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 1-10 of 39 results. Next

A279199 Number of reducible ways to split 1, 2, 3, ..., 3n into n arithmetic progressions each with 3 terms: a(n) = A104429(n) - A202705(n).

Original entry on oeis.org

0, 0, 1, 3, 9, 30, 117, 512, 2597, 14892, 99034, 721350, 5909324, 52578654, 516148082, 5422071091, 61889692290, 749456672155, 9767058240577, 134007989313530, 1958535749524107
Offset: 0

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Author

N. J. A. Sloane, Dec 15 2016

Keywords

References

  • R. K. Guy, Sedlacek's Conjecture on Disjoint Solutions of x+y= z, Univ. Calgary, Dept. Mathematics, Research Paper No. 129, 1971.
  • R. K. Guy, Sedlacek's Conjecture on Disjoint Solutions of x+y= z, in Proc. Conf. Number Theory. Pullman, WA, 1971, pp. 221-223.
  • R. K. Guy, Packing [1,n] with solutions of ax + by = cz; the unity of combinatorics, in Colloq. Internaz. Teorie Combinatorie. Rome, 1973, Atti Conv. Lincei. Vol. 17, Part II, pp. 173-179, 1976.

Crossrefs

All of A279197, A279198, A202705, A279199, A104429, A282615 are concerned with counting solutions to X+Y=2Z in various ways.
See also A002848, A002849.

Formula

a(n) = A104429(n)-A202705(n) = Sum_{i=1..n-1} A104429(i)*A202705(n-i). - Martin Fuller, Jul 08 2025

Extensions

Definition corrected by N. J. A. Sloane, Jan 09 2017 at the suggestion of Fausto A. C. Cariboni.
a(15)-a(17) from Fausto A. C. Cariboni, Feb 22 2017
a(18)-a(20) from Martin Fuller, Jul 08 2025

A000201 Lower Wythoff sequence (a Beatty sequence): a(n) = floor(n*phi), where phi = (1+sqrt(5))/2 = A001622.

Original entry on oeis.org

1, 3, 4, 6, 8, 9, 11, 12, 14, 16, 17, 19, 21, 22, 24, 25, 27, 29, 30, 32, 33, 35, 37, 38, 40, 42, 43, 45, 46, 48, 50, 51, 53, 55, 56, 58, 59, 61, 63, 64, 66, 67, 69, 71, 72, 74, 76, 77, 79, 80, 82, 84, 85, 87, 88, 90, 92, 93, 95, 97, 98, 100, 101, 103, 105, 106, 108, 110
Offset: 1

