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A181169
'ADPE(n,k)' triangle read by rows. ADPE(n,k) is the number of aperiodic k-double-palindromes of n up to cyclic equivalence.
4
0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 2, 2, 1, 0, 0, 2, 1, 2, 1, 0, 0, 3, 3, 3, 3, 1, 0, 0, 3, 3, 4, 3, 3, 1, 0, 0, 4, 3, 6, 6, 3, 4, 1, 0, 0, 4, 4, 8, 5, 8, 4, 4, 1, 0, 0, 5, 5, 10, 10, 10, 10, 5, 5, 1, 0, 0, 5, 4, 12, 10, 17, 10, 12, 4, 5, 1, 0, 0, 6, 6, 15, 15, 20, 20, 15, 15, 6, 6, 1, 0
OFFSET
1,12
COMMENTS
A k-composition of n is an ordered collection of k positive integers (parts) which sum to n. A k-composition is aperiodic (primitive) if its period is k, i.e., if it is not the concatenation of at least two smaller compositions. A palindrome is a word which is the same when written backwards.
A k-double-palindrome of n is a k-composition of n which is the concatenation of two palindromes, PP'=P|P', where both |P|, |P'|>=1.
For example 1123532=11|23532 is a 7-double-palindrome of 17 since both 11 and 23532 are palindromes. It is also aperiodic, and so it is an aperiodic 7-double-palindrome of 17. See sequence A181111.
Two k-compositions of n are cyclically equivalent if one can be obtained from the other by a cyclic permutation of its parts.
Let ADPE(n,k) denote the number of aperiodic k-double-palindromes of n up to cyclic equivalence, i.e., the number of cyclic equivalence classes containing at least one aperiodic k-double-palindrome.
This sequence is the `ADPE(n,k)' triangle read by rows.
REFERENCES
John P. McSorley: Counting k-compositions of n with palindromic and related structures. Preprint, 2010.
LINKS
FORMULA
T(n, 1) = 0; T(n, k) = A180424(n, k) for k > 1. - Andrew Howroyd, Sep 28 2019
EXAMPLE
The triangle begins:
0
0 0
0 1 0
0 1 1 0
0 2 2 1 0
0 2 1 2 1 0
0 3 3 3 3 1 0
0 3 3 4 3 3 1 0
0 4 3 6 6 3 4 1 0
0 4 4 8 5 8 4 4 1 0
...
For example, row 8 is: 0 3 3 4 3 3 1 0.
We have ADPE(8,3)=3 because the 6 aperiodic 3-double-palindromes of 8: 116, 611, 224, 422, 233, and 332 come in 3 cyclic equivalence classes: {116, 611, 161}, {224, 422, 242}, and {233, 323, 332}.
We have ADPE(8,4)=4 because there are 4 4-double-palindromes of 8 up to cyclic equivalence, the 4 classes are: {1115, 5111, 1511, 1151}, {1214, 4121, 1412, 2141}, {1133, 3113, 3311, 1331}, and {1232, 2123, 3212, 2321}.
PROG
(PARI) \\ here RE(n, k) is A119963(n, k).
RE(n, k) = binomial((n-k%2)\2, k\2);
T(n, k) = if(k<=1, 0, sumdiv(gcd(n, k), d, moebius(d)*RE(n/d, k/d))); \\ Andrew Howroyd, Sep 28 2019
CROSSREFS
Row sums are A181314.
If we remove the aperiodic requirement we get sequence A180918.
If we count the aperiodic k-double-palindromes of n (not the number of classes) we get sequence A181111 which is the 'ADP(n, k)' triangle read by rows, where ADP(n, k) is the number of aperiodic k-double-palindromes of n.
Cf. A180424.
Sequence in context: A287385 A191411 A133418 * A029390 A108040 A137566
KEYWORD
nonn,tabl
AUTHOR
John P. McSorley, Oct 07 2010
EXTENSIONS
Terms a(56) and beyond from Andrew Howroyd, Sep 27 2019
STATUS
approved