cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 11-16 of 16 results.

A181111 'ADP(n,k)' triangle read by rows. ADP(n,k) is the number of aperiodic k-double-palindromes of n.

Original entry on oeis.org

0, 0, 0, 0, 2, 0, 0, 2, 2, 0, 0, 4, 4, 4, 0, 0, 4, 2, 6, 4, 0, 0, 6, 6, 12, 12, 6, 0, 0, 6, 6, 14, 12, 16, 6, 0, 0, 8, 6, 24, 24, 18, 24, 8, 0, 0, 8, 8, 28, 20, 44, 24, 28, 8, 0, 0, 10, 10, 40, 40, 60, 60, 40, 40, 10, 0, 0, 10, 8, 44, 40, 94, 60, 88, 32, 46, 10, 0
Offset: 1

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Author

John P. McSorley, Oct 03 2010

Keywords

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. The 4-double-palindrome of 8 1313=131|3 is not aperiodic, so it is not an aperiodic 4-double-palindrome of 8.
Let ADP(n,k) denote the number of aperiodic k-double-palindromes of n.
This sequence is the 'ADP(n,k)' triangle read by rows.

Examples

			The triangle begins:
  0
  0 0
  0 2 0
  0 2 2  0
  0 4 4  4  0
  0 4 2  6  4  0
  0 6 6 12 12  6  0
  0 6 6 14 12 16  6  0
  0 8 6 24 24 18 24  8 0
  0 8 8 28 20 44 24 28 8 0
  ...
For example, row 8 is: 0 6 6 14 12 16 6 0.
We have ADP(8,3)=6 because there are 6 aperiodic 3-double-palindromes of 8: 116, 611, 224, 422, 233, and 332.
We have ADP(8,4)=14 because there are 14 4-double-palindromes of 8: 1115, 5111, 1511, 1151, 1214, 4121, 1412, 2141, 1133, 3311, 1232, 2123, 3212, and 2321.
		

References

  • John P. McSorley: Counting k-compositions of n with palindromic and related structures. Preprint, 2010.

Crossrefs

Row sums are A181135.
See sequence A180653 for the triangle whose (n, k) term gives the number of k-double-palindromes of n.
See sequence A179519 for the triangle whose (n, k) term gives the number of aperiodic k-palindromes (single-palindromes) of n.

Programs

  • PARI
    \\ here p(n,k) is A119963(n,k), q(n,k) is A051159(n-1, k-1).
    p(n, k) = { binomial((n-k%2)\2, k\2) }
    q(n, k) = { if(n%2==1&&k%2==0, 0, binomial((n-1)\2, (k-1)\2)) }
    T(n, k) = sumdiv(gcd(n, k), d, moebius(d) * (k*p(n/d, k/d) - q(n/d, k/d))); \\ Andrew Howroyd, Sep 27 2019

Formula

T(n,k) = A180279(n,k) - A179519(n,k). - Andrew Howroyd, Sep 27 2019

Extensions

a(37) corrected and terms a(56) and beyond from Andrew Howroyd, Sep 27 2019

A180171 Triangle read by rows: R(n,k) is the number of k-reverses of n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 4, 10, 5, 1, 1, 6, 9, 12, 15, 6, 1, 1, 7, 9, 19, 15, 21, 7, 1, 1, 8, 10, 24, 30, 20, 28, 8, 1, 1, 9, 12, 36, 26, 54, 28, 36, 9, 1, 1, 10, 15, 40, 50, 60, 70, 40, 45, 10, 1, 1, 11, 13, 53, 50, 108, 70, 106, 39, 55, 11, 1, 1, 12, 18, 60, 75, 120
Offset: 1

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Author

John P. McSorley, Aug 15 2010

Keywords

Comments

A k-composition of n is an ordered collection of k positive integers (parts) which sum to n.
Two k-compositions are cyclically equivalent if one can be obtained from the other by a cyclic permutation of its parts.
The reverse of a k-composition is the k-composition obtained by writing its parts in reverse.
For example the reverse of 123 is 321.
A k-reverse of n is a k-composition of n which is cyclically equivalent to its reverse.
For example 114 is a 3-reverse of 6 since its set of cyclic equivalents {114,411,141} contains its reverse 411. But 123 is not a 3-reverse of 6 since its set of cyclic equivalents {123,312,231} does not contain its reverse 321.

