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7, 31, 71, 127, 199, 287, 391, 511, 647, 799, 967, 1151, 1351, 1567, 1799, 2047, 2311, 2591, 2887, 3199, 3527, 3871, 4231, 4607, 4999, 5407, 5831, 6271, 6727, 7199, 7687, 8191, 8711, 9247, 9799, 10367, 10951, 11551, 12167, 12799, 13447, 14111, 14791
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| The identity (8*n^2-1)^2-(16*n^2-4)*(2*n)^2=1 can be written as a(n)^2-A158443(n)*A005843(n)^2=1.
Sequence found by reading the line from 7, in the direction 7, 31,..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Sep 03 2011
Bisection of A195605 (odd part). - Bruno Berselli, Sep 21 2011
The identity (8*n^2-1)^2-(64*n^2-16)*(n)^2 = 1 can be written as a(n)^2-A157913(n)*(n)^2 = 1. - Vincenzo Librandi, Feb 09 2012
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LINKS
| Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index to sequences with linear recurrences with constant coefficients, signature (3,-3,1).
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FORMULA
| a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).
G.f: x*(7+10*x-x^2)/(1-x)^3.
a(n) = A139098(n) - 1. - Omar E. Pol, Sep 03 2011
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MATHEMATICA
| Table[8n^2-1, {n, 50}]
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PROG
| (MAGMA) I:=[7, 31, 71]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]];
(PARI) a(n)=8*n^2-1 \\ Charles R Greathouse IV, Sep 03 2011
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CROSSREFS
| Cf. A157913, A158443, A005843.
Sequence in context: A163354 A105428 A050547 * A090684 A033199 A003550
Adjacent sequences: A157911 A157912 A157913 * A157915 A157916 A157917
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KEYWORD
| nonn,easy,changed
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AUTHOR
| Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Mar 09 2009
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