A000931 -id:A000931 - cp's OEIS frontend

A290821 Side length of largest equilateral triangle that can be made from n or fewer equilateral triangles with integer sides s_k, subject to gcd(s_1,s_2,...,s_n) = 1.

1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 39, 49

Offset: 1

Hugo Pfoertner, Aug 11 2017

No construction from 2, 3 or 5 equilateral triangles exists. The first difference from the Padovan numbers occurs for a(15)=39, where the corresponding term A000931(19)=37. a(16)=A000931(20)=49. a(n) >= A000931(n+3). From the growth behavior of A290697 it is conjectured that a(k) > A000931(k+3) for all k > 20.
a(19) is at least 130. This compares with A000931(23) = 114. It hints of growth behavior similar to sqrt(A014529) or sqrt(A001590). Ceiling(sqrt(A001590(n))) matches a(n) to n=14, then runs 38, 52, 70, 95, 128, ... . - Peter Munn, Mar 10 2018
From Peter Munn, Mar 14 2018 re monotonicity: (Start)
For n >= 6, a(n+1) > a(n).
Sketch of proof (inductive step) expressed in terms of tiling:
Given a triangle of side a(n) tiled with n equilateral triangular tiles. Let X, Y and Z be the tiles incident on its vertices, with X being not smaller than Y or Z.
Case 1: Y and Z have no vertices coincident. Remove Y and Z, thereby reducing the tiled area to a pentagon that has edges A and C that were previously internal to the area, and an edge B between A and C. Fit a new tile T against edge B, thereby extending edges A and C. Make the tiled area triangular by fitting a new tile against each of the extended edges.
Case 2: X, Y and Z have pairwise coincident vertices. It follows that these tiles are the same size. Remove Y and Z, thereby reducing the tiled area to a rhombus. Remove the tile at the rhombus vertex opposite X. The remaining area is a pentagon, since n >= 6. Extend the area by resiting Y against X, and Z against Y so that X and Z have external edges aligned. Make the area trapezoidal by fitting a new tile against the area's edge that includes an edge of Y. Fit another tile T against the smaller of the trapezoid's parallel edges.
In each case, we now have n+1 tiles, tiling an equilateral triangle with side length a(n) plus the side of T. As the sides of new and removed tiles can be calculated by adding sides of tiles that stayed in place, the GCD of the sides is unchanged.
(End)

			a(12) = 16:
                                  *
                                 / \
                                +   +
                               /     \
                              +       +
                             /         \
                            +           +
                           /             \
                          +               +
                         /                 \
                        +                   +
                       /                     \
                      +                       +
                     /                         \
                    +                           +
                   /                             \
                  +                               +
                 /                                 \
                *---+---*---+---+---+---+---+---+---*
               / \     / \                         / \
              +   +   +   +                       +   +
             /     \ /     \                     /     \
            *---*---*       +                   +       +
           / \ / \ /         \                 /         \
          +   *---*---+---+---*               +           +
         /     \             / \             /             \
        +       +           +   +           +               +
       /         \         /     \         /                 \
      +           +       +       +       +                   +
     /             \     /         \     /                     \
    +               +   +           +   +                       +
   /                 \ /             \ /                         \
  *---+---+---+---+---*---+---+---+---*---+---+---+---+---+---+---*

Stuart Anderson, An Introduction to Triangled Equilateral Triangles
Ales Drapal and Carlo Hamalainen, An enumeration of equilateral triangle dissections, arXiv:0910.5199 [math.CO], 2009-2010.
Hugo Pfoertner, Illustration of a(15)=39.

Cf. A000931, A001590, A167123, A290653, A290697, A290820.
A014529 gives greatest area of any convex polygon constructable from such triangles.
A089047 is this sequence's equivalent for squares.

Definition modified and 5 terms prepended by Peter Munn, Mar 14 2018

A005683 Numbers of Twopins positions.