Examples

			The triangle begins
  1
  1 1
  1 2 1
  1 3 3 1
  1 4 6 4 1
  1 5 4 10 5 1
  1 6 9 12 15 6 1
  1 7 9 19 15 21 7 1
  1 8 10 24 30 20 28 8 1
  1 9 12 36 26 54 28 36 9 1
For example row 8 is 1 7 9 19 15 21 7 1
We have R(8,3)=9 because there are 9 3-reverses of 8. In classes: {116,611,161} {224,422,242}, and {233,323,332}.
We have R(8,6)=21 because all 21 6-compositions of 8 are 6-reverses of 8.
		

References

  • John P. McSorley: Counting k-compositions of n with palindromic and related structures. Preprint, 2010.

Crossrefs

Row sums are A180249.

Programs

  • Mathematica
    f[n_Integer, k_Integer] := Block[{c = 0, j = 1, ip = IntegerPartitions[n, {k}]}, lmt = 1 + Length@ ip; While[j < lmt, c += g[ ip[[j]]]; j++ ]; c]; g[lst_List] := Block[{c = 0, len = Length@ lst, per = Permutations@ lst}, While[ Length@ per > 0, rl = Union[ RotateLeft[ per[[1]], # ] & /@ Range@ len]; If[ MemberQ[rl, Reverse@ per[[1]]], c += Length@ rl]; per = Complement[ per, rl]]; c]; Table[ f[n, k], {n, 13}, {k, n}] // Flatten (* Robert G. Wilson v, Aug 25 2010 *)
  • PARI
    \\ here p(n,k) is A119963, AR(n,k) is A180279.
    p(n,k) = binomial((n-k%2)\2, k\2);
    AR(n,k) = k*sumdiv(gcd(n,k), d, moebius(d) * p(n/d, k/d));
    T(n,k) = sumdiv(gcd(n,k), d, AR(n/d,k/d));
    for(n=1, 10, for(k=1, n, print1(T(n,k), ", ")); print) \\ Andrew Howroyd, Oct 08 2017

Formula

R(n,k) = Sum_{d|gcd(n,k)} A180279(n/d, k/d). - Andrew Howroyd, Oct 08 2017
From Petros Hadjicostas, Oct 21 2017: (Start)
For proofs of these formulae, see the links.
R(n,k) = Sum_{d|gcd(n,k)} phi^{(-1)}(d)*(k/d)*A119963(n/d, k/d), where phi^{(-1)}(d) = A023900(d) is the Dirichlet inverse function of Euler's totient function.
G.f.: Sum_{s >= 1} phi^{(-1)}(s)*g(x^s, y^s), where phi^{(-1)}(s) = A023900(s) and g(x,y) = (x*y+x+1)*(x*y-x+1)*(x+1)*x*y/(x^2*y^2+x^2-1)^2.
(End)

Extensions

a(56) onwards from Robert G. Wilson v, Aug 25 2010

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.

Original entry on oeis.org

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

Views

Author

John P. McSorley, Oct 07 2010

Keywords

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.

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.

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

  • 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

A292200 Number of Sommerville symmetrical cyclic compositions (on symmetric necklaces) of n that are Carlitz (adjacent parts on the circle are distinct).

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 5, 7, 10, 11, 16, 23, 27, 37, 51, 65, 86, 117, 148, 204, 267, 351, 461, 626, 803, 1088, 1419, 1899, 2473, 3341, 4319, 5840, 7583, 10202, 13263, 17889, 23191, 31295, 40627, 54752, 71094, 95878, 124388, 167790, 217781, 293617, 381153, 513989, 667029, 899589
Offset: 1

Views

Author

Petros Hadjicostas, Sep 11 2017

Keywords

Comments

We consider cyclic compositions (necklaces) as equivalence classes of compositions that can be obtained from each other by a cyclic shift. A cyclic composition is called Sommerville symmetrical (on a symmetric necklace) if its equivalence class contains at least one palindromic composition (type I) or a composition that becomes a palindromic composition if we remove the first part (type II). A composition with only one part is a palindromic composition of both types.
The equivalence class of each Sommerville symmetrical cyclic composition that is Carlitz contains exactly two type II palindromic Carlitz compositions (except in the case of a composition with only one part). For example, when n = 8, the equivalence class {(1,2,3,2), (2,3,2,1), (3,2,1,2), (2,1,2,3)} represents a Sommerville symmetrical cyclic composition of n = 8 that is Carlitz, but only two of the compositions in the set, i.e., (1,2,3,2) and (3,2,1,2), are type II palindromic.

