A087960 -id:A087960 - cp's OEIS frontend

A065941 T(n,k) = binomial(n-floor((k+1)/2), floor(k/2)). Triangle read by rows, for 0 <= k <= n.

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 4, 3, 3, 1, 1, 1, 5, 4, 6, 3, 1, 1, 1, 6, 5, 10, 6, 4, 1, 1, 1, 7, 6, 15, 10, 10, 4, 1, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1, 1, 1, 11, 10, 45, 36, 84, 56, 70, 35, 21, 6, 1

Offset: 0

Len Smiley, Nov 29 2001

Also the q-Stirling2 numbers at q = -1. - Peter Luschny, Mar 09 2020
Row sums give the Fibonacci sequence. So do the alternating row sums.
Triangle of coefficients of polynomials defined by p(-1,x) = p(0,x) = 1, p(n, x) = x*p(n-1, x) + p(n-2, x), for n >= 1. - Benoit Cloitre, May 08 2005 [rewritten with correct offset. - Wolfdieter Lang, Feb 18 2020]
Another version of triangle in A103631. - Philippe Deléham, Jan 01 2009
The T(n,k) coefficients appear in appendix 2 of Parks's remarkable article "A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov" if we assume that the b(n) coefficients are all equal to 1 and ignore the first column. The complete version of this triangle including the first column is A103631. - Johannes W. Meijer, Aug 11 2011
Signed ++--++..., the roots are chaotic using f(x) --> x^2 - 2 with cycle lengths shown in A003558 by n-th rows. Example: given row 3, x^3 + x^2 - 2x - 1; the roots are (a = 1.24697, ...; b = -0.445041, ...; c = -1.802937, ...). Then (say using seed b with x^2 - 2) we obtain the trajectory -0.445041, ... -> -1.80193, ... -> 1.24697, ...; matching the entry "3" in A003558(3). - Gary W. Adamson, Sep 06 2011
From Gary W. Adamson, Aug 25 2019: (Start)
Roots to the polynomials and terms in A003558 can all be obtained from the numbers below using a doubling series mod N procedure as follows: (more than one row may result). Any row ends when the trajectory produces a term already used. Then try the next higher odd term not used as the leftmost term, then repeat.
For example, for N = 11, we get: (1, 2, 4, 3, 5), showing that when confronted with two choices after the 4: (8 and -3), pick the smaller (abs) term, = 3. Then for the next row pick 7 (not used) and repeat the algorithm; succeeding only if the trajectory produces new terms. But 7 is also (-4) mod 11 and 4 was used. Therefore what I call the "r-t table" (for roots trajectory) has only one row: (1, 2, 4, 3, 5). Conjecture: The numbers of terms in the first row is equal to A003558 corresponding to N, i.e., 5 in this case with period 2.
Now for the roots to the polynomials. Pick N = 7. The polynomial is x^3 - x^2 - 2x + 1 = 0, with roots 1.8019..., -1.2469... and 0.445... corresponding to 2*cos(j*Pi/N), N = 7, and j = (1, 2, and 3). The terms (1, 2, 3) are the r-t terms for N = 7. For 11, the r-t terms are (1, 2, 4, 3, 5). This implies that given any roots of the corresponding polynomial, they are cyclic using f(x) --> x^2 - 2 with cycle lengths shown in A003558. The terms thus generated are 2*cos(j*Pi), with j = (1, 2, 4, 3, 5). Check: Begin with 2*j*Pi/N, with j = 1 (1.9189...). The other trajectory terms are: --> 1.6825..., --> 0.83083..., -1.3097...; 545...; (a 5 period and cyclic since we can begin with any of the constants). The r-t table for odd N begins as follows:
   3...............1
   5...............1, 2
   7...............1, 2, 3
   9...............1, 2, 4
    ...............3 (singleton terms reduce to "1") (9 has two rows)
  11...............1, 2, 4, 3, 5
  13...............1, 2, 4, 5, 3, 6
  15...............1, 2, 4, 7
   ................3, 6  (dividing through by the gcd gives (1, 2))
   ................5. (singleton terms reduce to "1")
The result is that 15 has 3 factors (since 3 rows), and the values of those factors are the previous terms "N", corresponding to the r-t terms in each row. Thus, the first row is new, the second (1, 2), corresponds to N = 5, and the "1" in row 3 corresponds to N = 3. The factors are those values apart from 15 and 1. Note that all of the unreduced r-t terms in all rows for N form a complete set of the terms 1 through (N-1)/2 without duplication. (End)
From Gary W. Adamson, Sep 30 2019: (Start)
The 3 factors of the 7th degree polynomial for 15: (x^7 - x^6 - 6x^5 + 5x^4 + 10x^3 - 6x^2 - 4x + 1) can be determined by getting the roots for 2*cos(j*Pi/1), j = (1, 2, 4, 7) and finding the corresponding polynomial, which is x^4 + x^3 - 4x^2 - 4x + 1. This is the minimal polynomial for N = 15 as shown in Table 2, p. 46 of (Lang). The degree of this polynomial is 4, corresponding to the entry in A003558 for 15, = 4. The trajectories (3, 6) and (5) are j values for 2*cos(j*Pi/15) which are roots to x^2 - x - 1 (relating to the pentagon), and (x - 1), relating to the triangle. (End)
From Gary W. Adamson, Aug 21 2019: (Start)
Matrices M of the form: (1's in the main diagonal, -1's in the subdiagonal, and the rest zeros) are chaotic if we replace (f(x) --> x^2 - 2) with f(x) --> M^2 - 2I, where I is the Identity matrix [1, 0, 0; 0, 1, 0; 0, 0, 1]. For example, with the 3 X 3 matrix M: [0, 0, 1; 0, 1, -1; 1, -1,  0]; the f(x) trajectory is:
  ....M^2 - 2I: [-1, -1, 0; -1, 0, -1; 0, -1, 0], then for the latter,
  ....M^2 - 2I: [0, 1, 1; 1, 0, 0; 1, 0, -1]. The cycle ends with period 3 since the next matrix is (-1) * the seed matrix. As in the case with f(x) --> x^2 - 2, the eigenvalues of the 3 chaotic matrices are (abs) 1.24697, 0.44504... and 1.80193, ... Also, the characteristic equations of the 3 matrices are the same as or variants of row 4 of the triangle below: (x^3 + x - 2x - 1) with different signs. (End)
Received from Herb Conn, Jan 2004: (Start)
Let x = 2*cos(2A) (A = Angle); then
sin(A)/sin A  = 1
sin(3A)/sin A = x + 1
sin(5A)/sin A = x^2 + x - 1
sin(7A)/sin A = x^3 + x - 2x - 1
sin(9A)/sin A = x^4 + x^3 - 3x^2 - 2x + 1
... (signed ++--++...). (End)
Or Pascal's triangle (A007318) with duplicated diagonals. Also triangle of coefficients of polynomials defined by P_0(x) = 1 and for n>=1, P_n(x) = F_n(x) + F_(n+1)(x), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 12 2012
The matrix inverse is given by
  1;
  1,  1;
  0, -1,  1;
  0,  1, -2,  1;
  0,  0,  1, -2,  1;
  0,  0, -1,  3, -3,  1;
  0,  0,  0, -1,  3, -3,  1;
  0,  0,  0,  1, -4,  6, -4,  1;
  0,  0,  0,  0,  1, -4,  6, -4,  1;
... apart from signs the same as A124645. - R. J. Mathar, Mar 12 2013

