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Search: id:A004771
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| A004771 |
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a(n) = 8n+7. Or, numbers n such that binary expansion ends 111. |
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+0
7
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| 7, 15, 23, 31, 39, 47, 55, 63, 71, 79, 87, 95, 103, 111, 119, 127, 135, 143, 151, 159, 167, 175, 183, 191, 199, 207, 215, 223, 231, 239, 247, 255, 263, 271, 279, 287, 295, 303, 311, 319, 327, 335, 343, 351, 359, 367, 375, 383, 391, 399, 407, 415
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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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
A056753(a(n)) = 7. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Aug 23 2009]
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LINKS
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Tanya Khovanova, Recursive Sequences
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 962
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FORMULA
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These numbers cannot be expressed as the sum of 3 squares - Artur Jasinski (grafix(AT)csl.pl), Nov 22 2006
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
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MAPLE
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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
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PROGRAM
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(Other) sage: [i+7 for i in range(415) if gcd(i, 8) == 8] # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 20 2009]
(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]
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CROSSREFS
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Sequence in context: A031490 A059562 A017149 this_sequence A133655 A029724 A056828
Adjacent sequences: A004768 A004769 A004770 this_sequence A004772 A004773 A004774
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KEYWORD
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nonn
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com).
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