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.

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A075195 Jablonski table T(n,k) read by antidiagonals: T(n,k) = number of necklaces with n beads of k colors.

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 5, 10, 11, 6, 1, 6, 15, 24, 24, 8, 1, 7, 21, 45, 70, 51, 14, 1, 8, 28, 76, 165, 208, 130, 20, 1, 9, 36, 119, 336, 629, 700, 315, 36, 1, 10, 45, 176, 616, 1560, 2635, 2344, 834, 60, 1, 11, 55, 249, 1044, 3367, 7826, 11165, 8230, 2195, 108, 1
Offset: 1

Views

Author

Christian G. Bower, Sep 07 2002

Keywords

Comments

From Richard L. Ollerton, May 07 2021: (Start)
Here, as in A000031, turning over is not allowed.
(1/n) * Dirichlet convolution of phi(n) and k^n. (End)

Examples

			The array T(n,k) for n >= 1, k >= 1 begins:
  1,  2,   3,    4,     5,     6,      7, ...
  1,  3,   6,   10,    15,    21,     28, ...
  1,  4,  11,   24,    45,    76,    119, ...
  1,  6,  24,   70,   165,   336,    616, ...
  1,  8,  51,  208,   629,  1560,   3367, ...
  1, 14, 130,  700,  2635,  7826,  19684, ...
  1, 20, 315, 2344, 11165, 39996, 117655, ...
From _Indranil Ghosh_, Mar 25 2017: (Start)
Triangle formed when the array is read by antidiagonals:
   1;
   2,  1;
   3,  3,   1;
   4,  6,   4,   1;
   5, 10,  11,   6,    1;
   6, 15,  24,  24,    8,    1;
   7, 21,  45,  70,   51,   14,    1;
   8, 28,  76, 165,  208,  130,   20,   1;
   9, 36, 119, 336,  629,  700,  315,  36,  1;
  10, 45, 176, 616, 1560, 2635, 2344, 834, 60, 1;
  ... (End)
		

References

  • F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-Like Structures, Cambridge, 1998, p. 86 (2.2.23).
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 496.
  • Louis Comtet, Analyse combinatoire, Tome 2, p. 104 #17, P.U.F., 1970.

Crossrefs

Main Diagonal: A056665. A054630 and A054631 are the upper and lower triangles.

Programs

  • Mathematica
    t[n_, k_] := (1/n)*Sum[EulerPhi[d]*k^(n/d), {d, Divisors[n]}]; Table[t[n-k+1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jan 20 2014, after Philippe Deléham *)
  • PARI
    T(n, k) = (1/n) * sumdiv(n, d, eulerphi(d)*k^(n/d));
    for(n=1, 15, for(k=1, n, print1(T(k, n - k + 1),", ");); print();) \\ Indranil Ghosh, Mar 25 2017
    
  • Python
    from sympy.ntheory import totient, divisors
    def T(n,k): return sum(totient(d)*k**(n//d) for d in divisors(n))//n
    for n in range(1, 16):
        print([T(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 25 2017

Formula

T(n,k) = (1/n)*Sum_{d | n} phi(d)*k^(n/d), where phi = Euler totient function A000010. - Philippe Deléham, Oct 08 2003
From Petros Hadjicostas, Feb 08 2021: (Start)
O.g.f. for column k >= 1: Sum_{n>=1} T(n,k)*x^n = -Sum_{j >= 1} (phi(j)/j) * log(1 - k*x^j).
Linear recurrence for row n >= 1: T(n,k) = Sum_{j=0..n} -binomial(j-n-1,j+1) * T(n,k-1-j) for k >= n + 2. (This recurrence is essentially due to Robert A. Russell, who contributed it in A321791.) (End)
From Richard L. Ollerton, May 07 2021: (Start)
T(n,k) = (1/n)*Sum_{i=1..n} k^gcd(n,i).
T(n,k) = (1/n)*Sum_{i=1..n} k^(n/gcd(n,i))*phi(gcd(n,i))/phi(n/gcd(n,i)).
T(n,k) = (1/n)*A185651(n,k) for n >= 1, k >= 1. (End)
Product_{n>=1} 1/(1 - x^n)^T(n,k) = Product_{n>=1} 1/(1 - k*x^n). - Seiichi Manyama, Apr 12 2025

Extensions

Additional references from Philippe Deléham, Oct 08 2003

A056293 Number of n-bead necklace structures using a maximum of five different colored beads.