			Triangle T(n, k) begins:
n\k 0  1  2  3   4   5  6   7  8  9 ...
---------------------------------------
[0] 1,
[1] 1, 1,
[2] 1, 1, 1,
[3] 1, 1, 2, 1,
[4] 1, 1, 3, 2,  1,
[5] 1, 1, 4, 3,  3,  1,
[6] 1, 1, 5, 4,  6,  3,  1,
[7] 1, 1, 6, 5, 10,  6,  4,  1,
[8] 1, 1, 7, 6, 15, 10, 10,  4,  1,
[9] 1, 1, 8, 7, 21, 15, 20, 10,  5, 1,
---------------------------------------
From _Gary W. Adamson_, Oct 23 2019: (Start)
Consider the roots of the polynomials corresponding to odd N such that for N=7 the polynomial is (x^3 + x^2 - 2x - 1) and the roots (a, b, c) are (-1.8019377..., 1.247697..., and -0.445041...). The discriminant of a polynomial derived from the roots is the square of the product of successive differences: ((a-b), (b-c), (c-a))^2 in this case, resulting in 49, matching the method derived from the coefficients of a cubic. For our purposes we use the product of the differences, not the square, resulting in (3.048...) * (1.69202...) * (1.35689...) = 7.0. Conjecture: for all polynomials in the set, the product of the differences of the roots = the corresponding N. For N = 7, we get x^3 - 7x + 7. It appears that for all prime N's, these resulting companion polynomials are monic (left coefficient is 1), and all other coefficients are N or multiples thereof, with the rightmost term = N. The companion polynomials for the first few primes are:
  N =  5:  x^2 - 5;
  N =  7:  x^3 - 7x + 7;
  N = 11:  x^5 - 11x^3 + 11x^2 + 11x - 11;
  N = 13:  x^6 - 13x^4 + 13x^3 + 26x^2 - 39x + 13;
  N = 17:  x^8 - 17x^6 + 17x^5 + 68x^4 - 119x^3 + 17x^2 + 51x - 17;
  N = 19:  x^9 - 19x^7 + 19x^6 + 95x^5 - 171x^4 - 19x^3 + 190x^2 - 114x + 19. (End)

Nathaniel Johnston, Rows 0..100, flattened
Leonard E. Dickson, The Inscription of Regular Polygons, The American Mathematical Monthly, Vol. 1, No. 9 (Sep., 1894), pp. 299-301.
Henry W. Gould, A Variant of Pascal's Triangle, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, pp. 257-271, with corrections.
A. F. Horadam, R. P. Loh and A. G. Shannon, Divisibility properties of some Fibonacci-type sequences, pp. 55-64 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979. See Table 4.
A. F. Horadam, R. P. Loh and A. G. Shannon, Divisibility properties of some Fibonacci-type sequences, pp. 55-64 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979. [Annotated scanned copy]
Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002; p. 490.
Jay Kappraff, S. Jablan, G. W. Adamson, and R. Sazdanovich, Golden Fields, Generalized Fibonacci Sequences, and Chaotic Matrices, FORMA, Vol. 19, No. 4, 2005.
Jay Kappraff and Gary W. Adamson, Polygons and Chaos, Journal of Dynamical Systems and Geometric Theories, Vol 2 (2004), p 65.
Thomas Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons, 2001 (Chapter 14)
E. Munarini and N. Z. Salvi, Binary strings without zigzags, Séminaire Lotharingien de Combinatoire, B49h (2004), 15 pp.
Wolfdieter Lang, The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon, arXiv:1210.1018 [math.GR], 2012-2017.
P.C. Parks, A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov , Math. Proc. of the Cambridge Philosophical Society, Vol. 58, Issue 04 (1962) p. 694-702.
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A065942 (central stalk sequence), A000045 (row sums), A108299.
Reflected version of A046854.
Some triangle sums (see A180662): A000045 (Fi1), A016116 (Kn21), A000295 (Kn23), A094967 (Fi2), A000931 (Ca2), A001519 (Gi3), A000930 (Ze3).
Cf. A003558, A182579, A059841, A333143.

a065941 n k = a065941_tabl !! n !! k
a065941_row n = a065941_tabl !! n
a065941_tabl = iterate (\row ->
   zipWith (+) ([0] ++ row) (zipWith (*) (row ++ [0]) a059841_list)) [1]
-- Reinhard Zumkeller, May 07 2012

[Binomial(n - Floor((k+1)/2), Floor(k/2)): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 10 2019

