A000931 -id:A000931 - cp's OEIS frontend

A034943 Binomial transform of Padovan sequence A000931.

1, 1, 1, 2, 5, 12, 28, 65, 151, 351, 816, 1897, 4410, 10252, 23833, 55405, 128801, 299426, 696081, 1618192, 3761840, 8745217, 20330163, 47261895, 109870576, 255418101, 593775046, 1380359512, 3208946545

Offset: 0

N. J. A. Sloane

Trisection of the Padovan sequence: a(n) = A000931(3n). - Paul Barry, Jul 06 2004
a(n+1) gives diagonal sums of Riordan array (1/(1-x),x/(1-x)^3). - Paul Barry, Oct 11 2005
a(n+2) is the sum, over all Boolean n-strings, of the product of the lengths of the runs of 1. For example, the Boolean 7-string (0,1,1,0,1,1,1) has two runs of 1s. Their lengths, 2 and 3, contribute a product of 6 to a(9). The 8 Boolean 3-strings contribute to a(5) as follows: 000 (empty product), 001, 010, 100, 101 all contribute 1, 011 and 110 contribute 2, 111 contributes 3. - David Callan, Nov 29 2007
[a(n), a(n+1), a(n+2)], n > 0, = [0,1,0; 0,0,1; 1,-2,3]^n * [1,1,1]. - Gary W. Adamson, Mar 27 2008
Without the initial 1 and 1: 1, 2, 5, 12, 28, this is also the transform of 1 by the T_{1,0} transformation; see Choulet link. - Richard Choulet, Apr 11 2009
Without the first 1: transform of 1 by T_{0,0} transformation (see Choulet link). - Richard Choulet, Apr 11 2009
Starting (1, 2, 5, 12, ...) = INVERT transform of (1, 1, 2, 3, 4, 5, ...) and row sums of triangle A159974. - Gary W. Adamson, Apr 28 2009
a(n+1) is also the number of 321-avoiding separable permutations. (A permutation is separable if it avoids both 2413 and 3142.) - Vince Vatter, Sep 21 2009
a(n+1) is an eigensequence of the sequence array for (1,1,2,3,4,5,...). - Paul Barry, Nov 03 2010
Equals the INVERTi transform of A055588: (1, 2, 4, 9, 22, 56, ...) - Gary W. Adamson, Apr 01 2011
The Ca3 sums, see A180662, of triangle A194005 equal the terms of this sequence without a(0) and a(1). - Johannes W. Meijer, Aug 16 2011
Without the initial 1, a(n) = row sums of A182097(n)*A007318(n,k); i.e., a Triangular array T(n,k) multiplying the binomial (Pascal's) triangle by the Padovan sequence where a(0) = 1, a(1) = 0 and a(2) = 1. - Bob Selcoe, Jun 28 2013
a(n+1) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 1; 0, 1, 1; 1, 0, 1] or [1, 1, 0; 1, 1, 1; 1, 0, 1] or [1, 1, 1; 1, 1, 0; 0, 1, 1] or [1, 0, 1; 1, 1, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
a(n) is the top left entry of the n-th power of the 3 X 3 matrix [1, 0, 1; 1, 1, 1; 0, 1, 1] or of the 3 X 3 matrix [1, 1, 0; 0, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
Number of sequences (e(1), ..., e(n-1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) != e(j) < e(k) and e(i) <= e(k). [Martinez and Savage, 2.8] - Eric M. Schmidt, Jul 17 2017
a(n+1) is the number of words of length n over the alphabet {0,1,2} that do not contain the substrings 01 or 12 and do not start with a 2 and do not end with a 0. - Yiseth K. Rodríguez C., Sep 11 2020

