A305540
Triangle read by rows: T(n,k) is the number of achiral loops (necklaces or bracelets) of length n using exactly k different colors.
Original entry on oeis.org
1, 1, 1, 1, 2, 1, 4, 3, 1, 6, 6, 1, 10, 21, 12, 1, 14, 36, 24, 1, 22, 93, 132, 60, 1, 30, 150, 240, 120, 1, 46, 345, 900, 960, 360, 1, 62, 540, 1560, 1800, 720, 1, 94, 1173, 4980, 9300, 7920, 2520, 1, 126, 1806, 8400, 16800, 15120, 5040, 1, 190, 3801, 24612, 71400, 103320, 73080, 20160, 1, 254, 5796, 40824, 126000, 191520, 141120, 40320
Offset: 1
The triangle begins with T(1,1):
1;
1, 1;
1, 2;
1, 4, 3;
1, 6, 6;
1, 10, 21, 12;
1, 14, 36, 24;
1, 22, 93, 132, 60;
1, 30, 150, 240, 120;
1, 46, 345, 900, 960, 360;
1, 62, 540, 1560, 1800, 720;
1, 94, 1173, 4980, 9300, 7920, 2520;
1, 126, 1806, 8400, 16800, 15120, 5040;
1, 190, 3801, 24612, 71400, 103320, 73080, 20160;
1, 254, 5796, 40824, 126000, 191520, 141120, 40320;
1, 382, 11973, 113652, 480060, 1048320, 1234800, 745920, 181440;
1, 510, 18150, 186480, 834120, 1905120, 2328480, 1451520, 362880;
For a(4,2)=4, the achiral loops are AAAB, AABB, ABAB, and ABBB.
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Table[(k!/2) (StirlingS2[Floor[(n + 1)/2], k] + StirlingS2[Ceiling[(n + 1)/2], k]), {n, 1, 15}, {k, 1, Ceiling[(n + 1)/2]}] // Flatten
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T(n, k) = (k!/2)*(stirling(floor((n+1)/2), k, 2)+stirling(ceil((n+1)/2), k, 2));
tabf(nn) = for(n=1, nn, for (k=1, ceil((n+1)/2), print1(T(n, k), ", ")); print); \\ Michel Marcus, Jul 02 2018
A056345
Number of bracelets of length n using exactly five different colored beads.
Original entry on oeis.org
0, 0, 0, 0, 12, 150, 1200, 7905, 46400, 255636, 1346700, 6901725, 34663020, 171786450, 843130688, 4110958530, 19951305240, 96528492700, 466073976900, 2247627076731, 10832193571460, 52194109216950
Offset: 1
For a(5)=12, pair up the 24 permutations of BCDE, each with its reverse, such as BCDE-EDCB. Precede the first of each pair with an A, such as ABCDE. These are the 12 arrangements, all chiral. If we precede the second of each pair with an A, such as AEDCB, we get the chiral partner of each. - _Robert A. Russell_, Sep 27 2018
- M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]
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t[n_, k_] := (For[t1 = 0; d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*k^(n/d)]]; If[EvenQ[n], (t1 + (n/2)*(1 + k)*k^(n/2))/(2*n), (t1 + n*k^((n + 1)/2))/(2*n)]);
T[n_, k_] := Sum[(-1)^i*Binomial[k, i]*t[n, k - i], {i, 0, k - 1}];
a[n_] := T[n, 5];
Array[a, 22] (* Jean-François Alcover, Nov 05 2017, after Andrew Howroyd *)
k=5; Table[k! DivisorSum[n, EulerPhi[#] StirlingS2[n/#,k]&]/(2n) + k!(StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k])/4, {n,1,30}] (* Robert A. Russell, Sep 27 2018 *)
A056501
Number of primitive (period n) periodic palindromes using exactly five different symbols.
Original entry on oeis.org
0, 0, 0, 0, 0, 0, 0, 60, 120, 960, 1800, 9300, 16800, 71400, 126000, 480000, 834120, 2968440, 5103000, 17354340, 29607600, 97566000, 165528000, 533264700, 901020120, 2854995360, 4809004080, 15050445900, 25292030400, 78417321240, 131542866000, 404936052000
Offset: 1
For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome.
- M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]
A305544
Number of chiral pairs of color loops of length n with exactly 5 different colors.
Original entry on oeis.org
0, 0, 0, 0, 12, 150, 1200, 7845, 46280, 254676, 1344900, 6892425, 34646220, 171715050, 843004688, 4110478470, 19950471120, 96525524140, 466068873900, 2247609721431, 10832163963860, 52194011649150, 251522234238000, 1212501695554920, 5848043487355752, 28223528190496380, 136307124614215660, 658800774340433025, 3186621527711606940
Offset: 1
For a(5)=12, the chiral pairs of color loops are ABCDE-AEDCB, ABCED-ADECB, ABDCE-AECDB, ABDEC-ACEDB, ABECD-ADCEB, ABEDC-ACDEB, ACBDE-AEDBC, ACBED-ADEBC, ACDBE-AEBCD, ACEDB-ABDEC, ADBCE-AECBD, ADBEC-ACEBD, and ADCBE-AEBCD.
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k=5; Table[(k!/(2n)) DivisorSum[n, EulerPhi[#] StirlingS2[n/#, k] &] - (k!/4) (StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k]), {n, 1, 40}]
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a(n) = my(k=5); -(k!/4)*(stirling(floor((n+1)/2),k,2) + stirling(ceil((n+1)/2),k,2)) + (k!/(2*n))*sumdiv(n, d, eulerphi(d)*stirling(n/d,k,2)); \\ Michel Marcus, Jun 06 2018
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