A089177
Triangle read by rows: T(n,k) (n >= 0, 0 <= k <= 1+log_2(floor(n))) giving number of non-squashing partitions of n into k parts.
Original entry on oeis.org
1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 4, 4, 1, 1, 5, 6, 2, 1, 6, 9, 4, 1, 7, 12, 6, 1, 8, 16, 10, 1, 1, 9, 20, 14, 2, 1, 10, 25, 20, 4, 1, 11, 30, 26, 6, 1, 12, 36, 35, 10, 1, 13, 42, 44, 14, 1, 14, 49, 56, 20, 1, 15, 56, 68, 26, 1, 16, 64, 84, 36, 1, 1, 17, 72, 100, 46, 2, 1, 18, 81, 120, 60, 4, 1
Offset: 0
Triangle begins:
1;
1, 1;
1, 2, 1;
1, 3, 2;
1, 4, 4, 1;
1, 5, 6, 2;
1, 6, 9, 4;
1, 7, 12, 6;
1, 8, 16, 10, 1;
-
T:= proc(n) option remember;
`if`(n=0, 1, zip((x, y)-> x+y, [T(n-1)], [0, T(floor(n/2))], 0)[])
end:
seq(T(n), n=0..25); # Alois P. Heinz, Apr 01 2012
-
row[0] = {1}; row[1] = {1, 1}; row[n_] := row[n] = Plus @@ PadRight[ {row[n-1], Join[{0}, row[Floor[n/2]]]} ]; Table[row[n], {n, 0, 25}] // Flatten (* Jean-François Alcover, Jan 31 2014 *)
A014557
Multiplicity of K_3 in K_n.
Original entry on oeis.org
0, 0, 0, 0, 0, 0, 2, 4, 8, 12, 20, 28, 40, 52, 70, 88, 112, 136, 168, 200, 240, 280, 330, 380, 440, 500, 572, 644, 728, 812, 910, 1008, 1120, 1232, 1360, 1488, 1632, 1776, 1938, 2100, 2280, 2460, 2660, 2860, 3080, 3300, 3542, 3784, 4048, 4312, 4600, 4888, 5200
Offset: 0
Any 2-coloring of the edges of K_6 produces at least two monochromatic triangles. Having colors induce K_3,3 and 2K_3 shows this is attained, so a(6) = 2.
- Alexander Adamchuk, Table of n, a(n) for n = 0..100
- R. Ehrenborg, Bounding monochromatic triangles using squares, Math. Magazine, 94 (2021), 383-386.
- A. W. Goodman, On Sets of Acquaintances and Strangers at Any Party, Amer. Math. Monthly 66, 778-783, 1959.
- L. Sauvé, On chromatic graphs, Amer. Math. Monthly, 68 (1961), 107-111.
- A. J. Schwenk, Acquaintance Party Problem, Amer. Math. Monthly 79 (1972), 1113-1117.
- V. Vijayalakshmi, Multiplicity of triangles in cocktail party graphs, Discrete Math., 206 (1999), 217-218.
- Eric Weisstein's World of Mathematics, Extremal Graph.
- Eric Weisstein's World of Mathematics, Monochromatic Forced Triangle.
- Index entries for linear recurrences with constant coefficients, signature (2,0,-2,2,-2,0,2,-1).
-
[n*(n-1)*(n-2)/6 - Floor((n/2)*Floor(((n-1)/2)^2)): n in [1..20]]; // G. C. Greubel, Oct 06 2017
-
A049322 := proc(n) local u; if n mod 2 = 0 then u := n/2; RETURN(u*(u-1)*(u-2)/3); elif n mod 4 = 1 then u := (n-1)/4; RETURN(u*(u-1)*(4*u+1)*2/3); else u := (n-3)/4; RETURN(u*(u+1)*(4*u-1)*2/3); fi; end;
-
Table[Binomial[n,3] - Floor[n/2*Floor[((n-1)/2)^2]],{n,0,100}] (* Alexander Adamchuk, Nov 29 2006 *)
-
x='x+O('x^99); concat(vector(6), Vec(2*x^6/((x-1)^4*(x+1)^2*(x^2+1)))) \\ Altug Alkan, Apr 08 2016
A088954
G.f.: 1/((1-x)^2*(1-x^2)*(1-x^4)*(1-x^8)*(1-x^16)).