			G.f. = 1 + x + x^2 + 2*x^3 + 5*x^4 + 12*x^5 + 28*x^6 + 65*x^7 + 151*x^8 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Miklos Bona and Rebecca Smith, Pattern avoidance in permutations and their squares, arXiv:1901.00026 [math.CO], 2018. See H(z), Ex. 4.1.
Richard Choulet, Curtz like Transformation
Michael Dairyko, Samantha Tyner, Lara Pudwell, and Casey Wynn, Non-contiguous pattern avoidance in binary trees, Electron. J. Combin. 19 (2012), no. 3, Paper 22, 21 pp. MR2967227. - From _N. J. A. Sloane_, Feb 01 2013
Stoyan Dimitrov, Sorting by shuffling methods and a queue, arXiv:2103.04332 [math.CO], 2021.
Phan Thuan Do, Thi Thu Huong Tran, and Vincent Vajnovszki, Exhaustive generation for permutations avoiding a (colored) regular sets of patterns, arXiv:1809.00742 [cs.DM], 2018.
Brian Hopkins and Hua Wang, Restricted Color n-color Compositions, arXiv:2003.05291 [math.CO], 2020.
Jia Huang and Erkko Lehtonen, Associative-commutative spectra for some varieties of groupoids, arXiv:2401.15786 [math.CO], 2024. See p. 18.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 904
H. Magnusson and H. Ulfarsson, Algorithms for discovering and proving theorems about permutation patterns, arXiv preprint arXiv:1211.7110 [math.CO], 2012.
Megan A. Martinez and Carla D. Savage, Patterns in Inversion Sequences II: Inversion Sequences Avoiding Triples of Relations, arXiv:1609.08106 [math.CO], 2016
Vincent Vatter, Finding regular insertion encodings for permutation classes, arXiv:0911.2683 [math.CO], 2009.
Chunyan Yan and Zhicong Lin, Inversion sequences avoiding pairs of patterns, arXiv:1912.03674 [math.CO], 2019.
Index entries for linear recurrences with constant coefficients, signature (3,-2,1).

First differences of A052921.
Cf. A000931, A007318, A055588, A095263, A097550.
Cf. A137531, A159974, A182097, A185963, A194005.

[n le 3 select 1 else 3*Self(n-1)-2*Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 14 2012

A034943 := proc(n): add(binomial(n+k-1, 3*k), k=0..floor(n/2)) end: seq(A034943(n), n=0..28); # Johannes W. Meijer, Aug 16 2011

LinearRecurrence[{3,-2,1},{1,1,1},30] (* Harvey P. Dale, Aug 11 2017 *)

{a(n) = if( n<1, n = 0-n; polcoeff( (1 - x + x^2) / (1 - 2*x + 3*x^2 - x^3) + x * O(x^n), n), n = n-1; polcoeff( (1 - x + x^2) / (1 - 3*x + 2*x^2 - x^3) + x * O(x^n), n))} /* Michael Somos, Mar 31 2012 */

@CachedFunction
def a(n): # a = A034943
    if (n<3): return 1
    else: return 3*a(n-1) - 2*a(n-2) + a(n-3)
[a(n) for n in range(51)] # G. C. Greubel, Apr 22 2023

a(n) = 3*a(n-1) - 2*a(n-2) + a(n-3).
a(n) = Sum_{k=0..floor(n/2)} binomial(n+k-1, 3*k). - Paul Barry, Jul 06 2004
G.f.: (1 - 2*x)/(1 - 3*x + 2*x^2 - x^3). - Paul Barry, Jul 06 2005
G.f.: 1 + x / (1 - x / (1 - x / (1 - x / (1 + x / (1 - x))))). - Michael Somos, Mar 31 2012
a(-1 - n) = A185963(n). - Michael Somos, Mar 31 2012
a(n) = A095263(n) - 2*A095263(n-1). - G. C. Greubel, Apr 22 2023

Edited by Charles R Greathouse IV, Apr 20 2010

A133034 First differences of Padovan sequence A000931.

Original entry on oeis.org

-1, 0, 1, -1, 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

Offset: 0

Omar E. Pol, Nov 05 2007

Yuksel Soykan, Vedat Irge, and Erkan Tasdemir, A Comprehensive Study of K-Circulant Matrices Derived from Generalized Padovan Numbers, Asian Journal of Probability and Statistics 26 (12):152-70, (2024). See p. 154.
Index entries for linear recurrences with constant coefficients, signature (0,1,1).

The following are basically all variants of the same sequence: A000931, A078027, A096231, A124745, A133034, A134816, A164001, A182097, A228361 and probably A020720. However, each one has its own special features and deserves its own entry.

Cf. A002026.

