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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A249580 List of quadruples (r,s,t,u): the matrix M = [[4,12,9][2,5,3][1,2,1]] is raised to successive negative powers, then (r,s,t,u) are the square roots of M[1,3], M[1,1], M[3,3], M[3,1] respectively.

Original entry on oeis.org

3, -1, -2, 1, -9, 4, 7, -3, 30, -13, -23, 10, -99, 43, 76, -33, 327, -142, -251, 109, -1080, 469, 829, -360, 3567, -1549, -2738, 1189, -11781, 5116, 9043, -3927, 38910, -16897, -29867, 12970, -128511, 55807, 98644, -42837
Offset: 1

Views

Author

Russell Walsmith, Nov 02 2014

Keywords

Comments

The sequence, in reverse order, comprises numbers to the left of a(0) in A249579, where the terms would be labeled a(-1), a(-2), a(-3), ... .
This sequence 'factors' into four sequences with alternating signs. Ignoring signage, they are A052906, A003688, A052924 and A006190 (listed as crossrefs below).

Examples

			M^-1 = [[1,-6,9][-1,5,-6][1,-4,4]]. sqrt(M[1,3]) = 3, sqrt(M[1,1]) = -1, sqrt(M[3,3]) = -2, sqrt(M[3,1]) = 1. Then r = 3; s = -1; t = -2; ; u = 1.
M^-2 = [[16,-72,81][-12,55,-63][9,-42,49]]. sqrt(M[1,3]) = -9, sqrt(M[1,1]) = 4, sqrt(M[3,3]) = 7, sqrt(M[3,1]) = -3. Then r = -9; s = 4; t = 7; ; u = -3.

Links

Crossrefs

Cf. A249579. Disregarding signage, a(4n) = A052906; a(4n+1) = A003688; a(4n+2) = A052924; a(4n+3) = A006190.

Programs

  • Mathematica
    m[e_] := MatrixPower[{{4, 12, 9}, {2, 5, 3}, {1, 2, 1}}, -e]; g[e_, x_, y_] := (-1)^If[x == y, e, e + 1] Sqrt@ m[e][[x, y]]; f[e_] := {g[e, 1, 3], g[e, 1, 1], g[e, 3, 3], g[e, 3, 1]}; Array[f, 10] // Flatten (* Robert G. Wilson v, Dec 19 2014 *)
  • PARI
    Vec(-x*(x^6+x^5+x^3-2*x^2-x+3)/(x^8-3*x^4-1) + O(x^100)) \\ Colin Barker, Nov 06 2014

Formula

a(n) = -3*a(n-4)+a(n-8). - Colin Barker, Nov 06 2014
G.f.: -x*(x^6+x^5+x^3-2*x^2-x+3) / (x^8-3*x^4-1). - Colin Barker, Nov 06 2014

A202209 Triangle T(n,k), read by rows, given by (2, 1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 2, 0, 5, 1, 0, 13, 5, 0, 0, 34, 19, 1, 0, 0, 89, 65, 8, 0, 0, 0, 233, 210, 42, 1, 0, 0, 0, 610, 654, 183, 11, 0, 0, 0, 0, 1597, 1985, 717, 74, 1, 0, 0, 0, 0, 4181, 5911, 2622, 394, 14, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Philippe Deléham, Dec 14 2011

Keywords

Comments

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

Examples

			Triangle begins :
1
2, 0
5, 1, 0
13, 5, 0, 0
34, 19, 1, 0, 0
89, 65, 8, 0, 0, 0
233, 210, 42, 1, 0, 0, 0

Crossrefs

Cf. A000045, A000079, A001519, A001870, A001906, A126124, A202207 (antidiagonal sums)

Formula

T(n,k) = 3*T(n-1,k) - T(n-2,k) + T(n-2,k-1).
G.f.: (1-x)/(1-3x+(1-y)*x^2).
Sum_{k, 0<=k<=n} T(n,k)*x^k = A057682(n+1), A000079(n), A122367(n), A025192(n), A052924(n), A104934(n), A202206(n), A122117(n), A197189(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6 respectively.
T(n,0) = A122367(n) = A000045(2n+1).

A332060 a(n) = 3*a(n-1) + a(n-2) after initial values a(0..5) = (0, 1, 2, 3, 5, 13).

Original entry on oeis.org

0, 1, 2, 3, 5, 13, 44, 145, 479, 1582, 5225, 17257, 56996, 188245, 621731, 2053438, 6782045, 22399573, 73980764, 244341865, 807006359, 2665360942, 8803089185, 29074628497, 96026974676, 317155552525, 1047493632251, 3459636449278, 11426402980085
Offset: 0

Views

Author

M. F. Hasler, Mar 04 2020

Keywords

Comments

These numbers arise as borders of intervals [a(n), a(n+1)] = [b(k)=k, b(m)=m] with m := b(k) + b(k-1) after the "holes" between the borders have been filled according to b(k+1) = b(k) + b(m) and b(m-1) = b(k+1) + b(n) for any such interval of length m - k > 2, i.e., starting from k = 5, m = 13.
The initial terms correspond to intervals of length <= 2 with only 0 or 1 "holes" to fill: In the first case we have the same recursion rule as for the Fibonacci sequence, and when there's one hole (between 3 and 5) the next Fibonacci number b(k+1) = b(k) + b(m) = 3 + 5 = 8 gets filled in there, and the next border is m = b(k) + b(k-1) = 5 + 8 = 13. See Example for more.

