

A003063


a(n) = 3^(n1)2^n.


14



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
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OFFSET

1,4


COMMENTS

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 nth 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


LINKS

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).


FORMULA

Let b(n) = 2*(3/2)^n  1. Then a(n) = b(1n)*3^(n1) for n > 0. A083313(n) = A064686(n) = b(n)*2^(n1) for n > 0.  Michael Somos, Aug 06 2006
a(n) = 5*a(n1)6*a(n2). G.f.: x*(4*x1) / ((2*x1)*(3*x1)).  Colin Barker, May 27 2013


EXAMPLE

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.


MATHEMATICA

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 *)


PROG

(PARI) a(n)=3^(n1)2^n \\ Charles R Greathouse IV, Oct 07 2015


CROSSREFS

Cf. A000918.
Sequence in context: A160671 A297521 A295420 * A124857 A302473 A126398
Adjacent sequences: A003060 A003061 A003062 * A003064 A003065 A003066


KEYWORD

sign,easy


AUTHOR

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


EXTENSIONS

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


STATUS

approved



