A052472 -id:A052472 - cp's OEIS frontend

A117651 A002415 and A052472 interlaced.

1, 0, 2, 1, 0, 6, 10, 20, 35, 50, 84, 105, 168, 196, 300, 336, 495, 540, 770, 825, 1144, 1210, 1638, 1716, 2275, 2366, 3080, 3185, 4080, 4200, 5304, 5440, 6783, 6936, 8550, 8721, 10640, 10830, 13090, 13300, 15939, 16170, 19228, 19481, 23000, 23276, 27300

Offset: 0

Roger L. Bagula, Apr 11 2006

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (1,4,-4,-6,6,4,-4,-1,1).

Cf. A002415, A052472, A018226.

R:=PowerSeriesRing(Integers(), 50); Coefficients(R!( (1-x -2*x^2+3*x^3-3*x^4+4*x^5+16*x^6-16*x^7 -14*x^8+14*x^9+4*x^10-4*x^11 )/( (1+x)^4*(1-x)^5) )); // G. C. Greubel, May 19 2019

f[n_]:= n*(n+1)*(n+2)*(n-3)/12; g[n_]:= n^2*(n^2 -1)/12; Table[{Abs[f[n]], g[n]}, {n, 1, 25}]//Flatten
LinearRecurrence[{1,4,-4,-6,6,4,-4,-1,1}, {1,0,2,1,0,6,10,20,35,50,84, 105}, 50] (* Harvey P. Dale, Mar 05 2016 *)

my(x='x+O('x^50)); Vec((1-x-2*x^2+3*x^3-3*x^4+4*x^5+16*x^6-16*x^7 -14*x^8+14*x^9+4*x^10-4*x^11 )/((1+x)^4*(1-x)^5)) \\ G. C. Greubel, May 19 2019

((1-x-2*x^2+3*x^3-3*x^4+4*x^5+16*x^6-16*x^7 -14*x^8+14*x^9+4*x^10 -4*x^11 )/((1+x)^4*(1-x)^5)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, May 19 2019

G.f.: (1 -x -2*x^2 +3*x^3 -3*x^4 +4*x^5 +16*x^6 -16*x^7 -14*x^8 +14*x^9 +4*x^10 -4*x^11 )/((1+x)^4*(1-x)^5). - Colin Barker, Mar 15 2013
a(n) = abs((2*n^4 +12*n^3 -2*n^2 -132*n -195 +(4*n^3 -6*n^2 -124*n -189)*(-1)^n))/384. - Luce ETIENNE, Jun 01 2015
a(n) = abs((-3*(65 +63*(-1)^n) -4*(33 +31*(-1)^n)*n -2*(1+3*(-1)^n)*n^2 +4*(3 +(-1)^n)*n^3 +2*n^4)/384). - Colin Barker, Jun 02 2015
a(n) = a(n-1) + 4*a(n-2) - 4*a(n-3) - 6*a(n-4) + 6*a(n-5) + 4*a(n-6) - 4*a(n-7) - a(n-8) + a(n-9) for n > 11. - Charles R Greathouse IV, Jun 02 2015

A192011 Let P(0,x) = -1, P(1,x) = 2x, and P(n,x) = xP(n-1,x) - P(n-2,x) for n > 1. This sequence is the triangle of polynomial coefficients in order of decreasing exponents.

Original entry on oeis.org

-1, 2, 0, 2, 0, 1, 2, 0, -1, 0, 2, 0, -3, 0, -1, 2, 0, -5, 0, 0, 0, 2, 0, -7, 0, 3, 0, 1, 2, 0, -9, 0, 8, 0, 1, 0, 2, 0, -11, 0, 15, 0, -2, 0, -1, 2, 0, -13, 0, 24, 0, -10, 0, -2, 0, 2, 0, -15, 0, 35, 0, -25, 0, 0, 0, 1, 2, 0, -17, 0, 48, 0, -49, 0, 10, 0, 3, 0, 2, 0, -19, 0, 63, 0, -84, 0, 35, 0, 3, 0, -1

Offset: 0

Paul Curtz, Jun 21 2011

			The first few rows are
  -1;
   2,   0;
   2,   0,   1;
   2,   0,  -1,   0;
   2,   0,  -3,   0,  -1;
   2,   0,  -5,   0,   0,   0;
   2,   0,  -7,   0,   3,   0,   1;
   2,   0,  -9,   0,   8,   0,   1,   0;
   2,   0, -11,   0,  15,   0,  -2,   0,  -1;
   2,   0, -13,   0,  24,   0, -10,   0,  -2,   0;
   2,   0, -15,   0,  35,   0, -25,   0,   0,   0,   1;

G. C. Greubel, Rows n = 0..30 of triangle, flattened

Left hand diagonals are: T(n,0) = [-1,2,2,2,2,2,...], T(n,2) = A165747(n), T(n,4) = A067998(n+1), T(n,6) = -A058373(n), T(n,8) = (-1)^(n+1) * A167387(n+2) (see also A052472(n)).

