A248156 -id:A248156 - cp's OEIS frontend

A248158 Expansion of (1 - 2*x^2)/(1 + x)^3. Second column of Riordan triangle A248156.

1, -3, 4, -4, 3, -1, -2, 6, -11, 17, -24, 32, -41, 51, -62, 74, -87, 101, -116, 132, -149, 167, -186, 206, -227, 249, -272, 296, -321, 347, -374, 402, -431, 461, -492, 524, -557, 591, -626, 662, -699, 737, -776, 816, -857, 899, -942, 986

Offset: 0

Wolfdieter Lang, Oct 05 2014

This is the column k=1 sequence of the Riordan triangle A248156 without a leading zero.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-3,-3,-1).

Cf. A046691, A148157, A248156(n+1,1).

[(-1)^n*(2+5*n-n^2)/2: n in [0..60]]; // G. C. Greubel, May 30 2025

Table[(-1)^n*(2+5*n-n^2)/2, {n,0,60}] (* G. C. Greubel, May 30 2025 *)

def A248158(n): return (-1)**n*(2+5*n-n**2)//2
print([A248158(n) for n in range(51)]) # G. C. Greubel, May 30 2025

O.g.f.: (1 - 2*x^2)/(1 + x)^3 = -2/(1 + x) + 4/(1 + x)^2 - 1/(1 + x)^3.
a(n) = (-1)^n*(4*(2*n+1) - (n+1)*(n+2))/2, n >= 0.
a(n) = -3*(a(n-1) + a(n-2)) - a(n-3), n >= 3 with a(0) = 1, a(1) = -3 and a(2) = 4.
From R. J. Mathar, Mar 13 2021: (Start)
a(n) = (-1)^(n+1)*A046691(n-5).
a(n) + a(n+1) = A248157(n+1). (End)
E.g.f.: (1/2)*(2 - 4*x - x^2)*exp(-x). - G. C. Greubel, May 30 2025

A248159 Expansion of (1 - 2*x^2)/(1 + x)^4. Third column of Riordan triangle A248156.

Original entry on oeis.org

1, -4, 8, -12, 15, -16, 14, -8, -3, 20, -44, 76, -117, 168, -230, 304, -391, 492, -608, 740, -889, 1056, -1242, 1448, -1675, 1924, -2196, 2492, -2813, 3160, -3534, 3936, -4367, 4828, -5320, 5844, -6401, 6992, -7618, 8280, -8979, 9716

Offset: 0

Wolfdieter Lang, Oct 07 2014

This is the column k=2 sequence of the Riordan triangle A248156 without the leading two zeros.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-4,-6,-4,-1).

Cf. A248156 (k=2).

Cf. A248157 (k=0), A248158 (k=1).

[(-1)^(n+1)*(n+1)*(n^2-7*n-6)/Factorial(3) : n in [0..50]]; // Wesley Ivan Hurt, Oct 07 2014

A248159:=n->(-1)^(n+1)*(n+1)*(n^2-7*n-6)/3!: seq(A248159(n),n=0..50); # Wesley Ivan Hurt, Oct 07 2014

Table[(-1)^(n + 1)*(n + 1)*(n^2 - 7*n - 6)/3!, {n, 0, 50}] (* Wesley Ivan Hurt, Oct 07 2014 *)

def A248159(n): return (-1)**n*(n+1)*(6+7*n-n**2)//6 # G. C. Greubel, May 30 2025

O.g.f.: (1 - 2*x^2)/(1 + x)^4 = -1/(1 + x)^4 + 4/(1 + x)^3 -2/(1 + x)^2.
a(n) = (-1)^n*(n+1)*(6 + 7*n - n^2)/3!, n >= 0.
a(n) = -4*(a(n-1) + a(n-3)) - 6*a(n-2) - a(n-4), n >= 4, with a(0) =1, a(1) = -4, a(2) = 8 and a(3) = -12.
a(n) + a(n+1) = A248158(n+1). - R. J. Mathar, Mar 13 2021
E.g.f.: (1/6)*(6 - 18*x + 3*x^2 + x^3)*exp(-x). - G. C. Greubel, May 30 2025

A248160 Expansion of (1 - 2*x^2)/(1 + x)^5. Fourth column of Riordan triangle A248156.

