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.

A077925 Expansion of 1/((1-x)*(1+2*x)).

Original entry on oeis.org

1, -1, 3, -5, 11, -21, 43, -85, 171, -341, 683, -1365, 2731, -5461, 10923, -21845, 43691, -87381, 174763, -349525, 699051, -1398101, 2796203, -5592405, 11184811, -22369621, 44739243, -89478485, 178956971, -357913941, 715827883, -1431655765, 2863311531, -5726623061
Offset: 0

Views

Author

N. J. A. Sloane, Nov 17 2002

Keywords

Comments

a(n+1) is the reflection of a(n) through a(n-1) on the numberline. - Floor van Lamoen, Aug 31 2004
If a zero is added as the (new) a(0) in front, the sequence represents the inverse binomial transform of A001045. Partial sums are in A077898. - R. J. Mathar, Aug 30 2008
a(n) = A077953(2*n+3). - Reinhard Zumkeller, Oct 07 2008
Related to the Fibonacci sequence by an INVERT transform: if A(x) = 1+x^2*g(x) is the generating function of the a(n) prefixed with 1, 0, then 1/A(x) = 2+(x+1)/(x^2-x+1) is the generating function of 1, 0, -1, 1, -2, 3, ..., the signed Fibonacci sequence A000045 prefixed with 1. - Gary W. Adamson, Jan 07 2011
Also: Gaussian binomial coefficients [n+1,1], or q-integers, for q=-2, diagonal k=1 in the triangular (or column r=1 in the square) array A015109. - M. F. Hasler, Nov 04 2012
With a leading zero, 0, 1, -1, 3, -5, 11, -21, 43, -85, 171, -341, 683, ... we obtain the Lucas U(-1,-2) sequence. - R. J. Mathar, Jan 08 2013
Let m = a(n). Then 18*m^2 - 12*m + 1 = A000225(2n+3). - Roderick MacPhee, Jan 17 2013

Examples

			G.f. = 1 - x + 3*x^2 - 5*x^3 + 11*x^4 - 21*x^5 + 43*x^6 - 85*x^7 + ...

Links

Crossrefs

Cf. A001045 (unsigned version).
Cf. A014983, A014985, A014986. - Zerinvary Lajos, Dec 16 2008
Cf. A232600, A232601, A232602.

Programs

Formula

G.f.: 1/(1+x-2*x^2).
a(n) = (1-(-2)^(n+1))/3. - Vladeta Jovovic, Apr 17 2003
a(n) = Sum_{k=0..n} (-2)^k. - Paul Barry, May 26 2003
a(n+1) - a(n) = A122803(n). - R. J. Mathar, Aug 30 2008
a(n) = Sum_{k=0..n} A112555(n,k)*(-2)^k. - Philippe Deléham, Sep 11 2009
a(n) = A082247(n+1) - 1. - Philippe Deléham, Oct 07 2009
G.f.: Q(0)/(3*x), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k + 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k - 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 22 2013
G.f.: Q(0)/2 , where Q(k) = 1 + 1/(1 - x*(4*k-1 + 2*x)/( x*(4*k+1 + 2*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 08 2013
E.g.f.: (2*exp(-2*x) + exp(x))/3. - Ilya Gutkovskiy, Nov 12 2016
a(n) = A086893(n+2) - A061547(n+3), n >= 0. - Yosu Yurramendi, Jan 16 2017
a(n) = (-1)^n*A001045(n+1). - M. F. Hasler, Feb 13 2020
a(n) - a(n-1) = a(n-1) - a(n+1) = (-2)^n, a(n+1) = - a(n) + 2*a(n-1) = 1 - 2*a(n). - Michael Somos, Feb 22 2023

A210801 Triangle of coefficients of polynomials u(n,x) jointly generated with A210802; see the Formula section.

Original entry on oeis.org

1, 3, 1, 6, 5, 2, 12, 15, 10, 3, 21, 39, 37, 19, 5, 39, 90, 111, 81, 35, 8, 66, 198, 300, 281, 171, 64, 13, 120, 414, 750, 855, 659, 346, 115, 21, 201, 846, 1776, 2391, 2230, 1474, 684, 205, 34, 363, 1683, 4044, 6255, 6828, 5441, 3170, 1323, 362, 55
Offset: 1

Views

Author

Clark Kimberling, Mar 27 2012

Keywords

Comments

Row n ends with F(n), where F=A000045 (Fibonacci numbers).
Row sums: A003462
Alternating row sums: 1,2,3,4,5,6,7,8,...
For a discussion and guide to related arrays, see A208510.

Examples

			First five rows:
1
3....1
6....5....2
12...15...10...3
21...39...37...19...5
First three polynomials u(n,x): 1, 3 + x, 6 + 5x + 2x^2.

Crossrefs

Cf. A210802, A208510.

Programs

  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + (x + j)*v[n - 1, x] + c;
d[x_] := h + x; e[x_] := p + x;
v[n_, x_] := d[x]*u[n - 1, x] + e[x]*v[n - 1, x] + f;
j = 1; c = 1; h = 2; p = -1; f = 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A210801 *)
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A210802 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]  (* A003462 *)
Table[v[n, x] /. x -> 1, {n, 1, z}]  (* A003462 *)
Table[u[n, x] /. x -> -1, {n, 1, z}] (* A000027 *)
Table[v[n, x] /. x -> -1, {n, 1, z}] (* A077898 *)

Formula

u(n,x)=u(n-1,x)+(x+1)*v(n-1,x)+1,
v(n,x)=(x+2)*u(n-1,x)+(x-1)*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.

