A181314 - OEIS

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

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OFFSET

1,4

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

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)

PROG

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

CROSSREFS

Row sums of A181169.