LinearRecurrence[{0,1,1},{-1,0,1},60] (* Harvey P. Dale, Dec 14 2013 *)

a(n+4) = A000931(n).
G.f.: ( 1-2*x^2 ) / ( -1+x^2+x^3 ). - R. J. Mathar, Sep 11 2011
a(n) = a(n-2) + a(n-3) with a(0) = -1, a(1) = 0, a(2) = 1. - Taras Goy, Mar 24 2019

A012781 Take every 5th term of Padovan sequence A000931, beginning with the second term.

Original entry on oeis.org

0, 1, 4, 16, 65, 265, 1081, 4410, 17991, 73396, 299426, 1221537, 4983377, 20330163, 82938844, 338356945, 1380359512, 5631308624, 22973462017, 93722435101, 382349636061, 1559831901918, 6363483400447, 25960439030624

Offset: 0

N. J. A. Sloane

Number of nonisomorphic graded posets with 0 and uniform hasse graph of rank n, with exactly 2 elements of each rank level above 0, for n > 0. (Uniform used in the sense of Retakh, Serconek and Wilson.) Here, we do not assume all maximal elements have maximal rank and thus use graded poset to mean: For every element x, all maximal chains among those with x as greatest element have the same finite length. - David Nacin, Feb 13 2012

R. Stanley, Enumerative combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1997, pp. 96-100.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
V. Retakh, S. Serconek, and R. Wilson, Hilbert Series of Algebras Associated to Directed Graphs and Order Homology, arXiv:1010.6295 [math.RA], 2010-2011.
Index entries for linear recurrences with constant coefficients, signature (5,-4,1).

I:=[0, 1, 4 ]; [n le 3 select I[n] else 5*Self(n-1)-4*Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 03 2012

LinearRecurrence[{5, -4, 1}, {0, 1, 4}, 25] (* Harvey P. Dale, Jan 10 2012 *)

def a(n, adict={0:0, 1:1, 2:4}):
    if n in adict:
        return adict[n]
    adict[n]=5*a(n-1) - 4*a(n-2) + a(n-3)
    return adict[n] # David Nacin, Feb 27 2012

a(n+3) = 5*a(n+2) - 4*a(n+1) + a(n).

G.f.: x*(1-x)/(1-5*x+4*x^2-x^3). - Colin Barker, Feb 03 2012

Initial term 0 added by Colin Barker, Feb 03 2012

A012814 Take every 5th term of Padovan sequence A000931, beginning with the third term.

Original entry on oeis.org

0, 1, 5, 21, 86, 351, 1432, 5842, 23833, 97229, 396655, 1618192, 6601569, 26931732, 109870576, 448227521, 1828587033, 7459895657, 30433357674, 124155792775, 506505428836, 2066337330754, 8429820731201, 34390259761825, 140298353215075, 572360547759276, 2334999585697905

Offset: 0

N. J. A. Sloane

			G.f. = x + 5*x^2 + 21*x^3 + 86*x^4 + 351*x^5 + 1432*x^6 + 5842*x^7 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Ulrich Brenner, Anna Hermann, and Jannik Silvanus, Constructing Depth-Optimum Circuits for Adders and AND-OR Paths, arXiv:2012.05550 [cs.DM], 2020.
Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 7.
Taras Goy and Mark Shattuck, Toeplitz-Hessenberg determinant formulas for the sequence F_n-1, Online J. Anal. Comb. 19 (2024), no. 19, Paper #1, 27 pp. See Theorem 3.1.
Index entries for linear recurrences with constant coefficients, signature (5,-4,1).

Cf. A000931, A012855.

I:=[0, 1, 5 ]; [n le 3 select I[n] else 5*Self(n-1)-4*Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 03 2012

LinearRecurrence[{5, -4, 1}, {0, 1, 5}, 25] (* Vincenzo Librandi, Feb 03 2012 *)

a(n+3) = 5*a(n+2) - 4*a(n+1) + a(n).
a(n) = A000931(5*n+2).
G.f.: x/(1-5*x+4*x^2-x^3). - Colin Barker, Feb 03 2012
a(n) = A012855(n+4) - A012855(n+3).

