A014552 -id:A014552 - cp's OEIS frontend

A014601 Numbers congruent to 0 or 3 mod 4.

0, 3, 4, 7, 8, 11, 12, 15, 16, 19, 20, 23, 24, 27, 28, 31, 32, 35, 36, 39, 40, 43, 44, 47, 48, 51, 52, 55, 56, 59, 60, 63, 64, 67, 68, 71, 72, 75, 76, 79, 80, 83, 84, 87, 88, 91, 92, 95, 96, 99, 100, 103, 104, 107, 108, 111, 112, 115, 116, 119, 120, 123, 124

Offset: 0

Eric Rains (rains(AT)caltech.edu)

Discriminants of orders in imaginary quadratic fields (negated). [Comment corrected by Christopher E. Thompson, Dec 11 2016]
Numbers such that Langford-Skolem problem has a solution - see A014552.
Complement of A042963. - Reinhard Zumkeller, Oct 04 2004
Also called skew amenable numbers; a number k is skew amenable if there exist a set {a(i)} of integers satisfying the relations k = Sum_{i=1..k} a(i) = -Product_{i=1..k} a(i). Thus we have 8 = 1 + 1 + 1 + 1 + 1 + 1 - 2 + 4 = -(1*1*1*1*1*1*(-2)*4). - Lekraj Beedassy, Jan 07 2005
Possible nonpositive discriminants of quadratic equation a*x^2 + b*x + c or discriminants of binary quadratic forms a*x^2 + b*x*y + c*y^2. - Artur Jasinski, Apr 28 2008
Also, disregarding the 0 term, positive integers m such that, equivalently,
  (i) +-1 +-2 +-... +-m is even for all choices of signs,
  (ii) +-1 +-2 +-... +-m = 0 for some choices of signs,
  (iii) for all -m <= k <= m, k = +-1 +-2 +-... +-(k-1) +-(k+1) +-(k+2) +-... +-m for at least one choice of signs. - Rick L. Shepherd, Oct 29 2008
A145768(a(n)) is even. - Reinhard Zumkeller, Jun 05 2012
Multiples of 4 interleaved with 1 less than multiples of 4. - Wesley Ivan Hurt, Nov 08 2013
((2*k+0) + (2*k+1) + ... + (2*k+m-1) + (2*k+m)) is even if and only if m = a(n) for some n where k is any nonnegative integer. - Gionata Neri, Jul 24 2015
Numbers whose binary reflected Gray code (A014550) ends with 0. - Amiram Eldar, May 17 2021

			G.f. = 3*x + 4*x^2 + 7*x^3 + 8*x^4 + 11*x^5 + 12*x^6 + 15*x^7 + 16*x^8 + ...

H. Cohen, Course in Computational Alg. No. Theory, Springer, 1993, pp. 514-5.
A. Scholz and B. Schoeneberg, Einführung in die Zahlentheorie, 5. Aufl., de Gruyter, Berlin, New York, 1973, p. 108.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
S. F. Barger, Solution to problem 10454: Amenable Numbers, Amer. Math. Monthly, Vol. 105, No. 4 (April 1998), p. 368.
Steven R. Finch, Class number theory [Cached copy, with permission of the author]
Heiko Harborth, Solution of Steinhaus's problem with plus and minus signs, Journal of Combinatorial Theory, Series A, Volume 12, Issue 2 (March 1972), Pages 253-259.
Mickaël Launay, Les routes de Numland, L’énigme maths du "Monde" n°2. In French.
Michael Penn, The most beautiful arrangement of numbers, YouTube video, 2024.
Rick L. Shepherd, Binary quadratic forms and genus theory, Master of Arts Thesis, University of North Carolina at Greensboro, 2013.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A004676, A014550, A042948, A079896.

