Search: a104429 -id:a104429
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A279199
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Number of reducible ways to split 1, 2, 3, ..., 3n into n arithmetic progressions each with 3 terms: a(n) = A104429(n) - A202705(n).
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+20
10
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0, 0, 1, 3, 9, 30, 117, 512, 2597, 14892, 99034, 721350, 5909324, 52578654, 516148082, 5422071091, 61889692290, 749456672155
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OFFSET
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0,4
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REFERENCES
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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.
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LINKS
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R. K. Guy, Letter to N. J. A. Sloane, June 24 1971: front, back [Annotated scanned copy, with permission] See sequence "L".
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A000201
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Lower Wythoff sequence (a Beatty sequence): a(n) = floor(n*phi), where phi = (1+sqrt(5))/2 = A001622.
(Formerly M2322 N0917)
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+10
311
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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
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OFFSET
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1,2
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COMMENTS
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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
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
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
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REFERENCES
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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.
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LINKS
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L. Carlitz, Richard Scoville, and V. E. Hoggatt, Jr., Fibonacci representations, Fib. Quart., Vol. 10, No. 1 (1972), pp. 1-28.
Eric Duchene, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, and Urban Larsson, Wythoff Wisdom, 43 pages, no date, unpublished.
Eric Duchene, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, and Urban Larsson, Wythoff Wisdom, unpublished, no date [Cached copy, with permission]
A. S. Fraenkel, Ratwyt, December 28 2011.
Clark Kimberling, Problem Proposals, The Fibonacci Quarterly, vol. 52 #5, 2015, p5-14.
Wolfdieter Lang, The Wythoff and the Zeckendorf representations of numbers are equivalent, in G. E. Bergum et al. (edts.) Application of Fibonacci numbers vol. 6, Kluwer, Dordrecht, 1996, pp. 319-337.[See A317208 for a link.]
D. J. Newman, Problem 3117, Amer. Math. Monthly, 34 (1927), 158-159.
D. J. Newman, Problem 5252, Amer. Math. Monthly, 72 (1965), 1144-1145.
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FORMULA
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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(Fibonacci(r-1)+j) = Fibonacci(r)+a(j) for 0 < j <= Fibonacci(r-2); 2 < r. - Paul Weisenhorn, Aug 18 2012
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EXAMPLE
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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)
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MAPLE
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Digits := 100; t := evalf((1+sqrt(5))/2); A000201 := n->floor(t*n);
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MATHEMATICA
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Table[Floor[N[n*(1+Sqrt[5])/2]], {n, 1, 75}]
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PROG
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(PARI) a(n)=floor(n*(sqrt(5)+1)/2)
(Maxima) makelist(floor(n*(1+sqrt(5))/2), n, 1, 60); /* Martin Ettl, Oct 17 2012 */
(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)
(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:]
(Python)
from math import isqrt
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CROSSREFS
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A185615 gives values n such that n divides A000201(n)^m for some integer m>0.
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
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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A001950
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Upper Wythoff sequence (a Beatty sequence): a(n) = floor(n*phi^2), where phi = (1+sqrt(5))/2.
(Formerly M1332 N0509)
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+10
253
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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
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OFFSET
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1,1
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COMMENTS
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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
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REFERENCES
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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.
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LINKS
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Eric Duchene, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, and Urban Larsson, Wythoff Wisdom, 43 pages, no date, unpublished.
Eric Duchene, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, and Urban Larsson, Wythoff Wisdom, unpublished, no date [Cached copy, with permission]
Aviezri S. Fraenkel, The Raleigh game, INTEGERS: Electronic Journal of Combinatorial Number Theory 7.2 (2007): A13, 10 pages. See Table 1.
Aviezri S. Fraenkel, Ratwyt, December 28 2011.
D. J. Newman, Problem 5252, Amer. Math. Monthly, Vol. 72, No. 10 (1965), pp. 1144-1145.
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FORMULA
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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
Append a 0 to the Zeckendorf expansion (cf. A035517) of n-th term of A000201.
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
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EXAMPLE
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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)
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MAPLE
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floor(n*(3+sqrt(5))/2) ;
end proc:
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MATHEMATICA
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Table[Floor[N[n*(1+Sqrt[5])^2/4]], {n, 1, 75}]
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PROG
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(PARI) a(n)=floor(n*(sqrt(5)+3)/2)
(Haskell)
(Magma) [Floor(n*((1+Sqrt(5))/2)^2): n in [1..80]]; // Vincenzo Librandi, Nov 19 2016
(Python)
from math import isqrt
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CROSSREFS
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a(n) = greatest k such that s(k) = n, where s = A026242.
