

A000931


Padovan sequence (or Padovan numbers): a(n) = a(n2) + a(n3) with a(0) = 1, a(1) = a(2) = 0.
(Formerly M0284 N0102)


242



1, 0, 0, 1, 0, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351, 465, 616, 816, 1081, 1432, 1897, 2513, 3329, 4410, 5842, 7739, 10252, 13581, 17991, 23833, 31572, 41824, 55405, 73396, 97229, 128801, 170625
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OFFSET

0,9


COMMENTS

Number of compositions of n into parts congruent to 2 mod 3 (offset 1).  Vladeta Jovovic, Feb 09 2005
a(n) is the number of compositions of n into parts that are odd and >= 3. Example: a(10)=3 counts 3+7, 5+5, 7+3.  David Callan, Jul 14 2006
Referred to as N0102 in R. K. Guy's "Anyone for Twopins?"  Rainer Rosenthal, Dec 05 2006
Zagier conjectures that a(n+3) is the maximum number of multiple zeta values of weight n > 1 which are linearly independent over the rationals.  Jonathan Sondow and Sergey Zlobin (sirg_zlobin(AT)mail.ru), Dec 20 2006
Starting with offset 6: (1, 1, 2, 2, 3, 4, 5, ...) = INVERT transform of A106510: (1, 1, 1, 0, 1, 1, 0, 1, 1, ...).  Gary W. Adamson, Oct 10 2008
Starting with offset 7, the sequence 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, ... is called the Fibonacci quilt sequence by Catral et al., in Fib. Q. 2017.  N. J. A. Sloane, Dec 24 2021
Triangle A145462: right border = A000931 starting with offset 6. Row sums = Padovan sequence starting with offset 7.  Gary W. Adamson, Oct 10 2008
Starting with offset 3 = row sums of triangle A146973 and INVERT transform of [1, 1, 2, 2, 3, 3, ...].  Gary W. Adamson, Nov 03 2008
a(n+5) corresponds to the diagonal sums of "triangle": 1; 1; 1,1; 1,1; 1,2,1; 1,2,1; 1,3,3,1; 1,3,3,1; 1,4,6,4,1; ..., rows of Pascal's triangle (A007318) repeated.  Philippe Deléham, Dec 12 2008
With offset 3: (1, 0, 1, 1, 1, 2, 2, ...) convolved with the tribonacci numbers prefaced with a "1": (1, 1, 1, 2, 4, 7, 13, ...) = the tribonacci numbers, A000073. (Cf. triangle A153462.)  Gary W. Adamson, Dec 27 2008
a(n) is also the number of strings of length (n8) from an alphabet {A, B} with no more than one A or 2 B's consecutively. (E.g., n = 4: {ABAB,ABBA,BABA,BABB,BBAB} and a(4+8) = 5.)  Toby Gottfried, Mar 02 2010
p(n):=A000931(n+3), n >= 1, is the number of partitions of the numbers {1,2,3,...,n} into lists of length two or three containing neighboring numbers. The 'or' is inclusive. For n=0 one takes p(0)=1. For details see the W. Lang link. There the explicit formula for p(n) (analog of the Binetde Moivre formula for Fibonacci numbers) is also given. Padovan sequences with different inputs are also considered there.  Wolfdieter Lang, Jun 15 2010
Equals the INVERTi transform of Fibonacci numbers prefaced with three 1's, i.e., (1 + x + x^2 + x^3 + x^4 + 2x^5 + 3x^6 + 5x^7 + 8x^8 + 13x^9 + ...).  Gary W. Adamson, Apr 01 2011
When run backwards gives (1)^n*A050935(n).
a(n) is the top left entry of the nth power of the 3 X 3 matrix [0, 0, 1; 1, 0, 1; 0, 1, 0] or of the 3 X 3 matrix [0, 1, 0; 0, 0, 1; 1, 1, 0].  R. J. Mathar, Feb 03 2014
Figure 4 of Brauchart et al., 2014, shows a way to "visualize the Padovan sequence as cuboid spirals, where the dimensions of each cuboid made up by the previous ones are given by three consecutive numbers in the sequence".  N. J. A. Sloane, Mar 26 2014
a(n) is the number of closed walks from a vertex of a unidirectional triangle containing an opposing directed edge (arc) between the second and third vertices. Equivalently the (1,1) entry of A^n where the adjacency matrix of digraph is A=(0,1,0;0,0,1;1,1,0).  David Neil McGrath, Dec 19 2014
Number of compositions of n3 (n >= 4) into 2's and 3's. Example: a(12)=5 because we have 333, 3222, 2322, 2232, and 2223.  Emeric Deutsch, Dec 28 2014
The Hoffman (2015) paper "offers significant evidence that the number of quantities needed to generate the weightn multiple harmonic sums mod p is" a(n).  N. J. A. Sloane, Jun 24 2016
a(n) gives the number of compositions of n5 into odd parts where the order of the 1's does not matter. For example, a(11)=4 counts the following compositions of 6: (5,1)=(1,5), (3,3), (3,1,1,1)=(1,3,1,1)=(1,1,3,1)=(1,1,1,3), (1,1,1,1,1,1).  Gregory L. Simay, Aug 04 2016
For n > 6, a(n) is the number of maximal matchings in the (n5)path graph, maximal independent vertex sets and minimal vertex covers in the (n6)path graph, and minimal edge covers in the (n5)pan graph and (n3)path graphs.  Eric W. Weisstein, Mar 30, Aug 03, and Aug 07 2017
a(2n + 5) + 2n  4, n > 2, is the number of maximal subsemigroups of the monoid of orderpreserving mappings on a set with n elements.
a(n + 6) + n  3, n > 3, is the number of maximal subsemigroups of the monoid of orderpreserving or reversing mappings on a set with n elements.
(End)
Has the property that the largest of any four consecutive terms equals the sum of the two smallest.  N. J. A. Sloane, Aug 29 2017. David Nacin points out that there are many sequences with this property, such as 1,1,1,2,1,1,1,2,1,1,1,2,... or 2,3,4,5,2,3,4,5,2,3,4,5,... or 2,2,1,3,3, 4,1,4, 5,5,1,6,6, 7,1,7, 8,8,1,9,9, 10,1,10, ... (spaces added for clarity), and a conjecture I made here in 2017 was simply wrong. I have deleted it.  N. J. A. Sloane, Oct 23 2018
a(n) is also the number of maximal cliques in the (n+6)path complement graph.  Eric W. Weisstein, Apr 12 2018
a(n+8) is the number of solus bitstrings of length n with no runs of 3 zeros.  Steven Finch, Mar 25 2020
Named after the architect Richard Padovan (b. 1935).  Amiram Eldar, Jun 08 2021
Shannon et al. (2006) credit a French architecture student Gérard Cordonnier with the discovery of these numbers.
For n >= 3, a(n) is the number of sequences of 0s and 1s of length (n2) that begin with a 0, end with a 0, contain no two consecutive 0s, and contain no three consecutive 1s.  Yifan Xie, Oct 20 2022


