A057077 -id:A057077 - cp's OEIS frontend

A244419 Coefficient triangle of polynomials related to the Dirichlet kernel. Rising powers. Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)).

1, 1, 2, -1, 2, 4, -1, -4, 4, 8, 1, -4, -12, 8, 16, 1, 6, -12, -32, 16, 32, -1, 6, 24, -32, -80, 32, 64, -1, -8, 24, 80, -80, -192, 64, 128, 1, -8, -40, 80, 240, -192, -448, 128, 256, 1, 10, -40, -160, 240, 672, -448, -1024, 256, 512, -1, 10, 60, -160, -560, 672, 1792, -1024, -2304, 512, 1024

Offset: 0

Wolfdieter Lang, Jul 29 2014

This is the row reversed version of A180870. See also A157751 and A228565.
The Dirichlet kernel is D(n,x) = Sum_{k=-n..n} exp(i*k*x) = 1 + 2*Sum_{k=1..n} T(n,x) = S(n, 2*y) + S(n-1, 2*y) = S(2*n, sqrt(2*(1+y))) with y = cos(x), n >= 0, with the Chebyshev polynomials T (A053120) and S (A049310). This triangle T(n, k) gives in row n the coefficients of the polynomial Dir(n,y) = D(n,x=arccos(y)) = Sum_{m=0..n} T(n,m)*y^m. See A180870, especially the Peter Bala comments and formulas.
This is the Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)) due to the o.g.f. for Dir(n,y) given by (1+z)/(1 - 2*y*z + z^2) = G(z)/(1 - y*F(z)) with G(z) = (1+z)/(1+z^2) and F(z) = 2*z/(1+z^2) (see the Peter Bala formula under A180870). For Riordan triangles and references see the W. Lang link 'Sheffer a- and z- sequences' under A006232.
The A- and Z- sequences of this Riordan triangle are (see the mentioned W. Lang link in the preceding comment also for the references): The A-sequence has o.g.f. 1+sqrt(1-x^2) and is given by A(2*k+1) = 0 and A(2*k) [2, -1/2, -1/8, -1/16, -5/128, -7/256, -21/1024, -33/2048, -429/32768, -715/65536, ...], k >= 0. (See A098597 and A046161.)
  The Z-sequence has o.g.f. sqrt((1-x)/(1+x)) and is given by
  [1, -1, 1/2, -1/2, 3/8, -3/8, 5/16, -5/16, 35/128, -35/128, ...]. (See A001790 and A046161.)
The column sequences are A057077, 2*(A004526 with even numbers signed), 4*A008805 (signed), 8*A058187 (signed), 16*A189976 (signed), 32*A189980 (signed) for m = 0, 1, ..., 5.
The row sums give A005408 (from the o.g.f. due to the Riordan property), and the alternating row sums give A033999.
The row polynomials Dir(n, x), n >= 0, give solutions to the diophantine equation (a + 1)*X^2 - (a - 1)*Y^2 = 2 by virtue of the identity (a + 1)*Dir(n, -a)^2 - (a - 1)*Dir(n, a)^2 = 2, which is easily proved inductively using the recurrence Dir(n, a) = (1 + a)*(-1)^(n-1)*Dir(n-1, -a) + a*Dir(n-1, a) given below by Wolfdieter Lang. - Peter Bala, May 08 2025

			The triangle T(n,m) begins:
  n\m  0   1   2    3    4    5    6     7     8    9    10 ...
  0:   1
  1:   1   2
  2:  -1   2   4
  3:  -1  -4   4    8
  4:   1  -4 -12    8   16
  5:   1   6 -12  -32   16   32
  6:  -1   6  24  -32  -80   32   64
  7:  -1  -8  24   80  -80 -192   64   128
  8:   1  -8 -40   80  240 -192 -448   128   256
  9:   1  10 -40 -160  240  672 -448 -1024   256  512
  10: -1  10  60 -160 -560  672 1792 -1024 -2304  512  1024
  ...
Example for A-sequence recurrence: T(3,1) = Sum_{j=0..2} A(j)*T(2,j) = 2*(-1) + 0*2 + (-1/2)*4 = -4. Example for Z-sequence recurrence: T(4,0) = Sum_{j=0..3} Z(j)*T(3,j) = 1*(-1) + (-1)*(-4) + (1/2)*4 + (-1/2)*8 = +1. (For the A- and Z-sequences see a comment above.)
Example for the alternate recurrence: T(4,2) = 2*T(3,1) - T(3,2) = 2*(-4) - 4 = -12. T(4,3) = 0*T(3,2) + T(3,3) = T(3,3) = 8. - _Wolfdieter Lang_, Jul 30 2014

Wikipedia, Dirichlet kernel.

Dir(n, x) : A005408 (x = 1), A002878 (x = 3/2), A001834 (x = 2), A030221 (x = 5/2), A002315 (x = 3), A033890 (x = 7/2), A057080 (x = 4), A057081 (x = 9/2), A054320 (x = 5), A077416 (x = 6), A028230 (x = 7), A159678 (x = 8), A049629 (x = 9), A083043 (x = 10),
(-1)^n * Dir(n, x): A122367 (x = -3/2); A079935 (x = -2), A004253 (x = -5/2), A001653 (x = -3), A049685 (x = -7/2), A070997 (x = -4), A070998 (x = -9/2), A072256(n+1) (x = -5).
Cf. A180870 (row reversed), A157751, A228565, A049310, A053120, A057077, A004526, A008805, A058187, A189976, A189980, A005408, A033999.