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Author

Keywords

Comments

This is the unique sequence a satisfying a'(n)=a(a(n))+1 for all n in the set N of natural numbers, where a' denotes the ordered complement (in N) of a. - Clark Kimberling, Feb 17 2003
This sequence and A001950 may be defined as follows. Consider the maps a -> ab, b -> a, starting from a(1) = a; then A000201 gives the indices of a, A001950 gives the indices of b. The sequence of letters in the infinite word begins a, b, a, a, b, a, b, a, a, b, a, ... Setting a = 0, b = 1 gives A003849 (offset 0); setting a = 1, b = 0 gives A005614 (offset 0). - Philippe Deléham, Feb 20 2004
These are the numbers whose lazy Fibonacci representation (see A095791) includes 1; the complementary sequence (the upper Wythoff sequence, A001950) are the numbers whose lazy Fibonacci representation includes 2 but not 1.
a(n) is the unique monotonic sequence satisfying a(1)=1 and the condition "if n is in the sequence then n+(rank of n) is not in the sequence" (e.g. a(4)=6 so 6+4=10 and 10 is not in the sequence) - Benoit Cloitre, Mar 31 2006
Write A for A000201 and B for A001950 (the upper Wythoff sequence, complement of A). Then the composite sequences AA, AB, BA, BB, AAA, AAB,...,BBB,... appear in many complementary equations having solution A000201 (or equivalently, A001950). Typical complementary equations: AB=A+B (=A003623), BB=A+2B (=A101864), BBB=3A+5B (=A134864). - Clark Kimberling, Nov 14 2007
Cumulative sum of A001468 terms. - Eric Angelini, Aug 19 2008
The lower Wythoff sequence also can be constructed by playing the so-called Mancala-game: n piles of total d(n) chips are standing in a row. The piles are numbered from left to right by 1, 2, 3, ... . The number of chips in a pile at the beginning of the game is equal to the number of the pile. One step of the game is described as follows: Distribute the pile on the very left one by one to the piles right of it. If chips are remaining, build piles out of one chip subsequently to the right. After f(n) steps the game ends in a constant row of piles. The lower Wythoff sequence is also given by n -> f(n). - Roland Schroeder (florola(AT)gmx.de), Jun 19 2010
With the exception of the first term, a(n) gives the number of iterations required to reverse the list {1,2,3,...,n} when using the mapping defined as follows: remove the first term of the list, z(1), and add 1 to each of the next z(1) terms (appending 1's if necessary) to get a new list. See A183110 where this mapping is used and other references given. This appears to be essentially the Mancala-type game interpretation given by R. Schroeder above. - John W. Layman, Feb 03 2011
Also row numbers of A213676 starting with an even number of zeros. - Reinhard Zumkeller, Mar 10 2013
From Jianing Song, Aug 18 2022: (Start)
Numbers k such that {k*phi} > phi^(-2), where {} denotes the fractional part.
Proof: Write m = floor(k*phi).
If {k*phi} > phi^(-2), take s = m-k+1. From m < k*phi < m+1 we have k < (m-k+1)*phi < k + phi, so floor(s*phi) = k or k+1. If floor(s*phi) = k+1, then (see A003622) floor((k+1)*phi) = floor(floor(s*phi)*phi) = floor(s*phi^2)-1 = s+floor(s*phi)-1 = m+1, but actually we have (k+1)*phi > m+phi+phi^(-2) = m+2, a contradiction. Hence floor(s*phi) = k.
If floor(s*phi) = k, suppose otherwise that k*phi - m <= phi^(-2), then m < (k+1)*phi <= m+2, so floor((k+1)*phi) = m+1. Suppose that A035513(p,q) = k for p,q >= 1, then A035513(p,q+1) = floor((k+1)*phi) - 1 = m = A035513(s,1). But it is impossible for one number (m) to occur twice in A035513. (End)
The formula from Jianing Song above is a direct consequence of an old result by Carlitz et al. (1972). Their Theorem 11 states that (a(n)) consists of the numbers k such that {k*phi^(-2)} < phi^(-1). One has {k*phi^(-2)} = {k*(2-phi)} = {-k*phi}. Using that 1-phi^(-1) = phi^(-2), the Jianing Song formula follows. - Michel Dekking, Oct 14 2023
In the Fokkink-Joshi paper, this sequence is the Cloitre (1,1,2,1)-hiccup sequence, i.e., a(1) = 1; for m < n, a(n) = a(n-1)+2 if a(m) = n-1, else a(n) = a(n-1)+1. - Michael De Vlieger, Jul 28 2025

Examples

			From Roland Schroeder (florola(AT)gmx.de), Jul 13 2010: (Start)
Example for n = 5; a(5) = 8;
(Start: [1,2,3,4,5]; 8 steps until [5,4,3,2,1]):
[1,2,3,4,5]; [3,3,4,5]; [4,5,6]; [6,7,1,1]; [8,2,2,1,1,1]: [3,3,2,2,2,1,1,1]; [4,3,3,2,1,1,1]; [4,4,3,2,1,1]; [5,4,3,2,1]. (End)
		

References

  • Eric Friedman, Scott M. Garrabrant, Ilona K. Phipps-Morgan, A. S. Landsberg and Urban Larsson, Geometric analysis of a generalized Wythoff game, in Games of no Chance 5, MSRI publ. Cambridge University Press, date?
  • M. Gardner, Penrose Tiles to Trapdoor Ciphers, W. H. Freeman, 1989; see p. 107.
  • 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).
  • I. M. Yaglom, Two games with matchsticks, pp. 1-7 of Qvant Selecta: Combinatorics I, Amer Math. Soc., 2001.

Crossrefs

a(n) = least k such that s(k) = n, where s = A026242. Complement of A001950. See also A058066.
The permutation A002251 maps between this sequence and A001950, in that A002251(a(n)) = A001950(n), A002251(A001950(n)) = a(n).
First differences give A014675. a(n) = A022342(n) + 1 = A005206(n) + n + 1. a(2n)-a(n)=A007067(n). a(a(a(n)))-a(n) = A026274(n-1). - Benoit Cloitre, Mar 08 2003
A185615 gives values n such that n divides A000201(n)^m for some integer m>0.
Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021
Bisections: A276854, A342279.

