

A000930


Narayana's cows sequence: a(0) = a(1) = a(2) = 1; thereafter a(n) = a(n1) + a(n3).
(Formerly M0571 N0207)


284



1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873, 2745, 4023, 5896, 8641, 12664, 18560, 27201, 39865, 58425, 85626, 125491, 183916, 269542, 395033, 578949, 848491, 1243524, 1822473, 2670964, 3914488, 5736961, 8407925
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OFFSET

0,4


COMMENTS

Named after a 14thcentury Indian mathematician. [The sequence first appeared in the book "Ganita Kaumudi" (1356) by the Indian mathematician Narayana Pandita (c. 1340  c. 1400).  Amiram Eldar, Apr 15 2021]
Number of compositions of n into parts 1 and 3.  Joerg Arndt, Jun 25 2011
A Lamé sequence of higher order.
Could have begun 1,0,0,1,1,1,2,3,4,6,9,... (A078012) but that would spoil many nice properties.
Number of tilings of a 3 X n rectangle with straight trominoes.
Number of ways to arrange n1 tatami mats in a 2 X (n1) room such that no 4 meet at a point. For example, there are 6 ways to cover a 2 X 5 room, described by 11111, 2111, 1211, 1121, 1112, 212.
Equivalently, number of compositions (ordered partitions) of n1 into parts 1 and 2 with no two 2's adjacent. E.g., there are 6 such ways to partition 5, namely 11111, 2111, 1211, 1121, 1112, 212, so a(6) = 6. [Minor edit by Keyang Li, Oct 10 2020]
This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n1) + a(nm), with a(n) = 1 for n = 0...m1. The generating function is 1/(1xx^m). Also a(n) = Sum_{i=0..floor(n/m)} binomial(n(m1)*i, i). This family of binomial summations or recurrences gives the number of ways to cover (without overlapping) a linear lattice of n sites with molecules that are m sites wide. Special case: m=1: A000079; m=4: A003269; m=5: A003520; m=6: A005708; m=7: A005709; m=8: A005710.
a(n+2) is the number of nbit 01 sequences that avoid both 00 and 010.  David Callan, Mar 25 2004 [This can easily be proved by the Cluster Method  see for example the NoonanZeilberger article.  N. J. A. Sloane, Aug 29 2013]
a(n4) is the number of nbit sequences that start and end with 0 but avoid both 00 and 010. For n >= 6, such a sequence necessarily starts 011 and ends 110; deleting these 6 bits is a bijection to the preceding item.  David Callan, Mar 25 2004
Also number of compositions of n+1 into parts congruent to 1 mod m. Here m=3, A003269 for m=4, etc.  Vladeta Jovovic, Feb 09 2005
Row sums of Riordan array (1/(1x^3), x/(1x^3)).  Paul Barry, Feb 25 2005
Row sums of Riordan array (1,x(1+x^2)).  Paul Barry, Jan 12 2006
The family a(n) = a(n1) + a(nm) with a(n)=1 for n=0..m1 can be generated by considering the sums (A102547):
1 1 1 1 1 1 1 1 1 1 1 1 1
1 2 3 4 5 6 7 8 9 10
1 3 6 10 15 21 28
1 4 10 20
1