T[n_, k_] := T[n, k] = Which[k == 0, (-1)^Quotient[n, 2], (0 <= n && n < k) || (n == -1 && k == 1), 0, True, 2 T[n-1, k-1] - T[n-2, k]];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 28 2019, from Sage *)

def T(n, k):
    if k == 0: return (-1)^(n//2)
    if (0 <= n and n < k) or (n == -1 and k == 1): return 0
    return 2*T(n-1, k-1) - T(n-2, k)
for n in range(11): [T(n,k) for k in (0..n)] # Peter Luschny, Jul 29 2014

T(n, m) = [y^m] Dir(n,y) for n >= m >= 0 and 0 otherwise, with the polynomials Dir(y) defined in a comment above.
T(n, m) = 2^m*(S(n,m) + S(n-1,m)) with the entries S(n,m) of A049310 given there explicitly.
O.g.f. for polynomials Dir(y) see a comment above (Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2))).
O.g.f. for column m: ((1 + x)/(1 + x^2))*(2*x/(1 + x^2))^m, m >= 0, (Riordan property).
Recurrence for the polynomials: Dir(n, y) = 2*y*Dir(n-1, y) - Dir(n-2, y), n >= 1, with input D(-1, y) = -1 and D(0, y) = 1.
Triangle three-term recurrence: T(n,m) = 2*T(n-1,m-1) - T(n-2,m) for n >= m >= 1 with T(n,m) = 0 if 0 <= n < m, T(0,0) = 1, T(-1,1) = 0 and T(n,0) = A057077(n) = (-1)^(floor(n/2)).
From Wolfdieter Lang, Jul 30 2014: (Start)
In analogy to A157751 one can derive a recurrence for the row polynomials Dir(n, y) = Sum_{m=0..n} T(n,m)*y^m also using a negative argument but only one recursive step: Dir(n,y) = (1+y)*(-1)^(n-1)*Dir(n-1,-y) + y*Dir(n-1,y), n >= 1, Dir(0,y) = 1 (Dir(-1,y) = -1). See also A180870 from where this formula can be derived by row reversion.
This entails another triangle recurrence T(n,m) = (1 + (-1)^(n-m))*T(n-1,m-1) - (-1)^(n-m)*T(n-1,m), for n >= m >= 1 with T(n,m) = 0 if n < m and T(n,0) = (-1)^floor(n/2). (End)
From Peter Bala, Aug 14 2022: (Start)
The row polynomials Dir(n,x), n >= 0, are related to the Chebyshev polynomials of the first kind T(n,x) by the binomial transform as follows:
(2^n)*(x - 1)^(n+1)*Dir(n,x) = (-1) * Sum_{k = 0..2*n+1} binomial(2*n+1,k)*T(k,-x).
Note that Sum_{k = 0..2*n} binomial(2*n,k)*T(k,x) = (2^n)*(1 + x)^n*T(n,x). (End)
From Peter Bala, May 04 2025: (Start)
For n >= 1, the n-th row polynomial Dir(n, x) = (-1)^n * (U(n, -x) - U(n-1, -x)) = U(2*n, sqrt((1+x)/2)), where U(n, x) denotes the n-th Chebyshev polynomial of the second kind.
For n >= 1 and x < 1, Dir(n, x) = (-1)^n * sqrt(2/(1 - x )) * T(2*n+1, sqrt((1 - x)/2)), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind.
Dir(n, x)^2 - 2*x*Dir(n, x)*Dir(n+1, x) + Dir(n+1, x)^2 = 2*(1 + x).
Dir(n, x) = (-1)^n * R(n, -2*(x+1)), where R(n, x) is the n-th row polynomial of the triangle A085478.
Dir(n, x) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n+k, 2*k) * (2*x + 2)^k. (End)

A143624 Decimal expansion of the negated constant cos(1) - sin(1) = -0.3011686789...

Original entry on oeis.org

3, 0, 1, 1, 6, 8, 6, 7, 8, 9, 3, 9, 7, 5, 6, 7, 8, 9, 2, 5, 1, 5, 6, 5, 7, 1, 4, 1, 8, 7, 3, 2, 2, 3, 9, 5, 8, 9, 0, 2, 5, 2, 6, 4, 0, 1, 8, 0, 4, 4, 8, 8, 3, 8, 0, 0, 2, 6, 5, 4, 4, 5, 4, 6, 1, 0, 8, 1, 0, 0, 0, 9, 6, 1, 6, 7, 6, 7, 9, 0, 4, 4, 3, 0, 6, 8, 7, 8, 8, 7, 4, 5, 5, 8, 6, 9, 6, 0, 6, 5

Offset: 0

Peter Bala, Aug 30 2008

cos(1) - sin(1) = Sum_{n>=0} (-1)^floor(n/2)*n/n! = 1/1! - 2/2! - 3/3! + 4/4! + 5/5! - 6/6! - 7/7! + + - - ... . Define E_2(k) = Sum_{n>=0} (-1)^floor(n/2)*n^k/n! for k = 0,1,2,... . Then E_2(1) = cos(1) - sin(1) and E_2(0) = cos(1) + sin(1). Furthermore, E_2(k) is an integral linear combination of E_2(0) and E_2(1) (a Dobinski-type relation). For example, E_2(2) = E_2(1) - E_2(0), E_2(3) = -3*E_2(0) and E_2(4) = -5*E_2(1) - 6*E_2(0). The precise result is E_2(k) = A121867(k) * E_2(0) - A121868(k) * E_2(1). The decimal expansion of the constant cos(1) + sin(1) is recorded in A143623. Compare with A143625.