Original entry on oeis.org

1, 2, 3, 7, 12, 42, 123, 503, 2008, 8720, 38365, 173609, 792828, 3662924, 17034381, 79703081, 374624254, 1767883444, 8370666417, 39751072847, 189262621864, 903220058756, 4319518316899, 20697040198889, 99343899144822, 477609477924308, 2299585449279713
Offset: 1

Views

Author

Keywords

Comments

Turning over the necklace is not allowed. Colors may be permuted without changing the necklace structure.

References

  • 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]

Crossrefs

Programs

  • Mathematica
    Adn[d_, n_] := Module[{ c, t1, t2}, t2 = 0; For[c = 1, c <= d, c++, If[Mod[d, c] == 0 , t2 = t2 + (x^c/c)*(E^(c*z) - 1)]]; t1 = E^t2; t1 = Series[t1, {z, 0, n+1}]; Coefficient[t1, z, n]*n!]; Pn[n_] := Module[{ d, e, t1}, t1 = 0; For[d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*Adn[d, n/d]/n]]; t1/(1 - x)]; Pnq[n_, q_] := Module[{t1}, t1 = Series[Pn[n], {x, 0, q+1}] ; Coefficient[t1, x, q]]; a[n_] := Pnq[n, 5]; Table[Print[an = a[n]]; an, {n, 1, 24}] (* Jean-François Alcover, Oct 04 2013, after N. J. A. Sloane's Maple code *)
    (* this Mathematica program uses Gilbert and Riordan's recurrence formula, which they recommend for calculations: *)
    Adn[d_, n_] := Adn[d, n] = If[1==n, DivisorSum[d, x^# &],
      Expand[Adn[d, 1] Adn[d, n-1] + D[Adn[d, n-1], x] x]];
    Table[SeriesCoefficient[DivisorSum[n, EulerPhi[#] Adn[#, n/#] &]
    /(n (1 - x)), {x, 0, 5}], {n, 1, 40}] (* Robert A. Russell, Feb 24 2018 *)
    From Robert A. Russell, May 29 2018: (Start)
    Table[(1/n) DivisorSum[n, EulerPhi[#] Which[Divisible[#, 60], 5 StirlingS2[n/#+4, 5] - 50 StirlingS2[n/#+3, 5] + 175 StirlingS2[n/#+2, 5] - 250 StirlingS2[n/#+1, 5] + 120 StirlingS2[n/#, 5], Divisible[#, 30], 4 StirlingS2[n/#+4, 5] - 41 StirlingS2[n/#+3, 5] + 149 StirlingS2[n/#+2, 5] - 226 StirlingS2[n/#+1, 5] + 120 StirlingS2[n/#, 5], Divisible[#, 20], 4 StirlingS2[n/#+4, 5] - 42 StirlingS2[n/#+3, 5] + 156 StirlingS2[n/#+2, 5] - 238 StirlingS2[n/#+1, 5] + 120 StirlingS2[n/#, 5], Divisible[#, 15], 3 StirlingS2[n/#+4, 5] - 33 StirlingS2[n/#+3, 5] + 129 