Original entry on oeis.org
1, 2, 4, 6, 10, 14, 20, 26, 36, 46, 60, 74, 94, 114, 140, 166, 202, 238, 284, 330, 390, 450, 524, 598, 692, 786, 900, 1014, 1154, 1294, 1460, 1626, 1827, 2028, 2264, 2500, 2780, 3060, 3384, 3708, 4088, 4468, 4904, 5340, 5844, 6348, 6920, 7492, 8148, 8804, 9544
Offset: 0
- Alois P. Heinz, Table of n, a(n) for n = 0..2000
- N. J. A. Sloane and J. A. Sellers, On non-squashing partitions, Discrete Math., 294 (2005), 259-274.
- Index entries for linear recurrences with constant coefficients, signature (2, 0, -2, 2, -2, 0, 2, 0, -2, 0, 2, -2, 2, 0, -2, 2, -2, 0, 2, -2, 2, 0, -2, 0, 2, 0, -2, 2, -2, 0, 2, -1).
-
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,6)
GFF := k->x^(2^(k-2))/((1-x)*mul((1-x^(2^j)),j=0..k-2)); # present g.f. is GFF(6)/x^16
a:= proc(n) local m, r; m:= iquo(n, 16, 'r'); r:= r+1; [1, 2, 4, 6, 10, 14, 20, 26, 36, 46, 60, 74, 94, 114, 140, 166][r] +(((((128/5*m +8*(15+r))*m +(228 +[0, 32, 68, 104, 144, 184, 228, 272, 320, 368, 420, 472, 528, 584, 644, 704][r]))*m +(172 +[0, 43, 98, 153, 223, 293, 378, 463, 566, 669, 790, 911, 1053, 1195, 1358, 1521][r]))*m +(247/5 +[0, 22, 55, 88, 138, 188, 255, 322, 415, 508, 627, 746, 900, 1054, 1243, 1432][r]))*m)/3 end: seq(a(n), n=0..60); # Alois P. Heinz, Apr 17 2009
-
CoefficientList[Series[1/((1-x)^2(1-x^2)(1-x^4)(1-x^8)(1-x^16)),{x,0,70}],x] (* or *) LinearRecurrence[{2,0,-2,2,-2,0,2,0,-2,0,2,-2,2,0,-2,2,-2,0,2,-2,2,0,-2,0,2,0,-2,2,-2,0,2,-1},{1,2,4,6,10,14,20,26,36,46,60,74,94,114,140,166,202,238,284,330,390,450,524,598,692,786,900,1014,1154,1294,1460,1626},70](* Harvey P. Dale, Feb 12 2013 *)
A008643
Molien series for group of 4 X 4 upper triangular matrices over GF(2).
Original entry on oeis.org
1, 1, 2, 2, 4, 4, 6, 6, 10, 10, 14, 14, 20, 20, 26, 26, 35, 35, 44, 44, 56, 56, 68, 68, 84, 84, 100, 100, 120, 120, 140, 140, 165, 165, 190, 190, 220, 220, 250, 250, 286, 286, 322, 322, 364, 364, 406, 406, 455, 455, 504, 504, 560, 560, 616, 616, 680, 680, 744
Offset: 0
- D. J. Benson, Polynomial Invariants of Finite Groups, Cambridge, 1993, p. 105.
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 233
- Index entries for Molien series
- Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1,-1,-1,1,1,-1,-1,1,-1,1,1,-1).
-
R:=PowerSeriesRing(Integers(), 65); Coefficients(R!( (&*[1/(1-x^(2^j)): j in [0..3]]) )); // G. C. Greubel, Feb 01 2020
-
a:= proc(n) local m, r; m := iquo(n, 8, 'r'); r:= iquo(r,2)+1; ([11, 17, 26, 35][r]+ (9+ 3*r+ 4*m) *m) *m/3+ [1, 2, 4, 6][r] end: seq(a(n), n=0..100); # Alois P. Heinz, Oct 06 2008
-
CoefficientList[1/((1-x)*(1-x^2)*(1-x^4)*(1-x^8)) + O[x]^65, x] (* Jean-François Alcover, May 29 2015 *)
LinearRecurrence[{1,1,-1,1,-1,-1,1,1,-1,-1,1,-1,1,1,-1}, {1,1,2,2,4,4,6,6,10,10,14,14,20,20,26}, 65] (* Ray Chandler, Jul 15 2015 *)
-
my(x='x+O('x^65)); Vec(1/((1-x)*(1-x^2)*(1-x^4)*(1-x^8))) \\ G. C. Greubel, Feb 01 2020
-
my(b(m) = (m^3 + 12*m^2 + (44 - 3*(m%2))*m + 48)\48); vector(59,n,b((n-1)\2)) \\ Hoang Xuan Thanh, Aug 14 2025
-
def A077952_list(prec):
P. = PowerSeriesRing(ZZ, prec)
return P( 1/((1-x)*(1-x^2)*(1-x^4)*(1-x^8)) ).list()
A077952_list(65) # G. C. Greubel, Feb 01 2020
A032250
"DHK[ n ](2n)" (bracelet, identity, unlabeled, n parts, evaluated at 2n) transform of 1,1,1,1,...