Cf. A274406. - Bruno Berselli, Jun 26 2016

a014601 n = a014601_list !! n
a014601_list = [x | x <- [0..], mod x 4 `elem` [0, 3]]
-- Reinhard Zumkeller, Jun 05 2012

[n: n in [0..200]|n mod 4 in {0,3}]; // Vincenzo Librandi, Dec 24 2010

A014601:=n->3*n-2*floor(n/2); seq(A014601(k), k=0..100); # Wesley Ivan Hurt, Nov 08 2013

aa = {}; Do[Do[Do[d = b^2 - 4 a c; If[d <= 0, AppendTo[aa, -d]], {a, 0, 50}], {b, 0, 50}], {c, 0, 50}]; Union[aa] (* Artur Jasinski, Apr 28 2008 *)
Select[Range[0, 124], Or[Mod[#, 4] == 0, Mod[#, 4] == 3] &] (* Ant King, Nov 18 2010 *)
CoefficientList[Series[2 x/(1 - x)^2 + (1/(1 - x) + 1/(1 + x)) x/2, {x, 0, 100}], x] (* Vincenzo Librandi, May 18 2014 *)
a[ n_] := 2 n + Mod[n, 2]; (* Michael Somos, Jul 24 2015 *)

{a(n) = 2*n + n%2}; /* Michael Somos, Dec 27 2010 */

a(n) = (n + 1)*2 + 1 - n mod 2. - Reinhard Zumkeller, Apr 21 2003
A014494(n) = A000217(a(n)). - Reinhard Zumkeller, Oct 04 2004
a(n) = Sum_{k=1..n} (2 - (-1)^k). - William A. Tedeschi, Mar 20 2008
A139131(a(n)) = A078636(a(n)). - Reinhard Zumkeller, Apr 10 2008
From R. J. Mathar, Sep 25 2009: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3) for n > 2.
G.f.: x*(3+x)/((1+x)*(x-1)^2). (End)
a(n) = 2*n + (n mod 2). - Paolo Valzasina (p.valzasina(AT)gmail.com), Nov 24 2009
a(n) = (4*n - (-1)^n + 1)/2. - Bruno Berselli, Oct 06 2010
a(n) = 4*n - a(n-1) - 1 (with a(0) = 0). - Vincenzo Librandi, Dec 24 2010
a(n) = -A042948(-n) for all n in Z. - Michael Somos, Dec 27 2010
G.f.: 2*x / (1 - x)^2 + (1 / (1 - x) + 1 / (1 + x)) * x/2. - Michael Somos, Dec 27 2010
a(n) = Sum_{k>=0} A030308(n,k)*b(k) with b(0) = 3 and b(k) = 2^(k+1) for k > 0. - Philippe Deléham, Oct 17 2011
a(n) = ceiling((4/3)*ceiling(3*n/2)). - Clark Kimberling, Jul 04 2012
a(n) = 3n - 2*floor(n/2). - Wesley Ivan Hurt, Nov 08 2013
a(n) = A042948(n+1) - 1 for all n in Z. - Michael Somos, Jul 24 2015
a(n) + a(n+1) = A004767(n) for all n in Z. - Michael Somos, Jul 24 2015
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*log(2)/4 - Pi/8. - Amiram Eldar, Dec 05 2021
E.g.f.: ((4*x + 1)*exp(x) - exp(-x))/2. - David Lovler, Aug 04 2022

A026272 a(n) = smallest k such that k=a(n-k-1) is the only appearance of k so far; if there is no such k, then a(n) = least positive integer that has not yet appeared.

Original entry on oeis.org

1, 2, 1, 3, 2, 4, 5, 3, 6, 7, 4, 8, 5, 9, 10, 6, 11, 7, 12, 13, 8, 14, 15, 9, 16, 10, 17, 18, 11, 19, 20, 12, 21, 13, 22, 23, 14, 24, 15, 25, 26, 16, 27, 28, 17, 29, 18, 30, 31, 19, 32, 20, 33, 34, 21, 35, 36, 22, 37, 23, 38, 39, 24, 40, 41, 25