StirlingS2[n/#+2, 5] - 210 StirlingS2[n/#+1, 5] + 120 StirlingS2[n/#, 5], Divisible[#, 12], 4 StirlingS2[n/#+4, 5] - 40 StirlingS2[n/#+3, 5] + 140 StirlingS2[n/#+2, 5] - 200 StirlingS2[n/#+1, 5] + 96 StirlingS2[n/#, 5], Divisible[#, 10], 3 StirlingS2[n/#+4, 5] - 33 StirlingS2[n/#+3, 5] + 130 StirlingS2[n/#+2, 5] - 214 StirlingS2[n/#+1, 5] + 120 StirlingS2[n/#, 5], Divisible[#, 6], 3 StirlingS2[n/#+4, 5] - 31 StirlingS2[n/#+3, 5] + 114 StirlingS2[n/#+2, 5] - 176 StirlingS2[n/#+1, 5] + 96 StirlingS2[n/#, 5], Divisible[#, 5], 2 StirlingS2[n/#+4, 5] - 23 StirlingS2[n/#+3, 5] + 95 StirlingS2[n/#+2, 5] - 165 StirlingS2[n/#+1, 5] + 100 StirlingS2[n/#, 5], Divisible[#, 4], 3 StirlingS2[n/#+4, 5] - 32 StirlingS2[n/#+3, 5] + 121 StirlingS2[n/#+2, 5] - 188 StirlingS2[n/#+1, 5] + 96 StirlingS2[n/#, 5], Divisible[#, 3], 2 StirlingS2[n/#+4, 5] - 23 StirlingS2[n/#+3, 5] + 94 StirlingS2[n/#+2, 5] - 160 StirlingS2[n/#+1, 5] + 96 StirlingS2[n/#, 5], Divisible[#, 2], 2 StirlingS2[n/#+4, 5] - 23 StirlingS2[n/#+3, 5] + 95 StirlingS2[n/#+2, 5] - 164 StirlingS2[n/#+1, 5] + 96 StirlingS2[n/#, 5], True, StirlingS2[n/#+4, 5] - 13 StirlingS2[n/#+3, 5] + 60 StirlingS2[n/#+2, 5] - 115 StirlingS2[n/#+1, 5] + 76 StirlingS2[n/#, 5]] &], {n, 1, 40}]
    mx = 40; Drop[CoefficientList[Series[1-Sum[(EulerPhi[d] / d) Which[ Divisible[d, 60], Log[1-5x^d], Divisible[d, 30], (3 Log[1-5x^d] + Log[1-x^d]) / 4, Divisible[d, 20], (2 Log[1-5x^d] + Log[1-2x^d]) / 3, Divisible[d, 15], (3 Log[1-5x^d] + 2 Log[1-3x^d] + 3 Log[1-x^d]) / 8, Divisible[d, 12], 4 Log[1-5x^d] / 5, Divisible[d, 10], (5 Log[1-5x^d] + 4 Log[1-2x^d] + 3 Log[1-x^d]) / 12, Divisible[d, 6], (11 Log[1-5x^d] + 5 Log[1-x^d]) / 20, Divisible[d, 5], (5 Log[1-5x^d] + 2 Log[1-3x^d] + 4 Log[1-2x^d] + 9 Log[1-x^d]) / 24, Divisible[d, 4], (7 Log[1-5x^d] + 5 Log[1-2x^d]) / 15, Divisible[d, 3], (7 Log[1-5x^d] + 10 Log[1-3x^d] + 15 Log[1-x^d]) / 40, Divisible[d, 2], (13 Log[1-5x^d] + 20 Log[1-2x^d] + 15 Log[1-x^d]) / 60, True, (Log[1-5x^d] + 10 Log[1-3x^d] + 20 Log[1-2x^d] + 45 Log[1-x^d]) / 120], {d, 1, mx}], {x, 0, mx}], x], 1]
    (End)