Initial term 0 added by Colin Barker, Feb 03 2012

A139038 Triangle read by rows: T(n,m) = A000931(m+6) if m <= floor(n/2), A000931(n+6-m) otherwise, for 0 <= m <= n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 3, 2, 2, 1, 1, 1, 1, 2, 2, 3, 3, 2, 2, 1, 1, 1, 1, 2, 2, 3, 4, 3, 2, 2, 1

Offset: 0

Roger L. Bagula and Gary W. Adamson, May 31 2008

The Padovan sequence is pushed back to a(-1)=1, so that the triangle is not almost all ones.

			The triangle begins:
  1;
  1, 1;
  1, 1, 1;,
  1, 1, 1, 1;
  1, 1, 2, 1, 1;
  1, 1, 2, 2, 1, 1;
  1, 1, 2, 2, 2, 1, 1;
  1, 1, 2, 2, 2, 2, 1, 1;
  1, 1, 2, 2, 3, 2, 2, 1, 1;
  1, 1, 2, 2, 3, 3, 2, 2, 1, 1;
  1, 1, 2, 2, 3, 4, 3, 2, 2, 1, 1;
  ...

Cf. A000931, A139147.

a[-1] = 1; a[0] = 1; a[1] = 1; a[n_] := a[n] = a[n - 2] + a[n - 3]; (* Padovan : A000931 *)
Table[If[m <= Floor[n/2], a[m], a[n - m]], {n, 0, 10}, {m, 0, n}]

Edited by N. J. A. Sloane, Feb 28 2009
Non-ASCII characters in %t line corrected by Wouter Meeussen, Feb 10 2013
Definition corrected and offset changed by Georg Fischer, May 16 2024

A144400 Triangle read by rows: row n (n > 0) gives the coefficients of x^k (0 <= k <= n - 1) in the expansion of Sum_{j=0..n} A000931(j+4)binomial(n, j)x^(j - 1)*(1 - x)^(n - j).

Original entry on oeis.org

1, 2, -1, 3, -3, 1, 4, -6, 4, 0, 5, -10, 10, 0, -3, 6, -15, 20, 0, -18, 10, 7, -21, 35, 0, -63, 70, -24, 8, -28, 56, 0, -168, 280, -192, 49, 9, -36, 84, 0, -378, 840, -864, 441, -89, 10, -45, 120, 0, -756, 2100, -2880, 2205, -890, 145, 11, -55, 165, 0

Offset: 1

Roger L. Bagula and Gary W. Adamson, Oct 03 2008

			Triangle begins:
    1;
    2,  -1;
    3,  -3,   1;
    4,  -6,   4, 0;
    5, -10,  10, 0,   -3;
    6, -15,  20, 0,  -18,   10;
    7, -21,  35, 0,  -63,   70,   -24;
    8, -28,  56, 0, -168,  280,  -192,   49;
    9, -36,  84, 0, -378,  840,  -864,  441,  -89;
   10, -45, 120, 0, -756, 2100, -2880, 2205, -890, 145;
     ... reformatted. - _Franck Maminirina Ramaharo_, Oct 22 2018

G. C. Greubel, Rows n = 1..50 of the triangle, flattened

Row sums: subsequence of A000931, A078027, A182097.

Cf. A122753, A123018, A123019, A123021, A123027, A123199, A123202, A123217, A123221, A141720, A144387, A174128.

a[n_]:= a[n]= If[n<3, Fibonacci[n], a[n-2] + a[n-3]];
p[x_, n_]:= Sum[a[k]*Binomial[n, k]*x^(k-1)*(1-x)^(n-k), {k, 0, n}];
Table[Coefficient[p[x, n], x, k], {n, 12}, {k, 0, n-1}]//Flatten

@CachedFunction
def f(n): return fibonacci(n) if (n<3) else f(n-2) + f(n-3)
def p(n,x): return sum( binomial(n,j)*f(j)*x^(j-1)*(1-x)^(n-j) for j in (0..n) )
def T(n): return ( p(n,x) ).full_simplify().coefficients(sparse=False)
[T(n) for n in (1..12)] # G. C. Greubel, Jul 14 2021

G.f.: (y - (1 - 2*x)*y^2)/(1 - 3*(1 - x)*y + (3 - 6*x + 2*x^2)*y^2 - (1 - 3*x + 2*x^2 + x^3)*y^3). - Franck Maminirina Ramaharo, Oct 22 2018

Edited, and new name by Franck Maminirina Ramaharo, Oct 22 2018

A100538 Volume of the 3-dimensional box of sides of length equal to consecutive Padovan numbers (A000931). These boxes form a spiral in three dimensions similar to the spiral of Fibonacci boxes in two dimensions.