Offset: 1

Clark Kimberling

From Daniel Joyce, Apr 13 2001: (Start)
This sequence displays every positive integer exactly twice and the gap between the two occurrences of n contains exactly n other values. The first occurrence of n precedes the first occurrence of n+1.
Also related to the Wythoff array (A035513) and the Para-Fibonacci sequence (A035513) where every positive integer is displayed exactly once in the whole array. Take any integer n in A026272 and let C = number of terms from the beginning of the sequence to the second occurrence of n. Then C = (2nd term after n in the applicable sequence for n in A035513).
Also in the second occurrence of n in A026272, let N=n ( - one term) = (first term value after n in the applicable sequence for n in A035513). In this format the second occurrence of n in A026272 will produce in A035513, n itself and two of the succeeding terms of n in the Wythoff array where every positive integer can only be displayed once.
In A026272 if |a(n)-a(n+1)| > 10 then phi ~ a(n)/|a(n)-a(n+1)|. When n -> infinity it will converge to phi. (End)
Or, put a copy of n in A000027 n places further along! - Zak Seidov, May 24 2008
Another version would prefix this sequence with two leading 0's (see the Angelini reference). If we use this form and write down the indices of the two 0's, the two 1's, the two 2's, the two 3's, etc., then we get A072061. - Jacques ALARDET, Jul 26 2008

Eric Angelini, "Jeux de suites", in Dossier Pour La Science, pp. 32-35, Volume 59 (Jeux math'), April/June 2008, Paris.

Zak Seidov, Table of n, a(n) for n = 1..1000
Jon Asier Bárcena-Petisco, Luis Martínez, María Merino, Juan Manuel Montoya, and Antonio Vera-López, Fibonacci-like partitions and their associated piecewise-defined permutations, arXiv:2503.19696 [math.CO], 2025. See p. 3.
Jeffrey Shallit, The Hurt-Sada Array and Zeckendorf Representations, arXiv:2501.08823 [math.NT], 2025. See p. 10.

Cf. A000027, A035513, A014552, A176127.

s=Range[1000];n=0;Do[n++;s=Insert[s,n,Position[s,n][[1]]+n+1],{500}];A026272=Take[s,1000] (* Zak Seidov, May 24 2008 *)

A026272=apply(t->t-1,A026242[3..-1]) \\ Use vecextract(A026242,"3..") in PARI versions < 2.7. - M. F. Hasler, Sep 17 2014

from collections import Counter
from itertools import count, islice
def agen(): # generator of terms
    aset, alst, k, mink, counts = set(), [0], 0, 1, Counter()
    for n in count(1):
        for k in range(1, len(alst)-1):
            if k == alst[n-k-1] and counts[alst[n-k-1]] == 1: an = k; break
        else: an = mink
        yield an; aset.add(an); alst.append(an); counts.update([an])
        while mink in aset: mink += 1
print(list(islice(agen(), 66))) # Michael S. Branicky, Jun 27 2022

a(n) = A026242(n+2) - 1 = A026350(n+3) - 2 = A026354(n+4) - 3.

Edited by Max Alekseyev, May 31 2011

A002047 Number of 3 X (2n+1) zero-sum arrays with entries -n,...,0,...,n.

Original entry on oeis.org

1, 2, 6, 28, 244, 2544, 35600, 659632, 15106128, 425802176, 14409526080, 577386122880

Offset: 0

N. J. A. Sloane

This can be interpreted as the number of ways to choose 2n+1 cells in a hexagonal grid of side n+1 such that no two are in the same row or left diagonal or right diagonal. - Alex Fink (a00(AT)shaw.ca), Mar 16 2005
Also the number of transversals of a partial Latin square L of order 2n+1 in which L_{ij} = i+j if n+1 < i+j < 3n+3 and L_{ij} is empty otherwise. [Cavenagh-Wanless]
Also the number of arrangements of the numbers n+1, n+1, ..., 3n+1, 3n+1 such that there are n numbers between the pair of n+1's, ..., 3n numbers between the pair of 3n+1's. For each of these arrangements and its mirror image, there is a bijection with a pair of the 3 X (2n+1) zero-sum arrays. - Stephen J Scattergood, Jul 19 2013
Also the number of sigma-permutations of length 2n+1 [Kotzig-Laufer]. - N. J. A. Sloane, Jul 27 2015
An (m,2n+1)-zero-sum array is an m X (2n+1) matrix whose m rows are permutations of the 2n+1 integers -n..n, the sum of each column is zero and the first row of the matrix is -n,-n+1,...,0,...,n-1,n. - Gheorghe Coserea, Dec 29 2016
a(n-1) is also number of ways of placing 2*n-1 nonattacking rooks on a hexagonal board with edge-length n in Glinski's hexagonal chess. - Vaclav Kotesovec, Aug 15 2019