Examples

			The initial a(0) is conventional. (One could also choose a(0) = 1 to have a(2) = a(1) + a(0) as for the next two terms, but this wouldn't correspond to b(m) = b(k) + b(k-1), either.)
We start with [a(1), a(2)] = [1, 2].
No gap or "hole" here to fill, so the next interval [a(2), a(3)] has upper bound a(3) = a(2) + 1 = 3, where 1 is the element just left to the right border a(2).
Again, no gap or hole to fill in [2, 3], so the next interval has upper bound a(4) = a(3) + 2 = 5, where 2 is the element just left to the right border a(3).
Now there's a hole in (3, 5), at position 4, which is filled with 3 + 5 = 8, so the next upper bound is a(5) = a(4) + 8 = 5 + 8 = 13: here the number 8 was the element left to the right border 5.
Now there are several holes in (5, 13). The leftmost one (position 6) is filled with 5 + 13 = 18, and the rightmost (position 12) is filled with 18 + 13 = 31. So the next interval [a(5), a(6)] has upper bound a(6) = a(5) + 31 = 44.

Links

Crossrefs

Cf. A000045 (Fibonacci numbers F(n+1) = F(n) + F(n-1)); A006190, A052924, A006497 (a(n+1) = 3*a(n) + a(n-1)); A000129 (Pell numbers a(n+1) = 2*a(n) + a(n-1)).

Programs

  • PARI
    for(n=1+#a=[0,1,2,3,5,13],#a=Vec(a,30),a[n]=a[n-1]*3+a[n-2]);a \\ Remove initial 0 to get a[1] = 1 etc.
apply( {A332060(n)=if(n>3,[5,13]*([0,1;1,3]^(n-4))[,1],n)}, [0..20])

Formula

G.f.: x*(1 - x - 4*x^2 - 6*x^3 - 5*x^4)/(1 - 3*x - x^2).

A107855 a(n) = 2*a(n-2)+4*a(n-4)+a(n-6), n>11.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 2, 3, 1, 10, 9, 33, 24, 109, 85, 360, 275, 1189, 914, 3927, 3013, 12970, 9957, 42837, 32880, 141481, 108601, 467280, 358679, 1543321, 1184642, 5097243, 3912601, 16835050, 12922449, 55602393, 42679944, 183642229, 140962285
Offset: 1

Views

Author

Roger L. Bagula, Jun 12 2005

Keywords

Links

Programs

  • Mathematica
    F[1] = 1; F[2] = 1; F[3] = 1; F[4] = 0; F[n__] := F[n] = If[ Mod[n, 2] == 0, -3*F[n - 1] + 3*F[n - 3] + F[n - 4], F[n - 1] + F[n - 2]] a = Table[Abs[F[n]], {n, 1, 50}]
Join[{1,1,1,0,1},LinearRecurrence[{0,2,0,4,0,1},{1,2,3,1,10,9},50]] (* Harvey P. Dale, Oct 18 2013 *)

Formula

lim_{ n->inf} a(n)/a(n-1) alternating between 0.767592... and 4.30278...
G.f.: x*(2*x^10+8*x^8+5*x^6+3*x^5+5*x^4+x^2-1+2*x^3-x)/( (x^2+1)*(x^4+3*x^2-1)). [ Sep 28 2009]
a(2n) = A006190(n-2), a(2n+1) = (1/3)*[A052924(n-2) - 4*(-1)^n], n>2.

Extensions

Definition replaced by recurrence by the Associate Editors of the OEIS, Sep 28 2009

A183189 Triangle T(n,k), read by rows, given by (2, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 2, 0, 6, 1, 0, 18, 5, 0, 0, 54, 21, 1, 0, 0, 162, 81, 8, 0, 0, 0, 486, 297, 45, 1, 0, 0, 0, 1458, 1053, 216, 11, 0, 0, 0, 0, 4374, 3645, 945, 78, 1, 0, 0, 0, 0, 13122, 12393, 3888, 450, 14, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Philippe Deléham, Dec 14 2011

Keywords

Comments

Riordan array ((1-x)/(1-3x), x^2/(1-3x)).
A skewed version of triangular array in A193723.
A202209*A007318 as infinite lower triangular matrices.

Examples

			Triangle begins:
  1
  2, 0
  6, 1, 0
  18, 5, 0, 0
  54, 21, 1, 0, 0
  162, 81, 8, 0, 0, 0
  486, 297, 45, 1, 0, 0, 0

Crossrefs

Cf. A000244, A025192, A081038, A183188 (antidiagonal sums).

Formula

G.f.: (1-x)/(1-3*x-y*x^2).
T(n,k) = Sum_{j, j>=0} T(n-2-j,k-1)*3^j.
T(n,k) = 3*T(n-1,k) + T(n-2,k-1).
Sum_{k, 0<=k<=n} T(n,k)*x^k = A057682(n+1), A000079(n), A122367(n), A025192(n), A052924(n), A104934(n), A202206(n), A122117(n), A197189(n) for x = -3, -2, -1, 0, 1, 2, 3, 4, 5 respectively.
Previous Showing 11-15 of 15 results.

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