A210802 Triangle of coefficients of polynomials v(n,x) jointly generated with A210801; see the Formula section.

Original entry on oeis.org

1, 2, 2, 5, 5, 3, 8, 16, 11, 5, 17, 34, 40, 22, 8, 26, 82, 107, 93, 43, 13, 53, 163, 287, 287, 201, 81, 21, 80, 352, 674, 862, 709, 419, 150, 34, 161, 676, 1592, 2272, 2326, 1641, 845, 273, 55, 242, 1378, 3482, 5878, 6797, 5863, 3638, 1666, 491, 89, 485
Offset: 1

Views

Author

Clark Kimberling, Mar 27 2012

Keywords

Comments

Row n ends with F(n+1), where F=A000045 (Fibonacci numbers).
Row sums: A003462
Alternating row sums: A077898
For a discussion and guide to related arrays, see A208510.

Examples

			First five rows:
1
2....2
5....5....3
8....16...11...5
17...34...40...22...8
First three polynomials v(n,x): 1, 2 + 2x, 5 + 5x + 3x^2

Crossrefs

Cf. A210801, A208510.

Programs

  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + (x + j)*v[n - 1, x] + c;
d[x_] := h + x; e[x_] := p + x;
v[n_, x_] := d[x]*u[n - 1, x] + e[x]*v[n - 1, x] + f;
j = 1; c = 1; h = 2; p = -1; f = 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A210801 *)
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A210802 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]  (* A003462 *)
Table[v[n, x] /. x -> 1, {n, 1, z}]  (* A003462 *)
Table[u[n, x] /. x -> -1, {n, 1, z}] (* A000027 *)
Table[v[n, x] /. x -> -1, {n, 1, z}] (* A077898 *)

Formula

u(n,x)=u(n-1,x)+(x+1)*v(n-1,x)+1,
v(n,x)=(x+2)*u(n-1,x)+(x-1)*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.

A366987 Triangle read by rows: T(n, k) = -(2^(n - k)*(-1)^n + 2^k + (-1)^k)/3.

Original entry on oeis.org

-1, 0, 0, -2, -1, -2, 2, 1, -1, -2, -6, -3, -3, -3, -6, 10, 5, 1, -1, -5, -10, -22, -11, -7, -5, -7, -11, -22, 42, 21, 9, 3, -3, -9, -21, -42, -86, -43, -23, -13, -11, -13, -23, -43, -86, 170, 85, 41, 19, 5, -5, -19, -41, -85, -170, -342, -171, -87, -45, -27, -21, -27, -45, -87, -171, -342
Offset: 0

Views

Author

Paul Curtz and Thomas Scheuerle, Oct 31 2023

Keywords

Examples

			Triangle T(n, k) starts:
   -1
    0   0
   -2  -1  -2
    2   1  -1  -2
   -6  -3  -3  -3  -6
   10   5   1  -1  -5 -10
  -22 -11  -7  -5  -7 -11 -22
   42  21   9   3  -3  -9 -21 -42
   ...
Note the symmetrical distribution of the absolute values of the terms in each row.

Links

Crossrefs

Rows sums: -A282577(n+2), if the conjectures from Colin Barker in A282577 are true.
First column: -(-1)^n * A078008(n).
Second column: (-1)^n * A001045(n).
Third column: -A140966(n).
Fourth column: (-1)^n * A155980(n+2).
Cf. A000079, A020989, A048573, A053088, A059570, A077898, A140966.

Programs

  • Maple
    T := (n, k) -> -(2^(n-k)*(-1)^n + 2^k + (-1)^k)/3:
seq(seq(T(n, k), k = 0..n), n = 0..10);  # Peter Luschny, Nov 02 2023
  • Mathematica
    A366987row[n_]:=Table[-(2^(n-k)(-1)^n+2^k+(-1)^k)/3,{k,0,n}];Array[A366987row,15,0] (* Paolo Xausa, Nov 07 2023 *)
  • PARI
    T(n, k) = (-2^(k+1) + 2*(-1)^(k+1) + (-1)^(n+1)*2^(1+n-k))/6 \\ Thomas Scheuerle, Nov 01 2023

Formula

T(n, 0) = -((-2)^n + 2)/3.
T(n, k+1) - T(n, k) = T(n-1, k) + (-1)^k.
T(2*n+1, n) = A001045(n).
T(2*n+1, n+1) = -A001045(n).
T(2*n, n+1) = -A048573(n-1), for n > 0.
Note that the definition of T extends to negative parameters:
T(2*n-2, n-1) = -A001045(n).
-2^n*Sum_{k=0..n} (-1)^k*T(-n, -k) = A059570(n+1).
-4^n*Sum_{k=0..2*n} T(-2*n, -k) = A020989(n).
-Sum_{k=0..n} (-1)^k*T(n, k) = A077898(n). See also A053088.
Sum_{k = 0..2*n} |T(2*n, k)| = (4^(n+1) - 1)/3.
Sum_{k = 0..2*n+1} |T(2*n+1, k)| = (1 + (-1)^n - 2^(2 + n) + 2^(1 + 2*n))/3.
G.f.: (-1 - x + x*y)/((1 - x)*(1 + 2*x)*(1 + x*y)*(1 - 2*x*y)). - Stefano Spezia, Nov 03 2023

Extensions

a(42) corrected by Paolo Xausa, Nov 07 2023
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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