Original entry on oeis.org

1, -5, 13, -25, 40, -56, 70, -78, 75, -55, 11, 65, -182, 350, -580, 884, -1275, 1767, -2375, 3115, -4004, 5060, -6302, 7750, -9425, 11349, -13545, 16037, -18850, 22010, -25544, 29480, -33847, 38675, -43995, 49839, -56240, 63232, -70850, 79130, -88109, 97825, -108317, 119625, -131790

Offset: 0

Wolfdieter Lang, Oct 09 2014

This is column k=3 of the Riordan triangle A248156 without the leading three zeros.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-5,-10,-10,-5,-1).

Cf. A248156 (column k=3).

Cf. A248157 (k=0), A248158 (k=1), A248159 (k=2).

[(-1)^n*(n + 1)*(n + 2)*(12 + 9*n - n^2)/24: n in [0..50]]; // G. C. Greubel, May 30 2025

A248160:=n->(-1)^n*(n+1)*(n+2)*(12 + 9*n - n^2)/4!: seq(A248160(n), n=0..30); # Wesley Ivan Hurt, Oct 09 2014

Table[(-1)^n*(n + 1)*(n + 2)*(12 + 9*n - n^2)/4!, {n, 0, 30}] (* Wesley Ivan Hurt, Oct 09 2014 *)
CoefficientList[Series[(1-2x^2)/(1+x)^5,{x,0,50}],x] (* or *) LinearRecurrence[ {-5,-10,-10,-5,-1},{1,-5,13,-25,40},50] (* Harvey P. Dale, Apr 13 2019 *)

Vec((1 - 2*x^2)/(1 + x)^5 + O(x^50)) \\ Michel Marcus, Oct 09 2014

def A248160(n): return (-1)**n*(n+1)*(n+2)*(12+9*n-n**2)//24 # G. C. Greubel, May 30 2025

O.g.f.: (1 - 2*x^2)/(1 + x)^5 = -2/(1 + x)^3 + 4/(1 + x)^4 - 1/(1 + x)^5.
a(n) = (-1)^n*(n+1)*(n+2)*(12 + 9*n - n^2)/4!.
a(n) =  -5*(a(n-1) + a(n-4)) - 10*(a(n-2) + a(n-3)) - a(n-5), n >= 5, with a(0) =1, a(1) = -5, a(2) = 13, a(3) = -25 and a(4) = 40.
E.g.f.: (1/4!)*(24 - 96*x + 48*x^2 - x^4)*exp(-x). - G. C. Greubel, May 30 2025

A248157 Expansion of (1 - 2*x^2)/(1 + x)^2.

Original entry on oeis.org

1, -2, 1, 0, -1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 14, -15, 16, -17, 18, -19, 20, -21, 22, -23, 24, -25, 26, -27, 28, -29, 30, -31, 32, -33, 34, -35, 36, -37, 38, -39, 40, -41, 42, -43, 44, -45, 46, -47, 48, -49, 50, -51, 52, -53, 54, -55, 56, -57

Offset: 0

Wolfdieter Lang, Oct 05 2014

First column of Riordan triangle A248156.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (-2,-1).

Cf. A038608, A097141, A248156 (row k=0).