Programs

  • Haskell
    a000201 n = a000201_list !! (n-1)
    a000201_list = f [1..] [1..] where
       f (x:xs) (y:ys) = y : f xs (delete (x + y) ys)
    -- Reinhard Zumkeller, Jul 02 2015, Mar 10 2013
    
  • Maple
    Digits := 100; t := evalf((1+sqrt(5))/2); A000201 := n->floor(t*n);
  • Mathematica
    Table[Floor[N[n*(1+Sqrt[5])/2]], {n, 1, 75}]
    Array[ Floor[ #*GoldenRatio] &, 68] (* Robert G. Wilson v, Apr 17 2010 *)
  • Maxima
    makelist(floor(n*(1+sqrt(5))/2),n,1,60); /* Martin Ettl, Oct 17 2012 */
    
  • PARI
    a(n)=floor(n*(sqrt(5)+1)/2)
    
  • PARI
    a(n)=(n+sqrtint(5*n^2))\2 \\ Charles R Greathouse IV, Feb 07 2013
    
  • Python
    def aupton(terms):
      alst, aset = [None, 1], {1}
      for n in range(1, terms):
        an = alst[n] + (1 if n not in aset else 2)
        alst.append(an); aset.add(an)
      return alst[1:]
    print(aupton(68)) # Michael S. Branicky, May 14 2021
    
  • Python
    from math import isqrt
    def A000201(n): return (n+isqrt(5*n**2))//2 # Chai Wah Wu, Jan 11 2022

Formula

Zeckendorf expansion of n (cf. A035517) ends with an even number of 0's.
Other properties: a(1)=1; for n>1, a(n) is taken to be the smallest integer greater than a(n-1) which is consistent with the condition "n is in the sequence if and only if a(n)+1 is not in the sequence".
a(1) = 1; for n>0, a(n+1) = a(n)+1 if n is not in the sequence, a(n+1) = a(n)+2 if n is in the sequence.
a(a(n)) = floor(n*phi^2) - 1 = A003622(n).
{a(k)} union {a(k)+1} = {1, 2, 3, 4, ...}. Hence a(1) = 1; for n>1, a(a(n)) = a(a(n)-1)+2, a(a(n)+1) = a(a(n))+1. - Benoit Cloitre, Mar 08 2003
{a(n)} is a solution to the recurrence a(a(n)+n) = 2*a(n)+n, a(1)=1 (see Barbeau et al.).
a(n) = A001950(n) - n. - Philippe Deléham, May 02 2004
a(0) = 0; a(n) = n + Max_{k : a(k) < n}. - Vladeta Jovovic, Jun 11 2004
a(Fibonacci(r-1)+j) = Fibonacci(r)+a(j) for 0 < j <= Fibonacci(r-2); 2 < r. - Paul Weisenhorn, Aug 18 2012
With 1 < k and A001950(k-1) < n <= A001950(k): a(n) = 2*n-k; A001950(n) = 3*n-k. - Paul Weisenhorn, Aug 21 2012

A001950 Upper Wythoff sequence (a Beatty sequence): a(n) = floor(n*phi^2), where phi = (1+sqrt(5))/2.

Original entry on oeis.org

2, 5, 7, 10, 13, 15, 18, 20, 23, 26, 28, 31, 34, 36, 39, 41, 44, 47, 49, 52, 54, 57, 60, 62, 65, 68, 70, 73, 75, 78, 81, 83, 86, 89, 91, 94, 96, 99, 102, 104, 107, 109, 112, 115, 117, 120, 123, 125, 128, 130, 133, 136, 138, 141, 143, 146, 149, 151, 154, 157
Offset: 1

