

A181510


Number of permutations of the multiset {1,1,2,2,3,3,...,n+1,n+1} avoiding the permutation patterns {132, 231, 2134}


7



6, 18, 34, 54, 78, 106, 138, 174, 214, 258, 306, 358, 414, 474, 538, 606, 678, 754, 834, 918, 1006, 1098, 1194, 1294, 1398, 1506, 1618, 1734, 1854, 1978
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OFFSET

1,1


COMMENTS

a(n) is also the surface ares of the nth solid in the following recursive construction:
The first solid is a unit cube (hence a(1)=6).
To form the nth solid from the (n1)st solid, construct a row of 2n1 cubes, then center the (n1)st solid on top of this row. (For example, the second solid is a row of 3 unit cubes, with a single unit cube centered on top of the middle cube. This construction has surface area a(2)=18.)
The sequence provides all nonnegative integers m such that 2*m+13 is a square.  Bruno Berselli, Mar 01 2013


LINKS

Table of n, a(n) for n=1..30.
Lara K. Pudwell, Stacking Blocks and Counting Permutations, Mathematics Magazine 83 (2010), 297302.
Index entries for linear recurrences with constant coefficients, signature (3,3,1).


FORMULA

a(n) = 2n^2+6n2.
From Bruno Berselli, Oct 29 2010: (Start)
G.f.: 2*x*(3x^2)/(1x)^3.
a(n) 3*a(n1) +3*a(n2) a(n3) = 0 for n>3.
a(n) = 2*A014209(n) = 2*A082111(n1)+4 = A051936(2n+2)+n+4. (End)


EXAMPLE

For n=1, the permutations of {1,1,2,2} avoiding the patterns {132, 231, 2134} are {1122, 1212, 1221, 2112, 2121, 2211}.
For n=2, the permutations of {1,1,2,2,3,3} avoiding the patterns {132, 231, 2134} are {112233, 121233, 122133, 211233, 212133, 221133, 311223, 312123, 312213, 321123, 321213, 322113, 331122, 331212, 331221, 332112, 332121, 332211}


PROG

(PARI) a(n)=2*n^2+6*n2 \\ Charles R Greathouse IV, Jun 17 2017


CROSSREFS

Sequence in context: A110671 A134078 A323148 * A301715 A269755 A270135
Adjacent sequences: A181507 A181508 A181509 * A181511 A181512 A181513


KEYWORD

nonn,easy


AUTHOR

Lara Pudwell, Oct 25 2010


STATUS

approved



