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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A001401 Number of partitions of n into at most 5 parts.

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 10, 13, 18, 23, 30, 37, 47, 57, 70, 84, 101, 119, 141, 164, 192, 221, 255, 291, 333, 377, 427, 480, 540, 603, 674, 748, 831, 918, 1014, 1115, 1226, 1342, 1469, 1602, 1747, 1898, 2062, 2233, 2418, 2611, 2818, 3034, 3266, 3507, 3765, 4033, 4319
Offset: 0

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Author

Keywords

Comments

a(n) = T_{r}(n) for r large, where T_{r}(n) = number of outcomes in which r indistinguishable dice yield a sum r+n-1.
a(n) = coefficient of q^n in the expansion of (m choose 5)_q as m goes to infinity. - Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
For n > 4: also number of partitions of n into parts <= 5: a(n) = A026820(n,5). - Reinhard Zumkeller, Jan 21 2010
Number of different distributions of n+15 identical balls in 5 boxes as x,y,z,p,q where 0 < x < y < z < p < q. - Ece Uslu and Esin Becenen, Jan 11 2016 [i.e., a(n) is the number of partitions of n+15 into 5 distinct parts. - R. J. Mathar, Feb 28 2021]
Tengely and Ulas prove that a(n) is a square only for n=1 and 2027. - Michel Marcus, Feb 11 2021

Examples

			(5 choose 5)_q = 1;
(6 choose 5)_q = q^5 + q^4 + q^3 + q^2 + q + 1;
(7 choose 5)_q = q^10 + q^9 + 2*q^8 + 2*q^7 + 3*q^6 + 3*q^5 + 3*q^4 + 2*q^3 + 2*q^2 + q + 1;
(8 choose 5)_q = q^15 + q^14 + 2*q^13 + 3*q^12 + 4*q^11 + 5*q^10 + 6*q^9 + 6*q^8 + 6*q^7 + 6*q^6 + 5*q^5 + 4*q^4 + 3*q^3 + 2*q^2 + q + 1;
so the coefficient of q^0 converges to 1, q^1 to 1, q^2 to 2 and so on.
a(3) = 3, i.e., {1,2,3,4,8}, {1,2,3,5,7}, {1,2,4,5,6}. Number of different distributions of 18 identical balls in 5 boxes as x,y,z,p,q where 0 < x < y < z < p < q. - _Ece Uslu_, Esin Becenen, Jan 11 2016
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 115, row m=5 of Q(m,n) table.
  • H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958, p. 2.
  • D. E. Knuth, The Art of Computer Programming, vol. 4, fascicle 3, Generating All Combinations and Partitions, Section 7.2.1.4., p. 56, exercise 31.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n) = A008284(n+5, 5), n >= 0.
Cf. A008619, A001400, A001399, A008667 (first differences), A008804.
First differences of A002622.

Programs

  • Maple
    with(combstruct):ZL6:=[S,{S=Set(Cycle(Z,card<6))}, unlabeled]:seq(count(ZL6,size=n),n=0..52); # Zerinvary Lajos, Sep 24 2007
    a:= n-> (Matrix(15, (i,j)-> if (i=j-1) then 1 elif j=1 then [1, 1, 0, 0, -1, -1, -1, 1, 1, 1, 0, 0, -1, -1, 1][i] else 0 fi)^n)[1,1]: seq(a(n), n=0..60); # Alois P. Heinz, Jul 31 2008
    B:=[S,{S = Set(Sequence(Z,1 <= card),card <=5)},unlabelled]: seq(combstruct[count](B, size=n), n=0..52); # Zerinvary Lajos, Mar 21 2009
  • Mathematica
    CoefficientList[ Series[ 1/((1 - x)*(1 - x^2)*(1 - x^3)*(1 - x^4)*(1 - x^5)), {x, 0, 60} ], x ]
    a[n_] := IntegerPartitions[n, 5] // Length; Table[a[n], {n, 0, 52}] (* Jean-François Alcover, Jul 13 2012 *)
    LinearRecurrence[{1,1,0,0,-1,-1,-1,1,1,1,0,0,-1,-1,1},{1,1,2,3,5,7,10,13,18,23,30,37,47,57,70},60] (* Harvey P. Dale, Jan 05 2019 *)
  • PARI
    a(n)=#partitions(n,,5) \\ Charles R Greathouse IV, Sep 15 2014
    
  • PARI
    a(n) = (n^4 + 30*n^3 + 310*n^2 + 1320*n - 90*n*(n%2) + 2880)\2880 \\ Hoang Xuan Thanh, Aug 12 2025

