A141124 -id:A141124 - cp's OEIS frontend

A100096 An inverse Chebyshev transform of the Jacobsthal numbers.

0, 1, 1, 6, 9, 36, 66, 218, 449, 1332, 2946, 8196, 18954, 50688, 120576, 314586, 761889, 1957092, 4794426, 12194828, 30093854, 76067256, 188595276, 474810276, 1180734234, 2965094536, 7387570516, 18521858088, 46203981924, 115721310552

Offset: 0

Paul Barry, Nov 03 2004

Image of x/(1-x-2*x^2) under the transform g(x)->(1/sqrt(1-4*x^2))*g(x*c(x^2)), where c(x) is the g.f. of the Catalan numbers A000108. This is the inverse of the Chebyshev transform which takes A(x) to ((1-x^2)/(1+x^2))*A(x/(1+x^2)).

Hankel transform is (-1)^n*n. Hankel transform of a(n+1) is A141124. - Paul Barry, Jun 05 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A100095, A100097.

CoefficientList[Series[x*Sqrt[1-4*x^2]*(3*Sqrt[1-4*x^2]+2*x+1)/(2*(4*x^2-1)*(10*x^2+x-2)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)

G.f.: x*sqrt(1-4*x^2)*(3*sqrt(1-4*x^2)+2*x+1)/(2*(4*x^2-1)*(10*x^2+x-2)).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*A001045(n-2*k).
Conjecture: 2*(-n+1)*a(n) +(n+3)*a(n-1) +18*(n-2)*a(n-2) +4*(-n-2)*a(n-3) +40*(-n+3)*a(n-4)=0. - R. J. Mathar, Nov 24 2012
a(n) ~ (5/2)^n / 3. - Vaclav Kotesovec, Feb 12 2014

A141125 Hankel transform of a transform of Fibonacci numbers.

Original entry on oeis.org

1, 4, -4, -16, 16, 64, -64, -256, 256, 1024, -1024, -4096, 4096, 16384, -16384, -65536, 65536, 262144, -262144, -1048576, 1048576, 4194304, -4194304, -16777216, 16777216, 67108864, -67108864, -268435456, 268435456, 1073741824, -1073741824, -4294967296, 4294967296

Offset: 0

Paul Barry, Jun 05 2008

Hankel transform of A100095(n+1).

Index entries for linear recurrences with constant coefficients, signature (0,-4).

Cf. A100095, A141124.

LinearRecurrence[{0,-4},{1,4},33] (* or *) CoefficientList[Series[(1+4x)/(1+4x^2),{x,0,32}],x] (* James C. McMahon, Jul 17 2025 *)

G.f.: (1+4x)/(1+4x^2).
a(n) = 2^n*(cos(Pi*n/2)+2*sin(Pi*n/2)).
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
E.g.f.: cos(2*x) + 2*sin(2*x). - Stefano Spezia, Jul 18 2025

More terms from James C. McMahon, Jul 18 2025

A216972 a(4n+2) = 2, otherwise a(n) = n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 2, 7, 8, 9, 2, 11, 12, 13, 2, 15, 16, 17, 2, 19, 20, 21, 2, 23, 24, 25, 2, 27, 28, 29, 2, 31, 32, 33, 2, 35, 36, 37, 2, 39, 40, 41, 2, 43, 44, 45, 2, 47, 48, 49, 2, 51, 52, 53, 2, 55, 56, 57, 2, 59, 60, 61, 2, 63, 64, 65, 2, 67, 68, 69, 2

Offset: 0

Paul Curtz, Sep 21 2012

For n>0, a(n) is the denominator of A214282(n)/(-A214283(n+1)):
1/1, 1/2, 1/3, 3/4, 3/5, 1/2, 3/7, 5/8, 5/9, ...
For n>0, a(n) is the denominator of A214283(n)/A214283(n+1):
0/1, 1/2, 2/3, 3/4, 2/5, 1/2, 4/7, 5/8, 4/9, ...
a(n), first and second differences:
0, 1, 2, 3,  4,  5,  2, 7,  8,  9,  2, 11,  12, ...
1, 1, 1, 1,  1, -3,  5, 1,  1, -7,  9,  1,   1, ...
0, 0, 0, 0, -4,  8, -4, 0, -8, 16, -8,  0, -12, ...

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Cf. A028233, A034684, A000265, A026741, A060819, A145979, A214682.

[n mod 4 eq 2 select 2 else n: n in [0..70]]; // Bruno Berselli, Sep 26 2012

a[n_] := If[Mod[n, 4] == 2, 2, n]; Table[a[n], {n, 0, 81}] (* Jean-François Alcover, Sep 25 2012 *)
LinearRecurrence[{0,0,0,2,0,0,0,-1},{0,1,2,3,4,5,2,7},80] (* Harvey P. Dale, Nov 06 2017 *)

makelist(expand(2+(4-(1+(-1)^n)*(1-%i^n))*(n-2)/4), n, 0, 70); /* Bruno Berselli, Sep 26 2012 */

def A216972(n): return 2 if n&3==2 else n # Chai Wah Wu, Jan 31 2024

a(n) = 2*a(n-4) - a(n-8).
a(n+4) - a(n) = 4*A152822(n).
a(2n) + a(2n+1) = |A141124(n)|.
a(4n) + a(4n+1) + a(4n+2) + a(4n+3) = 6*A005408(n) = A017593(n).
G.f.: (x+2*x^2+3*x^3+4*x^4+3*x^5-2*x^6+x^7) / (1-2*x^4+x^8). - Jean-François Alcover, Sep 25 2012
a(n) = 2+(4-(1+(-1)^n)*(1-i^n))*(n-2)/4, where i=sqrt(-1). - Bruno Berselli, Sep 26 2012
a(2n) = 2*|A009531(n)|, a(2n+1) = 2n+1. - Bruno Berselli, Sep 27 2012

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A100096 An inverse Chebyshev transform of the Jacobsthal numbers.

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A141125 Hankel transform of a transform of Fibonacci numbers.

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A216972 a(4n+2) = 2, otherwise a(n) = n.

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