Search: id:A001576
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%I A001576
%S A001576 3,7,21,73,273,1057,4161,16513,65793,262657,1049601,4196353,16781313,
%T A001576 67117057,268451841,1073774593,4295032833,17180000257,68719738881,
%U A001576 274878431233,1099512676353,4398048608257,17592190238721
%N A001576 1^n + 2^n + 4^n.
%C A001576 Equals A135576, except for the first member. [From Omar E. Pol (info(AT)polprimos.com), 
               Nov 18 2008]
%C A001576 Let n>1, if a(n)= 1^n+2^n+4^n is a prime number then n is the form 3^h. 
               Example, for h=1, n=3, a(n)= 1^3+2^3+4^3=73 (prime); h=2, n=9, a(n)= 
               1^9 + 2^9 + 4^9 = 262657 (prime); for h=3, n=27, a(n) is not prime. 
               [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Feb 08 2009]
%H A001576 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to 
               linear recurrences with constant coefficients</a>
%F A001576 a(n) = 6*a(n-1) - 8*a(n-2) +3.
%F A001576 O.g.f.: -1/(-1+x)-1/(-1+2*x)-1/(-1+4*x). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), 
               Feb 29 2008
%F A001576 E.g.f.: e^x+e^(2*x)+e^(4*x) [From Mohammad K. Azarian (azarian(AT)evansville.edu), 
               Dec 26 2008]
%F A001576 a(n)=6*a(n-1)-8*a(n-2)+3, (a(1)=3, a(2)=7. [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), 
               Feb 08 2009]
%F A001576 a(n) = A024088(n)/A000225(n). [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), 
               Feb 15 2009]
%t A001576 Table[1^n + 2^n + 4^n, {n, 0, 24}]
%o A001576 (Other) sage: [sigma(4,n)for n in xrange(0,23)] # [From Zerinvary Lajos 
               (zerinvarylajos(AT)yahoo.com), Jun 04 2009]
%Y A001576 Cf. A001550, A034513, A001579, A074501 - A074580. See also comments in 
               A051154.
%Y A001576 Cf. A135576, A135577.
%Y A001576 Sequence in context: A148678 A148679 A148680 this_sequence A075211 A075212 
               A049365
%Y A001576 Adjacent sequences: A001573 A001574 A001575 this_sequence A001577 A001578 
               A001579
%K A001576 easy,nonn
%O A001576 0,1
%A A001576 N. J. A. Sloane (njas(AT)research.att.com).

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