Formula

G.f.: 1/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)*(1-x^5)).
a(n) = 1 + (a(n-2) + a(n-3) + a(n-4)) - (a(n-6) + (2*a(n-7)) + a(n-8)) + (a(n-10) + a(n-11) + a(n-12)) - a(n-14). - Norman J. Meluch (norm(AT)iss.gm.com), Mar 09 2000
Let a1(n) = Sum_{i=0..floor(n/3)} (1 + ceiling((n-3*i-1)/2)), a2(n) = Sum_{i=0..floor(n/4)} (1 + ceiling((n-4*i-1)/2) + a1(n-4*i-3)), then a(n) = Sum_{i=0..floor(n/5)} (1 + ceiling((n-5*i-1)/2) + a1(n-5*i-3) + a2(n-5*i-4)). - Jon Perry, Jun 27 2003
(n choose 5)_q=(q^n-1)*(q^(n-1)-1)*(q^(n-2)-1)*(q^(n-3)-1)*(q^(n-4)-1)/((q^5-1)*(q^4-1)*(q^3-1)*(q^2-1)*(q-1)).
a(n) = round(((n+5)^4 + 10*((n+5)^3 + (n+5)^2) - 75*(n+5) - 45*(n+5)*(-1)^(n+5))/2880). - Washington Bomfim, Jul 03 2012
a(n) = a(n-1) + a(n-2) - a(n-5) - a(n-6) - a(n-7) + a(n-8) + a(n-9) + a(n-10) - a(n-13) - a(n-14) + a(n+15). - David Neil McGrath, Sep 13 2014
a(n+5) = a(n) + A001400(n) = A001400(n)+A026811(n). - Ece Uslu, Esin Becenen, Jan 11 2016
From Vladimír Modrák, Jul 13 2022: (Start)
a(n) = Sum_{k=0..floor(n/5)} Sum_{j=0..floor(n/4)} Sum_{i=0..floor(n/3)} ceiling((max(0, n + 1 - 3*i - 4*j - 5*k))/2).
a(n) = Sum_{j=0..floor(n/5)} Sum_{i=0..floor(n/4)} floor(((max(0, n + 3 - 4*i - 5*j))^2+4)/12). (End)
a(2n) = a(2n-1) + a(n) - a(n-8) = a(n) + Sum_{k=0..n-1} A008804(k). - David García Herrero, Aug 26 2024
a(n) = floor((n^4 + 30*n^3 + 310*n^2 + 1275*n + 45*n*(-1)^n+2880)/2880). - Hoang Xuan Thanh, Aug 12 2025

Extensions

Additional comments from Michael Somos and Branislav Kisacanin (branislav.kisacanin(AT)delphiauto.com)

A008610 Molien series of 4-dimensional representation of cyclic group of order 4 over GF(2) (not Cohen-Macaulay).

Original entry on oeis.org

1, 1, 3, 5, 10, 14, 22, 30, 43, 55, 73, 91, 116, 140, 172, 204, 245, 285, 335, 385, 446, 506, 578, 650, 735, 819, 917, 1015, 1128, 1240, 1368, 1496, 1641, 1785, 1947, 2109, 2290, 2470, 2670, 2870, 3091, 3311, 3553, 3795, 4060, 4324, 4612, 4900, 5213, 5525, 5863
Offset: 0

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Keywords

Comments

a(n) is the number of necklaces with 4 black beads and n white beads.
Also nonnegative integer 2 X 2 matrices with sum of elements equal to n, up to rotational symmetry.
The g.f. is Z(C_4,x), the 4-variate cycle index polynomial for the cyclic group C_4, with substitution x[i]->1/(1-x^i), i=1,...,4. Therefore by Polya enumeration a(n) is the number of cyclically inequivalent 4-necklaces whose 4 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_4,x). - Wolfdieter Lang, Feb 15 2005

Examples

			There are 10 inequivalent nonnegative integer 2 X 2 matrices with sum of elements equal to 4, up to rotational symmetry:
[0 0] [0 0] [0 0] [0 0] [0 1] [0 1] [0 1] [0 2] [0 2] [1 1]
[0 4] [1 3] [2 2] [3 1] [1 2] [2 1] [3 0] [1 1] [2 0] [1 1].
		

References

  • D. J. Benson, Polynomial Invariants of Finite Groups, Cambridge, 1993, p. 104.
  • E. V. McLaughlin, Numbers of factorizations in non-unique factorial domains, Senior Thesis, Allegeny College, Meadville, PA, April 2004.

Crossrefs

Row n=2 of A343874.
Column k=4 of A037306 and A047996.

Programs

  • GAP
    a:=[1,1,3,5,10,14,22,30];; for n in [9..50] do a[n]:=2*a[n-1]-2*a[n-3] +2*a[n-4]-2*a[n-5]+2*a[n-7]-a[n-1]; od; a; # G. C. Greubel, Jan 31 2020
  • Magma
    R:=PowerSeriesRing(Integers(), 50); Coefficients(R!( (1+2*x^3+x^4)/((1-x)*(1-x^2)^2*(1-x^4)) )); // G. C. Greubel, Jan 31 2020
    
  • Maple
    1/(1-x)/(1-x^2)^2/(1-x^4)*(1+2*x^3+x^4);
    seq(coeff(series(%, x, n+1), x, n), n=0..40);
  • Mathematica
    k = 4; Table[Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n, {n, k, 30}] (* Robert A. Russell, Sep 27 2004 *)
    LinearRecurrence[{2,0,-2,2,-2,0,2,-1}, {1,1,3,5,10,14,22,30}, 50] (* G. C. Greubel, Jan 31 2020 *)
  • PARI
    a(n)=if(n,([0,1,0,0,0,0,0,0; 0,0,1,0,0,0,0,0; 0,0,0,1,0,0,0,0; 0,0,0,0,1,0,0,0; 0,0,0,0,0,1,0,0; 0,0,0,0,0,0,1,0; 0,0,0,0,0,0,0,1; -1,2,0,-2,2,-2,0,2]^n*[1;1;3;5;10;14;22;30])[1,1],1) \\ Charles R Greathouse IV, Oct 22 2015
    
