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A079273
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Octo numbers (a polygonal sequence): 5n^2-6n+2, or (n-1)^2 + (2n-1)^2.
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3
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1, 10, 29, 58, 97, 146, 205, 274, 353, 442, 541, 650, 769, 898, 1037, 1186, 1345, 1514, 1693, 1882, 2081, 2290, 2509, 2738, 2977, 3226, 3485, 3754, 4033, 4322, 4621, 4930, 5249, 5578, 5917, 6266, 6625, 6994, 7373, 7762, 8161, 8570, 8989, 9418, 9857, 10306
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| a(n+1) = a(n)+10n-1 and n+a(n) is always congruent to 2 mod 10 (notice pattern of final digits). a(n)= the n-th hex number (3n^2-3n+1) added to the (2n-2)nd triangular number (2n^2-3n+1). The formula for the n-th octo number can be written as (2n-1)^2 + (n-1)^2; compare to formula for n-th octagonal number, n(3n-2)= (2n-1)^2 - (n-1)^2.
a(n+1) = 5n^2+4n+1 is also the number of ways of realizing the amount 10n using only coins with values 1, 2 and 5. [From Francois Brunault (brunault(AT)gmail.com), Nov 24 2009]
a(n) is the number of length 6 n-ary words beginning with the first character of the alphabet, that can be built by repeatedly inserting doublets into the initially empty word. - Alois P. Heinz, Sep 01 2011
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LINKS
| Eric Weisstein's World of Mathematics, Hex Number
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FORMULA
| a(n) = 10*n+a(n-1)-11 (with a(1)=1) [From Vincenzo Librandi, Aug 08 2010]
a(1) = 1, a(2) = 10, a(3) = 29, a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) [From Harvey P. Dale, May 03 2011]
G.f.: -(2*x^2+7*x+1)*x / (x-1)^3. - Alois P. Heinz, Sep 01 2011
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EXAMPLE
| a(4) = 58 because 58 dots can be arranged into a simple octagonal pattern with 4 dots on each side, its rows from top to bottom containing 4,5,6,7,7,7,7,6,5 and 4 dots respectively. The pattern is similar to the pattern for hex numbers (see link), with the exception that while the n-th hex figure has only 1 row of length 2n-1 dots (the maximum length) in the center, the n-th octo figure has n such rows.
a(4) = 58:
.. O O O O
. O O O O O
.O O O O O O
O O O O O O O
O O O O O O O
O O O O O O O
O O O O O O O
.O O O O O O
. O O O O O
.. O O O O
For n=2, a(2) = 10*2+1-11 = 10; n=3, a(3) = 10*3+10-11 = 29; n=4, a(4) = 10*4+29-11 = 58 [From Vincenzo Librandi, Aug 08 2010]
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MATHEMATICA
| Table[5n^2-6n+2, {n, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {1, 10, 29}, 150] (* From Harvey P. Dale, Apr 06 2011 & May 03 2011 *)
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CROSSREFS
| Cf. A000217 (triangular numbers), A000567 (octagonal numbers), A003215 (hex numbers).
Row n=3 of A183134. - Alois P. Heinz, Aug 31 2011
Sequence in context: A143190 A009771 A068197 * A048469 A048772 A055850
Adjacent sequences: A079270 A079271 A079272 * A079274 A079275 A079276
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KEYWORD
| easy,nonn,nice
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AUTHOR
| Matthew Vandermast (ghodges14(AT)comcast.net), Feb 06 2003
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