A248157:= func< n | n eq 0 select 1 else (-1)^n*(3-n) >; // G. C. Greubel, May 16 2025

CoefficientList[Series[(1-2x^2)/(1+x)^2,{x,0,60}],x] (* or *) LinearRecurrence[{-2,-1},{1,-2,1},60] (* Harvey P. Dale, Aug 25 2023 *)

Vec((1 - 2*x^2)/(1 + x)^2 + O(x^80)) \\ Michel Marcus, Oct 11 2014

def A248157(n): return (-1)^n*(3-n) - 2*int(n==0) # G. C. Greubel, May 16 2025

O.g.f.: (1 - 2*x^2)/(1 + x)^2 = -2 + 4/(1+x) - 1/(1+x)^2.
a(n) = -2*delta(n,0) + (-1)^n*(3 - n), n >= 0, with Kronecker delta(n,0) = 1 if n=0 else 0.
a(0) = 1, a(n) = -2*a(n-1) - a(n-2), n >= 3, with a(1) = -2, a(2) = 1.
a(n) = A038608(n-3) = A097141(n-1), n>=3.
E.g.f.: (3+x)*exp(-x) - 2. - G. C. Greubel, May 16 2025

A106513 A Pell-Pascal matrix.

Original entry on oeis.org

1, 2, 1, 5, 3, 1, 12, 8, 4, 1, 29, 20, 12, 5, 1, 70, 49, 32, 17, 6, 1, 169, 119, 81, 49, 23, 7, 1, 408, 288, 200, 130, 72, 30, 8, 1, 985, 696, 488, 330, 202, 102, 38, 9, 1, 2378, 1681, 1184, 818, 532, 304, 140, 47, 10, 1, 5741, 4059, 2865, 2002, 1350, 836, 444, 187, 57, 11, 1

Offset: 0

Paul Barry, May 05 2005

This triangle gives the iterated partial sums of the Pell sequence A000129(n+1), n>=0. - Wolfdieter Lang, Oct 05 2014

			The triangle T(n,k) begins:
n\k    0    1    2    3    4   5   6   7  8  9 10 ...
0:     1
1:     2    1
2:     5    3    1
3:    12    8    4    1
4:    29   20   12    5    1
5:    70   49   32   17    6   1
6:   169  119   81   49   23   7   1
7:   408  288  200  130   72  30   8   1
8:   985  696  488  330  202 102  38   9  1
9:  2378 1681 1184  818  532 304 140  47 10  1
10: 5741 4059 2865 2002 1350 836 444 187 57 11  1
... Reformatted and extended. - _Wolfdieter Lang_, Oct 05 2014
-----------------------------------------------------
Recurrence from the Z-sequence (see the formula above) for T(0,n) in terms of the entries of row n-1. For example, 29 = T(4,0) = 2*12 + 1*8 + (-1)*4 + 1*1 = 29. - _Wolfdieter Lang_, Oct 05 2014

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000129, A001333, A106514 (row sums), A106515 (antidiagonal sums), A248156.

[ (&+[Binomial(n+1, 2*j+k+1)*2^j: j in [0..Floor((n+1)/2)]]) : k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 05 2021

T[n_, k_]= Sum[Binomial[n+1, 2*j+k+1]*2^j, {j, 0, Floor[(n+1)/2]}];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Aug 05 2021 *)

@CachedFunction
def T(n, k):
    if (k<0 or k>n): return 0
    elif (k==0): return lucas_number1(n+1, 2, -1)
    else: return T(n-1,k-1) + T(n-1,k)
flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Aug 05 2021

Riordan array (1/(1-2*x-x^2), x/(1-x)).
Number triangle T(n,0) = A000129(n+1), T(n,k) = T(n-1,k-1) + T(n-1,k).
T(n,k) = Sum_{j=0..floor((n+1)/2)} binomial(n+1, 2*j+k+1)*2^j.
Sum_{k=0..n} T(n, k) = A106514(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A106515(n).
T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - 2*T(n-2,k-1) - T(n-3,k) - T(n-3,k-1), T(0,0)=1, T(1,0)=2, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 14 2014
From Wolfdieter Lang, Oct 05 2014: (Start)
O.g.f. for row polynomials R(n,x) = Sum_{k=0..n} T(n,k)*x^k: (1 - z)/((1 - 2*z - z^2)*(1 - (1+x)*z)).
O.g.f. column m: (1/(1 - 2*z - z^2))*(z/(1 - z))^m, m >= 0. (Riordan property).
The alternating row sums are shown in A001333.
A-sequence: [1, 1] (see the three term recurrence given above). Z-sequence has o.g.f. (2 + 3*x)/(1 + x), [2, 1, repeat(-1,1)] (unsigned A054977). See the W. Lang link under A006232 for Riordan A- and Z-sequences.
The inverse Riordan triangle is shown in A248156. (End)

A248161 Expansion of (2-x+x^2)/((1+x)(1-3x+x^2)).

