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%I #12 Jun 25 2022 21:43:27
%S 1,3,7,13,23,41,75,141,271,529,1043,2069,4119,8217,16411,32797,65567,
%T 131105,262179,524325,1048615,2097193,4194347,8388653,16777263,
%U 33554481,67108915,134217781,268435511,536870969,1073741883,2147483709
%N Start with Pascal's triangle; form a triangle by sliding down n steps from top on both sides and including the horizontal row, deleting the inner numbers; a(n) = sum of entries on perimeter of triangle.
%F For n > 1, a(n) = A061761(n-1). - _David Wasserman_, Jun 03 2004
%e The triangle pertaining to n = 4 is obtained from the solid triangle
%e 1
%e 1 1
%e 1 2 1
%e 1 3 3 1
%e giving
%e 1
%e 1 1
%e 1 1
%e 1 3 3 1
%e and the sum of all the numbers is 13, so a(4) = 13.
%p restart:a:= proc(n) option remember; if n=0 then 1 else add((binomial (n,j)+2), j=0..n-1) fi end: seq (a(n), n=0..31); # _Zerinvary Lajos_, Mar 29 2009
%Y Cf. A081495, A081496, A081497.
%Y First differences of A290707.
%K nonn
%O 1,2
%A _Amarnath Murthy_, Mar 25 2003
%E Corrected and extended by _David Wasserman_, Jun 03 2004