Formula

Use de Bruijn's generalization of Polya's enumeration theorem as discussed in reference.
From Robert A. Russell, May 29 2018: (Start)
a(n) = (1/n) * Sum_{d|n} phi(d) * ([d==0 mod 60] * (5*S2(n/d + 4, 5) - 50*S2(n/d + 3, 5) + 175*S2(n/d + 2, 5) - 250*S2(n/d + 1, 5) + 120*S2(n/d, 5)) + [d==30 mod 60] * (4*S2(n/d+4,5) - 41*S2(n/d+3,5) + 149*S2(n/d+2,5) - 226*S2(n/d + 1, 5) + 120*S2(n/d, 5)) + [d==20 mod 60 | d==40 mod 60] * (4*S2(n/d + 4, 5) - 42*S2(n/d + 3, 5) + 156*S2(n/d + 2, 5) - 238*S2(n/d + 1, 5) + 120*S2(n/d, 5)) + [d==15 mod 60 | d==45 mod 60] * (3*S2(n/d + 4, 5) - 33*S2(n/d + 3, 5) + 129*S2(n/d + 2, 5) - 210*S2(n/d + 1, 5) + 120*S2(n/d, 5)) + [d mod 60 in {12,24,36,48}] * (4*S2(n/d + 4, 5) - 40*S2(n/d + 3, 5) + 140*S2(n/d + 2, 5) - 200*S2(n/d+1, 5) + 96*S2(n/d, 5)) + [d=10 mod 60 | d==50 mod 60] * (3*S2(n/d + 4, 5) - 33*S2(n/d + 3, 5) + 130*S2(n/d + 2, 5) - 214*S2(n/d + 1, 5) + 120*S2(n/d, 5)) + [d mod 60 in {6,18,42,54}] * (3*S2(n/d + 4, 5) - 31*S2(n/d + 3, 5) + 114*S2(n/d + 2, 5) - 176*S2(n/d + 1, 5) + 96*S2(n/d, 5)) + [d mod 60 in {5,25,35,55}] * (2*S2(n/d + 4, 5) - 23*S2(n/d + 3, 5) + 95*S2(n/d + 2, 5) - 165*S2(n/d + 1, 5) + 100*S2(n/d, 5)) + [d mod 60 in {4,8,16,28,32,44,52,56}] * (3*S2(n/d + 4, 5) - 32*S2(n/d + 3, 5) + 121*S2(n/d + 2, 5) - 188*S2(n/d + 1, 5) + 96*S2(n/d, 5)) + [d mod 60 in {3,9,21,27,33,39,51,57}] * (2*S2(n/d + 4, 5) - 23*S2(n/d + 3, 5) + 94*S2(n/d + 2, 5) - 160*S2(n/d + 1, 5) + 96*S2(n/d, 5)) + [d mod 60 in {2,14,22,26,34,38,46,58}] * (2*S2(n/d + 4, 5) - 23*S2(n/d + 3, 5) + 95*S2(n/d + 2, 5) - 164*S2(n/d + 1, 5) + 96*S2(n/d, 5)) + [d mod 60 in {1,7,11,13,17,19,23,29,31,37,41,43,47,49,53,59}] * (S2[n/d + 4, 5) - 13*S2(n/d + 3, 5) + 60*S2(n/d + 2, 5) - 115*S2(n/d + 1, 5) + 76*S2(n/d, 5))), where S2(n,k) is the Stirling subset number, A008277.
G.f.: 1 - Sum_{d>0} (phi(d) / d) * ([d==0 mod 60] * log(1-5x^d) + [d==30 mod 60] * (3*log(1-5x^d) + log(1-x^d)) / 4 + [d==20 mod 60 | d==40 mod 60] * (2*log(1-5x^d) + log(1-2x^d)) / 3 + [d==15 mod 60 | d==45 mod 60] * (3*log(1-5x^d) + 2*log(1-3x^d) + 3*log(1-x^d)) / 8 + [d mod 60 in {12,24,36,48}] * 4*log(1-5x^d) / 5 + [d=10 mod 60 | d==50 mod 60] * (5*log(1-5x^d) + 4*log(1-2x^d) + 3*log(1-x^d)) / 12 + [d mod 60 in {6,18,42,54}] * (11*log(1-5x^d) + 5*log(1-x^d)) / 20 + [d mod 60 in {5,25,35,55}] * (5*log(1-5x^d) + 2*log(1-3x^d) + 4*log(1-2x^d) + 9*log(1-x^d)) / 24 + [d mod 60 in {4,8,16,28,32,44,52,56}] * (7*log(1-5x^d) + 5*log(1-2x^d)) / 15 + [d mod 60 in {3,9,21,27,33,39,51,57}] * (7*log(1-5x^d) + 10*log(1-3x^d) + 15*log(1-x^d)) / 40 + [d mod 60 in {2,14,22,26,34,38,46,58}] * (13*log(1-5x^d) + 20*log(1-2x^d) + 15*log(1-x^d)) / 60 +[d mod 60 in{1,7,11,13,17,19,23,29,31,37,41,43,47,49,53,59}] * (log(1-5x^d) + 10*log(1-3x^d) + 20*log(1-2x^d) + 45*log(1-x^d)) / 120).
(End)

A032276 Number of bracelets (turnover necklaces) with n beads of 5 colors.

Original entry on oeis.org

5, 15, 35, 120, 377, 1505, 5895, 25395, 110085, 493131, 2227275, 10196680, 46989185, 218102685, 1017448143, 4768969770, 22440372245, 105966797755, 501938733555, 2384200683816, 11353290089305
Offset: 1

Views

Author

Keywords

Comments

From Petros Hadjicostas, Sep 01 2018: (Start)
The DIK transform of the sequence (c(n): n >= 1), with g.f. C(x) = Sum_{n >= 1} c(n)*x^n, has g.f. -(1/2)*Sum_{m >= 1} (phi(m)/m))*log(1-C(x^m)) + (1 + C(x))^2/(4*(1-C(x^2))) - 1/4.
Here, c(1) = 5 and c(n) = 0 for n >= 2, and thus, C(x) = 5*x. Substituting this to the above g.f., we get that the g.f. of the current sequence is A(x) = Sum_{n >= 1} a(n)*x^n = -(1/2)*Sum_{m >= 1} (phi(m)/m))*log(1-5*x^m) + (1 + 5*x)^2/(4*(1-5*x^2)) - 1/4. This agrees with Herbert Kociemba's g.f. below except for an extra 1 because (1 + (1+5*x+10*x^2)/(1-5*x^2))/2 = 1 + (1 + 5*x)^2/(4*(1-5*x^2)) - 1/4.
(End)