			 -0.30116867893975678925156571418732239589025264018...

Eric Weisstein's World of Mathematics, Spherical Bessel Function of the First Kind

Cf. A049469, A049470, A057077, A121867, A121868, A143623, A143625, A174549.

RealDigits[Cos[1] - Sin[1], 10, 100][[1]] (* Amiram Eldar, Aug 07 2020 *)

cos(1)-sin(1) \\ Charles R Greathouse IV, Feb 04 2025

Equals sin(1-Pi/4)*sqrt(2). - Franklin T. Adams-Watters, Jun 27 2014
Equals j_1(1), where j_1(z) is the spherical Bessel function of the first kind. - Stanislav Sykora, Jan 11 2017
From Amiram Eldar, Aug 07 2020: (Start)
Equals -Integral_{x=0..1} x*sin(x) dx.
Equals Sum_{k>=1} (-1)^k/((2*k-1)! * (2*k+1)) = Sum_{k>=1} (-1)^k/A174549(k). (End)

Added sign in definition. Offset corrected by R. J. Mathar, Feb 05 2009

A203421 Reciprocal of Vandermonde determinant of (1,1/2,...,1/n).

Original entry on oeis.org

1, 1, -2, -18, 1152, 720000, -5598720000, -658683809280000, 1381360067999170560000, 59463021447701323327733760000, -59463021447701323327733760000000000000, -1542317635347398938581016812202229760000000000000

Offset: 0

Clark Kimberling, Jan 02 2012

Each term divides its successor, as in A000169.

G. C. Greubel, Table of n, a(n) for n = 0..35

Cf. A000169, A002109, A057077, A203422, A203424.

BarnesG:= func< n | (&*[Factorial(k): k in [0..n-2]]) >;
A203421:= func< n | (-1)^Binomial(n,2)*(Factorial(n))^n/BarnesG(n+2) >;
[A203421(n): n in [1..20]]; // G. C. Greubel, Dec 07 2023

(* First program *)
f[j_] := 1/j; z = 12;
v[n_] := Product[Product[f[k] - f[j], {j, 1, k - 1}], {k, 2, n}]
Table[v[n], {n, 1, z}]
1/%  (* A203421 *)
Table[v[n]/v[n + 1], {n, 1, z}]  (* A000169 signed *)
(* Additional programs *)
Table[(-1)^Floor[n/2]*Product[(k + 1)^k, {k, 0, n-1}], {n, 1, 10}] (* Vaclav Kotesovec, Oct 18 2015 *)
Table[(-1)^Binomial[n,2]*(n!)^n/BarnesG[n+2], {n, 20}] (* G. C. Greubel, Dec 07 2023 *)

a(n) = prod(i=2,n, (-i)^(i-1)); \\ Kevin Ryde, Apr 17 2022

def BarnesG(n): return product(factorial(k) for k in range(n-1))
def A203421(n): return (-1)^binomial(n, 2)*(gamma(n+1))^n/BarnesG(n+2)
[A203421(n) for n in range(1, 21)] # G. C. Greubel, Dec 07 2023

G.f.: G(0)/(2*x) -1/x, where G(k)= 1 + 1/(1 - x/(x + (2*k+1)/((2*k+1)^(2*k+1))/(1 + 1/(1 - x/(x - (2*k+2)/((2*k+2)^(2*k+2))/G(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, Jun 03 2013
a(n) = (-1)^floor(n/2) * hyperfactorial(n)/n! = A057077(n) * A002109(n)/n!. - Paul J. Harvey, Feb 08 2014
a(n) = Product_{i=2..n} (-i)^(i-1). - Kevin Ryde, Apr 17 2022
abs(a(n)) ~ A * n^(n*(n-1)/2 - 5/12) / (sqrt(2*Pi) * exp(n^2/4 - n)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Nov 20 2023
a(n) = (-1)^binomial(n,2) * (n!)^n / BarnesG(n+2). - G. C. Greubel, Dec 07 2023

a(0)=1 prepended by Alois P. Heinz, Apr 13 2024

A238442 Triangle read by rows demonstrating Euler's pentagonal theorem for the sum of divisors.