A192011 := proc(n,k)
        option remember;
        if k>n or k <0 or n<0 then
                0;
        elif n= 0 then
                -1;
        elif k=0 then
                2;
        else
                procname(n-1,k)-procname(n-2,k-2) ;
        end if;
end proc: # R. J. Mathar, Nov 03 2011

p[0, ] = -1; p[1, x] := 2x; p[n_, x_] := p[n, x] = x*p[n-1, x] - p[n-2, x]; row[n_] := CoefficientList[p[n, x], x]; Table[row[n] // Reverse, {n, 0, 9}] // Flatten (* Jean-François Alcover, Nov 26 2012 *)
T[n_,k_]:= If[k<0 || k>n, 0, If[n==0 && k==0, -1, If[k==0, 2, T[n-1,k] - T[n-2, k-2]]]]; Table[T[n,k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, May 19 2019 *)

{T(n,k) = if(k<0 || k>n, 0, if(n==0 && k==0, -1, if(k==0, 2, T(n-1,k) - T(n-2,k-2)))) };
for(n=0, 10, for(k=0, n, print1(T(n,k), ", "))) \\ G. C. Greubel, May 19 2019

def T(n,k):
    if (k<0 or k>n): return 0
    elif (n==0 and k==0): return -1
    elif (k==0): return 2
    else: return T(n-1,k) - T(n-2, k-2)
[[T(n,k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, May 19 2019

T(n, k) = T(n-1, k) - T(n-2, k-2), where T(0, 0) = -1, T(n, 0) = 2 and 0 <= k <= n, n >= 0. - G. C. Greubel, May 19 2019

A167387 a(n) = (-1)^(n+1) * n(n-1)(n-4)*(n+1)/12.

Original entry on oeis.org

1, -2, 0, 10, -35, 84, -168, 300, -495, 770, -1144, 1638, -2275, 3080, -4080, 5304, -6783, 8550, -10640, 13090, -15939, 19228, -23000, 27300, -32175, 37674, -43848, 50750, -58435, 66960, -76384, 86768, -98175, 110670, -124320, 139194, -155363, 172900

Offset: 2

Jamel Ghanouchi, Nov 02 2009

The coefficient of [x^4] of the Polynomial B_{2n}(x) defined in A137276.

Essentially the same as A052472.

G. C. Greubel, Table of n, a(n) for n = 2..1000
Index entries for linear recurrences with constant coefficients, signature (-5,-10,-10,-5,-1).

Cf. A052472, A137276, A135929, A138034.

List([2..40], n-> (-1)^(n+1)*(n-4)*Binomial(n+1,3)/2); # G. C. Greubel, May 19 2019

[(-1)^(n+1)*n*(n-1)*(n-4)*(n+1)/12: n in [2..40]]; // Vincenzo Librandi, Jun 13 2016

Table[(-1)^(n+1)*(n+1)*n*(n-1)*(n-4)/12, {n, 2, 40}] (* G. C. Greubel, Jun 12 2016 *)
LinearRecurrence[{-5, -10, -10, -5, -1}, {1, -2, 0, 10, -35}, 40] (* Vincenzo Librandi, Jun 13 2016 *)

vector(40, n, n++; (-1)^(n+1)*(n-4)*binomial(n+1,3)/2) \\ G. C. Greubel, May 19 2019

[(-1)^(n+1)*(n-4)*binomial(n+1,3)/2 for n in (2..40)] # G. C. Greubel, May 19 2019

a(n) = -5*a(n-1) -10*a(n-2) -10*a(n-3) -5*a(n-4) -a(n-5).
G.f.: x^2*(1+3*x)/(1+x)^5.
E.g.f.: x^2*(6 + 2*x - x^2)*exp(-x)/12. - G. C. Greubel, May 19 2019

A117652 a(n) = floor(n(n+2)(n+4)*(n-6)/192).

Original entry on oeis.org

0, -1, -1, -2, -2, -2, 0, 3, 10, 20, 35, 55, 84, 120, 168, 227, 300, 388, 495, 621, 770, 943, 1144, 1374, 1638, 1937, 2275, 2654, 3080, 3553, 4080, 4662, 5304, 6009, 6783, 7628, 8550, 9552, 10640, 11817, 13090, 14462, 15939, 17525, 19228, 21050, 23000, 25081

Offset: 0

Roger L. Bagula, Apr 11 2006

Quasipolynomial with period 16. - Charles R Greathouse IV, Sep 06 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-5,0,4,0,-4,0,4,0,-4,0,4,0,-4,0,5,-4,1).

Cf. A052472, A018226.