1 1 1 2 3 4 6 9 13 19 28 41 60
with (in this case 3) leading zeros added to each row.
(End)
Number of pairs of rabbits existing at period n generated by 1 pair. All pairs become fertile after 3 periods and generate thereafter a new pair at all following periods.  Carmine Suriano, Mar 20 2011
The compositions of n in which each natural number is colored by one of p different colors are called pcolored compositions of n. For n>=3, 2*a(n3) equals the number of 2colored compositions of n with all parts >= 3, such that no adjacent parts have the same color.  Milan Janjic, Nov 27 2011
Pisano period lengths of the sequence read mod m, m >= 1: 1, 7, 8, 14, 31, 56, 57, 28, 24, 217, 60, 56, 168, ... (A271953) If m=3, for example, the remainder sequence becomes 1, 1, 1, 2, 0, 1, 0, 0, 1, 1, 1, 2, 0, 1, 0, 0, 1, 1, 1, 2, 0, 1, 0, 0, 1, 1, 1, 2, 0, 1, 0, 0, 1, 1, ... with a period of length 8.  R. J. Mathar, Oct 18 2012
"In how many ways can a kangaroo jump through all points of the integer interval [1,n+1] starting at 1 and ending at n+1, while making hops that are restricted to {1,1,2}? (The OGF is the rational function 1/(1  z  z^3) corresponding to A000930.)" [Flajolet and Sedgewick, p. 373]  N. J. A. Sloane, Aug 29 2013
a(n) is the number of length n binary words in which the length of every maximal run of consecutive 0's is a multiple of 3. a(5) = 4 because we have: 00011, 10001, 11000, 11111.  Geoffrey Critzer, Jan 07 2014
a(n) is the top left entry of the nth power of the 3X3 matrix [1, 0, 1; 1, 0, 0; 0, 1, 0] or of the 3 X 3 matrix [1, 1, 0; 0, 0, 1; 1, 0, 0].  R. J. Mathar, Feb 03 2014
a(n3) is the top left entry of the nth power of any of the 3 X 3 matrices [0, 1, 0; 0, 1, 1; 1, 0, 0], [0, 0, 1; 1, 1, 0; 0, 1, 0], [0, 1, 0; 0, 0, 1; 1, 0, 1] or [0, 0, 1; 1, 0, 0; 0, 1, 1].  R. J. Mathar, Feb 03 2014
Counts closed walks of length (n+3) on a unidirectional triangle, containing a loop at one of remaining vertices.  David Neil McGrath, Sep 15 2014
a(n+2) equals the number of binary words of length n, having at least two zeros between every two successive ones.  Milan Janjic, Feb 07 2015
a(n+1)/a(n) tends to x = 1.465571... (decimal expansion given in A092526) in the limit n > infinity. This is the real solution of x^3  x^2 1 = 0. See also the formula by Benoit Cloitre, Nov 30 2002.  Wolfdieter Lang, Apr 24 2015
a(n+2) equals the number of subsets of {1,2,..,n} in which any two elements differ by at least 3.  Robert FERREOL, Feb 17 2016
Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*. Let g(n) be the set of nodes in the nth generation, so that g(0) = {0}, g(1) = {1}, g(2) = {2,x}, g(3) = {3,2x,x+1,x^2}, etc. Let T(r) be the tree obtained by substituting r for x. If a positive integer N such that r = N^(1/3) is not an integer, then the number of (not necessarily distinct) integers in g(n) is A000930(n), for n >= 1. (See A274142.)  Clark Kimberling, Jun 13 2016
a(n3) is the number of compositions of n excluding 1 and 2, n >= 3.  Gregory L. Simay, Jul 12 2016
a(n+1) is the number of multus bitstrings of length n with no runs of 3 ones.  Steven Finch, Mar 25 2020
Suppose we have a(n) samples, exactly one of which is positive. Assume the cost for testing a mix of k samples is 3 if one of the samples is positive (but you will not know which sample was positive if you test more than 1) and 1 if none of the samples is positive. Then the cheapest strategy for finding the positive sample is to have a(n3) undergo the first test and then continue with testing either a(n4) if none were positive or with a(n6) otherwise. The total cost of the tests will be n.  Ruediger Jehn, Dec 24 2020


REFERENCES

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, id. 8,80.
Crilly, Tony. "A supergolden rectangle." The Mathematical Gazette 78, No. 483 (1994): 320325. See page 234.
R. K. Guy, "Anyone for Twopins?" in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 215. [See p. 12, line 3]
H. Langman, Play Mathematics. Hafner, NY, 1962, p. 13.
David Sankoff and Lani Haque, Power Boosts for Cluster Tests, in Comparative Genomics, Lecture Notes in Computer Science, Volume 3678/2005, SpringerVerlag.  N. J. A. Sloane, Jul 09 2009
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).