Original entry on oeis.org

1, 1, 2, 3, 1, 4, 3, 7, 4, -5, 6, 7, -1, 12, 6, -3, -7, 8, 12, -4, -1, 15, 8, -7, -3, 13, 15, -6, -4, 18, 13, -12, -7, 12, 18, -8, -6, 12, 28, 12, -15, -12, 1, 14, 28, -13, -8, 3, 24, 14, -18, -15, 4, 15, 24, 24, -12, -13, 7, 1, 31, 24, -28, -18, 6, 3

Offset: 1

Omar E. Pol, Feb 26 2014

The law found by Leonhard Euler for the sum of divisors of n is that S(n) = S(n - 1) + S(n - 2) - S(n - 5) - S(n - 7) + S(n - 12) + S(n - 15) - S(n - 22) - S(n - 26) + S(n - 35) + S(n - 40) + ..., where the constants are the positive generalized pentagonal numbers, and S(0) = n, which is also a positive member of A001318.
Therefore column k lists A001318(k) together with the elements of A000203, starting at row A001318(k), but with all elements of column k multiplied by A057077(k-1).
The first element of column k is A057077(k-1)*A001318(k)which is also the last term of row A001318(k).
For Euler's pentagonal theorem for the partition numbers see A175003.
Note that both of Euler's pentagonal theorems refer to generalized pentagonal numbers (A001318), not to pentagonal numbers (A000326).

			Triangle begins:
   1;
   1,   2;
   3,   1;
   4,   3;
   7,   4,  -5;
   6,   7,  -1;
  12,   6,  -3,  -7;
   8,  12,  -4,  -1;
  15,   8,  -7,  -3;
  13,  15,  -6,  -4;
  18,  13, -12,  -7;
  12,  18,  -8,  -6,  12;
  28,  12, -15, -12,   1;
  14,  28, -13,  -8,   3;
  24,  14, -18, -15,   4,  15;
  24,  24, -12, -13,   7,   1;
  31,  24, -28, -18,   6,   3;
  18,  31, -14, -12,  12,   4;
  39,  18, -24, -28,   8,   7;
  20,  39, -24, -14,  15,   6;
  42,  20, -31, -24,  13,  12;
  32,  42, -18, -24,  18,   8, -22;
  36,  32, -39, -31,  12,  15,  -1;
  24,  36, -20, -18,  28,  13,  -3;
  60,  24, -42, -39,  14,  18,  -4;
  31,  60, -32, -20,  24,  12,  -7, -26;
  ...
For n = 21 the sum of divisors of 21 is 1 + 3 + 7 + 21 = 32. On the other hand, from Euler's Pentagonal Number Theorem we have that the sum of divisors of 21 is S_21 = S_20 + S_19 - S_16 - S_14 + S_9 + S_6, the same as the sum of the 21st row of triangle: 42 + 20 - 31 - 24 + 13 + 12 = 32, equaling the sum of divisors of 21.
For n = 22 the sum of divisors of 22 is 1 + 2 + 11 + 22 = 36. On the other hand, from Euler's Pentagonal Number Theorem we have that the sum of divisors of 22 is S_22 = S_21 + S_20 - S_17 - S_15 + S_10 + S_7 - S_0, the same as the sum of the 22nd row of triangle is 32 + 42 - 18 - 24 + 18 + 8 - 22 = 36, equaling the sum of divisors of 22. Note that S_0 = n, hence in this case S_0 = 22.

L. Euler, An observation on the sums of divisors, arXiv:math/0411587 [math.HO], 2004-2009, p. 8.
L. Euler, De mirabilibus proprietatibus numerorum pentagonalium
L. Euler, Découverte d'une loi tout extraordinaire des nombres par rapport à la somme de leurs diviseurs
L. Euler, Discovery of a most extraordinary law of numbers, relating to the sum of their divisors
L. Euler, Observatio de summis divisorum, p. 8.
L. Euler, On the remarkable properties of the pentagonal numbers, arXiv:math/0505373 [math.HO], 2005.
L. Euler, J. Bell, A demonstration of a theorem on the order observed in the sums of divisors, arXiv:math/0507201 [math.HO], 2005-2009.
Index entries for sequences related to sigma(n)

Row sums give A000203, the sum of divisors of n.
Row n has length A235963(n).
Cf. A001318, A027750, A057077, A175003, A196020, A237270, A237273.

rows = m = 18;
a057077[n_] := {1, 1, -1, -1}[[Mod[n, 4] + 1]];
a001318[n_] := (1/8)((2n + 1) Mod[n, 2] + 3n^2 + 2n);
a235963[n_] := Flatten[Table[k, {k, 0, m}, {(k+1)/(Mod[k, 2]+1)}]][[n+1]];
T[n_, k_] := If[n == a001318[k] && k == a235963[n], a001318[k] a057077[k - 1], a057077[k - 1] DivisorSigma[1, n - a001318[k]]];
Table[T[n, k], {n, 1, m}, {k, 1, a235963[n]}] // Flatten (* Jean-François Alcover, Nov 29 2018 *)

T(n,k) = A057077(k-1)*A001318(k), if n = A001318(k) and k = A235963(n). Otherwise T(n,k) = A057077(k-1)*A000203(n - A001318(k)), n >= 1, 1 <= k <= A235963(n).

A122046 Partial sums of floor(n^2/8).