Original entry on oeis.org

1, 2, 3, 5, 8, 13, 22, 37, 63, 108, 186, 322, 559, 973, 1697, 2964, 5183, 9071, 15886, 27835, 48790, 85545, 150021, 263136, 461596, 809812, 1420813, 2492945, 4374273, 7675598, 13468787, 23634817, 41474548, 72780553, 127718046, 224125677, 393308019, 690200668

Offset: 3

N. J. A. Sloane

nonn

Appears to be a bisection of A068930. - Ralf Stephan, Apr 20 2004
The Ze3 and Ze4 sums, see A180662 for their definitions, of Losanitsch's triangle A034851 lead to this sequence with a(1) = 1 and a(2) = 1; the recurrence relation below confirms these values and gives a(0) = 0. - Johannes W. Meijer, Jul 14 2011
The complete sequence by R. K. Guy in "Anyone for Twopins?" starts with a(0)=0, a(1)=1 and a(2)=1 and has g.f. x*(1-x-x^2)/(1-2*x+x^4+x^6). - Johannes W. Meijer, Aug 14 2011
a(n) is the number of equivalence classes of subsets of {1..n-2} without isolated elements up to reflection. The reflection of a subset is the set obtained by mapping each element i to n + 1 - i. For example, the a(6)=5 equivalence classes of subsets of {1..4} are {}, {1,2}/{3,4}, {2,3}, {1,2,3}/{2,3,4}, {1,2,3,4}. If reflections are not considered equivalent then A005251(n) gives the number of subsets of {1..n-2} without isolated elements. - Andrew Howroyd, Dec 24 2019

R. K. Guy, "Anyone for Twopins?", in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 2-15.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

R. K. Guy, Anyone for Twopins?, in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 2-15. [Annotated scanned copy, with permission]
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (2, 0, 0, -1, 0, -1).

Cf. A000931, A005251.

A005683:=-(-1+z**2+z**3+z**4+z**5)/(z**3-z**2+2*z-1)/(z**3+z**2-1); [Conjectured by Simon Plouffe in his 1992 dissertation.]

CoefficientList[Series[(1-x^2-x^3-x^4-x^5)/(1-2x+x^4+x^6),{x,0,40}],x] (* or *) LinearRecurrence[{2,0,0,-1,0,-1},{1,2,3,5,8,13},40] (* Harvey P. Dale, Jun 20 2011 *)

G.f.: x^3*(1-x^2-x^3-x^4-x^5)/(1-2*x+x^4+x^6). - Ralf Stephan, Apr 20 2004
a(3)=1, a(4)=2, a(5)=3, a(6)=5, a(7)=8, a(8)=13, a(n)=2*a(n-1)- a(n-4)- a(n-6). - Harvey P. Dale, Jun 20 2011
a(n) = (A005251(n) + A000931(n+4))/2. - Andrew Howroyd, Dec 24 2019

More terms from Harvey P. Dale, Jun 20 2011

A007307 a(n) = a(n-2) + a(n-3), with a(0) = 0, a(1) = 1, a(2) = 2.

Original entry on oeis.org

0, 1, 2, 1, 3, 3, 4, 6, 7, 10, 13, 17, 23, 30, 40, 53, 70, 93, 123, 163, 216, 286, 379, 502, 665, 881, 1167, 1546, 2048, 2713, 3594, 4761, 6307, 8355, 11068, 14662, 19423, 25730, 34085, 45153, 59815, 79238, 104968, 139053, 184206, 244021, 323259, 428227

Offset: 0

N. J. A. Sloane

Also the number of maximal matchings in the (n-2)-pan graph. - Eric W. Weisstein, Dec 30 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Matching
Eric Weisstein's World of Mathematics, Maximal Independent Edge Set
Eric Weisstein's World of Mathematics, Pan Graph
Index entries for linear recurrences with constant coefficients, signature (0,1,1).

Cf. A000931, A001608.

I:=[0,1,2]; [n le 3 select I[n] else Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Jun 09 2013

G(x):=(-1-x^3)/(-1+x^2+x^3): f[0]:=G(x): for n from 1 to 58 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n]/n!,n=1..43); # Zerinvary Lajos, Mar 27 2009
# second Maple program:
a:= n-> (<<0|1|0>, <0|0|1>, <1|1|0>>^n.<<($0..2)>>)[1$2]:
seq(a(n), n=0..60);  # Alois P. Heinz, Nov 06 2016

Table[- RootSum[-1 - # + #^3 &, -16 #^n - 13 #^(n + 1) + #^(n + 2) &]/23, {n, 20}] (* Eric W. Weisstein, Dec 30 2017 *)
LinearRecurrence[{0, 1, 1}, {1, 3, 3}, 20] (* Eric W. Weisstein, Dec 30 2017 *)
CoefficientList[Series[x (-1 - 3 x - 2 x^2)/(-1 + x^2 + x^3), {x, 0, 20}], x] (* Eric W. Weisstein, Dec 30 2017 *)

From Wolfdieter Lang, Jun 15 2010: (Start)
a(n) = p(n-1) + 2*p(n-2) = p(n+1) + p(n-2), with p(n):=A000931(n+3).
O.g.f: x*(1+2*x)/(1-x^2-x^3). (End)
a(n) = (A000931(n+1) + A001608(n+1))/2. - Elmo R. Oliveira, Dec 31 2022

A047328 Numbers that are congruent to {0, 3, 5, 6} mod 7.