A065941 := proc(n,k): binomial(n-floor((k+1)/2),floor(k/2)) end: seq(seq(A065941(n,k), k=0..n), n=0..15); # Johannes W. Meijer, Aug 11 2011
A065941 := proc(n,k) option remember: local j: if k=0 then 1 elif k=1 then 1: elif k>=2 then add(procname(j,k-2), j=k-2..n-2) fi: end: seq(seq(A065941(n,k), k=0..n), n=0..15);  # Johannes W. Meijer, Aug 11 2011
# The function qStirling2 is defined in A333143.
seq(print(seq(qStirling2(n, k, -1), k=0..n)), n=0..9);
# Peter Luschny, Mar 09 2020

Flatten[Table[Binomial[n-Floor[(k+1)/2],Floor[k/2]],{n,0,15},{k,0,n}]] (* Harvey P. Dale, Dec 11 2011 *)

T065941(n, k) = binomial(n-(k+1)\2, k\2); \\ Michel Marcus, Apr 28 2014

[[binomial(n - floor((k+1)/2), floor(k/2)) for k in (0..n)] for n in (0..15)] # G. C. Greubel, Jul 10 2019

T(n, k) = binomial(n-floor((k+1)/2), floor(k/2)).
As a square array read by antidiagonals, this is given by T1(n, k) = binomial(floor(n/2) + k, k). - Paul Barry, Mar 11 2003
Triangle is a reflection of that in A066170 (absolute values). - Gary W. Adamson, Feb 16 2004
Recurrences: T(k, 0) = 1, T(k, n) = T(k-1, n) + T(k-2, n-2), or T(k, n) = T(k-1, n) + T(k-1, n-1) if n even, T(k-1, n-1) if n odd. - Ralf Stephan, May 17 2004
G.f.: sum[n, sum[k, T(k, n)x^ky^n]] = (1+xy)/(1-y-x^2y^2). sum[n>=0, T(k, n)y^n] = y^k/(1-y)^[k/2]. - Ralf Stephan, May 17 2004
T(n, k) = A108299(n, k)*A087960(k) = abs(A108299(n, k)). - Reinhard Zumkeller, Jun 01 2005
From Johannes W. Meijer, Aug 11 2011: (Start)
  T(n,k) = A046854(n, n-k) = abs(A066170(n, n-k)).
  T(n+k, n-k) = A109223(n,k).
  T(n, k) = sum(T(j, k-2), j=k-2..n-2), 2 <= k <= n, n>=2;
  T(n, 0) =1, T(n+1, 1) = 1, n >= 0. (End)
For n > 1: T(n, k) = T(n-2, k) + T(n-1, k), 1 < k < n. - Reinhard Zumkeller, Apr 24 2013

A108299 Triangle read by rows, 0 <= k <= n: T(n,k) = binomial(n-[(k+1)/2],[k/2])*(-1)^[(k+1)/2].

Original entry on oeis.org

1, 1, -1, 1, -1, -1, 1, -1, -2, 1, 1, -1, -3, 2, 1, 1, -1, -4, 3, 3, -1, 1, -1, -5, 4, 6, -3, -1, 1, -1, -6, 5, 10, -6, -4, 1, 1, -1, -7, 6, 15, -10, -10, 4, 1, 1, -1, -8, 7, 21, -15, -20, 10, 5, -1, 1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1, 1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1, 1, -1, -11, 10, 45, -36, -84, 56, 70

Offset: 0

Reinhard Zumkeller, Jun 01 2005

Matrix inverse of A124645.
Let L(n,x) = Sum_{k=0..n} T(n,k)*x^(n-k) and Pi=3.14...:
L(n,x) = Product_{k=1..n} (x - 2*cos((2*k-1)*Pi/(2*n+1)));
Sum_{k=0..n} T(n,k) = L(n,1) = A010892(n+1);
Sum_{k=0..n} abs(T(n,k)) = A000045(n+2);
abs(T(n,k)) = A065941(n,k), T(n,k) = A065941(n,k)*A087960(k);
T(2*n,k) + T(2*n+1,k+1) = 0 for 0 <= k <= 2*n;
T(n,0) = A000012(n) = 1; T(n,1) = -1 for n > 0;
T(n,2) = -(n-1) for n > 1; T(n,3) = A000027(n)=n for n > 2;
T(n,4) = A000217(n-3) for n > 3; T(n,5) = -A000217(n-4) for n > 4;
T(n,6) = -A000292(n-5) for n > 5; T(n,7) = A000292(n-6) for n > 6;
T(n,n-3) = A058187(n-3)*(-1)^floor(n/2) for n > 2;
T(n,n-2) = A008805(n-2)*(-1)^floor((n+1)/2) for n > 1;
T(n,n-1) = A008619(n-1)*(-1)^floor(n/2) for n > 0;
T(n,n) = L(n,0) = (-1)^floor((n+1)/2);
L(n,1) = A010892(n+1); L(n,-1) = A061347(n+2);
L(n,2) = 1; L(n,-2) = A005408(n)*(-1)^n;
L(n,3) = A001519(n); L(n,-3) = A002878(n)*(-1)^n;
L(n,4) = A001835(n+1); L(n,-4) = A001834(n)*(-1)^n;
L(n,5) = A004253(n); L(n,-5) = A030221(n)*(-1)^n;
L(n,6) = A001653(n); L(n,-6) = A002315(n)*(-1)^n;
L(n,7) = A049685(n); L(n,-7) = A033890(n)*(-1)^n;
L(n,8) = A070997(n); L(n,-8) = A057080(n)*(-1)^n;
L(n,9) = A070998(n); L(n,-9) = A057081(n)*(-1)^n;
L(n,10) = A072256(n+1); L(n,-10) = A054320(n)*(-1)^n;
L(n,11) = A078922(n+1); L(n,-11) = A097783(n)*(-1)^n;
L(n,12) = A077417(n); L(n,-12) = A077416(n)*(-1)^n;
L(n,13) = A085260(n);
L(n,14) = A001570(n); L(n,-14) = A028230(n)*(-1)^n;
L(n,n) = A108366(n); L(n,-n) = A108367(n).
Row n of the matrix inverse (A124645) has g.f.: x^floor(n/2)*(1-x)^(n-floor(n/2)). - Paul D. Hanna, Jun 12 2005
From L. Edson Jeffery, Mar 12 2011: (Start)
Conjecture: Let N=2*n+1, with n > 2. Then T(n,k) (0 <= k <= n) gives the k-th coefficient in the characteristic function p_N(x)=0, of degree n in x, for the n X n tridiagonal unit-primitive matrix G_N (see [Jeffery]) of the form
  G_N=A_{N,1}=
  (0 1 0 ... 0)
  (1 0 1 0 ... 0)
  (0 1 0 1 0 ... 0)
  ...
  (0 ... 0 1 0 1)
  (0 ... 0 1 1),
with solutions phi_j = 2*cos((2*j-1)*Pi/N), j=1,2,...,n. For example, for n=3,
  G_7=A_{7,1}=
  (0 1 0)
  (1 0 1)
  (0 1 1).
We have {T(3,k)}=(1,-1,-2,1), while the characteristic function of G_7 is p(x) = x^3-x^2-2*x+1 = 0, with solutions phi_j = 2*cos((2*j-1)*Pi/7), j=1,2,3. (End)
The triangle sums, see A180662 for their definitions, link A108299 with several sequences, see the crossrefs. - Johannes W. Meijer, Aug 08 2011
The roots to the polynomials are chaotic using iterates of the operation (x^2 - 2), with cycle lengths L and initial seeds returning to the same term or (-1)* the seed. Periodic cycle lengths L are shown in A003558 such that for the polynomial represented by row r, the cycle length L is A003558(r-1). The matrices corresponding to the rows as characteristic polynomials are likewise chaotic [cf. Kappraff et al., 2005] with the same cycle lengths but substituting 2*I for the "2" in (x^2 - 2), where I = the Identity matrix. For example, the roots to x^3 - x^2 - 2x + 1 = 0 are 1.801937..., -1.246979..., and 0.445041... With 1.801937... as the initial seed and using (x^2 - 2), we obtain the 3-period trajectory of 8.801937... -> 1.246979... -> -0.445041... (returning to -1.801937...). We note that A003558(2) = 3. The corresponding matrix M is: [0,1,0; 1,0,1; 0,1,1,]. Using seed M with (x^2 - 2*I), we obtain the 3-period with the cycle completed at (-1)*M. - Gary W. Adamson, Feb 07 2012