Original entry on oeis.org
1, 1, 1, 2, 10, 29, 113, 368, 1316, 4490, 15907, 55866, 199550, 714601, 2583575, 9385280, 34311304, 126018592, 465044951, 1722987050, 6407739430, 23909854891, 89493459541, 335911158480, 1264104712300
Offset: 1
-
# This is a crude program that assumes the above conjecture is true (which was later proved in Hadjicostas (2019)). It is only valid for n >= 3 because of Bower's special definition of DHK[k] for the cases k=1 and k=2.
with(NumberTheory);
ff := proc (k, x) (1/2)*x^k*(add(phi(n)/(1-x^n)^(k/n), n in Divisors(k))/k-(x+1)/(1-x^2)^floor((1/2)*k+1)); end proc;
gg := proc (k, x) add(Moebius(d)*ff(k/d, x^d), d in Divisors(k)); end proc;
vv := proc (n) simplify(subs(x = 0, diff(gg(n, x), x$(2*n)))/factorial(2*n)); end proc;
for i from 3 to 100 do print(i, vv(i)); end do; # Petros Hadjicostas, Feb 24 2019
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
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.
- Colin Barker, Table of n, a(n) for n = 9..1000
- Hansraj Gupta, Enumeration of incongruent cyclic k-gons, Indian J. Pure and Appl. Math., 10 (1979), no. 8, 964-999.
- Petros Hadjicostas, The aperiodic version of Herbert Kociemba's formula for bracelets with no reflection symmetry, 2019.
- Arnold Knopfmacher and Neville Robbins, Some properties of dihedral compositions, Util. Math. 92 (2013), 207-220.
- Richard H. Reis, A formula for C(T) in Gupta's paper, Indian J. Pure and Appl. Math., 10 (1979), no. 8, 1000-1001.
- Frank Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc.
- Vladimir S. Shevelev, Necklaces and convex k-gons, Indian J. Pure and Appl. Math., 35 (2004), no. 5, 629-638.
- Vladimir S. Shevelev, Necklaces and convex k-gons, Indian J. Pure and Appl. Math., 35 (2004), no. 5, 629-638.
- Duncan M. Y. Sommerville, On certain periodic properties of cyclic compositions of numbers, Proc. London Math. Soc. S2-7(1) (1909), 263-313.
- Index entries for linear recurrences with constant coefficients, signature (2,1,-3,-1,1,4,-3,-3,4,1,-1,-3,1,2,-1).
-
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
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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
A308583
Triangle read by rows: T(n,k) = number of aperiodic chiral bracelets (turnover necklaces with no reflection symmetry and period n) with n beads, k of which are white and n - k are black, for n >= 1 and 1 <= k <= n.