Original entry on oeis.org

0, 0, 0, 1, 3, 6, 10, 16, 24, 34, 46, 61, 79, 100, 124, 152, 184, 220, 260, 305, 355, 410, 470, 536, 608, 686, 770, 861, 959, 1064, 1176, 1296, 1424, 1560, 1704, 1857, 2019, 2190, 2370, 2560, 2760, 2970, 3190, 3421, 3663, 3916, 4180, 4456, 4744, 5044, 5356, 5681, 6019, 6370

Offset: 0

Roger L. Bagula, Sep 13 2006

Degree of the polynomial P(n+1,x), defined by P(n,x) = [x^(n-1)*P(n-1,x)*P(n-4,x)+P(n-2,x)*P(n-3,x)]/P(n-5,x) with P(1,x)=P(0,x)=P(-1,x)=P(-2,x)=P(-3,x)=1.
Define the sequence b(n) = 1, 4, 10, 20, 36, 60,... for n>=0 with g.f. 1/((1+x)*(1+x^2)*(1-x)^5). Then a(n+3) = b(n)-b(n-1) and b(n)+b(n+1)+b(n+2)+b(n+3) = A052762(n+7)/24. - J. M. Bergot, Aug 21 2013
Maximum Wiener index of all maximal 4-degenerate graphs with n-1 vertices.  (A maximal 4-degenerate graph can be constructed from a 4-clique by iteratively adding a new 4-leaf (vertex of degree 4) adjacent to four existing vertices.)  The extremal graphs are 4th powers of paths, so the bound also applies to 4-trees. - Allan Bickle, Sep 15 2022

			a(6) = 10 = 0 + 0 + 0 + 1 + 2 + 3 + 4.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Allan Bickle and Zhongyuan Che, Wiener indices of maximal k-degenerate graphs, arXiv:1908.09202 [math.CO], 2019.
Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
A. N. W. Hone, Comments on A122046
A. N. W. Hone, Algebraic curves, integer sequences and a discrete Painlevé transcendent, Proceedings of SIDE 6, Helsinki, Finland, 2004; arXiv:0807.2538 [nlin.SI], 2008. [Set a(n)=d(n+3) on p. 8]
Brian O'Sullivan and Thomas Busch, Spontaneous emission in ultra-cold spin-polarised anisotropic Fermi seas, arXiv 0810.0231v1 [quant-ph], 2008. [Eq 10a, lambda=4]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1,1,-3,3,-1).

Cf. A014125, A052762, A057077, A090294, A122047, A144678, A190718.
Partial sums of A001972.
Maximum Wiener index of all maximal k-degenerate graphs for k=1..6: A000292, A002623, A014125, A122046 (this sequence), A122047, A175724.

[Round((2*n^3+3*n^2-8*n)/48): n in [0..60]]; // Vincenzo Librandi, Jun 25 2011

A122046 := proc(n) round((2*n^3+3*n^2-8*n)/48) ; end proc: # Mircea Merca

p[n_] := p[n] = Cancel[Simplify[ (x^(n - 1)p[n - 1]p[n - 4] + p[n - 2]*p[n - 3])/p[n - 5]]]; p[ -5] = 1;p[ -4] = 1;p[ -3] = 1;p[ -2] = 1;p[ -1] = 1; Table[Exponent[p[n], x], {n, 0, 20}]
Accumulate[Floor[Range[0,60]^2/8]] (* or *) LinearRecurrence[{3,-3,1,1,-3,3,-1},{0,0,0,1,3,6,10},60] (* Harvey P. Dale, Dec 23 2019 *)

a(n)=(2*n^3+3*n^2-8*n+3)\48 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = Sum_{k=0..n} floor(k^2/8).
a(n) = round((2*n^3 + 3*n^2 - 8*n)/48) = round((4*n^3 + 6*n^2 - 16*n - 9)/96) = floor((2*n^3 + 3*n^2 - 8*n + 3)/48) = ceiling((2*n^3 + 3*n^2 - 8*n - 12)/48). - Mircea Merca
a(n) = a(n-8) + (n-4)^2 + n, n > 8. - Mircea Merca
From Andrew Hone, Jul 15 2008: (Start)
a(n+1) = cos((2*n+1)*Pi/4)/(4*sqrt(2)) + (2*n+3)*(2*n^2 + 6*n - 5)/96 + (-1)^n/32.
a(n+1) = A057077(n+1)/8 + A090294(n-1)/32 + (-1)^n/32.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-4) - 3*a(n-5) + 3*a(n-6) - a(n-7). (End)
O.g.f.: x^3 / ( (1+x)*(x^2+1)*(x-1)^4 ). - R. J. Mathar, Jul 15 2008
From Johannes W. Meijer, May 20 2011: (Start)
a(n+3) = A144678(n) + A144678(n-1) + A144678(n-2) + A144678(n-3);
a(n+3) = Sum_{k=0..6} min(6-k+1,k+1)* A190718(n+k-6). (End)
a(n) = (4*n^3 + 6*n^2 - 16*n - 9 - 3*(-1)^n + 12*(-1)^((2*n - 1 + (-1)^n)/4))/96. - Luce ETIENNE, Mar 21 2014
E.g.f.: ((2*x^3 + 9*x^2 - 3*x - 6)*cosh(x) + 6*(cos(x) + sin(x)) + (2*x^3 + 9*x^2 - 3*x - 3)*sinh(x))/48. - Stefano Spezia, Apr 05 2023

Edited by N. J. A. Sloane, Sep 17 2006, Jul 11 2008, Jul 12 2008

More formulas and better name from Mircea Merca, Nov 19 2010

A098493 Triangle T(n,k) read by rows: difference between A098489 and A098490 at triangular rows.