			Triangle begins:
  1;
  1,  -1;
  1,  -1,  -1;
  1,  -1,  -2,   1;
  1,  -1,  -3,   2,   1;
  1,  -1,  -4,   3,   3,  -1;
  1,  -1,  -5,   4,   6,  -3,  -1;
  1,  -1,  -6,   5,  10,  -6,  -4,   1;
  1,  -1,  -7,   6,  15, -10, -10,   4,   1;
  1,  -1,  -8,   7,  21, -15, -20,  10,   5,  -1;
  1,  -1,  -9,   8,  28, -21, -35,  20,  15,  -5,  -1;
  1,  -1, -10,   9,  36, -28, -56,  35,  35, -15,  -6,   1;
  ...

Friedrich L. Bauer, 'De Moivre und Lagrange: Cosinus eines rationalen Vielfachen von Pi', Informatik Spektrum 28 (Springer, 2005).
Jay Kappraff, S. Jablan, G. Adamson, & R. Sazdonovich: "Golden Fields, Generalized Fibonacci Sequences, & Chaotic Matrices"; FORMA, Vol 19, No 4, (2005).

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Henry W. Gould, A Variant of Pascal's Triangle, Corrections, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, p. 257-271.
L. Edson Jeffery, Unit-primitive matrices.
Ju, Hyeong-Kwan On the sequence generated by a certain type of matrices. Honam Math. J. 39, No. 4, 665-675 (2017).
Michelle Rudolph-Lilith, On the Product Representation of Number Sequences, with Application to the Fibonacci Family, arXiv preprint arXiv:1508.07894 [math.NT], 2015.
Frank Ruskey and Carla Savage, Gray codes for set partitions and restricted growth tails, Australasian Journal of Combinatorics, Volume 10(1994), pp. 85-96. See Table 1 p. 95.

Cf. A049310, A039961, A124645 (matrix inverse).
Triangle sums (see the comments): A193884 (Kn11), A154955 (Kn21), A087960 (Kn22), A000007 (Kn3), A010892 (Fi1), A134668 (Fi2), A078031 (Ca2), A193669 (Gi1), A001519 (Gi3), A193885 (Ze1), A050935 (Ze3). - Johannes W. Meijer, Aug 08 2011
Cf. A003558.
Cf. A033999, A059841.

a108299 n k = a108299_tabl !! n !! k
a108299_row n = a108299_tabl !! n
a108299_tabl = [1] : iterate (\row ->
   zipWith (+) (zipWith (*) ([0] ++ row) a033999_list)
               (zipWith (*) (row ++ [0]) a059841_list)) [1,-1]
-- Reinhard Zumkeller, May 06 2012

A108299 := proc(n,k): binomial(n-floor((k+1)/2), floor(k/2))*(-1)^floor((k+1)/2) end: seq(seq(A108299 (n,k), k=0..n), n=0..11); # Johannes W. Meijer, Aug 08 2011

t[n_, k_?EvenQ] := I^k*Binomial[n-k/2, k/2]; t[n_, k_?OddQ] := -I^(k-1)*Binomial[n+(1-k)/2-1, (k-1)/2]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 16 2013 *)

{T(n,k)=polcoeff(polcoeff((1-x*y)/(1-x+x^2*y^2+x^2*O(x^n)),n,x)+y*O(y^k),k,y)} (Hanna)

T(n,k) = binomial(n-floor((k+1)/2),floor(k/2))*(-1)^floor((k+1)/2).
T(n+1, k) = if sign(T(n, k-1))=sign(T(n, k)) then T(n, k-1)+T(n, k) else -T(n, k-1) for 0 < k < n, T(n, 0) = 1, T(n, n) = (-1)^floor((n+1)/2).
G.f.: A(x, y) = (1 - x*y)/(1 - x + x^2*y^2). - Paul D. Hanna, Jun 12 2005
The generating polynomial (in z) of row n >= 0 is (u^(2*n+1) + v^(2*n+1))/(u + v), where u and v are defined by u^2 + v^2 = 1 and u*v = z. - Emeric Deutsch, Jun 16 2011
From Johannes W. Meijer, Aug 08 2011: (Start)
abs(T(n,k)) = A065941(n,k) = abs(A187660(n,n-k));
T(n,n-k) = A130777(n,k); abs(T(n,n-k)) = A046854(n,k) = abs(A066170(n,k)). (End)

Corrected and edited by Philippe Deléham, Oct 20 2008

A091337 a(n) = (2/n), where (k/n) is the Kronecker symbol.