Original entry on oeis.org

2, 3, 11, 26, 71, 183, 482, 1259, 3299, 8634, 22607, 59183, 154946, 405651, 1062011, 2780378, 7279127, 19056999, 49891874, 130618619, 341963987, 895273338, 2343856031, 6136294751, 16065028226, 42058789923, 110111341547

Offset: 0

Wolfdieter Lang, Nov 01 2014

The negative of this sequence provides the first component of the square of [F(n), F(n+1), F(n+2), F(n+3)], for n >= 0, where F(n) = A000045(n), in the Clifford algebra Cl_2 over Euclidean 2-space. The whole quartet of sequences for this square is [-a(n), A079472(n+1), A059929(n), A121801(n+1)]. See the Oct 15 2014 comment in A147973 where also a reference is given.

G. C. Greubel, Table of n, a(n) for n = 0..1000
R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.

Cf. A000045, A059929, A079472, A121801, A248156.

[-(Fibonacci(n)^2 +Fibonacci(n+1)^2 + Fibonacci(n+2)^2 - Fibonacci(n+3)^2): n in [0..30]]; // Vincenzo Librandi, Nov 01 2014

CoefficientList[Series[(2 - x + x^2)/((1 + x) (1 - 3 x + x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Nov 01 2014 *)
With[{F=Fibonacci}, Table[F[2*n+2] +F[n]*F[n+1] +(-1)^n, {n,0,40}]] (* G. C. Greubel, May 30 2025 *)

def A248161(n): return fibonacci(2*n+2) +fibonacci(n)*fibonacci(n+1) +(-1)^n
print([A248161(n) for n in range(41)]) # G. C. Greubel, May 30 2025

a(n) = F(n+3)^2 - (F(n)^2 + F(n+1)^2 + F(n+2)^2), F(n) = A000045(n).
a(n) = (6*F(2*n+2) + F(2*n) + 4*(-1)^n)/5, with the Fibonacci numbers F = A000045.
O.g.f.: (2-x+x^2)/((1+x)*(1-3*x+x^2)) = (4/(1+x) + (x+6)/(1-3*x+x^2))/5.
From G. C. Greubel, May 30 2025: (Start)
a(n) = Fibonacci(2*n+2) + Fibonacci(n)*Fibonacci(n+1) + (-1)^n.
E.g.f.: (1/5)*(exp(3*x/2)*(6*cosh(sqrt(5)*x/2) + 4*sqrt(5)*sinh(sqrt(5)*x/2)) + 4*exp(-x)). (End)

Typo in formula fixed by Vincenzo Librandi, Nov 01 2014

cp's OEIS Frontend

A248158 Expansion of (1 - 2*x^2)/(1 + x)^3. Second column of Riordan triangle A248156.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Python

Formula

A248159 Expansion of (1 - 2*x^2)/(1 + x)^4. Third column of Riordan triangle A248156.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Python

Formula

A248160 Expansion of (1 - 2*x^2)/(1 + x)^5. Fourth column of Riordan triangle A248156.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Python

Formula

A248157 Expansion of (1 - 2*x^2)/(1 + x)^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

A106513 A Pell-Pascal matrix.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Sage

Formula

A248161 Expansion of (2-x+x^2)/((1+x)*(1-3*x+x^2)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

Extensions

A248161 Expansion of (2-x+x^2)/((1+x)(1-3x+x^2)).