			a(2) = 6 corresponds to
..O.X.X.......X.X.O.......O.X.X.......X.O.X.......X.O.X.......X.X.O
.X.X.O.X.....X.O.X.X.....X.X.X.O.....X.X.X.O.....O.X.X.X.....O.X.X.X
X.X.X.X.O...O.X.X.X.X...X.O.X.X.X...O.X.X.X.X...X.X.X.X.O...X.X.X.O.X
.O.X.X.X.....X.X.X.O.....X.X.X.O.....X.O.X.X.....X.X.O.X.....O.X.X.X
..X.O.X.......X.O.X.......O.X.X.......X.X.O.......O.X.X.......X.X.O
The bijection with a pair of the 3 X (2n+1) zero-sum arrays:
n=1, a(1)=2 corresponds to
                           3 4 2 3 2 4
        and mirror image   4 2 3 2 4 3
element                  2  3  4  -(2n+1) --> -1  0  1
position, left element   3  1  2  -( n+1) -->  1 -1  0
position  in mirror      2  3  1  -( n+1) -->  0  1 -1
                          -------               -------
sum of column            7  7  7  -(4n+3)      0  0  0
Swapping rows 2,3 yields the other 3 X 3 zero sum array.
n=2, a(2)=6  an example and its mirror, so 2 of the 6 solutions:
                           5 6 7 3 4 5 3 6 4 7
            mirror image   7 4 6 3 5 4 3 7 6 5
            3  4  5  6  7  -(2n+1) --> -2 -1  0  1  2
            4  5  1  2  3  -( n+1) -->  1  2 -2 -1  0
            4  2  5  3  1  -( n+1) -->  1 -1  2  0 -2
            --------------              --------------
           11 11 11 11 11  -(4n+3) -->  0  0  0  0  0
Swapping rows 2,3 yields the other 3 X 5 zero sum array.

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

C. Bebeacua, T. Mansour, A. Postnikov and S. Severini, On the X-rays of permutations, arXiv:math/0506334 [math.CO], 2005.
B. T. Bennett and R. B. Potts, Arrays and brooks, J. Austral. Math. Soc., 7 (1967), 23-31.
B. T. Bennett and R. B. Potts, Arrays and brooks, J. Austral. Math. Soc., 7 (1967), 23-31. [Annotated scanned copy]
N. J. Cavenagh and I. M. Wanless, On the number of transversals in Cayley tables of cyclic groups, Disc. Appl. Math. 158 (2010), 136-146.
Gheorghe Coserea, Solutions for n=4.
Gheorghe Coserea, Solutions for n=5.
Gheorghe Coserea, MiniZinc model for generating solutions.
Diane Donovan, Asha Rao, Elif Üsküplü, and E. Ş. Yazıcı, QC-LDPC Codes from Difference Matrices and Difference Covering Arrays, arXiv:2205.00563 [math.CO], 2022.
A. Kotzig and P. J. Laufer, When are permutations additive?, Amer. Math. Monthly, 85 (1978), 364-365.
A. Kotzig and P. J. Laufer, When are permutations additive?, Amer. Math. Monthly, 85 (1978), 364-365. [Annotated by C. L. Mallows, scanned copy, together with letter from C. L. Mallows and N. J. A. Sloane to A. Kotzig, Jul 25 1978]
Wikipedia, Hexagonal chess - Gliński's hexagonal chess

Cf. A014552. A diagonal of the triangle in A260333.

Cf. A309260, A309746.

More terms from Alex Fink (a00(AT)shaw.ca), Mar 16 2005

a(10) and a(11) from Ian Wanless, Jul 30 2010, from the Cavenagh-Wanless paper

A004075 Number of Skolem sequences of order n.