Original entry on oeis.org
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 2, 2, 2, 0, 0, 0, 0, 0, 3, 4, 4, 3, 0, 0, 0, 0, 0, 4, 6, 10, 6, 4, 0, 0, 0, 0, 0, 5, 10, 16, 16, 10, 5, 0, 0, 0, 0, 0, 7, 14, 28, 29, 28, 14, 7, 0, 0, 0, 0, 0, 8, 20, 42, 56, 56, 42, 20, 8, 0, 0, 0, 0, 0, 10, 26, 64, 90, 113, 90, 64, 26, 10, 0, 0, 0, 0, 0, 12, 35, 90, 150, 197, 197, 150, 90, 35, 12, 0, 0, 0, 0, 0, 14, 44, 126, 222, 340, 368, 340, 222, 126, 44, 14, 0, 0, 0
Offset: 1
The triangle begins (with rows for n >= 1 and columns for k >= 1) as follows:
0;
0, 0;
0, 0, 0;
0, 0, 0, 0;
0, 0, 0, 0, 0;
0, 0, 1, 0, 0, 0;
0, 0, 1, 1, 0, 0, 0;
0, 0, 2, 2, 2, 0, 0, 0;
0, 0, 3, 4, 4, 3, 0, 0, 0;
0, 0, 4, 6, 10, 6, 4, 0, 0, 0;
0, 0, 5, 10, 16, 16, 10, 5, 0, 0, 0;
0, 0, 7, 14, 28, 29, 28, 14, 7, 0, 0, 0;
0, 0, 8, 20, 42, 56, 56, 42, 20, 8, 0, 0, 0;
0, 0, 10, 26, 64, 90, 113, 90, 64, 26, 10, 0, 0, 0;
...
Notice, for example, that T(14, 6) = 90 <> 91 = A180472(14, 6). Out of the 91 chiral bracelets with 6 W and 8 B beads, only WWBWBBBWWBWBBB is periodic.
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. Because chiral bracelets 2, 3, and 6 are aperiodic as well, we have T(9, 3) = 3 = T(9, 6).
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. In addition, chiral dihedral compositions 2, 3, and 6 are aperiodic as well, and so (again) T(9, 3) = 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. Chiral dihedral compositions 2, 3, and 6 are aperiodic as well, and hence T(9, 6) = 3.
A002985
Number of trees in an n-node wheel.
Original entry on oeis.org
1, 1, 1, 2, 3, 6, 11, 20, 36, 64, 108, 179, 292, 464, 727, 1124, 1714, 2585, 3866, 5724, 8418, 12290, 17830, 25713, 36898, 52664, 74837, 105873, 149178, 209364, 292793, 407990, 566668, 784521, 1082848, 1490197, 2045093, 2798895, 3820629, 5202085
Offset: 1
All trees that span a wheel on 5 nodes are equivalent to one of the following:
o o o
| | \ / \
o--o--o o--o o o--o o
| | /
o o o
- F. Harary, P. E. O'Neil, R. C. Read and A. J. Schwenk, The number of trees in a wheel, in D. J. A. Welsh and D. R. Woodall, editors, Combinatorics. Institute of Mathematics and Its Applications. Southend-on-Sea, England, 1972, pp. 155-163.
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
-
terms = 40;
A003293[n_] := SeriesCoefficient[Product[(1-x^k)^(-Ceiling[k/2]), {k, 1, terms}], {x, 0, n}];
A008804[n_] := SeriesCoefficient[1/((1-x)^4 (1+x)^2 (1+x^2)), {x, 0, n}];
a[n_] := A003293[n-1] - A008804[n-3];
Array[a, terms] (* Jean-François Alcover, Sep 02 2019 *)
-
\\ here b(n) is A003293 and d(n) is A008804.
b(n)={polcoeff( prod(k=1, n, (1-x^k+x*O(x^n))^-ceil(k/2)), n)}
d(n)={(84+12*(-1)^n+6*I*((-I)^n-I^n)+(85+3*(-1)^n)*n+24*n^2+2*n^3)/96}
a(n)=b(n-1)-d(n-3); \\ Andrew Howroyd, Oct 09 2017
A073463
Triangle of number of partitions of 2n into powers of 2 where the largest part is 2^k.
Original entry on oeis.org
1, 1, 1, 1, 2, 1, 1, 3, 2, 0, 1, 4, 4, 1, 0, 1, 5, 6, 2, 0, 0, 1, 6, 9, 4, 0, 0, 0, 1, 7, 12, 6, 0, 0, 0, 0, 1, 8, 16, 10, 1, 0, 0, 0, 0, 1, 9, 20, 14, 2, 0, 0, 0, 0, 0, 1, 10, 25, 20, 4, 0, 0, 0, 0, 0, 0, 1, 11, 30, 26, 6, 0, 0, 0, 0, 0, 0, 0, 1, 12, 36, 35, 10, 0, 0, 0, 0, 0, 0, 0, 0, 1, 13, 42, 44
Offset: 0
Rows start:
1;
1, 1;
1, 2, 1;
1, 3, 2, 0;
1, 4, 4, 1, 0;
1, 5, 6, 2, 0, 0;
...
Comments