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Keywords

Comments

Indices at which blocks (1;0) occur in infinite Fibonacci word; i.e., n such that A005614(n-2) = 0 and A005614(n-1) = 1. - Benoit Cloitre, Nov 15 2003
A000201 and this sequence may be defined as follows: Consider the maps a -> ab, b -> a, starting from a(1) = a; then A000201 gives the indices of a, A001950 gives the indices of b. The sequence of letters in the infinite word begins a, b, a, a, b, a, b, a, a, b, a, ... Setting a = 0, b = 1 gives A003849 (offset 0); setting a = 1, b = 0 gives A005614 (offset 0). - Philippe Deléham, Feb 20 2004
a(n) = n-th integer which is not equal to the floor of any multiple of phi, where phi = (1+sqrt(5))/2 = golden number. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), May 09 2007
Write A for A000201 and B for the present sequence (the upper Wythoff sequence, complement of A). Then the composite sequences AA, AB, BA, BB, AAA, AAB, ..., BBB, ... appear in many complementary equations having solution A000201 (or equivalently, the present sequence). Typical complementary equations: AB=A+B (=A003623), BB=A+2B (=A101864), BBB=3A+5B (=A134864). - Clark Kimberling, Nov 14 2007
Apart from the initial 0 in A090909, is this the same as that sequence? - Alec Mihailovs (alec(AT)mihailovs.com), Jul 23 2007
If we define a base-phi integer as a positive number whose representation in the golden ratio base consists only of nonnegative powers of phi, and if these base-phi integers are ordered in increasing order (beginning 1, phi, ...), then it appears that the difference between the n-th and (n-1)-th base-phi integer is phi-1 if and only if n belongs to this sequence, and the difference is 1 otherwise. Further, if each base-phi integer is written in linear form as a + b*phi (for example, phi^2 is written as 1 + phi), then it appears that there are exactly two base-phi integers with b=n if and only if n belongs to this sequence, and exactly three base-phi integers with b=n otherwise. - Geoffrey Caveney, Apr 17 2014
Numbers with an odd number of trailing zeros in their Zeckendorf representation (A014417). - Amiram Eldar, Feb 26 2021
Numbers missing from A066096. - Philippe Deléham, Jan 19 2023

Examples

			From _Paul Weisenhorn_, Aug 18 2012 and Aug 21 2012: (Start)
a(14) = floor(14*phi^2) = 36; a'(14) = floor(14*phi)=22;
with r=9 and j=1: a(13+1) = 34 + 2 = 36;
with r=8 and j=1: a'(13+1) = 21 + 1 = 22.
k=6 and a(5)=13 < n <= a(6)=15
a(14) = 3*14 - 6 = 36; a'(14) = 2*14 - 6 = 22;
a(15) = 3*15 - 6 = 39; a'(15) = 2*15 - 6 = 24. (End)
		

References

  • Claude Berge, Graphs and Hypergraphs, North-Holland, 1973; p. 324, Problem 2.
  • Eric Friedman, Scott M. Garrabrant, Ilona K. Phipps-Morgan, A. S. Landsberg and Urban Larsson, Geometric analysis of a generalized Wythoff game, in Games of no Chance 5, MSRI publ. Cambridge University Press, 2019.
  • Martin Gardner, Penrose Tiles to Trapdoor Ciphers, W. H. Freeman, 1989; see p. 107.
  • 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).
  • I. M. Yaglom, Two games with matchsticks, pp. 1-7 of Qvant Selecta: Combinatorics I, Amer Math. Soc., 2001.

Crossrefs

a(n) = greatest k such that s(k) = n, where s = A026242.
Complement of A000201 or A066096.
A002251 maps between A000201 and A001950, in that A002251(A000201(n)) = A001950(n), A002251(A001950(n)) = A000201(n).
Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864.
First differences give (essentially) A076662.
Bisections: A001962, A001966.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a001950 n = a000201 n + n  -- Reinhard Zumkeller, Mar 10 2013
    
  • Magma
    [Floor(n*((1+Sqrt(5))/2)^2): n in [1..80]]; // Vincenzo Librandi, Nov 19 2016
    
  • Maple
    A001950 := proc(n)
        floor(n*(3+sqrt(5))/2) ;
    end proc:
    seq(A001950(n),n=0..40) ; # R. J. Mathar, Jul 16 2024
  • Mathematica
    Table[Floor[N[n*(1+Sqrt[5])^2/4]], {n, 1, 75}]
    Array[ Floor[ #*GoldenRatio^2] &, 60] (* Robert G. Wilson v, Apr 17 2010 *)
  • PARI
    a(n)=floor(n*(sqrt(5)+3)/2)
    
  • PARI
    A001950(n)=(sqrtint(n^2*5)+n*3)\2 \\ M. F. Hasler, Sep 17 2014
    
  • Python
    from math import isqrt
    def A001950(n): return (n+isqrt(5*n**2)>>1)+n # Chai Wah Wu, Aug 10 2022