Examples

			For n=2, the 15 bracelets are AA, AB, AC, AD, AE, BB, BC, BD, BE, CC, CD, CE, DD, DE, and EE. - _Robert A. Russell_, Sep 24 2018
		

Crossrefs

Cf. A081720.
Column 5 of A051137.
Cf. A001869 (oriented), A056487 (achiral), A278641 (chiral).

Programs

  • Mathematica
    mx=40;CoefficientList[Series[(1-Sum[ EulerPhi[n]*Log[1-5*x^n]/n,{n,mx}]+(1+5 x+10 x^2)/(1-5 x^2))/2,{x,0,mx}],x] (* Herbert Kociemba, Nov 02 2016 *)
    k=5; Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) + (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}] (* Robert A. Russell, Sep 24 2018 *)

Formula

"DIK" (bracelet, indistinct, unlabeled) transform of 5, 0, 0, 0, ...
a(n) = A081720(n,5), n >= 1. - Wolfdieter Lang, Jun 03 2012
G.f.: (1 - Sum_{n>=1} phi(n)*log(1 - 5*x^n)/n + (1+5*x+10*x^2)/(1-5*x^2))/2. - Herbert Kociemba, Nov 02 2016
a(n) = (3/2)*5^(n/2) + (1/(2*n))*Sum_{d|n} phi(n/d)*5^d, if n is even, and = (1/2)*5^((n+1)/2) + (1/(2*n))*Sum_{d|n} phi(n/d)*5^d, if n is odd. - Petros Hadjicostas, Sep 01 2018
a(n) = (A001869(n) + A056487(n+1)) / 2 = A278641(n) + A056487(n+1) = A001869(n) - A278641(n). - Robert A. Russell, Oct 13 2018
a(n) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{d divides n} phi(d)*k^(n/d), where k=5 is the maximum number of colors. - Richard L. Ollerton, May 04 2021
a(n) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{i=1..n} k^gcd(n,i), where k=5 is the maximum number of colors. (See A051137.) - Richard L. Ollerton, May 04 2021

A056285 Number of n-bead necklaces with exactly five different colored beads.

Original entry on oeis.org

0, 0, 0, 0, 24, 300, 2400, 15750, 92680, 510312, 2691600, 13794150, 69309240, 343501500, 1686135376, 8221437000, 39901776360, 193054016840, 932142850800, 4495236798162, 21664357535320, 104388120866100, 503044634004000, 2425003924383900, 11696087875731624
Offset: 1

Views

Author

Keywords

Comments

Turning over the necklace is not allowed.

Examples

			For n=5, the 24 necklaces are A followed by the 24 permutations of BCDE.
		

References

  • 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]

Crossrefs

Column k=5 of A087854.

Programs

  • Mathematica
    k=5; Table[k!DivisorSum[n,EulerPhi[#]StirlingS2[n/#,k]&]/n,{n,1,30}] (* Robert A. Russell, Sep 26 2018 *)
  • PARI
    a(n) = my(k=5); k!*sumdiv(n, d, eulerphi(d)*stirling(n/d, k, 2))/n; \\ Michel Marcus, Sep 27 2018

Formula

a(n) = A001869(n) - 5*A001868(n) + 10*A001867(n) - 10*A000031(n) + 5.
From Robert A. Russell, Sep 26 2018: (Start)
a(n) = (k!/n) Sum_{d|n} phi(d) S2(n/d,k), where k=5 is the number of colors and S2 is the Stirling subset number A008277.
G.f.: -Sum_{d>0} (phi(d)/d) * Sum_{j} (-1)^(k-j) * C(k,j) * log(1-j x^d), where k=5 is the number of colors. (End)

A056286 Number of n-bead necklaces with exactly six different colored beads.