Original entry on oeis.org

1, 0, 0, 6, 10, 0, 0, 504, 2656, 0, 0, 455936, 3040560, 0, 0, 1400156768, 12248982496, 0, 0, 11435578798976, 123564928167168, 0, 0, 204776117691241344, 2634563519776965376, 0, 0, 7064747252076429464064, 105435171495207196553472, 0, 0

Offset: 1

N. J. A. Sloane

Number of permutations of the multiset {1,1,2,2,...,n,n} such that the distance between the elements i equals i for every i=1,2,...,n.

Number of super perfect rhythmic tilings of [0,2n-1] with pairs. See A285698 and A285527 for the definition and tilings of triples and quadruples. - Tony Reix, Apr 25 2017

CRC Handbook of Combinatorial Designs, 1996, p. 460.

Ali Assarpour, Amotz Bar-Noy, Ou Liuo, Counting the Number of Langford Skolem Pairings, arXiv:1507.00315 [cs.DM], 2015.
S. Burrill and L. Yen, Constructing Skolem sequences via generating trees, arXiv preprint arXiv:1301.6424 [math.CO], 2013.
J. E. Miller, Langford's Problem
G. Nordh, Perfect Skolem sequences, arXiv:math/0506155 [math.CO], 2005.

Cf. A014552, A059106, A176127, A268537.

(* Program not suitable to compute a large number of terms. *)
iter[n_] := Sequence @@ Table[{x[i], {-1, 1}}, {i, 1, 2n}];
a[n_] := 1/2^(2n) Sum[Product[x[i], {i, 1, 2n}] Product[Sum[x[k] x[k+i], {k, 1, 2n-i}], {i, 1, n}], iter[n] // Evaluate];
Table[Print[a[n]]; a[n], {n, 1, 10}] (* Jean-François Alcover, Sep 29 2018, from formula in Assarpour et al. *)

For n > 1, a(n) = A059106(n)*2 because A059106 ignores reflected solutions. - Martin Fuller, Mar 08 2007

More terms (via A059106) from Martin Fuller, Mar 08 2007
Extended using results from the Assarpour et al. (2015) paper by N. J. A. Sloane, Feb 22 2016 at the suggestion of William Rex Marshall
a(28)-a(31) from Assarpour et al. (2015), added by Max Alekseyev, Sep 24 2023

A059106 Number of solutions to Nickerson variant of Langford (or Langford-Skolem) problem.

Original entry on oeis.org

1, 0, 0, 3, 5, 0, 0, 252, 1328, 0, 0, 227968, 1520280, 0, 0, 700078384, 6124491248, 0, 0, 5717789399488, 61782464083584, 0, 0, 102388058845620672, 1317281759888482688, 0, 0, 3532373626038214732032, 52717585747603598276736, 0, 0

Offset: 1

N. J. A. Sloane, Feb 14 2001

How many ways are of arranging the numbers 1,1,2,2,3,3,...,n,n so that there are zero numbers between the two 1's, one number between the two 2's, ..., n-1 numbers between the two n's?
For n > 1, a(n) = A004075(n)/2 because A004075 also counts reflected solutions. - Martin Fuller, Mar 08 2007
Because of symmetry, is a(5) = 5 the largest prime in this sequence? - Jonathan Vos Post, Apr 02 2011

			For n=4 the a(4)=3 solutions, up to reversal of the order, are:
1 1 3 4 2 3 2 4
1 1 4 2 3 2 4 3
2 3 2 4 3 1 1 4
From _Gheorghe Coserea_, Aug 26 2017: (Start)
For n=5 the a(5)=5 solutions, up to reversal of the order, are:
1 1 3 4 5 3 2 4 2 5
1 1 5 2 4 2 3 5 4 3
2 3 2 5 3 4 1 1 5 4
2 4 2 3 5 4 3 1 1 5
3 5 2 3 2 4 5 1 1 4
(End)

Ali Assarpour, Amotz Bar-Noy, Ou Liuo, Counting the Number of Langford Skolem Pairings, arXiv:1507.00315 [cs.DM], 2015.
Gheorghe Coserea, Solutions for n=8.
Gheorghe Coserea, Solutions for n=9.
J. E. Miller, Langford's Problem
R. S. Nickerson and D. C. B. Marsh, E1845: A variant of Langford's Problem, American Math. Monthly, 1967, 74, 591-595.
Zan Pan, Conjectures on the number of Langford sequences, (2021).

