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%I #12 Sep 28 2019 22:15:33
%S 0,0,1,2,5,6,13,17,27,38,61,80,125,174,245,359,509,727,1021,1484,2029,
%T 3006,4093,6029,8183,12158,16351,24380,32765,48848,65533,97919,131005,
%U 196094,262121,392363,524285,785406,1048445,1571309,2097149,3143496,4194301,6288380,8388323
%N a(n) = ADPE(n) is the total number of aperiodic k-double-palindromes of n up to cyclic equivalence, where 1 <= k <= n.
%C a(n) = ADPE(n) is the total number of aperiodic k-double-palindromes of n up to cyclic equivalence. See sequence A181169 for the definitions of an aperiodic k-double-palindrome of n and of cyclic equivalence.
%C Sequence A181169 is the 'ADPE(n,k)' triangle read by rows where ADPE(n,k) is the number of aperiodic k-double-palindromes of n up to cyclic equivalence.
%C For example, we have a(6) = ADPE(6) = ADPE(6,1) + ADPE(6,2) + ADPE(6,3) + ADPE(6,4) + ADPE(6,5) + ADPE(6,6) = 0 + 2 + 1 + 2 + 1 + 0 = 6. The 6 aperiodic double-palindromes of 6 up to cyclic equivalence are: 15, 24, 114, 1113, 1122, 11112. They are the representatives of the cyclic equivalence classes: {15,51}, {24,42}, {114,411,141},{1113,3111,1311,1131}, {1122,2112,2211,1221} and {11112,21111,12111,11211,11121}.
%C Hence a(n) = ADPE(n) is the total number of cyclic equivalence classes of compositions of n containing at least one aperiodic double-palindrome of n.
%H Andrew Howroyd, <a href="/A181314/b181314.txt">Table of n, a(n) for n = 1..1000</a>
%F From _Andrew Howroyd_, Sep 28 2019: (Start)
%F a(n) = A056493(n) - 1 for n > 1.
%F G.f.: (x^2-2*x)/(1-x) + Sum_{k=1..n} mu(k)*x^k*(2 + 3*x^k)/(1 - 2*x^(2*k)).
%F (End)
%o (PARI) a(n)={sumdiv(n, d, moebius(n/d)*((3 + d%2)*2^(d\2-1) - 1)) - 1} \\ _Andrew Howroyd_, Sep 28 2019
%Y Row sums of A181169.
%Y If we remove the aperiodic requirement, we get sequence A027383, see the comment from McSorley there. Also see sequences A181111 and A181135.
%Y Cf. A056493.
%K nonn
%O 1,4
%A _John P. McSorley_, Oct 12 2010
%E Terms a(11) and beyond from _Andrew Howroyd_, Sep 27 2019
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