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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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9
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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, 423, 431
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,1
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COMMENTS
| These numbers cannot be expressed as the sum of 3 squares. [Artur Jasinski (grafix(AT)csl.pl), Nov 22 2006]
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]
a(n) is the set of numbers congruent to {7,15,23} mod 24. [From Gary Detlefs (gdetlefs(AT)aol.com), Mar 07 2010]
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LINKS
| Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Tanya Khovanova, Recursive Sequences
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 962
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FORMULA
| 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
a(n) = floor((24n-2)/3) with offset 1..a(1)=7 [From Gary Detlefs (gdetlefs(AT)aol.com), Mar 07 2010]
a(n) = 8*n + 7; a(n) = 2*a(n-1) - a(n-2); [Vincenzo Librandi, May 28 2011]
A056753(a(n)) = 7. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Aug 23 2009]
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MATHEMATICA
| Array[8*#+7&, 100, 0] [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Dec 14 2009]
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PROG
| (MAGMA) [8*n+7: n in [0..60]]; Vincenzo Librandi, May 28 2011
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CROSSREFS
| Cf. A008590, A017077, A017137.
Sequence in context: A059562 A017149 A133655 * A029724 A194400 A177768
Adjacent sequences: A004768 A004769 A004770 * A004772 A004773 A004774
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KEYWORD
| nonn
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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