Original entry on oeis.org

0, 0, 0, 0, 0, 120, 2160, 23940, 211680, 1643544, 11748240, 79419180, 516257280, 3262443120, 20193277104, 123071707080, 741419995680, 4427490147480, 26264144909520, 155018841055596, 911509010154720, 5344538384445120, 31272099902089200, 182707081122261480
Offset: 1

Views

Author

Keywords

Comments

Turning over the necklace is not allowed.

Examples

			For n=6, the 120 necklaces are A followed by the 120 permutations of BCDEF.
		

References

  • 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]

Crossrefs

Column k=6 of A087854.

Programs

  • Mathematica
    k=6; Table[k!DivisorSum[n,EulerPhi[#]StirlingS2[n/#,k]&]/n,{n,1,30}] (* Robert A. Russell, Sep 26 2018 *)

Formula

a(n) = A054625(n) - 6*A001869(n) + 15*A001868(n) - 20*A001867(n) + 15*A000031(n) - 6.
From Robert A. Russell, Sep 26 2018: (Start)
a(n) = (k!/n) Sum_{d|n} phi(d) S2(n/d,k), where k=6 is the number of colors and S2 is the Stirling subset number A008277.
G.f.: -Sum_{d>0} (phi(d)/d) * Sum_{j} (-1)^(k-j) * C(k,j) * log(1-j x^d), where k=6 is the number of colors. (End)

A054612 a(n) = Sum_{d|n} phi(d)*5^(n/d).

Original entry on oeis.org

0, 5, 30, 135, 660, 3145, 15810, 78155, 391320, 1953405, 9768870, 48828175, 244157820, 1220703185, 6103593930, 30517584915, 152588282640, 762939453205, 3814699250430, 19073486328215, 95367441415140, 476837158360185, 2384185839844050, 11920928955078235, 59604645020345640
Offset: 0

Views

Author

N. J. A. Sloane, Apr 16 2000

Keywords

Crossrefs

Column k=5 of A185651.
Cf. A001869.

Programs

  • Mathematica
    a[n_]:= Sum[5^GCD[n,k],{k,n}]; Array[a,30,0] (* Stefano Spezia, Sep 02 2025 *)
  • PARI
    a(n) = if (n, sumdiv(n, d, eulerphi(d)*5^(n/d)), 0); \\ Michel Marcus, Apr 16 2021

Formula

a(n) = Sum_{k=1..n} 5^gcd(n,k). - Ilya Gutkovskiy, Apr 16 2021

A184288 Table read by antidiagonals: T(n,k) = number of distinct n X k toroidal 0..4 arrays.

Original entry on oeis.org

5, 15, 15, 45, 175, 45, 165, 2635, 2635, 165, 629, 49075, 217125, 49075, 629, 2635, 976887, 20346485, 20346485, 976887, 2635, 11165, 20349075, 2034505661, 9536816875, 2034505661, 20349075, 11165, 48915, 435970995, 211927741375
Offset: 1

Views

Author

R. H. Hardin, Jan 10 2011

Keywords

Examples

			Table starts
      5        15           45           165           629         2635
     15       175         2635         49075        976887     20349075
     45      2635       217125      20346485    2034505661 211927741375
    165     49075     20346485    9536816875 4768372070757
    629    976887   2034505661 4768372070757
   2635  20349075 211927741375
  11165 435970995
  48915
		

Crossrefs

Columns 1-4 are A001869, A184286, A184287, A184288.

Programs

  • Mathematica
    T[n_, k_] := (1/(n*k))*Sum[Sum[EulerPhi[c] * EulerPhi[d] * 5^(n*k/LCM[c, d]), {d, Divisors[k]}], {c, Divisors[n]}];
    Table[T[n - k + 1, k], {n, 1, 9}, {k, 1, n}] // Flatten (* Jean-François Alcover, Oct 31 2017, after Andrew Howroyd *)
  • PARI
    T(n, k) = (1/(n*k)) * sumdiv(n, c, sumdiv(k, d, eulerphi(c) * eulerphi(d) * 5^(n*k/lcm(c,d)))); \\ Andrew Howroyd, Sep 27 2017

Formula

T(n,k) = (1/(n*k)) * Sum_{c|n} Sum_{d|k} phi(c) * phi(d) * 5^(n*k/lcm(c,d)). - Andrew Howroyd, Sep 27 2017

A106367 Number of necklaces with n beads of 5 colors, no 2 adjacent beads the same color.

