A187093 -id:A187093 - cp's OEIS frontend

A081654 a(n) = 2*4^n - 0^n.

1, 8, 32, 128, 512, 2048, 8192, 32768, 131072, 524288, 2097152, 8388608, 33554432, 134217728, 536870912, 2147483648, 8589934592, 34359738368, 137438953472, 549755813888, 2199023255552, 8796093022208, 35184372088832

Offset: 0

Paul Barry, Mar 26 2003

Binomial transform of A081632. Inverse binomial transform of A081655.

			a(0) = 2*4^0 - 0^0 = 2 - 1 = 1 (use 0^0 = 1).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (4).

Cf. A000244 (3^n), A187093.

Essentially the same as A004171.

[2*4^n-0^n: n in [0..30]]; // Vincenzo Librandi, Aug 10 2013

CoefficientList[Series[(1 + 4 x) / (1 - 4 x), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 10 2013 *)

a(n)=2*4^n-0^n \\ Charles R Greathouse IV, Apr 09 2012

x='x+O('x^100); Vec((1+4*x)/(1-4*x)) \\ Altug Alkan, Dec 14 2015

a(0)=1, a(n) = 2*4^n, n>0
G.f.: (1+4*x)/(1-4*x).
E.g.f. 2*exp(4*x)-1.
With interpolated zeros, this is 2^n - 0^n + (-2)^n. - Paul Barry, Sep 06 2003
a(n) = A081294(n+1), n>0. - R. J. Mathar, Sep 17 2008
For n>0, a(n) = 2 * (1 + 3^(n-1) + Sum{x=1..n-2}Sum{k=0..x-1}(binomial(x-1,k)*(3^(k+1) + 3^(n-x+k)))). - J. Conrad, Dec 10 2015

A215097 a(n) = n^3 - a(n-2) for n >= 2 and a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 8, 26, 56, 99, 160, 244, 352, 485, 648, 846, 1080, 1351, 1664, 2024, 2432, 2889, 3400, 3970, 4600, 5291, 6048, 6876, 7776, 8749, 9800, 10934, 12152, 13455, 14848, 16336, 17920, 19601, 21384, 23274, 25272, 27379, 29600, 31940, 34400, 36981, 39688, 42526

Offset: 0

Alex Ratushnyak, Aug 03 2012

Index entries for linear recurrences with constant coefficients, signature (4,-7,8,-7,4,-1).

Cf. A000217 (n^2 - a(n-1)).
Cf. A125577 (n^2 - a(n-1) with a(0)=1).
Cf. A011934 (n^3 - a(n-1)).
Cf. A153026 (n^3 - a(n-1) with a(1)=0).
Cf. A194274 (n^2 - a(n-2)).
Cf. A187093 (n^2 - a(n-2) with a(0)=a(1)=1, a(-1)=0).
Cf. A107386 ((n-2)^2 - a(n-1) with a(0)=0, a(1)=a(2)=1, a(3)=2).
Cf. A206481 ((n-1)^3 - a(n-2)).

RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == n^3 - a[n - 2]}, a[n], {n, 0, 43}] (* Bruno Berselli, Aug 07 2012 *)

prpr = 0
prev = 1
for n in range(2,77):
    print(prpr, end=',')
    curr = n*n*n - prpr
    prpr = prev
    prev = curr

G.f.: (x+4*x^2+x^3)/((-1+x)^4*(1+x^2)). - David Scambler, Aug 06 2012

a(n) = (n*(n^2-3)-(1-(-1)^n)*i^(n+1))/2, where i=sqrt(-1). - Bruno Berselli, Aug 07 2012

cp's OEIS Frontend

A081654 a(n) = 2*4^n - 0^n.

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