Original entry on oeis.org

0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1

Offset: 0

Eric W. Weisstein, Dec 30 2003

Sinh(1) in 'reflected factorial' base is 1.01010101010101010101010101010101010101010101... see A073097 for cosh(1). - Robert G. Wilson v, May 04 2005
A non-principal character for the Dirichlet L-series modulo 8, see arXiv:1008.2547 and L-values Sum_{n >= 1} a(n)/n^s in eq (318) by Jolley. - R. J. Mathar, Oct 06 2011 [The other two non-principal characters are A101455 = {(-4/n)} and A188510 = {(-2/n)}. - Jianing Song, Nov 14 2024]
Period 8: repeat [0, 1, 0, -1, 0, -1, 0, 1]. - Wesley Ivan Hurt, Sep 07 2015 [Adapted by Jianing Song, Nov 14 2024 to include a(0) = 0.]
a(n) = (2^(2i+1)/n), where (k/n) is the Kronecker symbol and i >= 0. - A.H.M. Smeets, Jan 23 2018

			G.f. = x - x^3 - x^5 + x^7 + x^9 - x^11 - x^13 + x^15 + x^17 - x^19 - x^21 + ...

L. B. W. Jolley, Summation of series, Dover (1961).

Jianing Song, Table of n, a(n) for n = 0..1000
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv:1105.3399 [math.GM], 2011.
R. J. Mathar, Table of Dirichlet L-series..., arXiv:1008.2547 [math.NT], 2010, 2015, L(m=8,r=2,s).
Michael Somos, Rational Function Multiplicative Coefficients
Eric Weisstein's World of Mathematics, Kronecker Symbol
Index entries for linear recurrences with constant coefficients, signature (0,0,0,-1).

Cf. A000129, A035185, A073097, A087960, A102587.

Kronecker symbols {(d/n)} where d is a fundamental discriminant with |d| <= 24: A109017 (d=-24), A011586 (d=-23), A289741 (d=-20), A011585 (d=-19), A316569 (d=-15), A011582 (d=-11), A188510 (d=-8), A175629 (d=-7), A101455 (d=-4), A102283 (d=-3), A080891 (d=5), this sequence (d=8), A110161 (d=12), A011583 (d=13), A011584 (d=17), A322829 (d=21), A322796 (d=24).

[(n mod 2) * (-1)^((n+1) div 4)  : n in [1..100]]; // Vincenzo Librandi, Oct 31 2014

A091337:= n -> [0, 1, 0, -1, 0, -1, 0, 1][(n mod 8)+1]: seq(A091337(n), n=1..100); # Wesley Ivan Hurt, Sep 07 2015

KroneckerSymbol[Range[100], 2] (* Alonso del Arte, Oct 30 2014 *)

{a(n) = (n%2) * (-1)^((n+1)\4)}; /* Michael Somos, Sep 10 2005 */

{a(n) = kronecker( 2, n)}; /* Michael Somos, Sep 10 2005 */

{a(n) = [0, 1, 0, -1, 0, -1, 0, 1][n%8 + 1]}; /* Michael Somos, Jul 17 2009 */

Euler transform of length 8 sequence [0, -1, 0, -1, 0, 0, 0, 1]. - Michael Somos, Jul 17 2009
a(n) is multiplicative with a(2^e) = 0^e, a(p^e) = 1 if p == 1, 7 (mod 8), a(p^e) = (-1)^e if p == 3, 5 (mod 8). - Michael Somos, Jul 17 2009
G.f.: x*(1 - x^2)/(1 + x^4). a(n) = -a(n + 4) = a(-n) for all n in Z. a(2*n) = 0. a(2*n + 1) = A087960(n). - Michael Somos, Apr 10 2011
Transform of Pell numbers A000129 by the Riordan array A102587. - Paul Barry, Jul 14 2005
a(n) = (2/n) = (n/2), Charles R Greathouse IV explained. - Alonso del Arte, Oct 31 2014
a(n) = (1 - (-1)^n)*(-1)^(n/4 - 1/8 - (-1)^n/8 + (-1)^((2*n + 1 - (-1)^n)/4)/4)/2. - Wesley Ivan Hurt, Sep 07 2015
From Jianing Song, Nov 14 2018: (Start)
a(n) = sqrt(2)*sin(Pi*n/2)*sin(Pi*n/4).
E.g.f.: sqrt(2)*cos(x/sqrt(2))*sinh(x/sqrt(2)).
Moebius transform of A035185.
a(n) = A101455(n)*A188510(n). (End)
a(n) = Sum_{i=1..n} (-1)^(i + floor((i-3)/4)). - Wesley Ivan Hurt, Apr 27 2020
Sum_{n>=1} a(n)/n = A196525. Sum_{n>=1} a(n)/n^2 = A328895. Sum_{n>=1} a(n)/n^3 = A329715. Sum_{n>=1} a(n)/n^4 = A346728. - R. J. Mathar, Dec 17 2024

a(0) prepended by Jianing Song, Nov 14 2024

A054440 Number of ordered pairs of partitions of n with no common parts.

Original entry on oeis.org

1, 0, 2, 4, 12, 16, 48, 60, 148, 220, 438, 618, 1302, 1740, 3216, 4788, 8170, 11512, 19862, 27570, 45448, 64600, 100808, 141724, 223080, 307512, 465736, 652518, 968180, 1334030, 1972164, 2691132, 3902432, 5347176, 7611484, 10358426, 14697028, 19790508, 27691500

Offset: 0

Herbert S. Wilf, May 13 2000

			a(3)=4 because of the 4 pairs of partitions of 3: (3,21),(3,111),(21,3),(111,3).