REFERENCES

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 47, ex. 4.
Minerva Catral, Pari L. Ford, Pamela E. Harris, Steven J. Miller, Dawn Nelson, Zhao Pan, and Huanzhong Xu, Legal Decompositions Arising from Nonpositive Linear Recurrences, Fib. Quart., 55:3 (2017), 252275. [Note that there is an earlier version of this paper, with only five authors, on the arXiv in 2016. Note to editors: do not merge these two citations.  N. J. A. Sloane, Dec 24 2021]
Richard K. Guy, "Anyone for Twopins?," in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 1011.
Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
A. G. Shannon, P. G. Anderson and A. F. Horadam, Properties of Cordonnier, Perrin and Van der Laan numbers, International Journal of Mathematical Education in Science and Technology, Volume 37:7 (2006), 825831. See P_n.
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).
Ian Stewart, L'univers des nombres, "La sculpture et les nombres", pp. 1920, BelinPour La Science, Paris, 2000.
Steven J. Tedford, Combinatorial identities for the Padovan numbers, Fib. Q., Vol. 57, No. 4 (2019), pp. 291298.
Hans van der Laan, Het plastische getal. XV lessen over de grondslagen van de architectonische ordonnantie. Leiden, E.J. Brill, 1967.
Don Zagier, Values of zeta functions and their applications, in First European Congress of Mathematics (Paris, 1992), Vol. II, A. Joseph et al. (eds.), Birkhäuser, Basel, 1994, pp. 497512.


