A165202 -id:A165202 - cp's OEIS frontend

A110666 Sequence is {a(1,n)}, where a(m,n) is defined at sequence A110665.

0, 1, 1, -2, -6, -6, 0, 7, 7, -2, -12, -12, 0, 13, 13, -2, -18, -18, 0, 19, 19, -2, -24, -24, 0, 25, 25, -2, -30, -30, 0, 31, 31, -2, -36, -36, 0, 37, 37, -2, -42, -42, 0, 43, 43, -2, -48, -48, 0, 49, 49, -2, -54, -54, 0, 55, 55, -2, -60, -60, 0, 61, 61, -2, -66, -66, 0, 67, 67, -2, -72, -72, 0, 73, 73, -2, -78, -78, 0, 79, 79, -2, -84

Offset: 0

Leroy Quet, Aug 02 2005

			a(0,n): 0,  1,  0, -3, -4, ...
a(1,n): 0,  1,  1, -2, -6, ...
a(2,n): 0,  1,  2,  0, -6, ...
a(3,n): 0,  1,  3,  3, -3, ...
a(4,n): 0,  1,  4,  7,  4, ...
Main diagonal of array is 0, 1, 2, 3, 4, ...

Michael De Vlieger, Table of n, a(n) for n = 0..1000

Cf. A110665, A110667, A110668, A110669, A110670, A110671, A110672.

A11066x := proc(mmax,nmax) local a,i,j ; a := array(0..mmax,0..nmax) ; a[0,0] := 0 ; for i from 1 to nmax do a[0,i] := i-sum(binomial(2*i-k-1,i-1)*a[0,k],k=0..i-1) : od ; for j from 1 to mmax do a[j,0] := 0 ; for i from 1 to nmax do a[j,i] := a[j-1,i]+a[j,i-1] ; od ; od ; RETURN(a) ; end : nmax := 100 : m := 1: a := A11066x(m,nmax) : for n from 0 to nmax do printf("%d,",a[m,n]) ; od ; # R. J. Mathar, Sep 01 2006

a[m_, 0] := 0; a[n_, n_] := n; a[0, n_] := n - Sum[Binomial[2*n - k - 1, n - 1]* a[0, k], {k, 0, n - 1}]; a[m_, n_] := a[m, n] = a[m - 1, n] + a[m, n - 1]; Table[a[1, n], {n, 0, 50}] (* G. C. Greubel, Sep 03 2017 *)

From R. J. Mathar, Oct 09 2013: (Start)
Conjecture: G.f. x*(-1+2*x) / ( (x-1)*(x^2-x+1)^2 ).
a(n) = -A010892(n-1) + A165202(n) -1. (End)

More terms from R. J. Mathar, Sep 01 2006

A165201 Expansion of 1/(1-x*c(x)^3), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 1, 4, 16, 65, 267, 1105, 4597, 19196, 80380, 337284, 1417582, 5965622, 25130844, 105954110, 447015744, 1886996681, 7969339643, 33670068133, 142301618265, 601586916703, 2543852427847, 10759094481491, 45513214057191

Offset: 0

Paul Barry, Sep 07 2009

Hankel transform is A165202. Essentially the same as A026674.

Vincenzo Librandi, Table of n, a(n) for n = 0..300

Cf. A000108, A026674, A165202.

List([0..30], n-> (1/2)*(2*0^n + Fibonacci(3*n-2) + Sum([0..n], j-> Binomial(2*j, j)*Fibonacci(3*(n-j)+1)/(1-2*j) ))); # G. C. Greubel, Jul 18 2019

R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( (1-3*x-2*x^2 + (1-x)*Sqrt(1-4*x))/(2*(1-4*x-x^2)) )); // G. C. Greubel, Jul 18 2019

CoefficientList[Series[(1-3*x-2*x^2+(1-x)*Sqrt[1-4*x])/(2*(1-4*x-x^2)), {x, 0, 30}], x] (* Vaclav Kotesovec, Oct 20 2012 *)

my(x='x+O('x^30)); Vec((1-3*x-2*x^2 + (1-x)*sqrt(1-4*x))/(2*(1-4*x-x^2))) \\ G. C. Greubel, Jul 18 2019

((1-3*x-2*x^2 + (1-x)*sqrt(1-4*x))/(2*(1-4*x-x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jul 18 2019

G.f.: (1-3*x-2*x^2 + (1-x)*sqrt(1-4*x))/(2*(1-4*x-x^2)).
a(n) = (1/2)*Sum_{k=0..n} C(2k,k)*F(3(n-k)+1)/(1-2k) + (1/2)*(F(3n-2) + 2*0^n).
Conjecture: n*(n-3)*a(n) +2*(-4*n^2+15*n-10)*a(n-1) +(15*n^2-69*n+80)*a(n-2) +2*(n-2)*(2*n-5)*a(n-3) =0. - R. J. Mathar, Nov 15 2011
a(n) ~ 1/10*(3*sqrt(5)-5)*(sqrt(5)+2)^n. - Vaclav Kotesovec, Oct 20 2012

A100050 A Chebyshev transform of n.