Reinhard Zumkeller, Table of n, a(n) for n = 0..5000
Sylvie Corteel, Carla D. Savage, Herbert S. Wilf, Doron Zeilberger, A pentagonal number sieve, J. Combin. Theory Ser. A 82 (1998), no. 2, 186-192.
Eric Weisstein's World of Mathematics, Pentagonal Number Theorem
Wikipedia, Pentagonal number theorem

Cf. A000041, A001255, A001318, A087960, A260672, A260664, A260669, A304873, A304877.

Main diagonal of A284592.

a054440 = sum . zipWith (*) a087960_list . map a001255 . a260672_row
-- Reinhard Zumkeller, Nov 15 2015

with(combinat): p1 := sum(numbpart(n)^2*x^n, n=0..500): it := p1*product((1-x^i), i=1..500): s := series(it, x, 500): for i from 0 to 100 do printf(`%d,`,coeff(s,x,i)) od:

nmax = 50; CoefficientList[Series[Sum[PartitionsP[k]^2*x^k, {k, 0, nmax}]/Sum[PartitionsP[k]*x^k, {k, 0, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 04 2016 *)

G.f.: Sum[p(n)^2*x^n]/Sum[p(n)*x^n], with p(n)=number of partitions of n.
a(n) ~ sqrt(3) * exp(Pi*sqrt(2*n)) / (64 * 2^(1/4) * n^(7/4)). - Vaclav Kotesovec, May 20 2018
a(n) = [(x*y)^n] Product_{k>=1} (1 + x^k / (1 - x^k) + y^k / (1 - y^k)). - Ilya Gutkovskiy, Apr 24 2025

Corrected and extended by James Sellers, May 23 2000

A097331 Expansion of 1 + 2x/(1 + sqrt(1 - 4x^2)).

Original entry on oeis.org

1, 1, 0, 1, 0, 2, 0, 5, 0, 14, 0, 42, 0, 132, 0, 429, 0, 1430, 0, 4862, 0, 16796, 0, 58786, 0, 208012, 0, 742900, 0, 2674440, 0, 9694845, 0, 35357670, 0, 129644790, 0, 477638700, 0, 1767263190, 0, 6564120420, 0, 24466267020, 0, 91482563640, 0, 343059613650, 0

Offset: 0

Paul Barry, Aug 05 2004

Binomial transform is A097332. Second binomial transform is A014318.
Essentially the same as A126120. - R. J. Mathar, Jun 15 2008
Hankel transform is A087960(n) = (-1)^binomial(n+1,2). - Paul Barry, Aug 10 2009

Michael De Vlieger, Table of n, a(n) for n = 0..3340
Jean-Luc Baril, Sergey Kirgizov, Armen Petrossian, Motzkin paths with a restricted first return decomposition, Integers (2019) Vol. 19, A46.

A097331_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
for w from 1 to n do a[w]:=a[w-1]-(-1)^w*add(a[j]*a[w-j-1],j=1..w-1) od; convert(a,list)end: A097331_list(48); # Peter Luschny, May 19 2011

a[0] = 1; a[n_?OddQ] := CatalanNumber[(n-1)/2]; a[] = 0; Table[a[n], {n, 0, 48}] (* _Jean-François Alcover, Jul 24 2013 *)

def A097331_list(n) :
    D = [0]*(n+2); D[1] = 1
    b = True; h = 1; R = []
    for i in range(2*n-1) :
        if b :
            for k in range(h,0,-1) : D[k] -= D[k-1]
            h += 1; R.append(abs(D[1]))
        else :
            for k in range(1,h, 1) : D[k] += D[k+1]
        b = not b
    return R
A097331_list(49) # Peter Luschny, Jun 03 2012

a(n) = 0^n + Catalan((n-1)/2)(1-(-1)^n)/2.
Unsigned version of A090192, A105523. - Philippe Deléham, Sep 29 2006
From Paul Barry, Aug 10 2009: (Start)
G.f.: 1+xc(x^2), c(x) the g.f. of A000108;
G.f.: 1/(1-x/(1+x/(1+x/(1-x/(1-x/(1+x/(1+x/(1-x/(1-x/(1+... (continued fraction);
G.f.: 1+x/(1-x^2/(1-x^2/(1-x^2/(1-x^2/(1-... (continued fraction). (End)
G.f.: 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1+2*x) (continued fraction); more generally g.f. C(x/(1+2*x)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
Conjecture: (n+1)*a(n) + n*a(n-1) + 4*(-n+2)*a(n-2) + 4*(-n+3)*a(n-3)=0. - R. J. Mathar, Dec 02 2012
Recurrence: (n+3)*a(n+2) = 4*n*a(n), a(0)=a(1)=1. For nonzero terms, a(n) ~ 2^(n+1)/((n+1)^(3/2)*sqrt(2*Pi)). - Fung Lam, Mar 17 2014

A014318 Convolution of Catalan numbers and powers of 2.

Original entry on oeis.org

1, 3, 8, 21, 56, 154, 440, 1309, 4048, 12958, 42712, 144210, 496432, 1735764, 6145968, 21986781, 79331232, 288307254, 1054253208, 3875769606, 14315659632, 53097586284, 197677736208, 738415086066

Offset: 0

N. J. A. Sloane

nonn

Binomial transform of A097332: (1, 2, 3, 5, 9, 18, 39, ...). - Gary W. Adamson, Aug 01 2011

Hankel transform is A087960. - Wathek Chammam, Dec 02 2011

Fung Lam, Table of n, a(n) for n = 0..1600
W. Chammam, F. Marcellán and R. Sfaxi, Orthogonal polynomials, Catalan numbers, and a general Hankel determinant evaluation, Linear Algebra and its Applications, Volume 436, Issue 7, 1 April 2012, Pages 2105-2116.

Cf. A000079, A000108, A087960, A097332, A106270.

