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 21-28 of 28 results.

A007998 Number of primitive solutions to x + y + z = 0 (mod n).

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 5, 7, 8, 9, 9, 15, 11, 17, 17, 18, 18, 31, 21, 33, 30, 33, 30, 53, 34, 48, 46, 53, 43, 87, 47, 73, 69, 79, 66, 110, 65, 100, 94, 111, 80, 159, 86, 135, 130, 135, 108, 202, 116, 181, 153, 175, 133, 255, 150, 212
Offset: 1

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Keywords

Comments

I can't reconstruct this. Possibly an erroneous version of A003050 or A048259. - N. J. A. Sloane

Crossrefs

Cf. A007997.

A032195 Number of necklaces with 10 black beads and n-10 white beads.

Original entry on oeis.org

1, 1, 6, 22, 73, 201, 504, 1144, 2438, 4862, 9252, 16796, 29414, 49742, 81752, 130752, 204347, 312455, 468754, 690690, 1001603, 1430715, 2016144, 2804880, 3856892, 5245128, 7060984, 9414328, 12440668, 16301164
Offset: 10

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The g.f. is Z(C_10,x)/x^10, the 10-variate cycle index polynomial for the cyclic group C_10, with substitution x[i]->1/(1-x^i), i=1,...,10. By Polya enumeration, a(n+10) is the number of cyclically inequivalent 10-necklaces whose 10 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_10,x). See the comment in A032191 on the equivalence of this problem with the one given in the `Name' line. - Wolfdieter Lang, Feb 15 2005

Crossrefs

Programs

  • Mathematica
    k = 10; Table[Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n, {n, k, 30}] (* Robert A. Russell, Sep 27 2004 *)

Formula

"CIK[ 10 ]" (necklace, indistinct, unlabeled, 10 parts) transform of 1, 1, 1, 1...
G.f.: (x^10)*(1-3*x+4*x^2+12*x^3-8*x^4-x^5+31*x^6-4*x^8+16*x^9 +11*x^10 +3*x^11+8*x^12+4*x^13+4*x^14+x^15+x^16) /((1-x)^4*(1-x^2)^4 *(1-x^5)*(1-x^10)).
G.f.: (1/10)*x^10*(1/(1 - x)^10 + 1/(1 - x^2)^5 + 4/(1 - x^5)^2 + 4/(1 - x^10)^1). - Herbert Kociemba, Oct 22 2016

A032196 Number of necklaces with 11 black beads and n-11 white beads.

Original entry on oeis.org

1, 1, 6, 26, 91, 273, 728, 1768, 3978, 8398, 16796, 32066, 58786, 104006, 178296, 297160, 482885, 766935, 1193010, 1820910, 2731365, 4032015, 5864750, 8414640, 11920740, 16689036, 23107896, 31666376, 42975796
Offset: 11

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The g.f. is Z(C_11,x)/x^11, the 11-variate cycle index polynomial for the cyclic group C_11, with substitution x[i]->1/(1-x^i), i=1..11. By Polya enumeration, a(n+11) is the number of cyclically inequivalent 11-necklaces whose 11 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_11,x). See the comment in A032191 on the equivalence of this problem with the one given in the `Name' line. - Wolfdieter Lang, Feb 15 2005

Crossrefs

Programs

  • Mathematica
    k = 11; Table[Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n, {n, k, 30}] (* Robert A. Russell, Sep 27 2004 *)
    DeleteCases[CoefficientList[Series[(x^11) (1 - 9 x + 41 x^2 - 109 x^3 + 191 x^4 - 229 x^5 + 191 x^6 - 109 x^7 + 41 x^8 - 9 x^9 + x^10)/((1 - x)^10 (1 - x^11)), {x, 0, 39}], x], 0] (* Michael De Vlieger, Oct 10 2016 *)

