A140323 -id:A140323 - cp's OEIS frontend

A272263 a(n) = numerator of A000032(n) - 1/2^n.

1, 1, 11, 31, 111, 351, 1151, 3711, 12031, 38911, 125951, 407551, 1318911, 4268031, 13811711, 44695551, 144637951, 468058111, 1514668031, 4901568511, 15861809151, 51329892351, 166107021311, 537533612031, 1739495309311, 5629125066751, 18216231370751

Offset: 0

Paul Curtz, Apr 24 2016

A000032(n), Lucas numbers, and 1/2^n are autosequences of the second kind.
Then a(n)/2^n is also an autosequence of the second kind.
Their corresponding autosequences of the first kind are A000045(n) and n/2^n, the Oresme numbers.
Difference table of A000032(n) - 1/2^n:
1, 1/2,  11/4,   31/8,  111/16,    351/32,  1151/64, ...
   9/4,   9/8,  49/16,  129/32,    449/64, 1409/128, ...
        31/16,  31/32,  191/64,   511/128, 1791/256, ...
               129/64, 129/128,   769/256, ...
                       511/256,   511/256, ...
                                2049/1024, ... .
The first upper diagonal is A140323(n)/A004171(n). The main diagonal is the double, i.e. A140323(n)/A000302(n). The inverse binomial transform is the signed sequence.
Quintisections from a(2):
     11,      31,      111,      351,      1151,
   3711,   12031,    38911,   125951,    407551,
1318911, 4268031, 13811711, 44695551, 144637951,
etc.

			Numerators of a(0) =2-1=1, a(1)=1-1/2=1/2, a(2)=3-1/4=11/4, a(3)=4-1/8=31/8, ... .

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,2,-4).

Cf. A000012, A000032, A000045, A000079, A000302, A004171, A085449, A087131, A140323.

CoefficientList[Series[(1 - 2*x + 6*x^2)/((1 - x)*(1 - 2*x - 4*x^2)), {x, 0, 30}], x] (* Robert Price, Apr 24 2016 *)
Table[Numerator[LucasL@ n - 1/2^n], {n, 0, 26}] (* Michael De Vlieger, Apr 24 2016 *)

Vec((1-2*x+6*x^2)/((1-x)*(1-2*x-4*x^2)) + O(x^50)) \\ Colin Barker, Apr 24 2016

a(n) = a(n-1) + 10*A085449(n), for n>0, a(0)=1.
a(n) = A087131(n) - 1.
From Colin Barker, Apr 24 2016: (Start)
a(n) = (-1+(1-sqrt(5))^n+(1+sqrt(5))^n).
a(n) = 3*a(n-1)+2*a(n-2)-4*a(n-3) for n>2.
G.f.: (1-2*x+6*x^2) / ((1-x)*(1-2*x-4*x^2)).
(End)

A321632 Expansion of e.g.f. (1 + sin(x))/exp(x).

Original entry on oeis.org

1, 0, -1, 1, 1, -5, 9, -9, 1, 15, -31, 31, 1, -65, 129, -129, 1, 255, -511, 511, 1, -1025, 2049, -2049, 1, 4095, -8191, 8191, 1, -16385, 32769, -32769, 1, 65535, -131071, 131071, 1, -262145, 524289, -524289, 1, 1048575, -2097151, 2097151, 1, -4194305, 8388609, -8388609

Offset: 0

Paolo P. Lava, Nov 16 2018

A140323(n) = |a(4*n-1)| = |a(4*n-2)|, A247281(n) = |a(4*n+1)|.

The absolute values of the coefficients of the expansion of the reciprocal of this function are listed in A186364.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-3,-4,-2).

Cf. A140323, A186364, A247281.

seq(factorial(n)*coeff(series((1+sin(x))/exp(x),x=0,48),x,n),n=0..47);

With[{nn=50},CoefficientList[Series[(1+Sin[x])/Exp[x],{x,0,nn}],x] Range[ 0,nn]!] (* or *) LinearRecurrence[{-3,-4,-2},{1,0,-1},50] (* Harvey P. Dale, Jul 21 2021 *)

Vec((1 + 3*x + 3*x^2) / ((1 + x)*(1 + 2*x + 2*x^2)) + O(x^40)) \\ Colin Barker, Nov 16 2018

a(4*k)   = 1;
a(4*k+1) = (-4)^k - 1;
a(4*k+2) = -2*a(4*k+1) - 1 = -2*(-4)^k + 1;
a(4*k+3) =  2*a(4*k+1) + 1 =  2*(-4)^k - 1.
From Colin Barker, Nov 16 2018: (Start)
G.f.: (1 + 3*x + 3*x^2) / ((1 + x)*(1 + 2*x + 2*x^2)).
a(n) = (-1)^n + i/2*((-1-i)^n - (-1+i)^n), where i=sqrt(-1).
a(n) = -3*a(n-1) - 4*a(n-2) - 2*a(n-3) for n>2. (End)

cp's OEIS Frontend

A272263 a(n) = numerator of A000032(n) - 1/2^n.

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