A014318:= func< n | (&+[2^(n-j)*Catalan(j): j in [0..n]]) >;
[A014318(n): n in [0..40]]; // G. C. Greubel, Jan 09 2023

a:=proc(n) options operator, arrow: sum(2^(n-j)*binomial(2*j,j)/(j+1), j=0..n) end proc: seq(a(n), n=0..23); # Emeric Deutsch, Oct 16 2008

a[n_]:= a[n]= Sum[2^(n-j)*CatalanNumber[j], {j,0,n}];
Table[a[n], {n,0,40}] (* G. C. Greubel, Jan 09 2023 *)

def A014318(n): return sum(2^(n-j)*catalan_number(j) for j in range(n+1))
[A014318(n) for n in range(41)] # G. C. Greubel, Jan 09 2023

From Emeric Deutsch, Oct 16 2008: (Start)
G.f.: (1-sqrt(1-4*z))/(2*z*(1-2*z)).
a(n) = Sum_{j=0..n} (2^(n-j) * binomial(2*j,j)/(j+1)). (End)
a(n) = Sum_{j=0..n} abs(A106270(n, j)) * A000079(j). - Gary W. Adamson, Apr 02 2009
Recurrence: (n+1)*a(n) = 32*(2*n-7)*a(n-5) + 48*(8-3*n)*a(n-4) + 8*(16*n-29)*a(n-3) + 4*(13-14*n)*a(n-2) + 12*n*a(n-1), n>=5. - Fung Lam, Mar 09 2014
Asymptotics: a(n) ~ 2^(2n+1)/n^(3/2)/sqrt(Pi). - Fung Lam, Mar 21 2014
G.f. A(x) satisfies: A(x) = 1 / (1 - 2*x) + x * (1 - 2*x) * A(x)^2. - Ilya Gutkovskiy, Nov 21 2021

A260664 Number of ordered triples of partitions of n with no common parts.

Original entry on oeis.org

1, 0, 6, 18, 90, 192, 864, 1710, 5970, 13110, 36810, 75984, 210546, 410130, 1003908, 2045808, 4616730, 8950176, 19746720, 37297710, 78247344, 147410640, 294299424, 543058032, 1067679540, 1925323308, 3653769792, 6555529158, 12129597486, 21348640230

Offset: 0

Reinhard Zumkeller, Nov 15 2015

nonn

			a(3) = 18 because of the 18 triples of partitions of 3: (3,3,21), (3,3,111), (3,21,3), (3,21,21), (3,21,111), (3,111,3), (3,111,21), (3,111,111), (21,3,3), (21,3,21), (21,3,111), (21,21,3), (21,111,3), (111,3,3), (111,3,21), (111,3,111), (111,21,3) and (111,111,3);
a(3) = A000041(3-A001318(0))^3 - A000041(3-A001318(1))^3 - A000041(3-A001318(2))^3 = 3^3 - 2^3 - 1^3 = 27 - 8 - 1 = 18.

Reinhard Zumkeller, Table of n, a(n) for n = 0..5000
Sylvie Corteel, Carla D. Savage, Herbert S. Wilf, Doron Zeilberger, A pentagonal number sieve, J. Combin. Theory Ser. A 82 (1998), no. 2, 186-192.
Eric Weisstein's World of Mathematics, Pentagonal Number Theorem
Wikipedia, Pentagonal number theorem

Cf. A000041, A133042, A001318, A087960, A260672, A054440, A304873, A304878.

a260664 = sum . zipWith (*) a087960_list . map a133042 . a260672_row

Table[Sum[(Cos[Pi*j/2] - Sin[Pi*j/2]) * PartitionsP[n - ((6*j^2 + 6*j + 1)/16 - (2*j + 1)*(-1)^j/16)]^3, {j, 0, Floor[Sqrt[8*n/3]]}], {n, 0, 30}] (* Vaclav Kotesovec, Jul 04 2016 *)
nmax = 50; CoefficientList[Series[Sum[PartitionsP[k]^3*x^k, {k, 0, nmax}] / Sum[PartitionsP[k]*x^k, {k, 0, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 04 2016 *)

a(n) = p(n)^3 - p(n-k(1))^3 - p(n-k(2))^3 + p(n-k(3))^3 + p(n-k(4))^3 - p(n-k(5))^3 - ..., with p=A000041 and k=A001318, see Wilf link: p. 2, (3).
G.f.: Sum[p(n)^3*x^n]/Sum[p(n)*x^n], with p(n)=number of partitions of n. - Vaclav Kotesovec, Jul 04 2016
a(n) ~ 2^(3/2) * exp(4*Pi*sqrt(n/3)) / (729 * 3^(1/4) * n^(11/4)). - Vaclav Kotesovec, May 20 2018

A112652 a(n) squared is congruent to a(n) (mod 12).

Original entry on oeis.org

0, 1, 4, 9, 12, 13, 16, 21, 24, 25, 28, 33, 36, 37, 40, 45, 48, 49, 52, 57, 60, 61, 64, 69, 72, 73, 76, 81, 84, 85, 88, 93, 96, 97, 100, 105, 108, 109, 112, 117, 120, 121, 124, 129, 132, 133, 136, 141, 144, 145, 148, 153, 156, 157, 160, 165, 168, 169, 172, 177, 180

Offset: 0

Jeremy Gardiner, Dec 28 2005

Numbers m such that A000217(3*m)/2 + A000217(2*m)/3 is an integer. - Bruno Berselli, Jul 01 2016

			a(3) = 9 because 9^2 = 81 = 6*12 + 9, hence 81 == 9 (mod 12).

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Cf. A000217, A087960.

m = 12 for n = 1 to 300 if n^2 mod m = n mod m then print n; next n

Select[Range[0, 180], Mod[#^2, 12] == Mod[#, 12] &] (* or *)
CoefficientList[Series[x (1 + 2 x + 3 x^2)/((x^2 + 1) (x - 1)^2), {x, 0, 60}], x] (* Michael De Vlieger, Jul 01 2016 *)

is(n)=(n^2-n)%12==0 \\ Charles R Greathouse IV, Oct 16 2015

From R. J. Mathar, Sep 25 2009: (Start)
G.f.: x*(1 + 2*x + 3*x^2)/((x^2 + 1)*(x - 1)^2).
a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - a(n-4).
a(n) = A087960(n) + 3*n - 1. (End)

A180666 Golden Triangle sums: a(n)=a(n-4)+A001654(n) with a(0)=0, a(1)=1, a(2)=2 and a(3)=6.