Original entry on oeis.org

0, 3, 5, 6, 7, 10, 12, 13, 14, 17, 19, 20, 21, 24, 26, 27, 28, 31, 33, 34, 35, 38, 40, 41, 42, 45, 47, 48, 49, 52, 54, 55, 56, 59, 61, 62, 63, 66, 68, 69, 70, 73, 75, 76, 77, 80, 82, 83, 84, 87, 89, 90, 91, 94, 96, 97, 98, 101, 103, 104, 105, 108, 110, 111

Offset: 1

N. J. A. Sloane

Indices of the odd numbers in the Padovan sequence (A000931). - Francesco Daddi, Jul 31 2011

Bruno Berselli, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A000931, A047280, A047382, A134719.

[ n: n in [0..111] | n mod 7 in [0,3,5,6] ]; // Bruno Berselli, Aug 01 2011

A047328:=n->(14*n-7+I^(2*n)-(1+3*I)*I^(-n)-(1-3*I)*I^n)/8: seq(A047328(n), n=1..100); # Wesley Ivan Hurt, May 31 2016

Table[(14n-7+I^(2n)-(1+3*I)*I^(-n)-(1-3*I)*I^n)/8, {n, 80}] (* Wesley Ivan Hurt, May 31 2016 *)

a(n)=n\4*7+[0,3,5,6][n%4+1] \\ Charles R Greathouse IV, Jul 31 2011

G.f.: x^2*(3+2x+x^2+x^3)/((1-x)^2*(1+x)*(1+x^2)). a(n) = A028762(n-2), 2R. J. Mathar, Oct 18 2008
a(n) = (1/8)*(14*n-5-(2-(-1)^n)*(1+2*(-1)^floor(n/2))). - Bruno Berselli, Aug 01 2011
From Wesley Ivan Hurt, May 31 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (14*n-7+i^(2*n)-(1+3*i)*i^(-n)-(1-3*i)*i^n)/8 where i=sqrt(-1).
a(2k) = A047280(k), a(2k-1) = A047382(k). (End)
E.g.f.: (4 - 3*sin(x) - cos(x) + (7*x - 4)*sinh(x) + (7*x - 3)*cosh(x))/4. - Ilya Gutkovskiy, May 31 2016

A068924 Number of ways to tile a 5 X 2n room with 1x2 Tatami mats. At most 3 Tatami mats may meet at a point.

Original entry on oeis.org

6, 3, 2, 2, 4, 4, 6, 8, 10, 14, 18, 24, 32, 42, 56, 74, 98, 130, 172, 228, 302, 400, 530, 702, 930, 1232, 1632, 2162, 2864, 3794, 5026, 6658, 8820, 11684, 15478, 20504, 27162, 35982, 47666, 63144, 83648, 110810, 146792, 194458, 257602, 341250

Offset: 1

Dean Hickerson, Mar 11 2002

R. J. Mathar, Paving rectangular regions with rectangular tiles,...., arXiv:1311.6135 [math.CO], Table 4.
F. Ruskey and J. Woodcock, Counting Fixed-Height Tatami Tilings, Electronic Journal of Combinatorics, Paper R126 (2009) 20 pages.
Index entries for linear recurrences with constant coefficients, signature (0,1,1).

Cf. A068930 for incongruent tilings, A068920 for more info. First column of A272474.

For n >= 6, a(n) = a(n-2) + a(n-3).
G.f.: x*(-6+x^4+7*x^3+4*x^2-3*x)/(-1+x^3+x^2). [Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009; checked and corrected by R. J. Mathar, Sep 16 2009]
a(n) = 2*A000931(n+3) for n>=3. - R. J. Mathar, Dec 06 2013

A077898 Expansion of (1 - x)^(-1)/(1 + x - 2*x^2).