Original entry on oeis.org

5, 10, 20, 70, 204, 700, 2340, 8230, 29140, 104968, 381300, 1398500, 5162220, 19175140, 71582940, 268439590, 1010580540, 3817763740, 14467258260, 54975633976, 209430787820, 799645010860, 3059510616420, 11728124734500
Offset: 1

Views

Author

Christian G. Bower, Apr 29 2005

Keywords

Crossrefs

Column 5 of A208535.

Programs

  • Mathematica
    a[n_] := If[n==1, 5, Sum[EulerPhi[n/d]*(4*(-1)^d+4^d), {d, Divisors[n]}]/n ];
    Array[a, 35] (* Jean-François Alcover, Jul 06 2018, after Andrew Howroyd *)
  • PARI
    a(n) = if(n==1, 5, sumdiv(n, d, eulerphi(n/d)*(4*(-1)^d + 4^d))/n); \\ Andrew Howroyd, Oct 14 2017

Formula

CycleBG transform of (5, 0, 0, 0, ...)
CycleBG transform T(A) = invMOEBIUS(invEULER(Carlitz(A)) + A(x^2) - A) + A.
Carlitz transform T(A(x)) has g.f. 1/(1-Sum_{k>0}(-1)^(k+1)*A(x^k)).
a(n) = (1/n) * Sum_{d | n} totient(n/d) * (4*(-1)^d + 4^d) for n > 1. - Andrew Howroyd, Mar 12 2017

A278641 Number of pairs of orientable necklaces with n beads and up to 5 colors; i.e., turning the necklace over does not leave it unchanged. The turned-over necklace is not included in the count.

Original entry on oeis.org

0, 0, 0, 10, 45, 252, 1130, 5270, 23520, 106960, 483756, 2211650, 10149805, 46911060, 217868310, 1017057518, 4767797895, 22438419120, 105960938380, 501928967930, 2384171386941, 11353241261180, 54185968572450, 259150507387910, 1241763071712930, 5960463867187752, 28656077411358180, 137973711706163210
Offset: 0

Views

Author

Herbert Kociemba, Nov 24 2016

Keywords

Comments

Number of chiral bracelets of n beads using up to five different colors.

Crossrefs

Column 5 of A293496.
Cf. A059076 (2 colors), A278639 (3 colors), A278640 (4 colors).
a(n) = (A001869(n) - A056487(n+1)) / 2 = A032276(n) - A056487(n+1).
Equals A001869 - A032276.

Programs

  • Mathematica
    mx=40;f[x_,k_]:=(1-Sum[EulerPhi[n]*Log[1-k*x^n]/n,{n,1,mx}]-Sum[Binomial[k,i]*x^i,{i,0,2}]/(1-k*x^2))/2;CoefficientList[Series[f[x,5],{x,0,mx}],x]
    k=5; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) - (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}], 0] (* Robert A. Russell, Sep 24 2018 *)

Formula

G.f.: k=5, (1 - Sum_{n>=1} phi(n)*log(1 - k*x^n)/n - Sum_{i=0..2} Binomial[k,i]*x^i / ( 1-k*x^2) )/2.
For n>0, a(n) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/2n)* Sum_{d|n} phi(d)*k^(n/d), where k=5 is the maximum number of colors. - Robert A. Russell, Sep 24 2018

A343467 a(n) = -(1/n) * Sum_{d|n} phi(n/d) * (-5)^d.

Original entry on oeis.org

5, -10, 45, -160, 629, -2590, 11165, -48910, 217045, -976258, 4438925, -20346440, 93900245, -435959830, 2034505661, -9536767660, 44878791365, -211927519090, 1003867701485, -4768372070128, 22706531350485, -108372079190350, 518301258916445, -2483526875847690, 11920928955078629
Offset: 1

Views

Author

Ilya Gutkovskiy, Apr 16 2021

Keywords

Crossrefs

Programs

  • Mathematica
    Table[-(1/n) Sum[EulerPhi[n/d] (-5)^d, {d, Divisors[n]}], {n, 1, 25}]
    nmax = 25; CoefficientList[Series[Sum[EulerPhi[k] Log[1 + 5 x^k]/k, {k, 1, nmax}], {x, 0, nmax}], x] // Rest

Formula

G.f.: Sum_{k>=1} phi(k) * log(1 + 5*x^k) / k.
a(n) = -(1/n) * Sum_{k=1..n} (-5)^gcd(n,k).
Product_{n>=1} 1 / (1 - x^n)^a(n) = g.f. for A261569.
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