  • PARI
    my(x='x+O('x^50)); Vec((1+2*x^3+x^4)/((1-x)*(1-x^2)^2*(1-x^4))) \\ G. C. Greubel, Jan 31 2020
    
  • Sage
    def A008610_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( (1+2*x^3+x^4)/((1-x)*(1-x^2)^2*(1-x^4)) ).list()
    A008610_list(50) # G. C. Greubel, Jan 31 2020
    

Formula

G.f.: (1+2*x^3+x^4)/((1-x)*(1-x^2)^2*(1-x^4)) = (1-x+x^2+x^3)/((1-x)^2*(1-x^2)*(1-x^4)).
a(n) = (1/48)*(2*n^3 + 3*(-1)^n*(n + 4) + 12*n^2 + 25*n + 24 + 12*cos(n*Pi/2)). - Ralf Stephan, Apr 29 2014
G.f.: (1/4)*(1/(1-x)^4 + 1/(1-x^2)^2 + 2/(1-x^4)). - Herbert Kociemba, Oct 22 2016
a(n) = -A032801(-n), per formulae of Colin Barker (A032801) and R. Stephan (above). Also, a(n) - A032801(n+4) = (1+(-1)^signum(n mod 4))/2, i.e., (1,0,0,0,1,0,0,0,...) repeating, (offset 0). - Gregory Gerard Wojnar, Jul 09 2022

Extensions

Comment and example from Vladeta Jovovic, May 18 2000

A001753 Expansion of 1/((1+x)*(1-x)^6).

Original entry on oeis.org

1, 5, 16, 40, 86, 166, 296, 496, 791, 1211, 1792, 2576, 3612, 4956, 6672, 8832, 11517, 14817, 18832, 23672, 29458, 36322, 44408, 53872, 64883, 77623, 92288, 109088, 128248, 150008, 174624, 202368, 233529
Offset: 0

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Comments

Number of symmetric nonnegative integer 5 X 5 matrices with sum of elements equal to 4*n under action of dihedral group D_4.
a(n) = A108561(n+6,n) for n>0. - Reinhard Zumkeller, Jun 10 2005

Examples

			There are 5 symmetric nonnegative integer 5 X 5 matrices with sum of elements equal to 4 under action of D_4:
[1 0 0 0 1] [0 0 1 0 0] [0 0 0 0 0] [0 0 0 0 0] [0 0 0 0 0]
[0 0 0 0 0] [0 0 0 0 0] [0 1 0 1 0] [0 0 1 0 0] [0 0 0 0 0]
[0 0 0 0 0] [1 0 0 0 1] [0 0 0 0 0] [0 1 0 1 0] [0 0 4 0 0]
[0 0 0 0 0] [0 0 0 0 0] [0 1 0 1 0] [0 0 1 0 0] [0 0 0 0 0]
[1 0 0 0 1] [0 0 1 0 0] [0 0 0 0 0] [0 0 0 0 0] [0 0 0 0 0].
		

Crossrefs

Cf. A000217, A002620, A008804, A038163, A054343, A001769 (partial sums), A001752 (first differences), A169793 (binomial transf).

Programs

  • Magma
    [(4*n^5+70*n^4+460*n^3+1400*n^2+1936*n+945)/960+(-1)^n/64: n in [0..40]]; // Vincenzo Librandi, Aug 15 2011
    
  • Mathematica
    CoefficientList[Series[1/((1+x)*(1-x)^6), {x, 0, 50}], x] (* G. C. Greubel, Nov 22 2017 *)
    LinearRecurrence[{5,-9,5,5,-9,5,-1},{1,5,16,40,86,166,296},40] (* Harvey P. Dale, Jun 05 2021 *)
  • PARI
    a(n)=(4*n^5+70*n^4+460*n^3+1400*n^2+1936*n)\/960+1 \\ Charles R Greathouse IV, Apr 17 2012

Formula

a(n) = Sum{k=0..n} (-1)^(n-k)*binomial(k+5, 5); a(n) = (4*n^5 + 70*n^4 + 460*n^3 + 1400*n^2 + 1936*n + 945)/960 + (-1)^n/64. - Paul Barry, Jul 01 2003
a(n) = a(n-2) + (n*(n + 1)*(n + 2)*(n - 1))/24, a(1) = 0, a(2) = 1; (15*(-1)^n - 15*(-1)^(2*n) + 96*n - 160*(-1)^(2*n)*n + 200*n^2 - 200*(-1)^(2*n)*n^2 + 140*n^3 - 80*(-1)^(2*n)*n^3 + 40*n^4 - 10*(-1)^(2*n)*n^4 + 4*n^5)/960. - Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 14 2004
a(n) + a(n+1) = A000389(n+6). - R. J. Mathar, Mar 14 2011

Extensions

Comment and example from Vladeta Jovovic, May 14 2000

A181322 Square array A(n,k), n>=0, k>=0, read by antidiagonals: A(n,k) is the number of partitions of 2*n into powers of 2 less than or equal to 2^k.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 4, 4, 1, 1, 2, 4, 6, 5, 1, 1, 2, 4, 6, 9, 6, 1, 1, 2, 4, 6, 10, 12, 7, 1, 1, 2, 4, 6, 10, 14, 16, 8, 1, 1, 2, 4, 6, 10, 14, 20, 20, 9, 1, 1, 2, 4, 6, 10, 14, 20, 26, 25, 10, 1, 1, 2, 4, 6, 10, 14, 20, 26, 35, 30, 11, 1, 1, 2, 4, 6, 10, 14, 20, 26, 36, 44, 36, 12, 1, 1, 2, 4, 6, 10, 14, 20, 26, 36, 46, 56, 42, 13, 1
Offset: 0

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Author

Alois P. Heinz, Jan 26 2011

Keywords

Comments

Column sequences converge towards A000123.