Original entry on oeis.org

1, 0, 3, -2, 9, -12, 31, -54, 117, -224, 459, -906, 1825, -3636, 7287, -14558, 29133, -58248, 116515, -233010, 466041, -932060, 1864143, -3728262, 7456549, -14913072, 29826171, -59652314, 119304657, -238609284, 477218599, -954437166, 1908874365, -3817748696, 7635497427

Offset: 0

N. J. A. Sloane, Nov 17 2002

Partial sums of A077925 (signed Jacobsthal numbers). - Paul Barry, Aug 26 2003

The generalized (3,-2)-Padovan sequence p(3,-2;n). See the W. Lang link under A000931 with (A,B)=(3,-2). - Wolfdieter Lang, Jun 28 2010

			(3,-2)-Padovan combinatorics from the (3,2)-Morse code with weights -2 and 3 for 3-lines -- and 2-lines -, respectively (see the W. Lang link under A000931). n=5: two codes - -- and -- - with the weights (3^1)*(-2)^1 and (-2)^1*3^1, respectively, adding up to 2*(3)(-2) = -12 = a(5). - _Wolfdieter Lang_, Jun 28 2010

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,-2).

Cf. A000931, A053088, A077925.

[(5+(-1)^n*2^(2+n)+3*n)/9 : n in [0..50]]; // Wesley Ivan Hurt, Apr 21 2016

A077898:=n->(5+(-1)^n*2^(2+n)+3*n)/9: seq(A077898(n), n=0..50); # Wesley Ivan Hurt, Apr 21 2016

CoefficientList[Series[(1 - x)^(-1)/(1 + x - 2*x^2), {x, 0, 40}], x] (* Wesley Ivan Hurt, Apr 21 2016 *)

Vec(1/(1-x)/(1+x-2*x^2)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

G.f.: (1-x)^(-1)/(1+x-2*x^2).
a(n) = Sum_{k=0..n} Sum_{j=0..k} Sum_{i=0..j} binomial(j, i)*(-3)^i. - Paul Barry, Aug 26 2003
a(n) = (-1)^n * A053088(n). - R. J. Mathar, Aug 30 2008
From Colin Barker, Apr 21 2016: (Start)
a(n) = 3*a(n-2) - 2*a(n-3) for n>2.
a(n) = (5+(-1)^n*2^(2+n)+3*n)/9. (End)
E.g.f.: (4*exp(-2*x) + (5 + 3*x)*exp(x))/9. - Ilya Gutkovskiy, Apr 21 2016
a(n) = Sum_{k=0..n} (n+1-k)*(-2)^k. - Bruno Berselli, May 15 2018

A103397 Semiprimes in A103377.

Original entry on oeis.org

4, 9, 15, 21, 33, 38, 58, 65, 86, 106, 121, 129, 265, 511, 8114, 8193, 16307, 16853, 17855, 19857, 31298, 68037, 104739, 124205, 131209, 134149, 140457, 152849, 252914, 259918, 265358, 274606, 417527, 2498871, 5291863, 8424051, 8743821

Offset: 1

Jonathan Vos Post, Feb 15 2005

			2071468241 is an element of A103377 and 2071468241= 17 * 121851073 which shows that it is a semiprime.

Cf. A001358, A000931, A079398, A103372-103381, A103377, A103397.

SemiprimeQ[n_]:=Plus@@FactorInteger[n][[All, 2]]?2; Clear[a]; k=9; Do[a[n]=1, {n, k+1}]; a[n_]:=a[n]=a[n-k]+a[n-k-1]; A103377=Array[a, 100] A103387=Union[Select[Array[a, 1000], PrimeQ]] A103397=Union[Select[Array[a, 300], SemiprimeQ]] N[Solve[x^10 - x - 1 == 0, x], 111][[2]]

Intersection of A103377 with A001358.

A104578 A Padovan convolution triangle.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 2, 0, 1, 1, 2, 3, 0, 1, 2, 3, 3, 4, 0, 1, 2, 6, 6, 4, 5, 0, 1, 3, 7, 12, 10, 5, 6, 0, 1, 4, 12, 16, 20, 15, 6, 7, 0, 1, 5, 17, 30, 30, 30, 21, 7, 8, 0, 1, 7, 24, 45, 60, 50, 42, 28, 8, 9, 0, 1, 9, 36, 70, 95, 105, 77, 56, 36, 9, 10, 0, 1, 12, 50, 111, 160, 175, 168, 112, 72