Formula

a(n) = n + floor(n*phi). In general, floor(n*phi^m) = Fibonacci(m-1)*n + floor(Fibonacci(m)*n*phi). - Benoit Cloitre, Mar 18 2003
a(n) = n + floor(n*phi) = n + A000201(n). - Paul Weisenhorn and Philippe Deléham
Append a 0 to the Zeckendorf expansion (cf. A035517) of n-th term of A000201.
a(n) = A003622(n) + 1. - Philippe Deléham, Apr 30 2004
a(n) = Min(m: A134409(m) = A006336(n)). - Reinhard Zumkeller, Oct 24 2007
If a'=A000201 is the ordered complement (in N) of {a(n)}, then a(Fib(r-2) + j) = Fib(r) + a(j) for 0 < j <= Fib(r-2), 3 < r; and a'(Fib(r-1) + j) = Fib(r) + a'(j) for 0 < j <= Fib(r-2), 2 < r. - Paul Weisenhorn, Aug 18 2012
With a(1)=2, a(2)=5, a'(1)=1, a'(2)=3 and 1 < k and a(k-1) < n <= a(k) one gets a(n)=3*n-k, a'(n)=2*n-k. - Paul Weisenhorn, Aug 21 2012

Extensions

Corrected by Michael Somos, Jun 07 2000

A104443 Square of P(n,t) read by antidiagonals. P(n,t) = number of ways to split [t*n] into n arithmetic progressions each with t terms.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 2, 15, 1, 1, 2, 5, 105, 1, 1, 2, 4, 15, 945, 1, 1, 2, 4, 11, 55, 10395, 1, 1, 2, 4, 10, 23, 232, 135135, 1, 1, 2, 4, 10, 21, 68, 1161, 2027025, 1, 1, 2, 4, 10, 20, 59, 161, 6643, 34459425, 1, 1, 2, 4, 10, 20, 57, 125, 488, 44566, 654729075, 1
Offset: 1

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Author

Jonas Wallgren, Mar 17 2005

Keywords

Examples

			Square array begins:
  1,        1,     1,    1,   1,   1,   1,   1,   1, ...
  1,        3,     2,    2,   2,   2,   2,   2,   2, ...
  1,       15,     5,    4,   4,   4,   4,   4,   4, ...
  1,      105,    15,   11,  10,  10,  10,  10,  10, ...
  1,      945,    55,   23,  21,  20,  20,  20,  20, ...
  1,    10395,   232,   68,  59,  57,  56,  56,  56, ...
  1,   135135,  1161,  161, 125, 119, 117, 116, 116, ...
  1,  2027025,  6643,  488, 349, 329, 323, 321, 320, ...
  1, 34459425, 44566, 1249, 848, 760, 745, 739, 737, ...
  ...
		

Crossrefs

Cf. A104429-A104442. P(1, )=P(, 1) = A000012, P(_, 2) = A001147.

Extensions

More terms from Alois P. Heinz, Nov 18 2020

A002849 Number of maximal collections of pairwise disjoint subsets {X,Y,Z} of {1, 2, ..., n}, each satisfying X + Y = Z.

Original entry on oeis.org

1, 1, 1, 2, 4, 6, 3, 10, 25, 12, 42, 8, 40, 204, 21, 135, 1002, 4228, 720, 5134, 29546, 4079, 35533, 3040, 28777, 281504, 20505, 212283, 2352469, 16907265, 1669221, 19424213, 167977344, 14708525, 191825926, 10567748, 149151774, 2102286756, 103372655, 1534969405
Offset: 1

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Keywords

Examples

			For n = 3, the unique solution is 1 + 2 = 3.
For n = 12, there are 8 solutions:
  1  5  6 | 1  5  6 | 2  5  7 | 1  6  7
  2  8 10 | 3  7 10 | 3  6  9 | 4  5  9
  4  7 11 | 2  9 11 | 1 10 11 | 3  8 11
  3  9 12 | 4  8 12 | 4  8 12 | 2 10 12
  --------+---------+---------+--------
  2  4  6 | 2  6  8 | 3  4  7 | 3  5  8
  1  9 10 | 4  5  9 | 1  8  9 | 2  7  9
  3  8 11 | 3  7 10 | 5  6 11 | 4  6 10
  5  7 12 | 1 11 12 | 2 10 12 | 1 11 12
		

References

  • R. K. Guy, "Sedlacek's Conjecture on Disjoint Solutions of x+y= z," in Proc. Conf. Number Theory. Pullman, WA, 1971, pp. 221-223.
  • R. K. Guy, "Packing [1,n] with solutions of ax + by = cz; the unity of combinatorics," in Colloq. Internaz. Teorie Combinatorie. Rome, 1973, Atti Conv. Lincei. Vol. 17, Part II, pp. 173-179, 1976.
  • 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).

