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Smallest positive integer for which n divides a(n)^6.
9

%I #37 Oct 27 2022 07:31:48

%S 1,2,3,2,5,6,7,2,3,10,11,6,13,14,15,2,17,6,19,10,21,22,23,6,5,26,3,14,

%T 29,30,31,2,33,34,35,6,37,38,39,10,41,42,43,22,15,46,47,6,7,10,51,26,

%U 53,6,55,14,57,58,59,30,61,62,21,2,65,66,67,34,69,70,71,6,73,74,15,38,77,78

%N Smallest positive integer for which n divides a(n)^6.

%C Differs from A007947 as follows: A007947(128)=2, a(128)=4; A007947(256)=2, a(256)=4; A007947(384)=6, a(384)=12; A007947(512)=2, a(512)=4; A007947(640)=10, a(640)=20, etc. - _R. J. Mathar_, Oct 28 2008

%H Harvey P. Dale, <a href="/A015053/b015053.txt">Table of n, a(n) for n = 1..1000</a>

%H Henry Bottomley, <a href="http://fs.gallup.unm.edu/Bottomley-Sm-Mult-Functions.htm">Some Smarandache-type multiplicative sequences</a>.

%H Kevin A. Broughan, <a href="https://doi.org/10.4064/aa101-2-2">Restricted divisor sums</a>, Acta Arithmetica, 101(2) (2002), 105-114.

%H Kevin A. Broughan, <a href="http://ijpam.eu/contents/2003-5-3/2/2.pdf">Relationship between the integer conductor and k-th root functions</a>, Int. J. Pure Appl. Math. 5(3) (2003), 253-275.

%H Kevin A. Broughan, <a href="https://www.thebookshelf.auckland.ac.nz/document.php?wid=2937">Relaxations of the ABC Conjecture using integer k'th roots</a>, New Zealand J. Math. 35(2) (2006), 121-136.

%H Henry Ibstedt, <a href="http://www.gallup.unm.edu/~smarandache/Ibstedt-surfing.pdf">Surfing on the Ocean of Numbers</a>, Erhus Univ. Press, Vail, 1997.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SmarandacheCeilFunction.html">Smarandache Ceil Function</a>.

%F Multiplicative with a(p^e) = p^ceiling(e/6). - _Christian G. Bower_, May 16 2005

%F Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(11)/2) * Product_{p prime} (1 - 1/p^2 + 1/p^3 - 1/p^4 + 1/p^5 - 1/p^6 + 1/p^7 - 1/p^8 + 1/p^9 - 1/p^10) = 0.3522558764... . - _Amiram Eldar_, Oct 27 2022

%t spi[n_]:=Module[{k=1},While[PowerMod[k,6,n]!=0,k++];k]; Array[spi,80] (* _Harvey P. Dale_, Feb 29 2020 *)

%t f[p_, e_] := p^Ceiling[e/6]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Sep 18 2020 *)

%o (PARI) a(n) = my(f=factor(n)); for (i=1, #f~, f[i,2] = ceil(f[i,2]/6)); factorback(f); \\ _Michel Marcus_, Feb 15 2015

%Y Cf. A000188 (inner square root), A019554 (outer square root), A053150 (inner 3rd root), A019555 (outer 3rd root), A053164 (inner 4th root), A053166 (outer 4th root), A015052 (5th outer root).

%Y Cf. A013669.

%K nonn,mult,easy

%O 1,2

%A R. Muller (Research37(AT)aol.com)

%E Corrected by _David W. Wilson_, Jun 04 2002