0,2

Number of moves to solve Chinese rings puzzle.

a(n-1) (with a(0):=0) enumerates all sequences of length m=1,2,...,floor(n/2) with nonzero integer entries n_i satisfying sum |n_i| <= n-m. Rephrasing K. A. Meissner's example in arXiv:gr-qc/0407052v1, p. 6. Example n=4: from length m=1: [1], [2], [3], each in 2 signed versions; from m=2: [1,1] in 2^2=4 signed versions. Hence a(3)=a(4-1)=3*2+1*4=10.

Richard I. Hess, Compendium of Over 7000 Wire Puzzles, privately printed, 1991.

Richard I. Hess, Analysis of Ring Puzzles, booklet distributed at 13-th International Puzzle Party, Amsterdam, Aug 20 1993.

Lee Hae-hwang, Illustration of initial terms in terms of rosemary plants

a(2k) = 2*a(2k-1), a(2k+1) = 2*a(2k)+2. - Peter Shor, Apr 11, 2002

For n>0: if n mod 2 = 0 then (2^(n+2)-4)/3 else (2^(n+2)-2)/3. - Richard Hess.

a(2n) = 2n-1 + Sum_{k = 1 to 2n-1} a(n); a(2n+1)= 2n-1 + Sum_{k=1 to 2n} a(n). - Lee Hae-hwang, Sep 17, 2002

a(n)=2n + 2*Sum_{k = 1 to n-2} a(n). - Lee Hae-hwang (mathmaniac(AT)empal.com), Sep 19 2002

G.f.: (1-x^2+2x^3)/((1-x)(1-x-2x^2)); a(n)=J(n+2)-1+0^n, where J(n)=A001045(n); a(n)=2a(n-1)+a(n-2)-2a(n-3); a(n)=0^n+sum{k=0..n, (2-2*0^(n-k))J(k+1)}; - Paul Barry (pbarry(AT)wit.ie), Oct 24 2007

f:=n-> if n mod 2 = 0 then (2^(n+2)-4)/3 else (2^(n+2)-2)/3; fi;

a(n) = T(n,0) + T(n,1) + ... + T(n,n), T given by A026637.

For n >= 1, equals twice A000975, also A001045 - 1.

For n >= 1, a(n+1) = a(n) + 2*b(n+1) + 4*b(n), where b(k) = A001045(k).

Sequence in context: A099413 A127392 A004647 this_sequence A026666 A121880 A094536

Adjacent sequences: A026641 A026642 A026643 this_sequence A026645 A026646 A026647

nonn

Clark Kimberling (ck6(AT)evansville.edu)

Recurrence in definition line found by Lee Hae-hwang (mathmaniac(AT)empal.com), Apr 03, 2002