Crossrefs

Programs

  • PARI
    nxyz(v,t)=local(n,r,x2); r=0; if(t==0,return(1)); for(i3=3*t,#v, n=v[i3]; for(i1=1,i3-2, x2=n-v[i1]; if(x2<=v[i1],break); for(i2=i1+1,i3-1, if(v[i2]>=x2, if(v[i2]==x2, r+=nxyz(vector(i3-3,k,v[if(kFranklin T. Adams-Watters

Extensions

Edited by N. J. A. Sloane, Feb 10 2010, based on posting to the Sequence Fans Mailing List by Franklin T. Adams-Watters, R. K. Guy, R. H. Hardin, Alois P. Heinz, Andrew Weimholt, Max Alekseyev and others
a(32)-a(39) from Max Alekseyev, Feb 23 2012
Definition corrected by Max Alekseyev, Nov 16 2012, Jul 06 2023
a(40)-a(41) from Fausto A. C. Cariboni, Feb 04 2017
a(42) from Fausto A. C. Cariboni, Mar 12 2017

A108235 Number of partitions of {1,2,...,3n} into n triples (X,Y,Z) each satisfying X+Y=Z.

Original entry on oeis.org

1, 1, 0, 0, 8, 21, 0, 0, 3040, 20505, 0, 0, 10567748, 103372655, 0, 0, 142664107305, 1836652173363, 0, 0
Offset: 0

Views

Author

N. J. A. Sloane, Feb 10 2010, based on posting to the Sequence Fans Mailing List by Franklin T. Adams-Watters, R. K. Guy, R. H. Hardin, Alois P. Heinz, Andrew Weimholt and others

Keywords

Comments

a(0)=1 by convention.

Examples

			For m = 1 the unique solution is 1 + 2 = 3.
For m = 4 there are 8 solutions:
  1  5  6 | 1  5  6 | 2  5  7 | 1  6  7
  2  8 10 | 3  7 10 | 3  6  9 | 4  5  9
  4  7 11 | 2  9 11 | 1 10 11 | 3  8 11
  3  9 12 | 4  8 12 | 4  8 12 | 2 10 12
  --------+---------+---------+--------
  2  4  6 | 2  6  8 | 3  4  7 | 3  5  8
  1  9 10 | 4  5  9 | 1  8  9 | 2  7  9
  3  8 11 | 3  7 10 | 5  6 11 | 4  6 10
  5  7 12 | 1 11 12 | 2 10 12 | 1 11 12
.
The 8 solutions for m = 4, one per line:
  (1,  5,  6), (2,  8, 10), (3,  9, 12), (4,  7, 11);
  (1,  5,  6), (2,  9, 11), (3,  7, 10), (4,  8, 12);
  (1, 10, 11), (2,  5,  7), (3,  6,  9), (4,  8, 12);
  (1,  6,  7), (2, 10, 12), (3,  8, 11), (4,  5,  9);
  (1,  9, 10), (2,  4,  6), (3,  8, 11), (5,  7, 12);
  (1, 11, 12), (2,  6,  8), (3,  7, 10), (4,  5,  9);
  (1,  8,  9), (2, 10, 12), (3,  4,  7), (5,  6, 11);
  (1, 11, 12), (2,  7,  9), (3,  5,  8), (4,  6, 10).
		

Crossrefs

Programs

  • Mathematica
    Table[Length[Select[Subsets[Select[Subsets[Range[3 n], {3}], #[[1]] + #[[2]] == #[[3]] &], {n}], Range[3 n] == Sort[Flatten[#]] &]], {n, 0,
    5}]  (* Suitable only for n<6. See Knuth's Dancing Links algorithm for n>5. *) (* Robert Price, Apr 03 2019 *)
  • Sage
    A = lambda n:sum(1 for t in DLXCPP([(a-1,b-1,a+b-1) for a in (1..3*n) for b in (1..min(3*n-a,a-1))])) # Tomas Boothby, Oct 11 2013

Formula

a(n) = 0 unless n == 0 or 1 (mod 4). For n == 0 or 1 (mod 4), a(n) = A002849(3n). See A002849 for references and further information.

