0.
Sums of antidiagonals are A045883(n).
Main diagonal: A192382(n).
First upper diagonal: A054881(n+1).
First subdiagonal: A003683(n+1).
Second subdiagonal: A246036(n).
Now consider the array from c(n) = (-1)^n*a(n) with its difference table:
4, 3, 5, 1, 9, -7, 25, -39, ... = c(n)
-1, 2, -4, 8, -16, 32, -64, 128, ... = -A122803(n)
3, -6, 12, -24, 48, -96, 192, -384, ... =
-9, 18, -36, 72, -144, 288, -576, 1152, ...
27, -54, 108, -216, 432, -864, 1728, -3456, ...
...
The first subdiagonal is -A000400(n). The second is A169604(n).

Links

Crossrefs

If a(0) = k then A001045 (k=0), A078008 (k=1), A140966 (k=2), A154879 (k=3), this sequence (k=4).
Essentially the same as A115335.
Cf. A000079, A002450, A340627.
Cf. A020714, A175805.
Cf. A045883, A192382, A003683, A246036.
Cf. A054881, A000400, A169604, A122803.
Cf. A132805, A082311, A082365.
Cf. A024495, A132804, A163868.

Programs

  • Maple
    a := proc(n) option remember; ifelse(n = 0, 4, 2^(n-1) - a(n-1)) end: # Peter Luschny, Mar 29 2022
A352691 := proc(n)
    (11*(-1)^n + 2^n)/3
end proc: # R. J. Mathar, Apr 26 2022
  • Mathematica
    LinearRecurrence[{1, 2}, {4, -3}, 40] (* Amiram Eldar, Mar 29 2022 *)
  • PARI
    a(n) = (11*(-1)^n + 2^n)/3; \\ Thomas Scheuerle, Mar 29 2022

Formula

abs(a(n)) = A115335(n-1) for n >= 1.
a(3*n) - (-1)^n*4 = A132805(n).
a(3*n+1) + (-1)^n*4 = A082311(n).
a(3*n+2) - (-1)^n*4 = A082365(n).
From Thomas Scheuerle, Mar 29 2022: (Start)
G.f.: (-4 + 7*x)/(-1 + x + 2*x^2).
Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*a(m + 2*n-k) = a(m)*2^n.
Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*a(1 + n-k) = -(-1)^n*A340627(n).
a(n) = (11*(-1)^n + 2^n)/3.
a(n + 2*m) = a(n) + A002450(m)*2^n.
a(2*n) = A192382(n+1) + (-1)^n*a(n).
a(n) = ( A045883(n) - Sum_{k=0..n-1}(-1)^k*a(k) )/n, for n > 0. (End)
a(n) = A001045(n) + 4*(-1)^n.
a(n+1) = 2*a(n) -11*(-1)^n.
a(n+2) = a(n) + 2^n.
a(n+4) = a(n) + A020714(n).
a(n+6) = a(n) + A175805(n).
a(2*n) = A163868(n).
a(2*n+1) = (2^(2*n+1) - 11)/3.

Extensions

Warning: The DATA is correct, but there may be errors in the COMMENTS, which should be rechecked. - Editors of OEIS, Apr 26 2022
Edited by M. F. Hasler, Apr 26 2022.
Showing 1-3 of 3 results.

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