LINKS

David Applegate, Marc LeBrun and N. J. A. Sloane, Dismal Arithmetic, J. Int. Seq. 14 (2011) # 11.9.8.
Richard 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. 90.
Steven J. Miller and Alexandra Newlon, The Fibonacci Quilt Game, arXiv preprint arXiv:1909.01938 [math.NT], 2019. Also Fib. Q., Vol. 58, No. 2 (2020), pp. 157168. (See Fig. 2, The "Fibonacci Quilt" sequence.)
Richard Padovan, Dom Hans van der Laan and the Plastic Number, Chapter 74, pp 407419, Volume II of K. Williams and M.J. Ostwald (eds.), Architecture and Mathematics from Antiquity to the Future, DOI 10.1007/9783319001432_27, Springer International Publishing Switzerland 2015.
Eric Weisstein's World of Mathematics, Pan Graph.
Eric Weisstein's World of Mathematics, Path Graph.


FORMULA

G.f.: (1x^2)/(1x^2x^3).
a(n) is asymptotic to r^n / (2*r+3) where r = 1.3247179572447... = A060006, the real root of x^3 = x + 1.  Philippe Deléham, Jan 13 2004
a(n)^2 + a(n+2)^2 + a(n+6)^2 = a(n+1)^2 + a(n+3)^2 + a(n+4)^2 + a(n+5)^2 (Barniville, Question 16884, Ed. Times 1911).
a(n+5) = a(0) + a(1) + ... + a(n).
a(n) = central and lower right terms in the (n3)th power of the 3 X 3 matrix M = [0 1 0 / 0 0 1 / 1 1 0]. E.g., a(13) = 7. M^10 = [3 5 4 / 4 7 5 / 5 9 7].  Gary W. Adamson, Feb 01 2004
G.f.: 1/(1  x^3  x^5  x^7  x^9  ...).  Jon Perry, Jul 04 2004
a(n+4) = Sum_{k=0..floor((n1)/2)} binomial(floor((n+k2)/3), k).  Paul Barry, Jul 06 2004
a(n+3) is diagonal sum of A026729 (as a number triangle), with formula a(n+3) = Sum_{k=0..floor(n/2)} Sum_{i=0..nk} (1)^(nk+i)*binomial(nk, i)*binomial(i+k, ik).  Paul Barry, Sep 23 2004
a(n+3) = Sum_{k=0..floor(n/2)} binomial((nk)/2, k)(1+(1)^(nk))/2.  Paul Barry, Sep 09 2005
The sequence 1/(1x^2x^3) (a(n+3)) is given by the diagonal sums of the Riordan array (1/(1x^3), x/(1x^3)). The row sums are A000930.  Paul Barry, Feb 25 2005
a(n+5) corresponds to the diagonal sums of A030528. The binomial transform of a(n+5) is A052921. a(n+5) = Sum_{k=0..floor(n/2)} Sum_{k=0..n} (1)^(nk+i)binomial(nk, i)binomial(i+k+1, 2k+1).  Paul Barry, Jun 21 2004
r^(n1) = (1/r)*a(n) + r*(n+1) + a(n+2), where r = 1.32471... is the real root of x^3  x  1 = 0. Example: r^8 = (1/r)*a(9) + r*a(10) + a(11) = ((1/r)*2 + r*3 + 4 = 9.483909...  Gary W. Adamson, Oct 22 2006
a(n) = (r^n)/(2r+3) + (s^n)/(2s+3) + (t^n)/(2t+3) where r, s, t are the three roots of x^3x1.  Keith Schneider (schneidk(AT)email.unc.edu), Sep 07 2007
a(n) = k*a(n1) + a(n2) + (k+1)a(n2) + k*a(n4), n > 3, for any value of k.  Gary Detlefs, Sep 13 2010
a(0) + a(2) + a(4) + a(6) + ... + a(2*n) = a(2*n+3).
a(0) + a(3) + a(6) + a(9) + ... + a(3*n) = a(3*n+2)+1.
a(0) + a(5) + a(10) + a(15) + ... + a(5*n) = a(5*n+1)+1.
a(0) + a(7) + a(14) + a(21) + ... + a(7*n) = (a(7*n) + a(7*n+1) + 1)/2. (End)
a(n+3) = Sum_{k=0..floor((n+1)/2)} binomial((n+k)/3,k), where binomial((n+k)/3,k)=0 for noninteger (n+k)/3.  Nikita Gogin, Dec 07 2012
a(n) = the kth difference of a(n+5k)  a(n+5k1), k>=1. For example, a(10)=3 => a(15)a(14) => 2nd difference of a(20)a(19) => 3rd difference of a(25)a(24)...  Bob Selcoe, Mar 18 2014
Construct the power matrix T(n,j) = [A^*j]*[S^*(j1)] where A=(0,0,1,0,1,0,1,...) and S=(0,1,0,0,...) or A063524. [* is convolution operation] Define S^*0=I with I=(1,0,0,...). Then a(n) = Sum_{j=1...n} T(n,j).  David Neil McGrath, Dec 19 2014
If x=a(n), y=a(n+1), z=a(n+2), then x^3 + 2*y*x^2  z^2*x  3*y*z*x + y^2*x + y^3  y^2*z + z^3 = 1.  Alexander Samokrutov, Jul 20 2015
For the sequence shifted by 6 terms, a(n) = Sum( binomial(k+1,3*kn), k=ceiling(n/3)..ceiling(n/2)) [DoslicZubac].  N. J. A. Sloane, Apr 23 2017
a(2n) = 2*a(n1)*a(n) + a(n)^2 + a(n+1)^2, for n > 8.
a(2n1) = 2*a(n)*a(n+1) + a(n1)^2, for n > 8.
a(2n+1) = 2*a(n+1)*a(n+2) + a(n)^2, for n > 7. (End)
0*a(0) + 1*a(1) + 2*a(2) + ... + n*a(n) = n*a(n+5)  a(n+9) + 2.  Greg Dresden and Zi Ye, Jul 02 2021
2*a(n) = a(n+2) + a(n5) for n >= 5.
3*a(n) = a(n+4)  a(n9) for n >= 9.
4*a(n) = a(n+5)  a(n9) for n >= 9. (End)