Examples

			A(3,2) = 6, because there are 6 partitions of 2*3=6 into powers of 2 less than or equal to 2^2=4: [4,2], [4,1,1], [2,2,2], [2,2,1,1], [2,1,1,1,1], [1,1,1,1,1,1].
Square array A(n,k) begins:
  1,  1,  1,  1,  1,  1,  ...
  1,  2,  2,  2,  2,  2,  ...
  1,  3,  4,  4,  4,  4,  ...
  1,  4,  6,  6,  6,  6,  ...
  1,  5,  9, 10, 10, 10,  ...
  1,  6, 12, 14, 14, 14,  ...
		

Crossrefs

Columns k=0-5 give: A000012, A000027(n+1), A002620(n+2), A008804, A088932, A088954.
Main diagonal gives A000123.
Cf. A145515.
See A262553 for another version of this array.
See A072170 for a related array (having the same limiting column).

Programs

  • Maple
    b:= proc(n, j) local nn, r;
          if n<0 then 0
        elif j=0 then 1
        elif j=1 then n+1
        elif n b(n/2^(k-1), k):
    seq(seq(A(n, d-n), n=0..d), d=0..13);
  • Mathematica
    b[n_, j_] := b[n, j] = Module[{nn, r}, Which[n<0, 0, j == 0, 1, j == 1, n+1, nJean-François Alcover, Jan 15 2014, translated from Maple *)
  • PARI
    A181322(n,k,r=1)={if(nA181322(n-1,k,0)+A181322(2*n,k-1,0),n-=r=1+n\1,(r-k)*binomial(r,k)*sum(i=0,min(k-1,k+n), binomial(k,i)/(r-k+i)*A181322(k-i+n,k,0) *(-1)^i))} \\ From Maple. - M. F. Hasler, Feb 19 2019

Formula

G.f. of column k: 1/(1-x) * 1/Product_{j=0..k-1} (1 - x^(2^j)).
A(n,k) = Sum_{i=0..k} A089177(n,i).
For n < 2^k, T(n,k) = A000123(k). T(n,0) = 1, T(n,1) = n+1. - M. F. Hasler, Feb 19 2019

A032246 "DHK[ 5 ]" (bracelet, identity, unlabeled, 5 parts) transform of 1,1,1,1,...

Original entry on oeis.org

2, 4, 10, 16, 28, 42, 64, 90, 126, 168, 224, 288, 370, 462, 576, 704, 858, 1030, 1232, 1456, 1716, 2002, 2330, 2688, 3094, 3536, 4032, 4570, 5168, 5814, 6528, 7296, 8140, 9044, 10032, 11088, 12236, 13460, 14784, 16192, 17710, 19320, 21050, 22880, 24840, 26910
Offset: 8

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Comments

a(n) is the number of bracelets with k = 5 black beads and n-k white beads which have no reflection symmetry. - Herbert Kociemba, Nov 27 2016
From Petros Hadjicostas, Feb 24 2019: (Start)
When k is odd >= 3, the DHK[k] transform of sequence c = (c(n): n >= 1), whose g.f. is C(x) = Sum_{n>=1} c(n)*x^n, has g.f. Sum_{n>=1} (DHK[k] c)n*x^n = (1/2)*Sum{d|k} mu(d)*((1/k)*C(x^d)^(k/d) - C(x^d)*C(x^(2*d))^((k/d) - 1)/2)).
For the current sequence we have k = 5 and c(n) = 1 for all n >= 1. Hence, C(x) = x/(1-x) and A(x) = Sum_{n>=1} a(n)*x^n = (x^k/2)*Sum_{d|k} mu(d)*((1/k)*(1-x^d)^(-k/d) - (1-x^d)^(-1)*(1-x^(2*d))^(-((k/d) - 1)/2)).
The latter g.f. agrees with Herbert Kociemba's formula found below only when k is an odd prime. The reason is that (DHK[k] c)_n, with c=(1,1,1,...), is the number of aperiodic bracelets without reflection symmetry with k black beads and n-k white beads, while Herbert Kociemba's formula counts all (periodic and aperiodic) bracelets without reflection symmetry with k black beads and n-k white beads. Hence, in the case k is an odd prime, the two formulas agree.
When k is even, the g.f. of the DHK[k] transform of sequence c = (c(n): n >= 1) is much more complicated.
Note that Herbert Kociemba's formula for counting all (periodic and aperiodic) bracelets with no reflection symmetry is still valid even when k is even; e.g., see sequence A008804 for the case k=4. For k = 4, all bracelets with 4 black beads and n-k = n-4 white beads that have no reflection symmetry are aperiodic, but this is not true anymore for k even >= 6.
(End)

Examples

			G.f. = 2*x^8 + 4*x^9 + 10*x^10 + 16*x^11 + 28*x^12 + 42*x^13 + 64*x^14 + ...
		