First differences give (essentially) A076662.
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
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A104443
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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.
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+10
20
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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
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OFFSET
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1,5
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LINKS
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EXAMPLE
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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, ...
...
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A002849
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Number of maximal collections of pairwise disjoint subsets {X,Y,Z} of {1, 2, ..., n}, each satisfying X + Y = Z.
(Formerly M0980 N0368)
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+10
11
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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
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OFFSET
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1,4
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REFERENCES
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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).
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LINKS
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R. K. Guy, Letter to N. J. A. Sloane, Jun 24 1971: front, back [Annotated scanned copy, with permission]
Richard K. Guy, The unity of combinatorics, in Proc. 25th Iran. Math. Conf., Tehran, (1994), Math. Appl. 329 (1994) 129-159, Kluwer Acad. Publ., Dordrecht, 1995.
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EXAMPLE
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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
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PROG
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(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(k<i1, k, if(k<i2-1, k+1, k+2))]), t-1)); break)))); r
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A108235
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Number of partitions of {1,2,...,3n} into n triples (X,Y,Z) each satisfying X+Y=Z.
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+10
10
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1, 1, 0, 0, 8, 21, 0, 0, 3040, 20505, 0, 0, 10567748, 103372655, 0, 0, 142664107305, 1836652173363, 0, 0
(list;
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OFFSET
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0,5
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COMMENTS
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a(0)=1 by convention.
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LINKS
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FORMULA
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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.
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EXAMPLE
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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).
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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a(12) confirmed and a(13) computed (using Knuth's dancing links algorithm) by Alois P. Heinz, Feb 11 2010
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STATUS
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approved
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A202705
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Number of irreducible ways to split 1, 2, 3, ..., 3n into n arithmetic progressions each with 3 terms.
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+10
10
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1, 1, 1, 2, 6, 25, 115, 649, 4046, 29674, 228030, 1987700, 18402704, 188255116, 2030067605, 23829298479, 293949166112, 3909410101509
(list;
graph;
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OFFSET
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0,4
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COMMENTS
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"Irreducible" means that there is no j such that the first j of the triples are a partition of 1, ..., 3j.
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REFERENCES
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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.
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LINKS
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R. K. Guy, Letter to N. J. A. Sloane, June 24 1971: front, back [Annotated scanned copy, with permission] See sequence "K".
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FORMULA
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G.f. = 1 - 1/g where g is g.f. for A104429.
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A279197
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Number of self-conjugate inseparable solutions of X + Y = 2Z (integer, disjoint triples from {1,2,3,...,3n}).
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+10
10
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1, 1, 2, 2, 11, 11, 55, 58, 486, 442, 4218, 3924, 45096, 42013, 538537, 505830, 7368091
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OFFSET
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1,3
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COMMENTS
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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
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)
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REFERENCES
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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.
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LINKS
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R. K. Guy, Letter to N. J. A. Sloane, June 24 1971: front, back [Annotated scanned copy, with permission] See sequence "I".
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EXAMPLE
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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)].
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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a(7) corrected and a(8)-a(13) added by Peter Kagey, Feb 14 2017
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STATUS
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approved
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A282615
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Number of self-conjugate separable solutions of X + Y = 2Z (integer, disjoint triples from {1,2,3,...,3n}).
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+10
10
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0, 1, 1, 3, 4, 9, 20, 35, 102, 160, 736, 930, 5972, 6766, 59017, 61814, 671651
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OFFSET
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1,4
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COMMENTS
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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 |
-------------------+-----------+-------------+---------+
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LINKS
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FORMULA
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EXAMPLE
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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).
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A104430
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Number of ways to split 1, 2, 3, ..., 4n into n arithmetic progressions each with 4 terms.
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+10
7
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1, 1, 2, 4, 11, 23, 68, 161, 488, 1249, 3771, 10388, 35725, 110449, 387057, 1411784, 5938390, 26054261, 129231034, 708657991
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graph;
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OFFSET
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0,3
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LINKS
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EXAMPLE
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{{{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.
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PROG
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(C) See Links section.
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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