LINKS

I. Amburg, K. Dasaratha, L. Flapan, T. Garrity, C. Lee, C. Mihailak, N. NeumannChun, S. Peluse, and M. Stoffregen, Stern Sequences for a Family of Multidimensional Continued Fractions: TRIPStern Sequences, arXiv:1509.05239v1 [math.CO] Sep 17 2015. See Conjecture 5.8.
M. Feinberg, New slants, Fib. Quart. 2 (1964), 223227.
R. K. Guy, Anyone for Twopins?, in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 215. [Annotated scanned copy, with permission]
Dov Jarden, Recurring Sequences, Riveon Lematematika, Jerusalem, 1966. [Annotated scanned copy] See p. 91.
Martin Küttler, Maksym Planeta, Jan Bierbaum, Carsten Weinhold, Hermann Härtig, Amnon Barak, and Torsten Hoefler, Corrected trees for reliable group communication, Proceedings of the 24th Symposium on Principles and Practice of Parallel Programming (PPoPP 2019), 287299.


FORMULA

a(n) = Sum_{i=0..floor(n/3)} binomial(n2*i, i).
a(n) = a(n2) + a(n3) + a(n4) for n>3.
a(n) = floor(d*c^n + 1/2) where c is the real root of x^3x^21 and d is the real root of 31*x^331*x^2+9*x1 (c = 1.465571... = A092526 and d = 0.611491991950812...).  Benoit Cloitre, Nov 30 2002
a(n) = Sum_{k=0..n} binomial(floor((n+2k2)/3), k).  Paul Barry, Jul 06 2004
a(n) = Sum_{k=0..n} binomial(k, floor((nk)/2))(1+(1)^(nk))/2.  Paul Barry, Jan 12 2006
a(n) = Sum_{k=0..n} binomial((n+2k)/3,(nk)/3)*(2*cos(2*Pi*(nk)/3)+1)/3.  Paul Barry, Dec 15 2006
a(n) = term (1,1) in matrix [1,1,0; 0,0,1; 1,0,0]^n.  Alois P. Heinz, Jun 20 2008
G.f.: exp( Sum_{n>=1} ((1+sqrt(1+4*x))^n + (1sqrt(1+4*x))^n)*(x/2)^n/n ).
For n >= 2, a(2*n1) = a(2*n2)+a(2*n4); a(2*n) = a(2*n1)+a(2*n3).  Vladimir Shevelev, Apr 12 2012
INVERT transform of (1,0,0,1,0,0,1,0,0,1,...) = (1, 1, 1, 2, 3, 4, 6, ...); but INVERT transform of (1,0,1,0,0,0,...) = (1, 1, 2, 3, 4, 6, ...).  Gary W. Adamson, Jul 05 2012
G.f.: 1/(G(0)x) where G(k) = 1  x^2/(1  x^2/(x^2  1/G(k+1) )); (continued fraction).  Sergei N. Gladkovskii, Dec 16 2012
G.f.: 1 + x/(G(0)x) where G(k) = 1  x^2*(2*k^2 + 3*k +2) + x^2*(k+1)^2*(1  x^2*(k^2 + 3*k +2))/G(k+1); (continued fraction).  Sergei N. Gladkovskii, Dec 27 2012
G.f.: Q(0)/2, where Q(k) = 1 + 1/(1  x*(4*k+1 + x^2)/( x*(4*k+3 + x^2) + 1/Q(k+1) )); (continued fraction).  Sergei N. Gladkovskii, Sep 08 2013
a(n) = v1*w1^n+v3*w2^n+v2*w3^n, where v1,2,3 are the roots of (1+9*x31*x^2+31*x^3): [v1=0.6114919920, v2=0.1942540040  0.1225496913*I, v3=conjugate(v2)] and w1,2,3 are the roots of (1x^2+x^3): [w1=1.4655712319, w2=0.2327856159  0.7925519925*I, w3=conjugate(w2)].  Gerry Martens, Jun 27 2015
a(n+6)^2 + a(n+1)^2 + a(n)^2 = a(n+5)^2 + a(n+4)^2 + 3*a(n+3)^2 + a(n+2)^2.  Greg Dresden, Jul 07 2021