Cf. A014552, A050998, A059107, A059108.

Cf. A004075, A268535.

a(20)-a(23) from Mike Godfrey (m.godfrey(AT)umist.ac.uk), Mar 14 2002
Extended using results from the Assarpour et al. (2015) paper by N. J. A. Sloane, Feb 22 2016 at the suggestion of William Rex Marshall
a(28)-a(31) from Assarpour et al. (2015), added by Max Alekseyev, Sep 24 2023

A176127 The number of permutations of {1,2,...,n,1,2,...,n} with the property that there are k numbers between the two k's in the set for k=1,...,n.

Original entry on oeis.org

0, 0, 2, 2, 0, 0, 52, 300, 0, 0, 35584, 216288, 0, 0, 79619280, 653443600, 0, 0, 513629782560, 5272675722400, 0, 0, 7598911885030976, 93690316113031872, 0, 0, 223367222197529806464, 3214766521218764786304, 0, 0

Offset: 1

Andrew McFarland, Apr 09 2010

			a(1)=0; a(2)=0; a(3)=a(4)=2 since {{2,3,1,2,1,3},{3,1,2,1,3,2}} and {{4,1,3,1,2,4,3,2},{2,3,4,2,1,3,1,4}} are the only ways to permute {1,2,3,1,2,3} and {1,2,3,4,1,2,3,4}, respectively, such that there is one number between the 1's, two numbers between the 2's,..., n numbers between the n's.

For references, see A014552.

Ali Assarpour, Amotz Bar-Noy, Ou Liuo, Counting the Number of Langford Skolem Pairings, arXiv:1507.00315 [cs.DM], 2015.

Cf. A014552, A059106, A004075, A264813, A268536.

a=lambda n:sum(1 for i in DLXCPP([(i-1,j+n,i+j+n+1)for i in[1..n]for j in[0..n+n-i-2]]+[(i,)for i in[n..n+n-1]]))if n%4 in[0,3] else 0
# Tomas Boothby, Jun 14 2013

a(n) = 2 * A014552(n).

Edited and more terms added from A014552 by Max Alekseyev, May 31 2011, May 19 2015

Corrected and extended using results from the Assarpour et al. (2015) paper by N. J. A. Sloane, Feb 22 2016 at the suggestion of William Rex Marshall.

A337226 Lexicographically earliest sequence of positive integers with the property that, for all k > 0, there is at most one j such that a(j) = a(j+k).

Original entry on oeis.org

1, 1, 2, 1, 3, 4, 2, 5, 1, 6, 3, 7, 8, 9, 4, 10, 2, 11, 5, 12, 1, 13, 6, 14, 15, 3, 16, 7, 17, 18, 8, 19, 20, 21, 22, 9, 23, 4, 24, 10, 25, 2, 26, 11, 27, 5, 28, 12, 29, 1, 30, 13, 31, 6, 32, 33, 14, 34, 15, 35, 36, 3, 37, 16, 38, 39, 40, 7, 41, 42, 17, 43, 18, 44, 45, 8

Offset: 1

Samuel B. Reid, Aug 19 2020

The sequence initially appears to be trivially fractal in that the removal of the first occurrence of each value seems to yield the original sequence. This pattern continues until a(121) where, if the sequence were fractal in this way, the value would be 72 or 1. The actual value is 13, so the pattern is broken.

Conjecture: For all k > 0, there is exactly one j such that a(j) = a(j+k). For 0 < k < 11911, this conjecture holds.

