1,2

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 François Brunault (brunault(AT)gmail.com), Nov 24 2009]

Eric Weisstein's World of Mathematics, Link to a section of The World Of Mathematics. (on hex numbers)

a(n)=10*n+a(n-1)-11 (with a(1)=1) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 10 2009]

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=1, a(1)=1; n=2, a(2)=10; [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Feb 08 2009]

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 (vincenzo.librandi(AT)tin.it), Nov 10 2009]

Cf. A000217 (triangular numbers), A000567 (octagonal numbers), A003215 (hex numbers).

Adjacent sequences: A079270 A079271 A079272 this_sequence A079274 A079275 A079276

easy,nonn,nice,new

Matthew Vandermast (ghodges14(AT)comcast.net), Feb 06 2003

First line of %F removed (wrong formula).Typo corrected in the next to last line of %e (wrong values) François Brunault (brunault(AT)gmail.com), Nov 24 2009