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.

A008577 Crystal ball sequence for planar net 4.8.8.

Original entry on oeis.org

1, 4, 9, 17, 28, 41, 57, 76, 97, 121, 148, 177, 209, 244, 281, 321, 364, 409, 457, 508, 561, 617, 676, 737, 801, 868, 937, 1009, 1084, 1161, 1241, 1324, 1409, 1497, 1588, 1681, 1777, 1876, 1977, 2081, 2188, 2297, 2409, 2524, 2641, 2761, 2884, 3009, 3137, 3268
Offset: 0

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Author

Keywords

Examples

			G.f. = 1 + 4*x + 9*x^2 + 17*x^3 + 28*x^4 + 41*x^5 + 67*x^6 + ... - _Michael Somos_, May 02 2020
		

Crossrefs

Partial sums of A008576.

Programs

Formula

G.f.: ((1+x)^2*(1+x^2)) / ((1-x)^3*(1+x+x^2)). - Ralf Stephan, Apr 24 2004
a(n) = 4*(n/3)*(n+1)+10/9+A099837(n+2)/9. - R. J. Mathar, Nov 20 2010
The above g.f. and formula were originally stated as conjectures, but I now have a proof. This also justifies the b-file. Details will be added later. - N. J. A. Sloane, Dec 29 2015
From Michael Somos, May 02 2020: (Start)
Euler transform of length 3 sequence [4, -1, 1, -1].
a(n) = a(-1-n) = floor((n^2+n+1)*4/3) for all n in Z.
a(n) - 2*a(n+1) + a(n+2) = A164359(n) unless n=0.
(End)

A167613 Array T(n,k) read by antidiagonals: the k-th term of the n-th difference of A131531.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 0, -1, -2, -3, 0, 0, 1, 3, 6, -1, -1, -1, -2, -5, -11, 0, 1, 2, 3, 5, 10, 21, 0, 0, -1, -3, -6, -11, -21, -42, 1, 1, 1, 2, 5, 11, 22, 43, 85, 0, -1, -2, -3, -5, -10, -21, -43, -86, -171, 0, 0, 1, 3, 6, 11, 21, 42, 85, 171, 342, -1, -1, -1, -2, -5, -11, -22, -43, -85, -170, -341, -683, 0, 1, 2, 3, 5, 10, 21, 43, 86, 171, 341, 682, 1365
Offset: 0

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Author

Paul Curtz, Nov 07 2009

Keywords

Comments

The array contains A131708(0) in diagonal 0, then -A024495(0..1) in diagonal 1, then A024493(0..2) in diagonal 2, then -A131708(0..3), then A024495(0..4), then -A024493(0..5).

Examples

			The table starts in row n=0 with columns k >= 0 as:
0, 0, 1, 0, 0, -1, 0, 0, 1, 0, 0, -1, 0, 0, 1, 0, 0, -1, 0, 0, 1, 0, 0 A131531
0, 1, -1, 0, -1, 1, 0, 1, -1, 0, -1, 1, 0, 1, -1, 0, -1, 1, 0, 1, -1 A092220
1, -2, 1, -1, 2, -1, 1, -2, 1, -1, 2, -1, 1, -2, 1, -1, 2, -1, 1, -2 A131556
-3, 3, -2, 3, -3, 2, -3, 3, -2, 3, -3, 2, -3, 3, -2, 3, -3, 2, -3 A164359
6, -5, 5, -6, 5, -5, 6, -5, 5, -6, 5, -5, 6, -5, 5, -6, 5, -5, 6, -5
-11, 10, -11, 11, -10, 11, -11, 10, -11, 11, -10, 11, -11, 10, -11
21, -21, 22, -21, 21, -22, 21, -21, 22, -21, 21, -22, 21, -21, 22
		

Crossrefs

Cf. A167617 (antidiagonal sums).

Programs

  • Maple
    A131531 := proc(n) op((n mod 6)+1,[0,0,1,0,0,-1]) ; end proc:
    A167613 := proc(n,k) option remember; if n= 0 then A131531(k); else procname(n-1,k+1)-procname(n-1,k) ; end if;end proc: # R. J. Mathar, Dec 17 2010
  • Mathematica
    nmax = 13;
    A131531 = Table[{0, 0, 1, 0, 0, -1}, {nmax}] // Flatten;
    T[n_] := T[n] = Differences[A131531, n];
    T[n_, k_] := T[n][[k]];
    Table[T[n-k, k], {n, 1, nmax}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Oct 20 2023 *)

Formula

T(0,k) = A131531(k). T(n,k) = T(n-1,k+1) - T(n-1,k), n > 0.
T(n,n) = A001045(n). T(n,n+1) = -A001045(n). T(n,n+2) = A078008(n).
T(n,0) = -T(n,3) = (-1)^(n+1)*A024495(n).
T(n,1) = (-1)^(n+1)*A131708(n).
T(n,2) = (-1)^n*A024493(n).
T(n,k+6) = T(n,k).
a(n) = A131708(0), -A024495(0,1), A024493(0,1,2), -A131708(0,1,2,3), A024495(0,1,2,3,4), -A024493(0,1,2,3,4,5).

A244893 a(n) = a(n-a(n-1)) with initial values 2,3,2.

Original entry on oeis.org

2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3
Offset: 1

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Author

N. J. A. Sloane, Jul 17 2014

Keywords

Comments

After first term, periodic with period {3,2,3}.

Programs

  • Maple
    f:=proc(n,o,S) option remember; # o = offset, S = init. terms
    L:=nops(S);
    if n < 0 then 0
    elif n <= o+L-1 then S[n-o+1]
    else f(n-f(n-1,o,S),o,S); fi; end;
    [seq(f(n,1,[2,3,2]),n=1..100)];
  • Mathematica
    Join[{2},LinearRecurrence[{0, 0, 1},{3, 2, 3},75]] (* Ray Chandler, Aug 25 2015 *)

Formula

a(n) = A164359(n), n>1. - R. J. Mathar, Jul 25 2014
Showing 1-3 of 3 results.