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.
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
Examples
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}.
References
- John P. McSorley: Counting k-compositions of n with palindromic and related structures. Preprint, 2010.
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..1275
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.
Programs
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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
Formula
T(n, 1) = 0; T(n, k) = A180424(n, k) for k > 1. - Andrew Howroyd, Sep 28 2019
Extensions
Terms a(56) and beyond from Andrew Howroyd, Sep 27 2019
Comments