Original entry on oeis.org

1, 0, -1, -1, -1, 1, -1, 1, 2, -1, 0, 3, 0, -3, 1, 1, 2, -5, -2, 4, -1, 1, -2, -7, 6, 5, -5, 1, 0, -5, 0, 15, -5, -9, 6, -1, -1, -3, 12, 9, -25, 1, 14, -7, 1, -1, 3, 15, -18, -29, 35, 7, -20, 8, -1, 0, 7, 0, -42, 14, 63, -42, -20, 27, -9, 1, 1, 4, -22, -24, 85, 14, -112, 42

Offset: 0

Ralf Stephan, Sep 12 2004

Also, coefficients of polynomials that have values in A098495 and A094954.

			Triangle begins:
   1;
   0, -1;
  -1, -1, 1;
  -1,  1, 2, -1;
   0,  3, 0, -3, 1;
  ...

Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114.

Columns include A010892, -A076118. Diagonals include A033999, A038608, (-1)^n*A000096. Row sums are in A057077.

Cf. A098494 (diagonal polynomials), A085478, A244419.

A098493 := proc (n, k)
add((-1)^(k+binomial(n-j+1,2))*binomial(floor((1/2)*n+(1/2)*j),j)* binomial(j,k), j = k..n);
end proc:
seq(seq(A098493(n, k), k = 0..n), n = 0..10); # Peter Bala, Jul 13 2021

T(n,k)=if(k>n||k<0||n<0,0,if(k>=n-1,(-1)^n*if(k==n,1,-k),if(n==1,0,if(k==0,T(n-1,0)-T(n-2,0),T(n-1,k)-T(n-2,k)-T(n-1,k-1)))))

T(n, k) = A098489[n(n+1)/2, k] - A098490[n(n+1)/2, k].
Recurrence: T(n, k) = T(n-1, k)-T(n-1, k-1)-T(n-2, k); T(n, k)=0 for n<0, k>n, k<0; T(n, n)=(-1)^n; T(n, n-1)=(-1)^n*(1-n).
G.f.: (1-x)/(1+(y-1)*x+x^2). [Vladeta Jovovic, Dec 14 2009]
From Peter Bala, Jul 13 2021: (Start)
Riordan array ( (1 - x)/(1 - x + x^2), -x/(1 - x + x^2) ).
T(n,k) = (-1)^k * the (n,k)-th entry of Q^(-1)*P = Sum_{j = k..n} (-1)^(k+binomial(n-j+1,2))*binomial(floor((1/2)*n+(1/2)*j),j)* binomial(j,k), where P denotes Pascal's triangle A007318 and Q denotes triangle A061554 (formed from P by sorting the rows into descending order). (End)
From Peter Bala, Jun 26 2025: (Start)
n-th row polynomial R(n, x) = Sum_{k = 0..n} (-1)^k * binomial(n+k, 2*k) * (1 + x)^k.
R(n, 2*x + 1) = (-1)^n * Dir(n, x), where Dir(n,x) denotes the n-th row polynomial of the triangle A244419.
R(n, -1 - x) = b(n, x), where b(n, x) denotes the n-th row polynomial of the triangle A085478. (End)

A108752 Numbers k such that 12 divides k*(k+1).

Original entry on oeis.org

0, 3, 8, 11, 12, 15, 20, 23, 24, 27, 32, 35, 36, 39, 44, 47, 48, 51, 56, 59, 60, 63, 68, 71, 72, 75, 80, 83, 84, 87, 92, 95, 96, 99, 104, 107, 108, 111, 116, 119, 120, 123, 128, 131, 132, 135, 140, 143, 144, 147, 152, 155, 156, 159, 164, 167, 168, 171, 176, 179, 180

Offset: 1

Robert Phillips (bobp(AT)usca.edu), Jun 23 2005

First differences are 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, ..., . - Robert G. Wilson v, May 31 2017

Numbers that are congruent to {0, 3, 8, 11} mod 12. - Amiram Eldar, Jul 26 2024

Vincenzo Librandi, Table of n, a(n) for n = 1..5000
Robert Phillips, Triangular numbers which are sums of two triangular numbers, 2006.
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Equals A112652-1, A218155-3, A174398-5, A072833+3.

Cf. A016777, A057077, A287765.

[3*n-2-(-1)^((2*n-3-(-1)^n) div 4): n in [1..80]]; // Vincenzo Librandi, May 04 2017

a:= proc(n) if is(n*(n+1)/12, integer) then n fi end: seq(a(n), n=0..200); # Emeric Deutsch, Jun 25 2005

Select[ Range[0, 182], Mod[ #(# + 1), 12] == 0 &] (* Robert G. Wilson v, Jun 25 2005 *)
LinearRecurrence[{2, -2, 2, -1}, {0, 3, 8, 11}, 200] (* Vincenzo Librandi, Jun 04 2017 *)

From R. J. Mathar, Jan 07 2009: (Start)
a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - a(n-4) = A016777(n) - A057077(n).
G.f.: x*(3 + 2*x + x^2)/((1 + x^2)*(1 - x)^2). (End)
a(n) = 3*n - 2 - (-1)^((2*n-3-(-1)^n)/4). - Luce ETIENNE, Apr 04 2015
Sum_{n>=2} 1/a(n) = log(2)/2 + arccoth(sqrt(3))/(2*sqrt(3)) - Pi*(3+2*sqrt(3))/72. - Amiram Eldar, Jul 26 2024