			  1 1 2 1 3 4 2
   (1)1 2 1 3 4   k = 1
      1(1)2 1 3   k = 2
       (1)1 2 1   k = 3
          1 1(2)  k = 4
            1 1   k = 5
              1   k = 6
Coincidences are circled. There can only be one coincidence per row.
a(3) cannot be 1 because that would result in two coincidences for k = 1.
a(5) cannot be 1 or 2 because those values would result in two coincidences for k = 1 and k = 2, respectively.
a(7) cannot be 1, 3, or 4 because those values would result in two coincidences for k = 3, k = 2, and k = 1, respectively. It can, however, be 2 because this results in no double coincidences.

Samuel B. Reid, Table of n, a(n) for n = 1..10000
Samuel B. Reid, Python program for A337226

Cf. A003602, A014552, A026272.

Python
```
# See Links section.
```

A059108 Number of solutions to variant of triples version of Langford (or Langford-Skolem) problem.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 0, 0, 0, 9, 20, 33, 0, 0, 0, 0, 0, 0, 200343, 869006, 4247790, 0, 0, 0, 0, 0, 0

Offset: 0

N. J. A. Sloane, Feb 14 2001

How many ways are of arranging the numbers 1,1,1,2,2,2,3,3,3,...,n,n,n so that there are zero numbers between the first and second 1's and zero numbers between the second and third 1's; one number between the first and second 2's and one number between the second and third 2's; ... n-1 numbers between the first and second n's and n-1 numbers between the second and third n's?

a(n)=0 for n mod 9 not in {0,1,2}. - Gheorghe Coserea, Aug 23 2017

			From _Gheorghe Coserea_, Jul 14 2017: (Start)
For n=9 the a(9)=9 solutions, up to reversal of the order, are:
2 4 2 8 2 4 6 7 9 4 3 8 6 3 7 5 3 9 6 8 5 7 1 1 1 5 9
2 4 2 9 2 4 5 6 7 4 8 5 9 6 3 7 5 3 8 6 3 9 7 1 1 1 8
4 2 5 2 4 2 9 5 4 7 8 3 5 6 3 9 7 3 8 6 1 1 1 7 9 6 8
5 1 1 1 7 5 8 6 9 3 5 7 3 6 8 3 4 9 7 6 4 2 8 2 4 2 9
5 6 1 1 1 5 8 6 9 3 5 7 3 6 8 3 4 9 7 2 4 2 8 2 4 7 9
6 7 9 2 5 2 6 2 7 5 8 9 6 3 5 7 3 4 8 3 9 4 1 1 1 4 8
6 7 9 2 5 2 6 2 7 5 8 9 6 4 5 7 3 4 8 3 9 4 3 1 1 1 8
7 4 2 8 2 4 2 7 9 4 3 8 6 3 7 5 3 9 6 8 5 1 1 1 6 5 9
7 5 3 6 9 3 5 7 3 6 8 5 4 9 7 6 4 2 8 2 4 2 9 1 1 1 8
(End)

Gheorghe Coserea, Solutions for n=10.
Gheorghe Coserea, Solutions for n=11.
J. E. Miller, Langford's Problem

Cf. A014552, A050998, A059106, A059107.

Fausto A. C. Cariboni has confirmed the values a(1) to a(20). - N. J. A. Sloane, Mar 27 2017

a(21) from Fausto A. C. Cariboni, Mar 28 2017

A050998 Inequivalent solutions to Langford (or Langford-Skolem) problem of arranging the numbers 1,1,2,2,3,3,...,n,n so that there is one number between the two 1's, two numbers between the two 2's, ..., n numbers between the two n's, listed by length and lexicographic order.

Original entry on oeis.org

231213, 23421314, 14156742352637, 14167345236275, 15146735423627, 15163745326427, 15167245236473, 15173465324726, 16135743625427, 16172452634753, 17125623475364, 17126425374635, 23627345161475