Crossrefs

Cf. A001399, A008804, A032248. Column k = 5 of A180472.

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( 2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5)) )); // G. C. Greubel, Feb 25 2019
    
  • Mathematica
    gf[x_,k_]:=x^k/2 (1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])-(1+x)/(1-x^2)^Floor[k/2+1]); CoefficientList[Series[gf[x,5],{x,0,50}],x] (* Herbert Kociemba, Nov 27 2016 *)
    Drop[CoefficientList[Series[2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5)), {x,0,50}], x], 8] (* G. C. Greubel, Feb 25 2019 *)
  • PARI
    {a(n) = if( n<0, n=5-n); polcoeff( 2 * x^8 / ((1-x)^2*(1-x^2)^2*(1-x^5)) + x * O(x^n), n)}; /* Michael Somos, Nov 28 2016 */
    
  • PARI
    Vec(2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5)) + O(x^40)) \\ Colin Barker, Mar 13 2019
    
  • Sage
    a=(2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5))).series(x, 50).coefficients(x, sparse=False); a[8:] # G. C. Greubel, Feb 25 2019

Formula

G.f.: 2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5)).
From Herbert Kociemba, Nov 27 2016: (Start)
More generally gf(k) is the g.f. for the number of bracelets without reflection symmetry with k black beads and n-k white beads.
gf(k): x^k/2 * ((1/k)*Sum_{n|k} phi(n)/(1 - x^n)^(k/n) - (1 + x)/(1-x^2)^floor(k/2 + 1)). (End)
a(n) = a(5-n) for all n in Z. - Michael Somos, Nov 28 2016
0 = a(n) - 2*a(n+1) - a(n+2) + 4*a(n+3) - a(n+4) - 3*a(n+5) + 3*a(n+6) + a(n+7) - 4*a(n+8) + a(n+9) + 2*a(n+10) - a(n+11) for all n in Z. - Michael Somos, Nov 28 2016

A032248 "DHK[ 7 ]" (bracelet, identity, unlabeled, 7 parts) transform of 1,1,1,1,...

Original entry on oeis.org

4, 10, 28, 56, 113, 197, 340, 544, 856, 1284, 1896, 2709, 3816, 5247, 7128, 9504, 12540, 16302, 21001, 26728, 33748, 42185, 52364, 64448, 78832, 95725, 115600, 138720, 165648, 196707, 232560, 273600, 320601, 374034, 434796, 503448, 581020, 668173, 766084
Offset: 10

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Comments

From Petros Hadjicostas, Feb 24 2019: (Start)
When k is odd >= 3, the DHK[k] transform of sequence c = (c(n): n >= 1), whose g.f. is C(x) = Sum_{n>=1} c(n)*x^n, has g.f. Sum_{n>=1} (DHK[k] c)n*x^n = (1/2)*Sum{d|k} mu(d)*((1/k)*C(x^d)^(k/d) - C(x^d)*C(x^(2*d))^((k/d) - 1)/2)).
For the current sequence we have k = 7 and c(n) = 1 for all n >= 1. Hence, C(x) = x/(1-x) and A(x) = Sum_{n>=1} a(n)*x^n = (x^k/2)*Sum_{d|k} mu(d)*((1/k)*(1-x^d)^(-k/d) - (1-x^d)^(-1)*(1-x^(2*d))^(-((k/d) - 1)/2)).
The latter g.f. agrees with Herbert Kociemba's formula found in the documentations of sequences and A008804 and A032246 only when k is an odd prime. The reason is that (DHK[k] c)_n, with c=(1,1,1,...), is the number of aperiodic bracelets without reflection symmetry with k black beads and n-k white beads, while Herbert Kociemba's formula (in the documentations of sequences and A008804 and A032246) counts all (periodic and aperiodic) bracelets without reflection symmetry with k black beads and n-k white beads. Hence, in the case k is an odd prime, the two formulas agree.
When k is even, the g.f. of the DHK[k] transform of sequence c = (c(n): n >= 1) is much more complicated.
Note that Herbert Kociemba's formula for counting all (periodic and aperiodic) bracelets with no reflection symmetry is still valid even when k is even; e.g., see sequence A008804 for the case k=4. For k = 4, all bracelets with 4 black beads and n-k = n-4 white beads that have no reflection symmetry are aperiodic, but this is not true anymore for k even >= 6.
(End)

Crossrefs

Cf. A001399, A008804, A032246, A032247, A032250. Column k = 7 of A180472.

Programs

  • Mathematica
    LinearRecurrence[{3,0,-8,6,6,-8,1,0,-1,8,-6,-6,8,0,-3,1},{4,10,28,56,113,197,340,544,856,1284,1896,2709,3816,5247,7128,9504},40] (* Harvey P. Dale, Jul 08 2024 *)
  • PARI
    Vec(x^10*(4 - 2*x - 2*x^2 + 4*x^3 + x^4 - 2*x^5 + x^6) / ((1 - x)^7*(1 + x)^3*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)) + O(x^40)) \\ Colin Barker, Feb 25 2019

Formula

G.f.: x^7*(1/(14*(1 - x)^7) - 1/((2*(1 - x))*(1 - x^2)^3) + 3/(7*(1 - x^7))). - Petros Hadjicostas, Feb 24 2019
a(n) = 3*a(n-1) - 8*a(n-3) + 6*a(n-4) + 6*a(n-5) - 8*a(n-6) + a(n-7) - a(n-9) + 8*a(n-10) - 6*a(n-11) - 6*a(n-12) + 8*a(n-13) - 3*a(n-15) + a(n-16) for n>25. - Colin Barker, Feb 25 2019

A032247 "DHK[ 6 ]" (bracelet, identity, unlabeled, 6 parts) transform of 1,1,1,1,...