Offset: 0

Paul Barry, Mar 16 2005

A Padovan convolution triangle. See A000931 for the Padovan sequence.
Row sums are tribonacci numbers A000073(n+2). Antidiagonal sums are A008346. The first columns are A000931(n+3), A228577.
From Wolfdieter Lang, Oct 30 2018: (Start)
The alternating row sums give A001057(n+1), for n >= 0.
The inverse of this Riordan triangle is given in A319203.
The row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k, with R(-1, x) = 0, appear in the Cayley-Hamilton formula for nonnegative powers of a 3 X 3 matrix with Det M = sigma(3;3) = x1*x2*x3 = +1, sigma(3; 2) := x1*x2 + x1*x*3 + x2*x^3  = -1 and Tr M = sigma(3; 1) = x1 + x2 = x, where x1, x2, and  x3, are the eigenvalues of M, and sigma the elementary symmetric functions, as M^n = R(n-2, x)*M^2  + (R(n-3, x) + R(n-4, x))*M + R(n-3, x)*1_3, for n >= 3, where M^0 = 1_3 is the 3 X 3 unit matrix.
For the Cayley-Hamilton formula for 3 X 3 matrices with Det M = +1, sigma(3,2) = +1 and Tr(M) = x see A321196.
(End)

			From _Wolfdieter Lang_, Oct 30 2018: (Start)
The triangle T begins:
    n\k   0  1  2  3  4  5  6  7  8  9 10 ...
    --------------------------------------
    0:    1
    1:    0  1
    2:    1  0  1
    3:    1  2  0  1
    4:    1  2  3  0  1
    5:    2  3  3  4  0  1
    6:    2  6  6  4  5  0  1
    7:    3  7 12 10  5  6  0  1
    8:    4 12 16 20 15  6  7  0  1
    9:    5 17 30 30 30 21  7  8  0  1
   10:    7 24 45 60 50 42 28  8  9  0  1
   ...
Cayley-Hamilton formula for the tribonacci Q-matrix TQ(x) =[[x,1,1], [1,0,0], [0,1,0]] with Det(TQ) = +1, sigma(3, 2) = -1, and Tr(TQ) = x. For n = 3: TQ(x)^3 = R(1, x)*TQ(x)^2  + (R(0 x) + R(-1, x))*TQ(x) + R(0, x)*1_3 = x*TQ(x)^2 + TQ(x) + 1_3. For x = 1 see also A058265 (powers of the tribonacci constant).
Recurrence: T(6, 2) = T(5, 1) + T(4, 2) + T(3, 2) = 3 + 3 + 0 = 6.
Z- and A- recurrence with A319202 = {1, 0, 1, 1, -1, -3, 0, ...}:
  T(5, 0) = 0*1 + 1*2 + 1*3 + (-1)*0 + (-3)*1 = 2; T(5,2) = 1*2 + 0*3 + 1*0 + 1*1 = 3.
Boas-Buck type recurrence with b = {0, 2, 3, ...}: T(5, 2) = ((1+2)/(5-2)) * (3*1 + 2*0 + 0*3) = 1*3 = 3.
(End)

Tomislav Došlic and Luka Podrug, Tilings of a Honeycomb Strip and Higher Order Fibonacci Numbers, arXiv:2203.11761 [math.CO], 2022.
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Cf. A000073, A000931, A001057, A001608, A008346, A058265, A228577, A319202, A319203 (inverse), A321196.

T[n_, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + T[n-2, k] + T[n-3, k]; T[0, 0] = 1; T[, ] = 0; Table[T[n, k], {n, 0, 12}, {k, 0, n}] (* Jean-François Alcover, Jun 11 2019 *)

# uses[riordan_array from A256893]
riordan_array( 1/(1 - x^2 - x^3), x/(1 - x^2 - x^3), 8) # Peter Luschny, Nov 09 2018

Riordan array (1/(1 - x^2 - x^3), x/(1 - x^2 - x^3)).
T(n,k) = T(n-1,k-1) + T(n-2,k) + T(n-3,k), T(0,0)=1, T(n,k)=0 if k > n or if k < n. - Philippe Deléham, Jan 08 2014
From Wolfdieter Lang, Oct 30 2018: (Start)
The Riordan property  T = (G(x), x*G(x)) with G(x)= 1/(1-x^2-x^3) implies the following.
G.f. of row polynomials R(n, x) is G(x,z) = 1/(1- x*z - z^2 - z^3).
G.f. of column sequence k:  x^k/(1 - x^2 - x^3)^(k+1), k >= 0.
Boas-Buck recurrence (see the Aug 10 2017 remark in A046521, also for the reference):
T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} b(n-1-j)*T(j, k), for n >= 1, k = 0,1, ..., n-1, and input T(n,n) = 1, for n >= 0. Here b(n) = [x^n]*(d/dx)log(G(x)) = A001608(n+1), for n >= 0.
Recurrences from the A- and Z- sequences (see the W. Lang link under A006232 with references), which are A(n) = A319202(n) and Z(n) = A(n+1).
  T(0, 0) = 1, T(n, k) = 0 for n < k, and
  T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and
  T(n, k) = Sum_{j=0..n-k} A(j)*T(n-1, k-1+j), for n >= m >= 1.
(End)