EXAMPLE

The number of compositions of 11 without any 1's and 2's is a(113) = a(8) = 13. The compositions are (11), (8,3), (3,8), (7,4), (4,7), (6,5), (5,6), (5,3,3), (3,5,3), (3,3,5), (4,4,3), (4,3,4), (3,4,4).  Gregory L. Simay, Jul 12 2016
The compositions from the above example may be mapped to the a(8) compositions of 8 into 1's and 3's using this (more generally applicable) method: replace all numbers greater than 3 with a 3 followed by 1's to make the same total, then remove the initial 3 from the composition. Maintaining the example's order, they become (1,1,1,1,1,1,1,1), (1,1,1,1,1,3), (3,1,1,1,1,1), (1,1,1,1,3,1), (1,3,1,1,1,1), (1,1,1,3,1,1), (1,1,3,1,1,1), (1,1,3,3), (3,1,1,3), (3,3,1,1), (1,3,1,3), (1,3,3,1), (3,1,3,1).  Peter Munn, May 31 2017


MAPLE

f := proc(r) local t1, i; t1 := []; for i from 1 to r do t1 := [op(t1), 0]; od: for i from 1 to r+1 do t1 := [op(t1), 1]; od: for i from 2*r+2 to 50 do t1 := [op(t1), t1[i1]+t1[i1r]]; od: t1; end; # set r = order
with(combstruct): SeqSetU := [S, {S=Sequence(U), U=Set(Z, card > 2)}, unlabeled]: seq(count(SeqSetU, size=j), j=3..40); # Zerinvary Lajos, Oct 10 2006
add(binomial(n2*k, k), k=0..floor(n/3)) ;
a:= n> (Matrix([[1, 1, 0], [0, 0, 1], [1, 0, 0]])^n)[1, 1]: seq(a(n), n=0..50); # Alois P. Heinz, Jun 20 2008


MATHEMATICA

a[0] = 1; a[1] = a[2] = 1; a[n_] := a[n] = a[n  1] + a[n  3]; Table[ a[n], {n, 0, 40} ]
CoefficientList[Series[1/(1xx^3), {x, 0, 45}], x] (* Zerinvary Lajos, Mar 22 2007 *)
a[n_] := HypergeometricPFQ[{(1n)/3, (2n)/3, n/3}, {(1n)/ 2, n/2}, 27/4]; Table[a[n], {n, 0, 43}] (* JeanFrançois Alcover, Feb 26 2013 *)


PROG

(PARI) a(n)=polcoeff(exp(sum(m=1, n, ((1+sqrt(1+4*x))^m + (1sqrt(1+4*x))^m)*(x/2)^m/m)+x*O(x^n)), n) \\ Paul D. Hanna, Oct 08 2009
(PARI) x='x+O('x^66); Vec(1/(1(x+x^3))) \\ Joerg Arndt, May 24 2011
(Maxima) makelist(sum(binomial(n2*k, k), k, 0, n/3), n, 0, 18); \\ Emanuele Munarini, May 24 2011
(Haskell)
a000930 n = a000930_list !! n
a000930_list = 1 : 1 : 1 : zipWith (+) a000930_list (drop 2 a000930_list)
(Magma) [1, 1] cat [ n le 3 select n else Self(n1)+Self(n3): n in [1..50] ]; // Vincenzo Librandi, Apr 25 2015
(GAP) a:=[1, 1, 1];; for n in [4..50] do a[n]:=a[n1]+a[n3]; od; a; # Muniru A Asiru, Aug 13 2018
(Python)
from itertools import islice
def A000930_gen(): # generator of terms
blist = [1]*3
while True:
yield blist[0]
blist = blist[1:]+[blist[0]+blist[2]]
(SageMath)
@CachedFunction
if (n<3): return 1
else: return a(n1) + a(n3)


CROSSREFS

A120562 has the same recurrence for odd n.


KEYWORD

nonn,easy,nice


AUTHOR



EXTENSIONS



STATUS

approved



