Search: id:A004771
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%I A004771
%S A004771 7,15,23,31,39,47,55,63,71,79,87,95,103,111,119,127,135,143,151,159,
%T A004771 167,175,183,191,199,207,215,223,231,239,247,255,263,271,279,287,
%U A004771 295,303,311,319,327,335,343,351,359,367,375,383,391,399,407,415
%N A004771 a(n) = 8n+7. Or, numbers n such that binary expansion ends 111.
%C A004771 These numbers cannot be perfect squares. Proof: Assume x^2 = 8k+7. Then
x is odd of the form 2m+1. So (2m+1)^2 - 7 = 8k 4m^2+4m - 6 = 8k
2m^2+2m - 3 = 4k or odd = even a contradiction. So the assumption
that x^2 = 8k+7 is false. - Cino Hilliard (hillcino368(AT)gmail.com),
Sep 03 2006
%C A004771 A056753(a(n)) = 7. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com),
Aug 23 2009]
%H A004771 Tanya Khovanova, Recursive Sequences
%H A004771 INRIA Algorithms Project,
Encyclopedia of Combinatorial Structures 962
%F A004771 These numbers cannot be expressed as the sum of 3 squares - Artur Jasinski
(grafix(AT)csl.pl), Nov 22 2006
%F A004771 O.g.f: (7+x)/(-1+x)^2 = 8/(-1+x)^2+1/(-1+x) . - R. J. Mathar (mathar(AT)strw.leidenuniv.nl),
Nov 30 2007
%p A004771 with(finance):seq(add(cashflows([2,2,4], 0 ),k=1..n)-1,n=1..54); - Zerinvary
Lajos (zerinvarylajos(AT)yahoo.com), Jun 21 2008
%o A004771 (Other) sage: [i+7 for i in range(415) if gcd(i,8) == 8] # [From Zerinvary
Lajos (zerinvarylajos(AT)yahoo.com), May 20 2009]
%o A004771 (Other) sage: [crt(2,n,4,3 )+crt(1,n,4,3 ) for n in xrange(2, 54)]# [From
Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 07 2009]
%Y A004771 Sequence in context: A031490 A059562 A017149 this_sequence A133655 A029724
A056828
%Y A004771 Adjacent sequences: A004768 A004769 A004770 this_sequence A004772 A004773
A004774
%K A004771 nonn
%O A004771 0,1
%A A004771 N. J. A. Sloane (njas(AT)research.att.com).
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