login
A143721
Aliquot sequence starting at 38.
2
38, 22, 14, 10, 8, 7, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
OFFSET
0,1
COMMENTS
From Michal Paulovic, Dec 31 2017: (Start)
The only integer with the sum of its proper divisors equal to 38 is 1369.
The sequence's pattern: 6+2^5, 6+2^4, 6+2^3, 6+2^2, 6+2^1, 6+2^0, 1, 0, 0, 0, ...
(End)
Part of the larger aliquot sequence: 12135617199, 5556030801, 2107460079, 1016939121, 496646799, 231221769, 85406391, 55582041, 19742503, 1794785, 366535, 107225, 25765, 5159, 1369, 38, 22, 14, 10, 8, 7, 1, 0, ..., . - Robert G. Wilson v, Mar 05 2018
LINKS
Eric Weisstein's MathWorld, Aliquot Sequence
FORMULA
a(n+1) = A001065(a(n)). - R. J. Mathar, Oct 11 2017
From Iain Fox, Dec 31 2017: (Start)
G.f.: 38 + 22*x + 14*x^2 + 10*x^3 + 8*x^4 + 7*x^5 + x^6.
E.g.f.: (27360 + 15840*x + 5040*x^2 + 1200*x^3 + 240*x^4 + 42*x^5 + x^6)/720.
(End)
MATHEMATICA
NestList[If[# == 0, 0, DivisorSigma[1, #] - #] &, 38, 100] (* Michael De Vlieger, Dec 31 2017 *)
PROG
(PARI) x=1369; while(x, x=sigma(x)-x; if(x, print1(x ", "), print1(x ", " x ", " x ", ..."))) \\ Michal Paulovic, Dec 31 2017
CROSSREFS
Sequence in context: A160147 A033358 A033974 * A333850 A070725 A249280
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Nov 30 2008
STATUS
approved