Examples

			For n = 7, there are exactly a(7) = 5 Sommerville symmetrical cyclic compositions (symmetric necklaces) of 7 that are Carlitz: 7, 1+6, 2+5, 3+4, 2+1+3+1. (Note that 1+6 is the same as 6+1, 3+1+2+1 is the same as 2+1+3+1, and so on, because in each case one composition can be obtained from the other by a cyclic shift.)
		

Crossrefs

Formula

a(n) = (A291941(n) + 1)/2.
G.f.: x/(1 - x) - A(x)/2 + B(x)^2/(2*(1 - A(x))), where A(x) = Sum_{n >= 1} x^(2*n)/(1 + x^(2*n)) and B(x) = Sum_{n >= 1} x^n/(1 + x^(2*n)).

Extensions

More terms from Altug Alkan, Sep 18 2017

A180424 "ARE(n,k)" triangle read by rows. ARE(n,k) is the number of aperiodic k-reverses of n up to cyclic equivalence.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 3, 3, 3, 3, 1, 0, 1, 3, 3, 4, 3, 3, 1, 0, 1, 4, 3, 6, 6, 3, 4, 1, 0, 1, 4, 4, 8, 5, 8, 4, 4, 1, 0, 1, 5, 5, 10, 10, 10, 10, 5, 5, 1, 0, 1, 5, 4, 12, 10, 17, 10, 12, 4, 5, 1, 0
Offset: 1

Views

Author

John P. McSorley, Sep 03 2010

Keywords

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, or if it is not the concatenation of at least two smaller compositions.
Two k-compositions are cyclically equivalent if one can be obtained from the other by a cyclic permutation of its parts.
The reverse of a k-composition is the k-composition obtained by writing its parts in reverse. For example, the reverse of 123 is 321.
A k-reverse of n is a k-composition of n which is cyclically equivalent to its reverse. And an aperiodic k-reverse of n is a k-reverse of n which is aperiodic.
For example, 114 is an aperiodic 3-reverse of 6 since it is aperiodic and its set of cyclic equivalents {114,411,141} contains its reverse 411.
But 123 is not an aperiodic 3-reverse of 6 since, even though it is aperiodic, its set of cyclic equivalents {123,312,231} does not contain its reverse 321.
Let AR(n,k) denote the number of aperiodic k-reverses of n, then sequence A180279 is the 'AR(n,k)' triangle read by rows.
For the above sequence we count the aperiodic k-reverses of n up to cyclic equivalence, ARE(n,k), in other words, the number of equivalence classes under cyclic permutation which contain at least one aperiodic k-reverse of n.

Examples

			The triangle begins
  1
  1 0
  1 1 0
  1 1 1 0
  1 2 2 1 0
  1 2 1 2 1 0
  1 3 3 3 3 1 0
  1 3 3 4 3 3 1 0
  1 4 3 6 6 3 4 1 0
  1 4 4 8 5 8 4 4 1 0
For example, row 8 is 1 3 3 4 3 3 1 0.
We have ARE(8,3)=3 because there are 9 aperiodic 3-reverses of 8 in the 3 classes {116,611,161}, {224,422,242}, and {233,323,332}, and so there are ARE(8,3)=3 aperiodic 3-reverses of 8 up to cyclic equivalence.
We have ARE(8,6)=3 because there are 3 aperiodic 6-reverses of 8 up to cyclic equivalence. The representatives of the 3 classes are 111113, 111122, and 111212.
		

References

  • John P. McSorley: Counting k-compositions of n with palindromic and related structures. Preprint, 2010.