Original entry on oeis.org

0, 1, 2, 0, -4, -5, 0, 7, 8, 0, -10, -11, 0, 13, 14, 0, -16, -17, 0, 19, 20, 0, -22, -23, 0, 25, 26, 0, -28, -29, 0, 31, 32, 0, -34, -35, 0, 37, 38, 0, -40, -41, 0, 43, 44, 0, -46, -47, 0, 49, 50, 0, -52, -53, 0, 55, 56, 0, -58, -59, 0, 61, 62, 0, -64, -65, 0, 67, 68, 0, -70, -71, 0, 73, 74, 0, -76, -77, 0, 79, 80, 0, -82, -83, 0, 85, 86, 0

Offset: 0

Paul Barry, Oct 31 2004

A Chebyshev transform of x/(1-x)^2: if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))A(x/(1+x^2)).

			x + 2*x^2 - 4*x^4 - 5*x^5 + 7*x^7 + 8*x^8 - 10*x^10 - 11*x^11 + 13*x^13 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Michael Somos, Rational Function Multiplicative Coefficients
Index entries for linear recurrences with constant coefficients, signature (2,-3,2,-1).

Cf. A165202 (partial sums).

Cf. A099837, A099443, A011655, A100047, A100048, A100051, A091684.

LinearRecurrence[{2, -3, 2, -1}, {0, 1, 2, 0},50] (* G. C. Greubel, Aug 08 2017 *)

{a(n) = n * (-1)^(n\3) * sign( n%3)} /* Michael Somos, Mar 19 2011 */

{a(n) = local(A, p, e); if( abs(n)<1, 0, A = factor(abs(n)); prod( k=1, matsize(A)[1], if( p=A[k,1], e=A[k,2]; if( p==2, -(-2)^e, (kronecker( -12, p) * p)^e))))} /* Michael Somos, Mar 19 2011 */

[lucas_number1(n,2,1)*lucas_number1(n,1,1) for n in range(0,88)] # Zerinvary Lajos, Jul 06 2008

Euler transform of length 6 sequence [ 2, -3, -2, 0, 0, 2]. - Michael Somos, Mar 19 2011
a(n) is multiplicative with a(2^e) = -(-2)^e if e>0, a(3^e) = 0^e, a(p^e) = p^e if p == 1 (mod 6), a(p^e) = (-p)^e if p == 5 (mod 6). - Michael Somos, Mar 19 2011
G.f.: x*(1 - x^2)^3 *(1 - x^3)^2 / ((1 - x)^2 *(1 - x^6)^2) = x *(1 + x)^2 *(1 - x^2) / (1 + x^3)^2. - Michael Somos, Mar 19 2011
a(3*n) = 0, a(3*n + 1) = (-1)^n * (3*n + 1), a(3*n + 2) = (-1)^n * (3*n + 2). a(-n) = a(n). - Michael Somos, Mar 19 2011
G.f.: x(1-x^2)/(1-x+x^2)^2.
a(n) = 2*a(n-1) -3*a(n-2) +2*a(n-3) -a(n-4).
a(n) = n*Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k,k)*(n-2k)/(n-k).

A122765 Triangle read by rows: Let p(k, x) = x*p(k-1, x) - p(k-2, x). Then T(k,x) = dp(k,x)/dx.

Original entry on oeis.org

1, -1, 2, -2, -2, 3, 2, -6, -3, 4, 3, 6, -12, -4, 5, -3, 12, 12, -20, -5, 6, -4, -12, 30, 20, -30, -6, 7, 4, -20, -30, 60, 30, -42, -7, 8, 5, 20, -60, -60, 105, 42, -56, -8, 9, -5, 30, 60, -140, -105, 168, 56, -72, -9, 10

Offset: 1

Roger L. Bagula and Gary W. Adamson, Sep 22 2006

Based on the coefficients of derivatives of the polynomials in A130777.

			Triangle begins as:
   1;
  -1,   2;
  -2,  -2,   3;
   2,  -6,  -3,   4;
   3,   6, -12,  -4,   5;
  -3,  12,  12, -20,  -5,   6;
  -4, -12,  30,  20, -30,  -6,   7;
   4, -20, -30,  60,  30, -42,  -7,   8;
   5,  20, -60, -60, 105,  42, -56,  -8,  9;

G. C. Greubel, Rows n = 1..50 of the triangle, flattened
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.

Cf. A000217, A026741, A046854, A066170, A076118, A122766, A130777, A165202.