[Floor( n*(n+2)*(n+4)*(n-6)/192): n in [0..50]]; // Vincenzo Librandi, Sep 06 2011

Table[Floor[n*(n+1)*(n+2)*(n-3)/12], {n, 0, 25, 1/2}]
LinearRecurrence[{4,-5,0,4,0,-4,0,4,0,-4,0,4,0,-4,0,5,-4,1},{0,-1,-1,-2,-2,-2,0,3,10,20,35,55,84,120,168,227,300,388},50] (* Harvey P. Dale, Nov 02 2024 *)

a(n)=n*(n+2)*(n+4)*(n-6)\192 \\ Charles R Greathouse IV, Sep 06 2011

[floor(n*(n+2)*(n+4)*(n-6)/192) for n in (0..50)] # G. C. Greubel, May 20 2019

a(n) = floor( n*(n+2)*(n+4)*(n-6)/192).

a(n) = 4*a(n-1) - 5*a(n-2) + 4*a(n-4) - 4*a(n-6) + 4*a(n-8) - 4*a(n-10) + 4*a(n-12) - 4*a(n-14) + 5*a(n-16) - 4*a(n-17) + a(n-18).

More precise description, converted to a more regular signed sequence - the Assoc. Eds. of the OEIS, Jun 27 2010

A128890 Triangle T(n,k) related to walks in the positive quadrant.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 0, 5, 0, 1, 10, 0, 9, 0, 1, 0, 35, 0, 14, 0, 1, 70, 0, 84, 0, 20, 0, 1, 0, 294, 0, 168, 0, 27, 0, 1, 588, 0, 840, 0, 300, 0, 35, 0, 1, 0, 2772, 0, 1980, 0, 495, 0, 44, 0, 1

Offset: 0

Philippe Deléham, Apr 20 2007

			Triangle begins:
    1;
    0,    1;
    2,    0,   1;
    0,    5,   0,    1;
   10,    0,   9,    0,   1;
    0,   35,   0,   14,   0,   1;
   70,    0,  84,    0,  20,   0,  1;
    0,  294,   0,  168,   0,  27,  0,  1;
  588,    0, 840,    0, 300,   0, 35,  0,  1;
    0, 2772,   0, 1980,   0, 495,  0, 44,  0,  1;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A000012, A000096, A052472, A005568, A000356.

T[n_, k_]:= If[k==0 && EvenQ[n], 4*Binomial[n,n/2]*Binomial[n+2,(n+2)/2 ]/((n+2)*(n+4)), If[EvenQ[n+k], Binomial[n, (n+k)/2]*Binomial[n+2, (n - k)/2] - Binomial[n+2, (n+k+2)/2]*Binomial[n, (n-k-2)/2], 0]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten

{ T(n,k) = if(k==0 && n%2==0, 4*binomial(n,n/2)*binomial(n+2, (n+2)/2)/((n+2)*(n+4)), if((n+k)%2==0, binomial(n, (n+k)/2)*binomial(n + 2, (n-k)/2) - binomial(n+2, (n+k+2)/2)*binomial(n, (n-k-2)/2), 0)) }; \\ G. C. Greubel, May 20 2019

def T(n, k):
    if (k==0 and n%2==0): return 4*binomial(n,n/2)*binomial(n+2, (n+2)/2)/((n+2)*(n+4))
    elif ((n+k)%2==0): return binomial(n, (n+k)/2)*binomial(n + 2, (n-k)/2) - binomial(n+2, (n+k+2)/2)*binomial(n, (n-k-2)/2)
    else: return 0
[[T(n, k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, May 20 2019

T(n,k) = binomial(n, r)*binomial(n+2, s) - binomial(n+2, r+1)*binomial(n, s-1) with r=(n+k)/2 and s=(n-k)/2, if n+k is even otherwise T(n,k)=0. Also T(2*n,0) = A000108(n)*A000108(n+1) = A005568(n).

cp's OEIS Frontend

A117651 A002415 and A052472 interlaced.

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A192011 Let P(0,x) = -1, P(1,x) = 2*x, and P(n,x) = x*P(n-1,x) - P(n-2,x) for n > 1. This sequence is the triangle of polynomial coefficients in order of decreasing exponents.

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A167387 a(n) = (-1)^(n+1) * n*(n-1)*(n-4)*(n+1)/12.

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A117652 a(n) = floor(n*(n+2)*(n+4)*(n-6)/192).

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A128890 Triangle T(n,k) related to walks in the positive quadrant.

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Author

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Crossrefs

Programs

Mathematica

PARI

Sage

Formula

A192011 Let P(0,x) = -1, P(1,x) = 2x, and P(n,x) = xP(n-1,x) - P(n-2,x) for n > 1. This sequence is the triangle of polynomial coefficients in order of decreasing exponents.

A167387 a(n) = (-1)^(n+1) * n(n-1)(n-4)*(n+1)/12.

A117652 a(n) = floor(n(n+2)(n+4)*(n-6)/192).