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 A003063 a(n) = 3^(n-1)-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 (list; graph; refs; listen; history; text; internal format)
 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 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 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(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 a(n) = 5*a(n-1)-6*a(n-2). G.f.: x*(4*x-1) / ((2*x-1)*(3*x-1)). - 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^(n-1)-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

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Last modified December 6 16:24 EST 2019. Contains 329808 sequences. (Running on oeis4.)