EXAMPLE

G.f. = 1 + x^3 + x^5 + x^6 + x^7 + 2*x^8 + 2*x^9 + 3*x^10 + 4*x^11 + ...


MAPLE

A000931 := proc(n) option remember; if n = 0 then 1 elif n <= 2 then 0 else procname(n2)+procname(n3); fi; end;
A000931:=(1+z)/(1+z^2+z^3); # Simon Plouffe in his 1992 dissertation; gives sequence without five leading terms
a[0]:=1; a[1]:=0; a[2]:=0; for n from 3 to 50 do a[n]:=a[n2]+a[n3]; end do; # Francesco Daddi, Aug 04 2011


MATHEMATICA

CoefficientList[Series[(1x^2)/(1x^2x^3), {x, 0, 50}], x]
a[0]=1; a[1]=a[2]=0; a[n_]:= a[n]= a[n2] + a[n3]; Table[a[n], {n, 0, 50}] (* Robert G. Wilson v, May 04 2006 *)
LinearRecurrence[{0, 1, 1}, {1, 0, 0}, 50] (* Harvey P. Dale, Jan 10 2012 *)
Table[RootSum[1 # +#^3 &, 5#^n 6#^(n+1) +4#^(n+2) &]/23, {n, 0, 50}] (* Eric W. Weisstein, Nov 09 2017 *)


PROG

(Haskell)
a000931 n = a000931_list !! n
a000931_list = 1 : 0 : 0 : zipWith (+) a000931_list (tail a000931_list)
(PARI) {a(n) = if( n<0, polcoeff(1/(1+xx^3) + x * O(x^n), n), polcoeff( (1  x^2)/(1x^2x^3) + x * O(x^n), n))}; /* Michael Somos, Sep 18 2012 */
(Magma) I:=[1, 0, 0]; [n le 3 select I[n] else Self(n2) + Self(n3): n in [1..60]]; // Vincenzo Librandi, Jul 21 2015
(Sage)
P.<x> = PowerSeriesRing(ZZ, prec)
return P( (1x^2)/(1x^2x^3) ).list()
(GAP) a:=[1, 0, 0];; for n in [4..50] do a[n]:=a[n2]+a[n3]; od; a; # G. C. Greubel, Dec 30 2019
(Python)
def aupton(nn):
alst = [1, 0, 0]
for n in range(3, nn+1): alst.append(alst[n2]+alst[n3])
return alst


CROSSREFS



KEYWORD

nonn,easy,nice


AUTHOR



EXTENSIONS

Deleted certain dangerous or potentially dangerous links.  N. J. A. Sloane, Jan 30 2021


STATUS

approved