Extensions

a(12) from R. H. Hardin, Feb 11 2010
a(12) confirmed and a(13) computed (using Knuth's dancing links algorithm) by Alois P. Heinz, Feb 11 2010
a(13) confirmed by Tomas Boothby, Oct 11 2013
a(16) from Frank Niedermeyer, Apr 19 2020
a(17)-a(19) from Frank Niedermeyer, May 02 2020

A202705 Number of irreducible ways to split 1, 2, 3, ..., 3n into n arithmetic progressions each with 3 terms.

Original entry on oeis.org

1, 1, 1, 2, 6, 25, 115, 649, 4046, 29674, 228030, 1987700, 18402704, 188255116, 2030067605, 23829298479, 293949166112, 3909410101509, 54360507919179, 806312701922676
Offset: 0

Views

Author

N. J. A. Sloane, Dec 26 2011

Keywords

Comments

"Irreducible" means that there is no j such that the first j of the triples are a partition of 1, ..., 3j.

References

  • R. K. Guy, Sedlacek's Conjecture on Disjoint Solutions of x+y= z, Univ. Calgary, Dept. Mathematics, Research Paper No. 129, 1971.
  • R. K. Guy, Sedlacek's Conjecture on Disjoint Solutions of x+y= z, in Proc. Conf. Number Theory. Pullman, WA, 1971, pp. 221-223.
  • R. K. Guy, Packing [1,n] with solutions of ax + by = cz; the unity of combinatorics, in Colloq. Internaz. Teorie Combinatorie. Rome, 1973, Atti Conv. Lincei. Vol. 17, Part II, pp. 173-179, 1976.

Crossrefs

All of A279197, A279198, A202705, A279199, A104429, A282615 are concerned with counting solutions to X+Y=2Z in various ways.
See also A002848, A002849.

Formula

G.f.: 2 - 1/g where g is g.f. for A104429. [corrected by Martin Fuller, Jul 08 2025]
a(n) = A279197(n) + 2*A279198(n) for n>0.

Extensions

a(11)-a(14) from Alois P. Heinz, Dec 28 2011
a(15)-a(17) from Fausto A. C. Cariboni, Feb 22 2017
a(18)-a(19) from Martin Fuller, Jul 08 2025

A279197 Number of self-conjugate inseparable solutions of X + Y = 2Z (integer, disjoint triples from {1,2,3,...,3n}).

Original entry on oeis.org

1, 1, 2, 2, 11, 11, 55, 58, 486, 442, 4218, 3924, 45096, 42013, 538537, 505830, 7368091, 6959545, 111877294, 105723374, 1886636688, 1763443165, 34585786729, 32401780965, 687085545694, 642233156868, 14691047314846, 13788837896728, 340221989868538, 317342350394678, 8327884506579315
Offset: 1

Views

Author

N. J. A. Sloane, Dec 15 2016

Keywords

Comments

In Richard Guy's letter, the term 50 is marked with a question mark. Peter Kagey has shown that the value should be 55. - N. J. A. Sloane, Feb 15 2017
From Peter Kagey, Feb 14 2017: (Start)
An inseparable solution is one in which "there is no j such that the first j of the triples are a partition of 1, ..., 3j" (See A202705.)
A self-conjugate solution is one in which for every triple (a, b, c) in the partition there exists a "conjugate" triple (m-a, m-b, m-c) or (m-b, m-a, m-c) where m = 3n+1.
(End)

Examples

			Examples of solutions X,Y,Z for n=5:
  2,4,3
  5,7,6
  1,15,8
  9,11,10
  12,14,13
and in his letter Richard Guy has drawn links pairing the first and fifth solutions, and the second and fourth solutions.
For n = 2 the a(2) = 1 solution is
  [(2,6,4),(1,5,3)].
For n = 3 the a(3) = 2 solutions are
  [(1,7,4),(3,9,6),(2,8,5)] and
  [(2,4,3),(6,8,7),(1,9,5)].
		

References

  • R. K. Guy, Sedlacek's Conjecture on Disjoint Solutions of x+y= z, Univ. Calgary, Dept. Mathematics, Research Paper No. 129, 1971.
  • R. K. Guy, Sedlacek's Conjecture on Disjoint Solutions of x+y= z, in Proc. Conf. Number Theory. Pullman, WA, 1971, pp. 221-223.
  • R. K. Guy, Packing [1,n] with solutions of ax + by = cz; the unity of combinatorics, in Colloq. Internaz. Teorie Combinatorie. Rome, 1973, Atti Conv. Lincei. Vol. 17, Part II, pp. 173-179, 1976.

