A102699 Number of strings of length n, using as symbols numbers from the set {1, 2, ..., n}, in which consecutive symbols differ by exactly 1.

1, 1, 2, 6, 16, 42, 104, 252, 592, 1370, 3112, 6996, 15536, 34244, 74832, 162616, 351136, 754938, 1615208, 3443940, 7314928, 15493676, 32714992, 68918856, 144815456, 303703972, 635554064, 1327816392, 2769049312, 5766417480, 11989472672, 24897569648

Offset

0,3

Comments

Equally, number of different n-digit numbers, using only the digits 1 through n, where consecutive digits differ by 1. It is assumed that there are n different digits available even when n > 9.

Number of endomorphisms of a path P_n. - N. J. A. Sloane, Sep 20 2009

a(n) is also the number of distinct paths of length n starting from the bottom row of an n X n chess board and ending at the top row, such that all the n squares traversed in the path are of the same color. - Kiran Ananthpur Bacche, Oct 25 2022

Links

Alois P. Heinz, Table of n, a(n) for n = 0..3311 (terms n = 1..300 from T. D. Noe)

Sr. Arworn, An algorithm for the number of endomorphisms on paths, Disc. Math., 309 (2009), 94-103 (see p. 95).

Zhicong Lin and Jiang Zeng, On the number of congruence classes of paths, arXiv preprint arXiv:1112.4026 [math.CO], 2011.

M. A. Michels and U. Knauer, The congruence classes of paths and cycles, Discr. Math., 309 (2009), 5352-5359. See p. 5356. [From N. J. A. Sloane, Sep 20 2009]

Joseph Myers, BMO 2008-2009 Round 1 Problem 1-Generalisation

Formula

It appears that the limit of a(n)/a(n-1) is decreasing towards 2. - Ben Paul Thurston, Oct 04 2006

a(n) = (n+1)2^(n-1) - 4(n-1)binomial(n-2,(n-2)/2) for n even, a(n) = (n+1)2^(n-1) - (2n-1)binomial(n-1,(n-1)/2) for n odd. - Joseph Myers, Dec 23 2008

a(n) = 2 * Sum_{k=1..n-1} k*A110971(n,k). - N. J. A. Sloane, Sep 20 2009

G.f.: x * (2*(1 - x) - sqrt(1 - 4*x^2)) / (1 - 2*x)^2. - Michael Somos, Mar 17 2014

0 = a(n) * 8*n^2 - a(n+1) * 4*(n^2 - 2*n - 1) - a(n+2) * 2*(n^2 + 3*n - 2) + a(n+3) * (n-1)*(n+2) for n>0. - Michael Somos, Mar 17 2014

0 = a(n) * (16*a(n+1) - 16*a(n+2) + 4*a(n+3)) + a(n+1) * (-16*a(n+1) + 20*a(n+2) - 4*a(n+3)) + a(n+2) * (-4*a(n+2) + a(n+3)) for n>0. - Michael Somos, Mar 17 2014

Example

For example, a(4)=16: the 16 strings are 1212, 1232, 1234, 2121, 2123, 2321, 2323, 2343, 3212, 3232, 3234, 3432, 3434, 4321, 4323, 4343.
G.f. = x + 2*x^2 + 6*x^3 + 16*x^4 + 42*x^5 + 104*x^6 + 252*x^7 + 592*x^8 + ...

Maple

p:= 0; paths := proc(m, n, s, t) global p; if(((t+1) <= m) and s <= (n)) then paths(m, n, s+1, t+1); end if; if(((t-1) > 0) and s <= (n)) then paths(m, n, s+1, t-1); end if; if(s = n) then p:=p+1; end if; end proc; sumpaths:=proc(j) global p; p:=0; sp:=0; for h from 1 to j do p:=0; paths(j, j, 1, h); sp:=sp+ p ; end do; sp; end proc; for l from 1 to 50 do sumpaths(l); end do; # Ben Paul Thurston, Oct 04 2006
# second Maple program:
a:= proc(n) option remember;
      `if`(n<5, [1, 1, 2, 6, 16][n+1], ((2*n^2-6*n-4) *a(n-1)
      +(56-32*n+4*n^2) *a(n-2) -8*(n-3)^2 *a(n-3))/ ((n-1)*(n-4)))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 23 2012

Mathematica

a[n_] := a[n] = If[n <= 4, n*((n-3)*n+4)/2, ((2*n^2 - 6*n - 4)*a[n-1] + (4*n^2 - 32*n + 56)*a[n-2] - 8*(n-3)^2*a[n-3])/((n-1)*(n-4))]; Table[ a[n], {n, 1, 30}] (* Jean-François Alcover, Nov 10 2015, after Alois P. Heinz *)

Prog

(PARI) x='x+O('x^55); Vec(x*(2*(1-x)-sqrt(1-4*x^2))/(1-2*x)^2) \\ Altug Alkan, Nov 10 2015

Crossrefs

Cf. A110971, A152086.

Main diagonal of A220062. - Alois P. Heinz, Dec 03 2012

Sequence in context: A178438 A365548 A143123 * A266124 A217194 A304662

Adjacent sequences: A102696 A102697 A102698 * A102700 A102701 A102702

Keyword

nonn

Author

Don Rogers (donrogers42(AT)aol.com), Feb 07 2005

Extensions

More terms from Ben Paul Thurston, Oct 04 2006

a(20) onwards from David Wasserman, Apr 26 2008

Edited by N. J. A. Sloane, Jan 03 2009 and Sep 23 2010

a(0)=1 prepended by Alois P. Heinz, Apr 17 2017

Status

approved