A124788 Triangle read by rows: expansion of (1+xy)/(1-x^2y^2-x^3*y^2).

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 0, 1, 3, 1, 0, 0, 0, 0, 0, 0, 3, 3, 1, 0, 0, 0, 0, 0, 0, 1, 3, 4, 1, 0, 0, 0, 0, 0, 0, 0, 1, 6, 4, 1, 0, 0, 0, 0, 0, 0, 0, 0, 4, 6, 5, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 10, 5, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 10, 10, 6

Offset: 0

Paul Barry, Nov 07 2006

Row sums give A000931(n+5). Diagonal sums are A124789.

			Triangle begins
1,
0, 1,
0, 0, 1,
0, 0, 1, 1,
0, 0, 0, 1, 1,
0, 0, 0, 0, 2, 1,
0, 0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 0, 1, 3, 1,
0, 0, 0, 0, 0, 0, 3, 3, 1,
0, 0, 0, 0, 0, 0, 1, 3, 4, 1,
0, 0, 0, 0, 0, 0, 0, 1, 6, 4, 1

Cf. A124745.

A124788 := proc(n,k) binomial(floor(k/2),n-k) ; end: for n from 0 to 20 do for k from 0 to n do printf("%d, ",A124788(n,k)) ; od ; od ; # R. J. Mathar, Feb 10 2007

Number triangle T(n,k) = binomial(floor(k/2),n-k).

Column k has g.f. x^k*(1+x)^floor(k/2). - Paul Barry, Feb 01 2007

More terms from R. J. Mathar, Feb 10 2007

A185963 Row sums of number triangle A185962.

Original entry on oeis.org

1, 0, -2, -3, 0, 7, 11, 1, -24, -40, -7, 82, 145, 37, -279, -524, -174, 945, 1888, 767, -3185, -6783, -3244, 10676, 24301, 13330, -35567, -86823, -53615, 117672, 309366, 212101, -386224, -1099385, -827997, 1255937, 3896480, 3197152, -4039199, -13773374

Offset: 0

Paul Barry, Feb 07 2011

			G.f. = 1 - 2*x^2 - 3*x^3 + 7*x^5 + 11*x^6 + x^7 - 24*x^8 - 40*x^9 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Taras Goy and Mark Shattuck, Toeplitz-Hessenberg determinant formulas for the sequence F_n-1, Online J. Anal. Comb. (2025) Vol. 19, Paper 1, 1-26. See p. 12.
Index entries for linear recurrences with constant coefficients, signature (2,-3,1).

Cf. A000931.

a := n -> hypergeom([(n+1)/2, n/2+1, -n], [1/3, 2/3], 4/27):
seq(simplify(a(n)), n=0..39); # Peter Luschny, Nov 03 2017

LinearRecurrence[{2,-3,1},{1,0,-2},50] (* Vincenzo Librandi, Feb 18 2012 *)

x='x+O('x^50); Vec((1-x)^2/(1-2*x+3*x^2-x^3)) \\ G. C. Greubel, Jul 23 2017

G.f.: (1-x)^2/(1-2x+3x^2-x^3).
a(n) = Sum_{k=0..n} Sum_{i=0..(2k+2)} C(2k+2,i)*Sum_{j=0..(n-k-i)} C(k+j,j)*C(j,n-k-i-j)*(-1)^(n-k-j).
a(n) = Sum_{k=0..n} binomial(n+2k,3k)*(-1)^k = Sum_{k=0..n} A109955(n,k)*(-1)^k. - Philippe Deléham, Feb 18 2012
a(n) = A000931(-3*n). - Michael Somos, Sep 18 2012
a(n) = hypergeom([(n+1)/2, n/2+1, -n], [1/3, 2/3], 4/27). - Peter Luschny, Nov 03 2017

More terms from Philippe Deléham, Feb 07 2012

cp's OEIS Frontend

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