A122765:= func< n,k | k*(-1)^Binomial(n-k+1, 2)*Binomial(Floor((n+k)/2), k) >;
[A122765(n,k): k in [1..n], n in [1..14]]; // G. C. Greubel, Dec 30 2022

(* First program *)
p[0,x]=1; p[1,x]=x-1; p[k_,x_]:= p[k, x]= x*p[k-1,x] -p[k-2,x]; a = Table[Expand[p[n, x]], {n, 0, 10}]; Table[CoefficientList[D[a[[n]], x], x], {n, 2, 10}]//Flatten
(* Second program *)
T[n_, k_]:= k*(-1)^Binomial[n-k+1,2]*Binomial[Floor[(n+k)/2], k];
Table[T[n, k], {n,14}, {k,n}]//Flatten (* G. C. Greubel, Dec 30 2022 *)

tpol(n) = if (n<=0, 1, if (n==1, x-1, x*tpol(n-1) -tpol(n-2)));
lista(nn) = {for(n=0, nn, pol = deriv(tpol(n)); for (k=0, poldegree(pol), print1(polcoeff(pol, k), ", ");););} \\ Michel Marcus, Feb 07 2014

def A122765(n, k): return k*(-1)^binomial(n-k+1, 2)*binomial(((n+k)//2), k)
flatten( [[A122765(n,k) for k in range(1,n+1)] for n in range(1,15)] ) # G. C. Greubel, Dec 30 2022

From G. C. Greubel, Dec 30 2022: (Start)
T(n, k) = coefficient [x^k]( p(n, x) ), where p(n,x) = (2/(x^2-4))*((n+1)*chebyshev_T(n+1,x/2) -n*chebyshev_T(n,x/2) - (x/2)*(chebyshev_U(n,x/2) - chebyshev_U(n-1,x/2))).
T(n, k) = k*(-1)^binomial(n-k+1, 2)*binomial(floor((n+k)/2), k).
T(n, n) = n.
T(n, n-1) = -(n-1).
T(n, n-2) = -2*A000217(n-2).
T(n, n-3) = 2*A000217(n-3).
T(n, 1) = (-1)^binomial(n, 2)*floor((n+1)/2).
T(n, 2) = 2*(-1)^binomial(n-1, 2)*binomial(floor((n+2)/2), 2).
Sum_{k=1..n} T(n, k) = A076118(n).
Sum_{k=1..n} (-1)^k*T(n, k) = (-1)^(n-1)*A165202(n).
Sum_{k=1..floor((n+1)/2)} T(n-k+1, k) = [n=1] - [n=2].
Sum_{k=1..floor((n+1)/2)} (-1)^k*T(n-k+1, k) = (-1)^binomial(n+1, 2)*b(n), where b(n) = 4^floor(n/4)*A026741(n/2) if n is even and b(n) = 4^floor((n-1)/4)*A026741((n-1)/4) if n is odd. (End)

Name corrected and more terms from Michel Marcus, Feb 07 2014

A213172 Floor of the Euclidean distance of a point on the (1, 2, 3; 4, 5, 6) 3D walk.

Original entry on oeis.org

0, 1, 2, 3, 6, 9, 12, 16, 21, 26, 32, 38, 45, 52, 61, 69, 78, 88, 99, 110, 121, 133, 146, 159, 173, 188, 203, 218, 234, 251, 268, 286, 305, 324, 343, 364, 384, 406, 428, 450, 473, 497, 521, 546, 571, 597, 624, 651, 679, 707, 736, 765, 795, 826, 857

Offset: 0

Jon Perry, Apr 14 2013

nonn

Consider a standard 3-dimensional Euclidean lattice. We take 1 step along the positive x-axis, 2 along the positive y-axis, 3 along the positive z-axis, 4 along the positive x-axis, and so on. This sequence gives the floor of the Euclidean distance to the origin after n steps.

The (x,y,z) coordinates are (1,0,0), (1,2,0), (1,2,3), (5,2,3), (5,7,3), (5,7,9), (12,7,9) etc, where the x values run through A000326, the y-values through A005449, and the z-values through A045943. The squared Euclidean distances are s(n) = 1, 5, 14, 38, 83, 155, 274, 450,..., which obey the recurrence s(n) = 3*s(n-1) -3*s(n-2) +3*s(n-3) -6*s(n-4) +6*s(n-5) -3*s(n-6) +3*s(n-7) -3*s(n-8) +s(n-9), s(n) = (3*n^2+9*n+10)^2/108 +4*A099837(n+3)/27 -2*(-1)^n*A165202(n)/9, with a = floor(sqrt(s(n))). - R. J. Mathar, May 02 2013

			For a(4) we are at [5,2,3], so a(n) = floor(sqrt(25+4+9)) = 6.

Cf. A054925, A224985, A225215.

p=new Array(0,0,0);
for (a=0;a<100;a++) {
p[a%3]+=a;
document.write(Math.floor(Math.sqrt(p[0]*p[0]+p[1]*p[1]+p[2]*p[2]))+", ");
}

a(n) ~ n^2 sqrt(3)/6. - Charles R Greathouse IV, May 02 2013

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A110666 Sequence is {a(1,n)}, where a(m,n) is defined at sequence A110665.

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A165201 Expansion of 1/(1-x*c(x)^3), c(x) the g.f. of A000108.

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A100050 A Chebyshev transform of n.

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A122765 Triangle read by rows: Let p(k, x) = x*p(k-1, x) - p(k-2, x). Then T(k,x) = dp(k,x)/dx.

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A213172 Floor of the Euclidean distance of a point on the (1, 2, 3; 4, 5, 6) 3D walk.

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