More terms from Robert G. Wilson v and Emeric Deutsch, Jun 25 2005

A380853 Number of ways to place six distinct positive integers on a triangle, three on the corners and three on the sides such that the sum of the three values on each side is n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 3, 5, 13, 14, 25, 37, 47, 58, 89, 98, 126, 159, 188, 219, 276, 303, 362, 423, 478, 536, 633, 688, 781, 881, 973, 1068, 1211, 1301, 1443, 1589, 1724, 1866, 2066, 2202, 2396, 2598, 2790, 2986, 3250, 3439, 3699, 3967, 4219, 4480, 4819, 5071

Offset: 1

Derek Holton and Alex Holton, Feb 06 2025

Solutions differing by only rotation or reflections are not counted separately.

If the numbers do not need to be distinct and rotations and reflections are counted separately we get A019298(n-2). If the numbers do not need to be distinct but rotations and reflections do not count separately we get A006918(n-2). If the six numbers must be distinct and reflections and rotations count separately we get 6*a(n). - R. J. Mathar, Feb 27 2025

			The a(9) = 1 solution is:
       1
     5   6
   3   4   2

R. J. Mathar, Illustrations - Examples (2025)
Index entries for linear recurrences with constant coefficients, signature (-1,0,2,3,2,0,-3,-4,-3,0,2,3,2,0,-1,-1).

Cf. A342467, A380105.

A380853 := proc(n)
    -3412+2353*n+30*n^3-480*n^2+(135*n-900)*(-1)^n ;
    %-144*(2*A010891(n)+A010891(n-1)+2*A010891(n-2)) ;
    %-160*(17*A049347(n)+8*A049347(n-1)) ;
    %-360*A057077(n) ;
    %+480*(-1)^n*(A099254(n)-A099254(n-1)) ;
    %/720 ;
end proc:
seq(A380853(n),n=1..40) ; # R. J. Mathar, Feb 27 2025

a(n) = my(c=0, t); for(x=3, n-5, t=n-x; for(y=2, min(x-1, t-1), for(z=1, y-1, if(#Set([x, y, z, t-y, t-z, n-y-z])==6, c++)))); c; \\ Jinyuan Wang, Feb 07 2025

G.f.: x^9*(1 + 4*x + 8*x^2 + 16*x^3 + 18*x^4 + 18*x^5 + 15*x^6 + 10*x^7)/((1 - x)^4*(1 + 2*x + 2*x^2 + x^3)^2*(1 + x + 2*x^2 + 2*x^3 + 2*x^4 + x^5 + x^6)). - Stefano Spezia, Feb 08 2025

A380105(n) = a(n)-a(n-3). - R. J. Mathar, Mar 13 2025

A089928 a(n) = 2a(n-1) + 2a(n-3) + a(n-4), with a(0)=1, a(1)=2, a(3)=4, a(4)=10.

Original entry on oeis.org

1, 2, 4, 10, 25, 60, 144, 348, 841, 2030, 4900, 11830, 28561, 68952, 166464, 401880, 970225, 2342330, 5654884, 13652098, 32959081, 79570260, 192099600, 463769460, 1119638521, 2703046502, 6525731524, 15754509550, 38034750625, 91824010800

Offset: 0

Paul Barry, Nov 15 2003

a(n) is the number of tilings of an n-board (a board of size n X 1) using white squares, black squares, and white (1,1)-fences. A (1,1)-fence is a tile composed of two squares separated by a gap of width 1. - Michael A. Allen, Mar 12 2021

a(n) is the number of tilings of an n-board using white squares, black squares, white trominoes, black trominoes, and white tetrominoes. - Michael A. Allen, Mar 12 2021

G. C. Greubel, Table of n, a(n) for n = 0..1000
Kenneth Edwards and Michael A. Allen, New combinatorial interpretations of the Fibonacci numbers squared, golden rectangle numbers, and Jacobsthal numbers using two types of tile, J. Int. Seq. 24 (2021) Article 21.3.8.
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (2,0,2,1).

Cf. A000129, A001333, A006498, A057077, A079291, A114620.

[(Evaluate(DicksonFirst(n+2,-1), 2) + 2*(-1)^Binomial(n,2))/8: n in [0..40]]; // G. C. Greubel, Aug 18 2022

CoefficientList[Series[1/(1-2x-2x^3-x^4),{x,0,30}],x] (* Michael A. Allen, Mar 12 2021 *)
LinearRecurrence[{2,0,2,1}, {1,2,4,10}, 41] (* G. C. Greubel, Aug 18 2022 *)
nxt[{a_,b_,c_,d_}]:={b,c,d,2d+2b+a}; NestList[nxt,{1,2,4,10},30][[;;,1]] (* Harvey P. Dale, Jul 18 2024 *)