Original entry on oeis.org

3, 6, 16, 29, 56, 90, 150, 222, 336, 474, 672, 908, 1233, 1612, 2112, 2693, 3432, 4282, 5340, 6542, 8008, 9666, 11648, 13874, 16503, 19432, 22848, 26639, 31008, 35832, 41346, 47400, 54264, 61776, 70224, 79434, 89733, 100914
Offset: 9

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Author

Keywords

Comments

Here, a(n) is the number of aperiodic bracelets with k = 6 black beads and n-k = n-6 white beads that have no reflection symmetry. We conjecture that we can use Herbert Kociemba's formula from the documentation of sequences A008804 and A032246 to derive the g.f. of (a(n): n >= 1). See below for more details. - Petros Hadjicostas, Feb 24 2019

Crossrefs

Formula

From Petros Hadjicostas, Feb 24 2019: (Start)
Let gf(k, x) = x^k/2 * ((1/k) * Sum_{n|k} phi(n)/(1 - x^n)^(k/n) - (1 + x)/(1 -x^2)^floor(k/2 + 1)) be Herbert Kociemba's formula for the g.f. of the number of all bracelets with k black beads and n-k white beads that have no reflection symmetry.
We prove in the note that g.f. = Sum_{n>=1} a(n)*x^n = gf(6, x) - gf(3, x^2).
(End)
From Colin Barker, Feb 25 2019: (Start)
G.f.: x^9*(3 + 4*x^2 + 4*x^4 + 2*x^6 - 2*x^7 + x^8) / ((1 - x)^6*(1 + x)^3*(1 - x + x^2)*(1 + x^2)*(1 + x + x^2)^2).
a(n) = 2*a(n-1) - a(n-3) - 2*a(n-5) + 3*a(n-6) - 2*a(n-7) + a(n-8) + a(n-9) - 2*a(n-10) + 3*a(n-11) - 2*a(n-12) - a(n-14) + 2*a(n-16) - a(n-17) for n > 25.
(End)
From Petros Hadjicostas, May 25 2019: (Start)
G.f.: (x^k/(2*k)) * Sum_{d|k} mu(d)*(1/(1 - x^d)^(k/d) - k*(1 + x^d)/(1 - x^(2*d))^floor(k/(2*d) + 1)) with k = 6.
a(n) = (1/12)* Sum_{d|gcd(n, 6)} mu(d) * (binomial((n/d) - 1, (6/d) - 1) - 6*binomial(floor(b(n,d)/2), floor(3/d))) for n >= 6, where b(n,d) = n/d + ((-1)^(6/d) - 1)/2. (Thus, b(n,d) = n/d for d = 1, 3, and b(n, d) = n/d - 1 for d = 2, 6.) (End)

A180472 Triangle T(n, k) = OC(n, k; not -1), read by rows, where OC(n, k; not -1) is the number of k-subsets of Z_n without -1 as a multiplier, up to congruency.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 2, 2, 2, 0, 0, 0, 0, 0, 0, 3, 4, 4, 3, 0, 0, 0, 0, 0, 0, 4, 6, 10, 6, 4, 0, 0, 0, 0, 0, 0, 5, 10, 16, 16, 10, 5, 0, 0, 0, 0, 0, 0, 7, 14, 28, 30, 28, 14, 7, 0, 0, 0, 0, 0, 0, 8, 20, 42, 56, 56, 42, 20, 8, 0, 0, 0, 0, 0, 0, 10, 26, 64, 91, 113, 91, 64, 26, 10, 0, 0, 0, 0, 0, 0, 12, 35, 90, 150, 197, 197, 150, 90, 35, 12, 0, 0, 0, 0, 0, 0, 14, 44, 126, 224, 340, 370, 340, 224, 126, 44, 14, 0, 0, 0, 0, 0, 0, 16, 56, 168, 336, 544, 680, 680, 544, 336, 168, 56, 16, 0, 0, 0
Offset: 0