Formula

"CIK[ 11 ]" (necklace, indistinct, unlabeled, 11 parts) transform of 1, 1, 1, 1...
G.f.: (x^11) * (1 - 9*x + 41*x^2 - 109*x^3 + 191*x^4 - 229*x^5 + 191*x^6 - 109*x^7 + 41*x^8 - 9*x^9 + x^10) / ((1-x)^10 * (1-x^11)).
a(n) = ceiling(binomial(n, 11)/n) (conjecture Wolfdieter Lang).
From Herbert Kociemba, Oct 11 2016: (Start)
This conjecture indeed is true.
Sketch of proof:
There are binomial(n,11) ways to place the 11 black beads in the necklace with n beads. If n is not divisible by 11 there are no necklaces with a rotational symmetry. So exactly n necklaces are equivalent up to rotation and there are binomial(n,11)/n = ceiling(binomial(n,11)/n) equivalence classes.
If n is divisible by 11 the only way to get a necklace with rotational symmetry is to space out the 11 black beads evenly. There are n/11 ways to do this and all ways are equivalent up to rotation. So there are binomial(n,11) - n/11 unsymmetric necklaces which give binomial(n,11)/n - 1/11 equivalence classes. If we add the single symmetric equivalence class we get Binomial(n,11)/n - 1/11 + 1 which also is ceiling(binomial(n,11)/n). (End)
G.f.: (10/(1 - x^11) + 1/(1 - x)^11)*x^11/11. - Herbert Kociemba, Oct 16 2016

A032197 Number of necklaces with 12 black beads and n-12 white beads.

Original entry on oeis.org

1, 1, 7, 31, 116, 364, 1038, 2652, 6310, 14000, 29414, 58786, 112720, 208012, 371516, 643856, 1086601, 1789515, 2883289, 4552275, 7056280, 10752060, 16128424, 23841480, 34769374, 50067108, 71250060, 100276894, 139672312
Offset: 12

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The g.f. is Z(C_12,x)/x^12, the 12-variate cycle index polynomial for the cyclic group C_12, with substitution x[i]->1/(1-x^i), i=1,...,12. Therefore by Polya enumeration a(n+12) is the number of cyclically inequivalent 12-necklaces whose 12 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_12,x). See the comment in A032191 on the equivalence of this problem with the one given in the `Name' line. - Wolfdieter Lang, Feb 15 2005

Crossrefs

Programs

  • Mathematica
    k = 12; Table[Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n, {n, k, 30}] (* Robert A. Russell, Sep 27 2004 *)

Formula

"CIK[ 12 ]" (necklace, indistinct, unlabeled, 12 parts) transform of 1, 1, 1, 1...
G.f.: (x^12)*(1-3*x+7*x^2+9*x^3+18*x^4+38*x^5+72*x^6+92*x^7+168*x^8+160*x^9+238*x^10+230*x^11+296*x^12+234*x^13+330*x^14+248*x^15+284*x^16+238*x^17+230*x^18+166*x^19+172*x^20+78*x^21+80*x^22+38*x^23+21*x^24+7*x^25+3*x^26+x^27) /((1+x)*(1-x)*(1-x^2)*(1-x^3)*(1-x)^5*(1+x+x^2)*(1-x^4)^2*(1-x^6)*(1-x^12)). - Wolfdieter Lang, Feb 15 2005 (see comment)
G.f.: 1/12 x^12 ((1 - x)^-12 + (1 - x^2)^-6 + 2 (1 - x^3)^-4 + 2 (1 - x^4)^-3 + 2 (1 - x^6)^-2 + 4 (1 - x^12)^-1). - Herbert Kociemba, Oct 22 2016

A058936 Decomposition of Stirling's S(n,2) based on associated numeric partitions.

Original entry on oeis.org

0, 1, 3, 8, 3, 30, 20, 144, 90, 40, 840, 504, 420, 5760, 3360, 2688, 1260, 45360, 25920, 20160, 18144, 403200, 226800, 172800, 151200, 72576, 3991680, 2217600, 1663200, 1425600, 1330560, 43545600, 23950080, 17740800, 14968800, 13685760, 6652800, 518918400
Offset: 1

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Author

Alford Arnold, Jan 11 2001

Keywords

Comments

These values also appear in a wider context when counting elements of finite groups by cycle structure. For example, the alternating group on four symbols has 12 elements; eight associated with the partition 3+1, three associated with 2+2 and the identity associated with 1+1+1+1. The cross-referenced sequences are all associated with similar numeric partitions and "M2" weights.

Examples

			Triangle begins:
  0;
  1;
  3;
  8, 3;
  30, 20;
  144, 90, 40;
  840, 504, 420;
  ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 831.