[(lucas_number2(n+2,2,-1) +2*(-1)^binomial(n,2))/8 for n in (0..40)] # G. C. Greubel, Aug 18 2022

a(n) = ( (1+sqrt(2))^(n+2) + (1-sqrt(2))^(n+2) + 2*(-1)^floor(n/2) )/8.
a(n) = (-i)^n*Sum_{k=0..floor(n/2)} U(n-2*k, i) with i^2 = -1.
a(n) + a(n+2) = A000129(n+3). - Alex Ratushnyak, Aug 06 2012
G.f.: 1/ ( (1+2*x)*(1-2*x-x^2) ). - R. J. Mathar, Apr 26 2013
4*a(n) = A057077(n) + A001333(n+2). - R. J. Mathar, Apr 26 2013
a(2*n) = (A000129(n+1))^2 = A079291(n+1). - Michael A. Allen, Mar 12 2021
a(2*n+1) = A000129(n+1)*A000129(n+2) = A114620(n+1). - Michael A. Allen, Mar 12 2021

Formula corrected by Max Alekseyev, Aug 22 2013

A097332 Expansion of (1/(1-x))(1+2x/(1-x+sqrt(1-2x-3x^2))).

Original entry on oeis.org

1, 2, 3, 5, 9, 18, 39, 90, 217, 540, 1375, 3563, 9361, 24872, 66707, 180341, 490913, 1344380, 3701159, 10237541, 28436825, 79288844, 221836403, 622599626, 1752360041, 4945087838, 13988490339, 39658308815, 112666081617

Offset: 0

Paul Barry, Aug 05 2004

Binomial transform of A097331. Binomial transform is A014318. Partial sums of 1+2x/(1-x+sqrt(1-2x-3x^2)) or (1+x+sqrt(1-2x-3x^2))/(1-x+sqrt(1-2x-3x^2)), which is A001006 with an extra leading 1.
Apparently the Motzkin transform of 1, 2, bar(1, -1, -1, 1), where bar() denotes a periodically continued series, as in A057077. - R. J. Mathar, Dec 11 2008
Starting with offset 1 = iterates of M * [1,1,0,0,0,...] where M = a tridiagonal matrix with [1,1,1,...] in the main and superdiagonals and [0,1,1,1,...] in the subdiagonal. - Gary W. Adamson, Jan 08 2009
Hankel transform is A087960(n) = (-1)^binomial(n+1,2). - Paul Barry, Aug 10 2009

			G.f. = 1 + 2*x + 3*x^2 + 5*x^3 + 9*x^4 + 18*x^5 + 39*x^6 + 90*x^7 + 217*x^8 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Emeric Deutsch and Bruce E. Sagan, Congruences for Catalan and Motzkin numbers and related sequences, arXiv:math/0407326 [math.CO], 2004; J. Num. Theory 117 (2006), 191-215. [See S_n on page 7.]

CoefficientList[Series[1/(1-x)*(1+(2x)/(1-x+Sqrt[1-2x-3x^2])),{x,0,40}],x] (* Harvey P. Dale, May 03 2012 *)
a[ n_] := SeriesCoefficient[ (1 + x - Sqrt[1 - 2 x - 3 x^2]) / (2 x (1 - x)), {x, 0, n}]; (* Michael Somos, May 19 2014 *)

{a(n) = if( n<0, 0, polcoeff( (1 + x - sqrt(1 - 2*x - 3*x^2 + x^2 * O(x^n))) / (2 * x * (1 - x)), n))}; /* Michael Somos, May 19 2014 */

a(n) = Sum_{k=0..n} (-1)^(n+k)*binomial(n, k)*Sum_{i=0..k} Catalan(k-i)*2^i.
G.f.: 1/(1-x-x/(1+x/(1-x+x/(1-x/(1-x-x/(1+x/(1-x+x/(1-x/(1-x-x/(1+... (continued fraction). - Paul Barry, Aug 10 2009
Conjecture D-finite with recurrence: (n+1)*a(n) - 3*n*a(n-1) + (-n+5)*a(n-2) + 3*(n-2)*a(n-3) = 0. - R. J. Mathar, Nov 26 2012
a(n) ~ 3^(n+3/2) / (4 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Feb 13 2014
0 = a(n)*(9*a(n+1) + 6*a(n+2) - 27*a(n+3) + 12*a(n+4)) + a(n+1)*(-12*a(n+1) + 10*a(n+2) + 12*a(n+3) - 7*a(n+4)) + a(n+2)*(-12*a(n+2) + 14*a(n+3) - 6*a(n+4)) + a(n+3)*(a(n+4)). - Michael Somos, May 19 2014

cp's OEIS Frontend

A244419 Coefficient triangle of polynomials related to the Dirichlet kernel. Rising powers. Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Sage

Formula

A143624 Decimal expansion of the negated constant cos(1) - sin(1) = -0.3011686789...

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A203421 Reciprocal of Vandermonde determinant of (1,1/2,...,1/n).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

Extensions

A238442 Triangle read by rows demonstrating Euler's pentagonal theorem for the sum of divisors.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A122046 Partial sums of floor(n^2/8).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

A098493 Triangle T(n,k) read by rows: difference between A098489 and A098490 at triangular rows.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

Formula

A108752 Numbers k such that 12 divides k*(k+1).

Views

A089928 a(n) = 2a(n-1) + 2a(n-3) + a(n-4), with a(0)=1, a(1)=2, a(3)=4, a(4)=10.