Offset: 1

Eric W. Weisstein

Entries are indexed by numbers n == -1 or 0 mod 4 (A014601).
More precisely, for each given n = (3, 4, 7, 8, ...) in A014601, all of the A014552(n) inequivalent solutions are listed in lexicographic order. For example,  a(1), a(2) and a(3) correspond to n=3, 4 and 7, but a(4) is not the first solution for n=8 but the second solution for n=7. - M. F. Hasler, Nov 12 2015
"Inequivalent" means that for two solutions related by symmetry (reading the digits backwards), only the (lexicographic) smaller one is listed. - M. F. Hasler, Nov 15 2015
It is unclear how the sequence goes on after the first 1+1+26+150 terms, with the solutions for n >= 11. Will a solution s=(s[1],...,s[n]) be coded again by Sum_{i=1..n} s[i]*b^(n-i) in base b=10, or in some larger base b >= n+1? Maybe using as many decimal digits as needed, i.e., b=100 for 11 <= n <= 99? - M. F. Hasler, Nov 16 2015

			The first n which allows a solution (A014552(n) > 0; n in A014601) is n=3, the solutions are a(1) = 231213 and the same read backwards, 312132.
The next solutions are given for n=4, again there is only A014552(4)=1 solution a(2) = 23421314 up to reversal (41312432, not listed).
Then follow the A014552(7)=26 (inequivalent) solutions for n=7, viz. a(3)-a(28).

M. Gardner, Mathematical Magic Show, New York: Vintage, pp. 70 and 77-78, 1978.

Seiichi Manyama, Table of n, a(n) for n = 1..178
R. K. Guy, The unity of combinatorics, Proc. 25th Iranian Math. Conf, Tehran, (1994), Math. Appl 329 129-159, Kluwer Dordrecht 1995, Math. Rev. 96k:05001.
C. D. Langford, Problem, Math. Gaz., 1958, vol. 42, p. 228.
J. Miller, Langford's problem.
Eric Weisstein's World of Mathematics, Langford's Problem.

See A014552 (the main entry for this problem) for number of solutions.

Definition clarified by M. F. Hasler, Nov 15 2015

A059107 Number of solutions to triples version of Langford (or Langford-Skolem) problem.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 3, 5, 0, 0, 0, 0, 0, 0, 13440, 54947, 249280, 0, 0, 0, 0, 0, 0

Offset: 1

N. J. A. Sloane, Feb 14 2001

How many ways are there of arranging the numbers 1,1,1,2,2,2,3,3,3, ...,n,n,n so that there is one number between the first and second 1's and one number between the second and third 1's; two numbers between the first and second 2's and two numbers between the second and third 2's; ... n numbers between the first and second n's and n numbers between the second and third n's?

a(n)=0 for n mod 9 not in {-1,0,1}. - Gheorghe Coserea, Aug 23 2017

			For n=9 the a(9)=3 solutions, up to reversal of the order, are:
1 8 1 9 1 5 2 6 7 2 8 5 2 9 6 4 7 5 3 8 4 6 3 9 7 4 3
1 9 1 2 1 8 2 4 6 2 7 9 4 5 8 6 3 4 7 5 3 9 6 8 3 5 7
1 9 1 6 1 8 2 5 7 2 6 9 2 5 8 4 7 6 3 5 4 9 3 8 7 4 3
From _Gheorghe Coserea_, Aug 26 2017: (Start)
For n=10 the a(10)=5 solutions, up to reversal of the order, are:
1 3 1 10 1 3 4 9 6 3 8 4 5 7 10 6 4 9 5 8 2 7 6 2 5 10 2 9 8 7
1 10 1 2 1 4 2 9 7 2 4 8 10 5 6 4 7 9 3 5 8 6 3 10 7 5 3 9 6 8
1 10 1 6 1 7 9 3 5 8 6 3 10 7 5 3 9 6 8 4 5 7 2 10 4 2 9 8 2 4
4 10 1 7 1 4 1 8 9 3 4 7 10 3 5 6 8 3 9 7 5 2 6 10 2 8 5 2 9 6
5 2 7 9 2 10 5 2 6 4 7 8 5 9 4 6 10 3 7 4 8 3 6 9 1 3 1 10 1 8
(End)

F. S. Gillespie and W. R. Utz, A generalized Langford Problem, Fibonacci Quart., 1966, v4, 184-186.
J. E. Miller, Langford's Problem

Cf. A014552, A050998, A059106, A059108.

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A176127 The number of permutations of {1,2,...,n,1,2,...,n} with the property that there are k numbers between the two k's in the set for k=1,...,n.

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