Original entry on oeis.org

1, 2, 4, 12, 24, 60, 140, 315, 756, 1728, 4032, 9408, 21756, 50764, 117845, 273910, 637260, 1480404, 3442800, 8003000, 18603000, 43251975, 100540440, 233735040, 543371136, 1263161472, 2936540824, 6826574552, 15869878969, 36893076570

Offset: 1

John Lien, Nov 27 2004

nonn

a(n)^(1/3) rounded to the nearest integer equals A000931(n+5). - Peter M. Chema, Apr 24 2017

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
I. Stewart, Tales of a Neglected Number
Ian Stewart, Tales of a Neglected Number, Mathematical Recreations, Scientific American, Vol. 274, No. 6 (1996), pp. 102-103.
Index entries for linear recurrences with constant coefficients, signature (1,2,3,-2,4,-4,-1,-1,0,-1).

Cf. A000931.

LinearRecurrence[{1, 2, 3, -2, 4, -4, -1, -1, 0, -1}, {1, 2, 4, 12, 24, 60, 140, 315, 756, 1728}, 50] (* Vincenzo Librandi, Apr 24 2017 *)

For large n a(n+1) -> a(n) * p^3 where p is the plastic number = 1.324718... a(n+1) = a(n)+ (a(n)/P(n))*P(n+1 ) where P are the Padovan numbers (A000931) starting 1, 1, 1, 2, 2, 3, 4, 5, 7, etc.

a(n) = +a(n-1) +2*a(n-2) +3*a(n-3) -2*a(n-4) +4*a(n-5) -4*a(n-6) -a(n-7) -a(n-8) -a(n-10) = A000931(n+4)*A000931(n+5)*A000931(n+6). G.f.: x*(1+x+x^3) / ( (x-1)*(x^3-2*x^2+3*x-1)*(x^6+3*x^5+5*x^4+5*x^3+5*x^2+3*x+1) ). - R. J. Mathar, Sep 14 2010

More terms from R. J. Mathar, Sep 14 2010

A133037 a(n) = A000931(n)^2.

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 1, 1, 4, 4, 9, 16, 25, 49, 81, 144, 256, 441, 784, 1369, 2401, 4225, 7396, 12996, 22801, 40000, 70225, 123201, 216225, 379456, 665856, 1168561, 2050624, 3598609, 6315169, 11082241, 19448100, 34128964, 59892121, 105103504, 184443561, 323676081

Offset: 0

Omar E. Pol, Nov 02 2007

a(n+3) is the number of tilings of an n-board (a board with dimensions n X 1) with (1/2,1/2;2)-combs and (1/2,1/2;3)-combs. A (w,g;m)-comb is a tile composed of m pieces of dimensions w X 1 separated horizontally by gaps of width g. - Michael A. Allen, Sep 25 2024

			a(10)=9 because Padovan(10)=3 and 3^2=9.

Michael A. Allen and Kenneth Edwards, Connections between two classes of generalized Fibonacci numbers squared and permanents of (0,1) Toeplitz matrices, Lin. Multilin. Alg. 72:13 (2024) 2091-2103.
Index entries for linear recurrences with constant coefficients, signature (1,1,1,-1,1,-1).

Cf. A000290, A001248, A007598. Padovan sequence: A000931.

a[0] = a[3] = a[5] = a[6] = 1; a[1] = a[2] = a[4] = 0; a[n_Integer] := a[n] = 2*a[n - 2] + 2*a[n - 3] - a[n - 7]; Table[a[i], {i, 0, 40}] (* Olivier Gérard, Jul 05 2011 *)
Table[RootSum[-1 - # + #^3 &, #^n (5 - 6 # + 4 #^2) &]^2/529, {n, 0,
40}] (* Eric W. Weisstein, Apr 16 2018 *)
LinearRecurrence[{1, 1, 1, -1, 1, -1}, {1, 0, 0, 1, 0, 1}, 40] (* Eric W. Weisstein, Apr 16 2018 *)

