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A114830 Each term is previous term plus ceiling of geometric mean of all previous terms. 1

%I #6 Jun 23 2014 16:57:59

%S 1,2,4,6,9,13,18,24,31,39,48,59,71,85,101,119,139,162,187,215,246,280,

%T 318,359,404,453,507,565,628,697,771,851,937,1029,1128,1234,1348,1470,

%U 1600,1738,1885,2042,2209,2386,2574,2773,2984,3207,3443,3692,3955,4232,4524,4831,5154,5494,5851,6226,6620

%N Each term is previous term plus ceiling of geometric mean of all previous terms.

%C What is this sequence, asymptotically? a(n) is prime for n = 2, 6, 9, 12, 13, 15, 17, 24, ... are there an infinite number of prime values?

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

%F a(1) = 1, a(n+1) = a(n) + ceiling(GeometricMean[a(1),a(2),...,a(n)]). a(n+1) = a(n) + ceiling[((a(1)*a(2)*,...,*a(n))^(1/n)].

%e a(2) = 1 + ceiling(1^(1/1)) = 1 + 1 = 2.

%e a(3) = 2 + ceiling[(1*2)^(1/2)] = 2 + ceiling[sqrt(2)] = 2 + 2 = 4.

%e a(4) = 4 + ceiling[(1*2*4)^(1/3)] = 4 + ceiling[CubeRoot(8)] = 4 + 2 = 6.

%e a(5) = 6 + ceiling[(1*2*4*6)^(1/4)] = 6 + floor[4thRoot(48)] = 6 + 3 = 9.

%e a(6) = 9 + ceiling[(1*2*4*6*9)^(1/5)] = 9 + ceiling[5thRoot(432)] = 9 + 4 = 13.

%e a(7) = 13 + ceiling[(1*2*4*6*9*13)^(1/6)] = 6 + floor[6thRoot(5616)] = 13 + 5 = 18.

%e a(25) = 359 + ceiling[(1 * 2 * 4 * 6 * 9 * 13 * 18 * 24 * 31 * 39 * 48 * 59 * 71 * 85 * 101 * 119 * 139 * 162 * 187 * 215 * 246 * 280 * 318 * 359)^(1/24)] = 359 + ceiling[44.8074289] = 359 + 45 = 404.

%p A114830 := proc(n)

%p option remember;

%p if n= 1 then

%p 1;

%p else

%p mul(procname(i),i=1..n-1) ;

%p procname(n-1)+ceil(root[n-1](%)) ;

%p end if;

%p end proc:

%p seq(A114830(n),n=1..60) ; # _R. J. Mathar_, Jun 23 2014

%Y Cf. A065094, A065095.

%K easy,nonn

%O 1,2

%A _Jonathan Vos Post_, Feb 19 2006

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Last modified March 28 10:55 EDT 2024. Contains 371241 sequences. (Running on oeis4.)