Crossrefs

As mentioned above, if we don't count the classes, but rather the elements in the classes, we get sequence A180279.
If we remove the aperiodic requirement, see sequence A119963.
The row sums of the "ARE(n, k)" triangle above give sequence A056493 (except for the first term). See also sequence A056498.

Programs

  • Mathematica
    Table[DivisorSum[GCD[n, k], MoebiusMu[#]*Binomial[Floor[((n/#) - Boole[OddQ[k/#]])/2], Floor[k/(2 #)]] &], {n, 12}, {k, n}] // Flatten (* Michael De Vlieger, Oct 31 2021 *)
  • PARI
    \\ here RE(n,k) is A119963(n,k).
    RE(n,k) = binomial((n-k%2)\2, k\2);
    T(n,k) = sumdiv(gcd(n,k), d, moebius(d)*RE(n/d,k/d));
    for(n=1, 10, for(k=1, n, print1(T(n,k), ", ")); print) \\ Andrew Howroyd, Oct 07 2017

Formula

ARE(n,k) = Sum_{d|gcd(n,k)} mu(d) * A119963(n/d,k/d). - Andrew Howroyd, Oct 07 2017

Extensions

Terms a(56) and beyond from Andrew Howroyd, Oct 07 2017

A308401 Number of bracelets (turnover necklaces) of length n that have no reflection symmetry and consist of 6 white beads and n-6 black beads.

Original entry on oeis.org

3, 6, 16, 30, 56, 91, 150, 224, 336, 477, 672, 912, 1233, 1617, 2112, 2700, 3432, 4290, 5340, 6552, 8008, 9678, 11648, 13888, 16503, 19448, 22848, 26658, 31008, 35853, 41346, 47424, 54264, 61803, 70224, 79464, 89733, 100947, 113344, 126840, 141680, 157780, 175416, 194480, 215280, 237708
Offset: 9

Views

Author

Petros Hadjicostas, May 24 2019

Keywords

Comments

Bracelets that have no reflection symmetry are also known as chiral bracelets.
Here, for n >= 6, a(n) is also the number of dihedral compositions of n with 6 parts that have no reflection symmetry. Taking the MacMahon conjugates of these dihedral compositions, we see that a(n) is also the number of dihedral compositions of n into n-6 parts that have no reflection symmetry.
A cyclic composition b_1 + b_2 + ... + b_k of n into k parts is an equivalent class of (linear) compositions of n into k parts (placed on a circle) such that two such (linear) compositions are equivalent iff one can be obtained from the other by a rotation. Such compositions were first studied extensively by Sommerville (1909).
A dihedral composition b_1 + b_2 + ... + b_k of n into k parts is an equivalent class of (linear) compositions of n into k parts (placed on a circle) such that two such (linear) compositions are equivalent iff one can be obtained from the other by a rotation or a reversal of order. Such compositions were studied, for example, by Knopfmacher and Robbins (2013).
Given a bracelet of length n with k white beads and n-k black beads, we may get the corresponding dihedral composition using MacMahon's correspondence: start with a white bead and count that bead and the black beads that follow (in one direction), and call that b_1; then start with the next white bead and count that one and the black beads that follow, and call that b_2; repeat this process until you reach the k-th white bead and count that one and the black beads that follow, and call that b_k. The corresponding dihedral composition is b_1 + b_2 + ... + b_k.
If in the previous paragraph (given a bracelet of length n with k white beads and n-k black beads), we replace the white beads with black beads and the black beads with white beads, we get a dihedral composition of n into n-k parts: c_1 + c_2 + ... + c_{n-k}. These two dihedral compositions (which correspond to the same bracelet) are called "conjugate" compositions. See p. 273 in Sommerville (1909) for an explanation of "conjugate" compositions in the context of cyclic compositions.
Symmetric cyclic compositions of a positive integer n were first studied by Sommerville (1909, pp. 301-304). It can be proved that the study of necklaces with reflection symmetry using beads of two colors is equivalent to the study of symmetric cyclic compositions of a positive integer. Clearly all the necklaces with reflection symmetry are all the bracelets (turnover necklaces) with reflection symmetry. See also the comments for sequences A119963, A292200, and A295925.

