A369490 -id:A369490 - cp's OEIS frontend

-1, -1, 1, 11, 49, 179, 601, 1931, 6049, 18659, 57001, 173051, 523249, 1577939, 4750201, 14283371, 42915649, 128878019, 386896201, 1161212891, 3484687249, 10456158899, 31372671001, 94126401611, 282395982049, 847221500579, 2541731610601, 7625329049531, 22876255584049

Offset: 1

Henrik Johansson (Henrik.Johansson(AT)Nexus.SE)

Binomial transform of A000918: (-1, 0, 2, 6, 14, 30, ...). - Gary W. Adamson, Mar 23 2012

This sequence demonstrates 2^n as a loose lower bound for g(n) in Waring's problem. Since 3^n > 2(2^n) for all n > 2, the number 2^(n + 1) - 1 requires 2^n n-th powers for its representation since 3^n is not available for use in the sum: the gulf between the relevant powers of 2 and 3 widens considerably as n gets progressively larger. - Alonso del Arte, Feb 01 2013

			a(3) = 1 because 3^2 - 2^3 = 9 - 8 = 1.
a(4) = 11 because 3^3 - 2^4 = 27 - 16 = 11.
a(5) = 49 because 3^4 - 2^5 = 81 - 32 = 49.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
D. Knuth, Letter to N. J. A. Sloane, date unknown
Index entries for linear recurrences with constant coefficients, signature (5,-6).

Cf. A000918, A056182 (first differences), A064686, A083313, A214091, A369490.
Cf. A363024 (prime terms).
From the third term onward the first differences of A005173.
Difference between two leftmost columns of A090888.
A diagonal in A254027.
Right edge of irregular triangle A252750.

[3^(n-1) -2^n: n in [1..30]]; // G. C. Greubel, Nov 03 2022

Table[3^(n-1) - 2^n, {n, 25}] (* Alonso del Arte, Feb 01 2013 *)
LinearRecurrence[{5,-6},{-1,-1},30] (* Harvey P. Dale, Feb 02 2015 *)

a(n)=3^(n-1)-2^n \\ Charles R Greathouse IV, Oct 07 2015

[3^(n-1) -2^n for n in range(1,31)] # G. C. Greubel, Nov 03 2022

Let b(n) = 2*(3/2)^n - 1. Then a(n) = -b(1-n)*3^(n-1) for n > 0. A083313(n) = A064686(n) = b(n)*2^(n-1) for n > 0. - Michael Somos, Aug 06 2006
From Colin Barker, May 27 2013: (Start)
a(n) = 5*a(n-1) - 6*a(n-2).
G.f.: -x*(1-4*x) / ((1-2*x)*(1-3*x)). (End)
E.g.f.: (1/3)*(2 - 3*exp(2*x) + exp(3*x)). - G. C. Greubel, Nov 03 2022

A few more terms from Alonso del Arte, Feb 01 2013

cp's OEIS Frontend

A003063 a(n) = 3^(n-1) - 2^n.

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