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Author

John P. McSorley, Sep 06 2010

Keywords

Comments

Let Z_n = {0,1,...,n-1} denote the integers mod n.
Let S be a k-subset of Z_n.
Then S has multiplier -1 iff there is a z in Z_n for which S = -S + z. Otherwise, S doesn't have multiplier -1.
For example in Z_7 the set S = {0,1,2} has multiplier -1 since -S = {0,-1,-2} = {0,5,6} and then {0,1,2} = {0,5,6} + 2, so S = -S + 2. But S={0,1,3} doesn't have multiplier -1.
Let S and S' be two k-subsets of Z_n.
Define an equivalence relation on the set of k-subsets as follows: S is congruent to S' iff S=S'+z or S = -S' + z for some z in Z_n.
Then define OC(n, k) to be the number of such congruence classes.
And define OC(n, k; not -1) to be the number of such congruence classes in which the representative doesn't have -1 as a multiplier.
Then this sequence is the 'OC(n,k; not -1)' triangle read by rows.
For convenience we start the triangle at n = 0, and we have 0 <= k <= n.
See the McSorley and Schoen (2013) reference below for equivalent definitions of this sequence in terms of (n,k)-Ovals and k-compositions of n.
From Petros Hadjicostas, May 29 2019: (Start)
Here, T(n, k) is the number of bracelets (turnover necklaces) of length n that have no reflection symmetry and consist of k white beads and n - k black beads. (Bracelets that have no reflection symmetry are also known as chiral bracelets.)
It is also the number of dihedral compositions of n into k parts with no reflection symmetry. It is also the number of dihedral compositions of n into n - k parts with no reflection symmetry. (For a definition of a dihedral composition, see Knopfmacher and Robbins (2013) in the references.)
For MacMahon's method for transforming a cyclic composition into a necklace and vice versa, see the comments for sequence A308401. See also p. 273 in Sommerville (1909).
(End)

Examples

			The triangle begins (with rows for n >= 0 and columns for k >= 0) as follows:
  0
  0  0
  0  0  0
  0  0  0  0
  0  0  0  0  0
  0  0  0  0  0  0
  0  0  0  1  0  0  0
  0  0  0  1  1  0  0   0
  0  0  0  2  2  2  0   0  0
  0  0  0  3  4  4  3   0  0  0
  0  0  0  4  6 10  6   4  0  0  0
  0  0  0  5 10 16 16  10  5  0  0  0
  0  0  0  7 14 28 30  28 14  7  0  0  0
  0  0  0  8 20 42 56  56 42 20  8  0  0  0
  0  0  0 10 26 64 91 113 91 64 26 10  0  0  0  0
  ...
For example the row which corresponds to Z_7 is: 0 0 0 1 1 0 0 0.
The first '1' here corresponds to the 3-subsets of Z_7.
There are 4 congruence classes of the 3-subsets of Z_7, their representatives are {0,1,2}, {0,2,4}, {0,1,4} and {0,1,3}. The first 3 representatives have multiplier -1, but the last doesn't. Hence there is just one 3-subset of Z_7 without multiplier -1, up to congruency.
		

Crossrefs

This sequence is A052307-A119963. The sequence A052307 is formed from the triangle whose (n, k)-term is the number of k-subsets of Z_n up to congruence, and the sequence A119963 is formed from the triangle whose (n, k)-term is the number of k-subsets of Z_n with multiplier -1 up to congruence.
The row sums of the 'OC(n, k, not -1)' triangle above give sequence A059076.
Cf. A001399 (column k = 3 with different offset), A008804 (column k = 4 with different offset), A032246 (column k = 5), A308401 (column k = 6), A032248 (column k = 7).

Programs

  • PARI
    T(n,k) = if ((n==0) && (k==0), 0, -binomial(floor(n/2) - (k % 2) * (1 - n % 2), floor(k/2)) / 2 + sumdiv(gcd(n,k), d, (eulerphi(d)*binomial(n/d, k/d))) / (2*n));tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print); \\ Michel Marcus, May 30 2019

Formula

From Petros Hadjicostas, May 29 2019: (Start)
T(n,k) = -binomial(floor(n/2) - (k mod 2) * (1 - (n mod 2)), floor(k/2)) / 2 + Sum_{d|n, d|k} (phi(d)*binomial(n/d, k/d)) / (2*n) for n >= 1 and 0 <= k <= n. (This is a modification of formulas due to Gupta (1979), Shevelev (2004), and W. Bomfim in sequence A052307.)
T(n, k) = A052307(n, k) - A119963(n,k) for 0 <= k <= n. (See the comments in CROSSREFS by J. P. McSorley.)
T(n, k) = T(n, n - k) for 0 <= k <= n.
G.f. for column k >= 1: (x^k/2) * (-(1 + x)/(1 - x^2)^floor((k/2) + 1) + (1/k) * Sum_{m|k} phi(m)/(1 - x^m)^(k/m)). (This formula is due to Herbert Kociemba.)
(End)
Bivariate g.f.: Sum_{n,k >= 0} T(n, k)*x^n*y^k = (1/2) * (1 - (1 + x) * (1 + x*y) / (1 - x^2 * (1 + y^2)) - Sum_{d >= 1} (phi(d) / d) * log(1 - x^d * (1 + y^d))). - Petros Hadjicostas, Jun 15 2019

Extensions

Name edited by Petros Hadjicostas, May 29 2019
Offset corrected by Andrew Howroyd, Sep 27 2019

A032249 "DHK[ 8 ]" (bracelet, identity, unlabeled, 8 parts) transform of 1,1,1,1,...