Crossrefs

All of A279197, A279198, A202705, A279199, A104429, A282615 are concerned with counting solutions to X+Y=2Z in various ways.
See also A002848, A002849.

Formula

a(n) = A282616(n) - A282615(n). - Martin Fuller, Jul 15 2025

Extensions

a(7) corrected and a(8)-a(13) added by Peter Kagey, Feb 14 2017
a(14)-a(16) from Fausto A. C. Cariboni, Feb 27 2017
a(17) from Fausto A. C. Cariboni, Mar 22 2017
a(18)-a(24) from Bert Dobbelaere, May 29 2025
a(25)-a(31) from Martin Fuller, Jul 15 2025

A282615 Number of self-conjugate separable solutions of X + Y = 2Z (integer, disjoint triples from {1,2,3,...,3n}).

Original entry on oeis.org

0, 1, 1, 3, 4, 9, 20, 35, 102, 160, 736, 930, 5972, 6766, 59017, 61814, 671651, 675379, 8844028, 8675583, 130880467, 126385830, 2163551657, 2049560059, 39112954305, 36883483406, 768337929193, 720918897940, 16279025598443, 15303083773040, 373743187469167, 349148771223261, 9095126347788632
Offset: 1

Views

Author

Peter Kagey, Feb 19 2017

Keywords

Comments

An inseparable solution is one in which "there is no j such that the first j of the triples are a partition of 1, ..., 3j" (see A202705).
A self-conjugate solution is one in which for every triple (a, b, c) in the partition there exists a "conjugate" triple (m-a, m-b, m-c) or (m-b, m-a, m-c) where m = 3n+1.
| separable | inseparable | either |
-------------------+-----------+-------------+---------+
self-conjugate | A282615 | A279197 | A282616 |
non-self-conjugate | A282618 | A282617 | A282619 |
either | A279199 | A202705 | A104429 |

Examples

			For n = 4 the a(4) = 3 solutions are:
  (10,12,11),(7,9,8),(4,6,5),(1,3,2),
  (10,12,11),(5,9,7),(4,8,6),(1,3,2), and
  (8,12,10),(7,11,9),(2,6,4),(1,5,3).
		

Crossrefs

All of A279197, A279198, A202705, A279199, A104429, A282615 are concerned with counting solutions to X+Y=2Z in various ways.

Formula

a(n) = A282616(n) - A279197(n).
a(n) = A279199(n) - A282618(n).
a(n) = Sum_{i=1..floor(n/2)} A202705(i) * (A282616(n-2*i) if n>2*i else 1) = Sum_{i=1..floor(n/2)} A104429(i) * (A279197(n-2*i) if n>2*i else 1). - Martin Fuller, Jul 15 2025

Extensions

a(11)-a(16) from Fausto A. C. Cariboni, Feb 27 2017
a(17) from Fausto A. C. Cariboni, Mar 22 2017
a(18)-a(24) from Bert Dobbelaere, May 29 2025
a(25)-a(33) from Martin Fuller, Jul 15 2025

A104430 Number of ways to split 1, 2, 3, ..., 4n into n arithmetic progressions each with 4 terms.

Original entry on oeis.org

1, 1, 2, 4, 11, 23, 68, 161, 488, 1249, 3771, 10388, 35725, 110449, 387057, 1411784, 5938390, 26054261, 129231034, 708657991
Offset: 0

Views

Author

Jonas Wallgren, Mar 17 2005

Keywords

Examples

			{{{1,2,3,4},{5,6,7,8},{9,10,11,12}}, {{1,2,3,4},{5,7,9,11},{6,8,10,12}}, {{1,3,5,7},{2,4,6,8},{9,10,11,12}}, {{1,4,7,10},{2,5,8,11},{3,6,9,12}}} are the 4 ways to split 1, 2, 3, ..., 12 into 3 arithmetic progressions each with 4 terms. Thus a(3)=4.
		

Crossrefs

Programs

  • C
    See Links section.

Extensions

a(11)-a(17) from Alois P. Heinz, Dec 28 2011
a(0)=1 prepended by Alois P. Heinz, Nov 18 2020
a(18)-a(19) from Rémy Sigrist, Feb 07 2022
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