A181314 a(n) = ADPE(n) is the total number of aperiodic k-double-palindromes of n up to cyclic equivalence, where 1 <= k <= n.

0, 0, 1, 2, 5, 6, 13, 17, 27, 38, 61, 80, 125, 174, 245, 359, 509, 727, 1021, 1484, 2029, 3006, 4093, 6029, 8183, 12158, 16351, 24380, 32765, 48848, 65533, 97919, 131005, 196094, 262121, 392363, 524285, 785406, 1048445, 1571309, 2097149, 3143496, 4194301, 6288380, 8388323

Offset: 1

Author

John P. McSorley, Oct 12 2010

Keywords

• nonn

Comments

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.

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.

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}.

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.

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1000

Crossrefs

Row sums of A181169.

If we remove the aperiodic requirement, we get sequence A027383, see the comment from McSorley there. Also see sequences A181111 and A181135.

Cf. A056493.

Programs

PARI

a(n)={sumdiv(n, d, moebius(n/d)*((3 + d%2)*2^(d\2-1) - 1)) - 1} \\ Andrew Howroyd, Sep 28 2019

Formula

From Andrew Howroyd, Sep 28 2019: (Start)

a(n) = A056493(n) - 1 for n > 1.

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)).

(End)

Extensions

Terms a(11) and beyond from Andrew Howroyd, Sep 27 2019