Vec(O(x^20)+(1-x-x^2-x^5)/(1-x-x^2-x^3+x^4-x^5+x^6)) \\ Charles R Greathouse IV, Jul 05 2011

a(n) = A000931(n)^2.
a(n) = a(n-1) + a(n-2) + a(n-3) - a(n-4) + a(n-5) - a(n-6).
G.f.: (x^5+x^2+x-1)/(-x^6+x^5-x^4+x^3+x^2+x-1).
a(n) = a(n-2) + a(n-3) + 2*Sum_{r=8..n} ( A000930(r-8)*a(n+3-r) ) for n >= 3. - Michael A. Allen, Sep 25 2024

A134732 Concatenation of first n members of the Padovan sequence A000931, starting at (1, 1, 1, 2,).

Original entry on oeis.org

1, 11, 111, 1112, 11122, 111223, 1112234, 11122345, 111223457, 1112234579, 111223457912, 11122345791216, 1112234579121621, 111223457912162128, 11122345791216212837, 1112234579121621283749, 111223457912162128374965

Offset: 1

Omar E. Pol, Nov 10 2007

Harvey P. Dale, Table of n, a(n) for n = 1..125

Cf. A000931, A007908, A019523, A102397. See A132347 for another version.

Module[{nn=20,padseq},padseq=LinearRecurrence[{0,1,1},{1,1,1,2},nn];Table[ FromDigits[ Flatten[IntegerDigits/@Take[padseq,n]]],{n,nn}]] (* Harvey P. Dale, Feb 18 2023 *)

A329227 Products of consecutive terms of the Padovan sequence A000931.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 2, 4, 6, 12, 20, 35, 63, 108, 192, 336, 588, 1036, 1813, 3185, 5590, 9804, 17214, 30200, 53000, 93015, 163215, 286440, 502656, 882096, 1547992, 2716504, 4767161, 8365777, 14680890, 25763220, 45211238, 79340228, 139232412, 244335771

Offset: 0

David Nacin, Nov 08 2019

nonn

			For n=5, a(5) = A000931(5)*A000931(6) = 1*1.

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A000931, A133037.

Times@@@Partition[LinearRecurrence[{0,1,1},{1,0,0},50],2,1] (* Harvey P. Dale, Jul 05 2021 *)

p = lambda x:[1,0,0][x] if x<3 else p(x-2)+p(x-3)
a = lambda x:p(x)*p(x+1)

a(n) = A000931(n)*A000931(n+1).
a(n+2) = Sum_{i=0..n} A000931(i)*A000931(i+2).
a(n) - a(n-2) - a(n-3) - a(n-4) = A133037(n-2) + A133037(n-3) for n>3.
G.f.: x^5 / ((1 - 2*x + x^2 - x^3)*(1 + x - x^3)) (conjectured). - Colin Barker, Nov 08 2019

cp's OEIS Frontend

A034943 Binomial transform of Padovan sequence A000931.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

SageMath

Formula

Extensions

A133034 First differences of Padovan sequence A000931.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A012781 Take every 5th term of Padovan sequence A000931, beginning with the second term.

Views

Author

Keywords

Comments

References

Links

Programs

Magma

Mathematica

Python

Formula

Extensions

A012814 Take every 5th term of Padovan sequence A000931, beginning with the third term.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Formula

Extensions

A139038 Triangle read by rows: T(n,m) = A000931(m+6) if m <= floor(n/2), A000931(n+6-m) otherwise, for 0 <= m <= n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Extensions

A144400 Triangle read by rows: row n (n > 0) gives the coefficients of x^k (0 <= k <= n - 1) in the expansion of Sum_{j=0..n} A000931(j+4)*binomial(n, j)*x^(j - 1)*(1 - x)^(n - j).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Sage

Formula

Extensions

A100538 Volume of the 3-dimensional box of sides of length equal to consecutive Padovan numbers (A000931). These boxes form a spiral in three dimensions similar to the spiral of Fibonacci boxes in two dimensions.

Views

Author

Keywords

Comments

Links

Crossrefs

A144400 Triangle read by rows: row n (n > 0) gives the coefficients of x^k (0 <= k <= n - 1) in the expansion of Sum_{j=0..n} A000931(j+4)binomial(n, j)x^(j - 1)*(1 - x)^(n - j).