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Numerator of 1 - (3/4)^n - frac((3/2)^n), where frac(x) = x - floor(x).
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%I #21 Aug 18 2024 19:04:11

%S 0,-1,3,13,159,173,1767,12789,17759,126237,292183,1930245,3724303,

%T 23940141,14206087,99585429,640559295,12562430525,7042526903,

%U 43417422885,813747135599,494896655693,3000760993767,18098709141429,249612172740383

%N Numerator of 1 - (3/4)^n - frac((3/2)^n), where frac(x) = x - floor(x).

%C The presumption that the fraction is positive for n > 1 underlies the presumed solution to Waring's problem.

%H Harry J. Smith, <a href="/A065622/b065622.txt">Table of n, a(n) for n = 0..200</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/WaringsProblem.html">Waring's Problem</a>

%F a(n) = 4^n*(1 + floor((3/2)^n)) - 3^n - 6^n = A005061(n) - A002380(n)*A000079(n) = A000302(n)*(1 + A002379(n)) - A000244(n) - A000400(n).

%e a(3) = 13 since 1 - (3/4)^3 - frac((3/2)^3) = 1 - 27/64 - frac(27/8) = 1 - 27/64 - 3/8 = (64 - 27 - 24)/64 = 13/64.

%t Table[1 - (3/4)^n - FractionalPart[(3/2)^n], {n, 0, 24}] // Numerator (* _Jean-François Alcover_, Apr 26 2016 *)

%o (PARI) { for (n=0, 200, a=numerator(1 - (3/4)^n - frac((3/2)^n)); write("b065622.txt", n, " ", a) ) } \\ _Harry J. Smith_, Oct 24 2009

%Y Denominator is A000302. Cf. A002804.

%K frac,sign

%O 0,3

%A _Henry Bottomley_, Dec 03 2001