Original entry on oeis.org

0, 1, 2, 6, 15, 41, 106, 279, 729, 1911, 5001, 13095, 34281, 89752, 234971, 615165, 1610520, 4216400, 11038675, 28899630, 75660210, 198081006, 518582802, 1357667406, 3554419410, 9305590831, 24362353076, 63781468404

Offset: 0

Johannes W. Meijer, Sep 21 2010

The a(n) are the Gi2 sums of the Golden Triangle A180662. See A180662 for information about these giraffe and other chess sums.

Index entries for linear recurrences with constant coefficients, signature (2,2,-1,1,-2,-2,1).

Cf. A064831, A180664, A180665, A115730, A180666.

nmax:=27: with(combinat): for n from 0 to nmax do A001654(n):=fibonacci(n)*fibonacci(n+1) od: a(0):=0: a(1):=1: a(2):=2: a(3):=6: for n from 4 to nmax do a(n):=a(n-4)+A001654(n) od: seq(a(n),n=0..nmax);
A180666 := proc(n)
    option remember;
    if n <=3 then
        op(n+1,[0,1,2,6]) ;
    else
        procname(n-4)+A001654(n) ;
    end if;
end proc:
seq(A180666(n),n=0..100 ) ; # R. J. Mathar, Aug 18 2016

Take[Total@{#, PadLeft[Drop[#, -4], Length@ #]}, Length@ # - 4] &@ Table[Times @@ Fibonacci@ {n, n + 1}, {n, 0, 31}] (* or *)
CoefficientList[Series[(-x)/((x^2 - 3 x + 1) (x - 1) (x + 1)^2 (x^2 + 1)), {x, 0, 27}], x] (* Michael De Vlieger, Aug 18 2016 *)

a(n) = a(n-4)+A001654(n) with a(0)=0, a(1)=1, a(2)=2 and a(3)=6.
G.f.: (-x)/((x^2-3*x+1)*(x-1)*(x+1)^2*(x^2+1)).
a(n) = Sum_{k=0..floor(n/4)} A180662(n-3*k,n-4*k).
120*a(n) = 8*A001519(n) -10*A087960(n) -9*(-1)^n -15 -6*(n+1)*(-1)^n. - R. J. Mathar, Aug 18 2016

A186423 Partial sums of A186421.

Original entry on oeis.org

0, 1, 3, 4, 8, 11, 17, 20, 28, 33, 43, 48, 60, 67, 81, 88, 104, 113, 131, 140, 160, 171, 193, 204, 228, 241, 267, 280, 308, 323, 353, 368, 400, 417, 451, 468, 504, 523, 561, 580, 620, 641, 683, 704, 748, 771, 817, 840, 888, 913, 963, 988, 1040, 1067, 1121, 1148, 1204, 1233, 1291, 1320

Offset: 0

Reinhard Zumkeller, Feb 21 2011

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

A062717 is the subsequence of even terms.

A186424 is the subsequence of odd terms.

List([0..65], n-> (6*n^2 +6*n +3 +(-1)^n*(2*n+1) -4*(-1)^Binomial(n+1, 2))/16); # G. C. Greubel, Oct 09 2019

a186423 n = a186423_list !! n
a186423_list = scanl1 (+) a186421_list

[(6*n^2 +6*n +3 +(-1)^n*(2*n+1) -4*(-1)^Binomial(n+1, 2))/16: n in [0..65]]; // G. C. Greubel, Oct 09 2019

A087960 := proc(n) op((n mod 4)+1,[1,-1,-1,1]) ; end proc:
A186423 := proc(n) 3*n*(n+1)/8 +3/16 +(-1)^n*(2*n+1)/16 -A087960(n)/4 ; end proc: # R. J. Mathar, Feb 28 2011

CoefficientList[Series[x(1+2x+2x^3+x^4)/((1-x)^3(1+x)^2(1+x^2)),{x, 0, 65}],x]  (* Harvey P. Dale, Mar 13 2011 *)
Table[(6*n^2 +6*n +3 +(-1)^n*(2*n+1) -4*(-1)^Binomial[n+1, 2])/16, {n, 0, 65}] (* G. C. Greubel, Oct 09 2019 *)

vector(66, n, my(m=n-1); (6*m^2 +6*m +3 +(-1)^m*(2*m+1) -4*(-1)^binomial(m+1, 2))/16) \\ G. C. Greubel, Oct 09 2019

def A186423(n): return (6*n*(n+1)+3+(-2*n-1 if n&1 else 2*n+1)+(4 if n+1&2 else -4))>>4 # Chai Wah Wu, Jan 31 2023

[(6*n^2 +6*n +3 +(-1)^n*(2*n+1) -4*(-1)^binomial(n+1, 2))/16 for n in (0..65)] # G. C. Greubel, Oct 09 2019

From R. J. Mathar, Feb 28 2011: (Start)
G.f.: x*(1 + 2*x + 2*x^3 + x^4)/( (1+x^2)*(1+x)^2*(1-x)^3 ).
a(n) = (6*n*(n+1) + 3 + (-1)^n*(2*n+1) - 4*A087960(n))/16. (End)
E.g.f.: ((2 + 5*x + 3*x^2)*cosh(x) + (1 + 7*x + 3*x^2)*sinh(x) + 2*sin(x) - 2*cos(x))/8. - G. C. Greubel, Oct 09 2019

More terms added by G. C. Greubel, Oct 09 2019

cp's OEIS Frontend

A065941 T(n,k) = binomial(n-floor((k+1)/2), floor(k/2)). Triangle read by rows, for 0 <= k <= n.

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A108299 Triangle read by rows, 0 <= k <= n: T(n,k) = binomial(n-[(k+1)/2],[k/2])*(-1)^[(k+1)/2].

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A091337 a(n) = (2/n), where (k/n) is the Kronecker symbol.

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Magma

Maple

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PARI

PARI

PARI

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A054440 Number of ordered pairs of partitions of n with no common parts.

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Haskell

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A097331 Expansion of 1 + 2x/(1 + sqrt(1 - 4x^2)).

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Maple

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Sage

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A014318 Convolution of Catalan numbers and powers of 2.

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