Original entry on oeis.org

5, 14, 42, 90, 197, 368, 680, 1152, 1926, 3044, 4740, 7100, 10494, 15072, 21384, 29680, 40755, 54994, 73502, 96854, 126555, 163424, 209456, 265792, 335036, 418728, 520200, 641496, 786828, 958848, 1162800, 1402080
Offset: 11

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Author

Keywords

Comments

Here, a(n) is the number of aperiodic bracelets with k = 8 black beads and n-k = n-8 white beads that have no reflection symmetry. We conjecture that we can use Herbert Kociemba's formula from the documentation of sequences A008804 and A032246 to derive the g.f. of (a(n): n >= 1). See below for more details. - Petros Hadjicostas, Feb 24 2019

Crossrefs

Formula

From Petros Hadjicostas, Feb 24 2019, proven in Hadjicostas (2019): (Start)
Let gf(k, x) = x^k/2 * ( (1/k)*Sum_{n|k} phi(n)/(1 - x^n)^(k/n) - (1 + x)/(1 -x^2)^floor(k/2 + 1) ) be Herbert Kociemba's formula for the g.f. of the number of all bracelets with k black beads and n-k white beads that have no reflection symmetry.
We conjecture that g.f. = Sum_{n>=1} a(n)*x^n = gf(8,x) - gf(4, x^2).
(End)
G.f.: (x^k/(2*k)) * Sum_{d|k} mu(d) * (1/(1 - x^d)^(k/d) - k*(1 + x^d)/(1 - x^(2*d))^floor(k/(2*d) + 1)) with k = 8. - Petros Hadjicostas, May 24 2019
a(n) = (1/16)* Sum_{d|gcd(n, 8)} mu(d) * (binomial((n/d) - 1, (8/d) - 1) - 8 * binomial(floor(b(n,d)/2), floor(4/d))) for n >= 11, where b(n,d) = n/d + ((-1)^(8/d) - 1)/2. (Thus, b(n,d) = n/d for d = 1, 2, 4, and b(n, d) = n/d - 1 for d = 8.) - Petros Hadjicostas, May 27 2019

A088932 G.f.: 1/((1-x)^2*(1-x^2)*(1-x^4)*(1-x^8)).

Original entry on oeis.org

1, 2, 4, 6, 10, 14, 20, 26, 36, 46, 60, 74, 94, 114, 140, 166, 201, 236, 280, 324, 380, 436, 504, 572, 656, 740, 840, 940, 1060, 1180, 1320, 1460, 1625, 1790, 1980, 2170, 2390, 2610, 2860, 3110, 3396, 3682, 4004, 4326, 4690, 5054, 5460, 5866, 6321, 6776, 7280, 7784
Offset: 0

Views

Author

N. J. A. Sloane, Dec 02 2003

Keywords

Comments

a(n) is the number of partitions of 2*n into powers of 2 less than or equal to 2^4. First differs from A000123 at n=16. - Alois P. Heinz, Apr 02 2012

Crossrefs

See A000027, A002620, A008804, A088954, A000123 for similar sequences.
Column k=4 of A181322.
Cf. A010873.

Programs

  • Maple
    f := proc(n,k) option remember; if k > n then RETURN(0); fi; if k= 0 then if n=0 then RETURN(1) else RETURN(0); fi; fi; if k = 1 then RETURN(1); fi; if n mod 2 = 1 then RETURN(f(n-1,k)); fi; f(n-1,k)+f(n/2,k-1); end; # present sequence is f(2m,5)
    GFF := k->x^(2^(k-2))/((1-x)*mul((1-x^(2^j)),j=0..k-2)); # present g.f. is GFF(5)/x^8
    a:= proc(n) local m, r; m := iquo(n, 8, 'r'); r:= r+1; [1, 2, 4, 6, 10, 14, 20, 26][r]+ (((8/3*m +(4*r +28)/3)*m +[0, 4, 9, 14, 20, 26, 33, 40][r] +43/3)*m +[22, 33, 50, 67, 93, 119, 154, 189][r]/3)*m end: seq(a(n), n=0..60); # Alois P. Heinz, Apr 17 2009
  • Mathematica
    CoefficientList[Series[1/((1-x)^2(1-x^2)(1-x^4)(1-x^8)), {x,0,60}], x]  (* Harvey P. Dale, Apr 22 2011 *)
    Table[1 + 1237*n/1536 + 17*n^2/96 + 13*n^3/768 + n^4/1536 + (5/32 + n/32) * Floor[n/4] + (81/256 + 3*n/32 + n^2/128) * Floor[n/2] - Floor[(n+1)/8]/4 - (n+3) * Floor[(n+1)/4]/32 - Floor[(n+2)/8]/4, {n, 0, 100}] (* Vaclav Kotesovec, May 02 2018 *)
    Table[Simplify[1023/1024 + 85*n/96 + 341*n^2/1536 + n^3/48 + n^4/1536 + (-1)^n*(113/1024 + n/32 + n^2/512) - (1 + Sqrt[2])*Cos[Pi*n/4]/16 + Cos[Pi*n/2]/64 + (Sqrt[2] - 1) * Cos[3*Pi*n/4]/16 + (1/8 + n/64)*Sin[Pi*n/2]], {n, 0, 100}] (* Vaclav Kotesovec, May 02 2018 *)
  • PARI
    Vec(1/((1-x)^2*(1-x^2)*(1-x^4)*(1-x^8))+O(x^99)) \\ Charles R Greathouse IV, Sep 03 2011

Formula

a(n) = (8*floor(n/4)^4 + 8*(m+8)*floor(n/4)^3 - 2*(m^3 - 6*m^2 - 19*m - 86)*floor(n/4)^2 -8*(m^3 - 6*m^2 - 6*m - 22)*floor(n/4) - 7*m^3 + 42*m^2 + 13*m + 54 - (m^3 - 6*m^2 + 5*m + 6)*(-1)^floor(n/4))/48 where m = n mod 4. - Luce ETIENNE, Apr 07 2018
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