Examples

			Using Frank Ruskey's website (listed above) to generate bracelets of fixed content (6, 3) with string length n = 9 and alphabet size 2, we get the following A005513(n = 9) = 7 bracelets: (1) WWWWWWBBB, (2) WWWWWBWBB, (3) WWWWBWWBB, (4) WWWWBWBWB, (5) WWWBWWWBB, (6) WWWBWWBWB, and (7) WWBWWBWWB. From these, bracelets 1, 4, 5, and 7 have reflection symmetry, while bracelets 2, 3 and 6 have no reflection symmetry (and thus, a(9) = 3).
Starting with a black bead, we count that bead and how many white beads follow (in one direction), and continue this process until we count all beads around the circle. We thus use MacMahon's correspondence to get the following dihedral compositions of n = 9 into 3 parts: (1) 1 + 7 + 1, (2) 1 + 2 + 6, (3) 1 + 3 + 5, (4) 2 + 5 + 2, (5) 4 + 1 + 4, (6) 2 + 3 + 4, and (7) 3 + 3 + 3. Again, dihedral compositions 1, 4, 5, and 7 are symmetric (have reflection symmetry), while dihedral compositions 2, 3, and 6 are not symmetric (and thus, a(9) = 3).
We may also start with a white bead and count that bead and how many black beads follow (in one direction), and continue this process until we count all beads around the circle. We thus use MacMahon's correspondence again to get the following (conjugate) dihedral compositions of n = 9 into 6 parts: (1) 1 + 1 + 1 + 1 + 1 + 4, (2) 1 + 1 + 1 + 1 + 2 + 3, (3) 1 + 1 + 1 + 2 + 1 + 3, (4) 1 + 1 + 1 + 2 + 2 + 2, (5) 1 + 1 + 2 + 1 + 1 + 3, (6) 1 + 1 + 2 + 1 + 2 + 2, and (7) 1 + 2 + 1 + 2 + 1 + 2. Again, dihedral compositions 1, 4, 5, and 7 have reflection symmetries, while dihedral compositions 2, 3, and 6 do not have reflection symmetries (and thus, a(9) = 3). For example, dihedral composition 1 is symmetric because we can draw an axis of symmetry through one of the 1s and 4. In addition, dihedral composition 5 is symmetric because we may draw an axis of symmetry through the numbers 2 and 3.
		

Crossrefs

Programs

  • PARI
    a(n) = (1/12)* (sumdiv(gcd(n, 6), d,  eulerphi(d)*binomial((n/d) - 1, (6/d) - 1))) - (1/2)*binomial(floor(n/2), 3); \\ Michel Marcus, May 28 2019
    
  • PARI
    Vec(x^9*(3 + x^2 + x^3 + x^4) / ((1 - x)^6*(1 + x)^3*(1 - x + x^2)*(1 + x + x^2)^2) + O(x^50)) \\ Colin Barker, Jun 02 2019

Formula

G.f.: (x^k/2) * (-(1 + x)/(1 - x^2)^floor((k/2) + 1) + (1/k) * Sum_{m|k} phi(m)/(1 - x^m)^(k/m)) with k = 6. (This formula is due to Herbert Kociemba.)
a(n) = A005513(n) - A058187(n-6) = A005513(n) - binomial(floor(n/2), 3) for n >= 6.
a(n) = -(1/2)*binomial(floor(n/2), 3) + (1/12)* Sum_{d|gcd(n, 6)} phi(d)*binomial((n/d) - 1, (6/d) - 1) for n >= 6. (This is a modification of formulas found in Gupta (1979) and Shevelev (2004).)
From Colin Barker, May 26 2019: (Start)
G.f.: x^9*(3 + x^2 + x^3 + x^4) / ((1 - x)^6*(1 + x)^3*(1 - x + x^2)*(1 + x + x^2)^2).
a(n) = 2*a(n-1) + a(n-2) - 3*a(n-3) - a(n-4) + a(n-5) + 4*a(n-6) - 3*a(n-7) - 3*a(n-8) + 4*a(n-9) + a(n-10) - a(n-11) - 3*a(n-12) + a(n-13) + 2*a(n-14) - a(n-15) for n > 23. (End)
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