Crossrefs

Formula

From Sean A. Irvine, Sep 05 2022: (Start)
T(1,1) = 0.
T(n,k) = n! / (k * (n-k)) for 1 <= k < n/2.
T(2n,n) = (2*n)! / (2*n^2).
(End)

Extensions

More terms from Sean A. Irvine, Sep 05 2022

A195798 Number of triangular n X n X n 0..1 arrays with all rows and diagonals having the same length having the same sum, with corners zero.

Original entry on oeis.org

1, 1, 2, 8, 16, 64, 1184, 5300, 130324, 14748808, 421963232, 54990266540
Offset: 1

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Author

R. H. Hardin, Sep 23 2011

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Comments

From Pontus von Brömssen, Sep 08 2022: (Start)
For n <= 6, all solutions are rotationally symmetric, which implies that a(n) = 2^(A007997(n+4)-1). But for n >= 7 there exist asymmetric solutions like
0
0 0
0 0 1
0 1 0 0
1 0 0 0 0
0 0 0 0 1 0
0 0 0 1 0 0 0.
(End)

Examples

			Some solutions for n=5
......0..........0..........0..........0..........0..........0..........0
.....0.0........0.0........0.0........0.1........0.1........0.1........0.1
....0.1.0......1.0.1......1.1.1......0.0.0......0.1.0......1.0.1......1.1.1
...0.1.1.0....0.0.0.0....0.1.1.0....1.0.0.0....1.1.1.0....1.0.0.0....1.1.1.0
..0.0.0.0.0..0.0.1.0.0..0.0.1.0.0..0.0.0.1.0..0.0.0.1.0..0.0.1.1.0..0.0.1.1.0
		

Crossrefs

Column 1 of A195805.
Cf. A007997.

A332050 Number of ways to arrange Palago tiles in a triangle of side length n, up to rotation, reflection, and swapping colors.

Original entry on oeis.org

1, 1, 7, 129, 9882, 2391930, 1743402771, 3812799008214, 25015772571200361, 492385451093553791610, 29074868501520453489499806, 5150525730438768829942800034449, 2737200544710109691113626131721984885, 4363981784043856212945753449232929426200329
Offset: 0

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Author

Peter Kagey, Feb 06 2020

Keywords

Comments

A Palago tile is a hexagonal tile with four regions of alternating colors. See links for illustrations.

Crossrefs

Programs

  • Mathematica
    a[n_] = (3^Binomial[n + 1, 2] +
        3*3^((Binomial[n + 1, 2] - Ceiling[n/2])/2) +
        If[Mod[n, 3] == 1, 0, 2*3^(Binomial[n + 1, 2]/3)])/6

Formula

a(n) = (3^A000217(n) + 3*3^A002620(n) + 2*3^A007997(n+4))/6 if n = 1 (mod 3), and
a(n) = (3^A000217(n) + 3*3^A002620(n))/6 otherwise.

A349216 Number of ternary triples (u,v,w) with 1 <= u < v < w <= n.

Original entry on oeis.org

0, 0, 1, 2, 4, 8, 13, 20, 30, 40, 53, 70, 88, 110, 137, 166, 200, 240, 281, 328, 382, 438, 501, 572, 646, 728, 819, 910, 1010, 1120, 1233, 1356, 1490, 1628, 1777, 1938, 2100, 2274, 2461, 2652, 2856, 3074, 3297, 3534, 3786, 4040, 4309, 4594, 4884, 5190, 5513, 5842, 6188, 6552, 6917
Offset: 1

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Author

Peter J. Dukes, Nov 10 2021

Keywords

Comments

A triple of integers (u,v,w) is a ternary triple if in the ternary expansions of u,v,w, all three disagree at the least significant position at which any two disagree.
Equivalently, (u,v,w) is a ternary triple if the highest power of three dividing 2w-u-v is greater than the highest power of three dividing gcd(w-u,w-v).

Examples

			For n = 7 the 13 ternary triples are (1, 2, 3), (2, 3, 4), (1, 3, 5), (3, 4, 5), (1, 2, 6), (2, 4, 6), (1, 5, 6), (4, 5, 6), (2, 3, 7), (1, 4, 7), (3, 5, 7), (2, 6, 7), (5, 6, 7).
		

Crossrefs

Programs

  • Mathematica
    Array[Sum[Sum[Sum[Boole[IntegerExponent[w + w - u - v, 3] > IntegerExponent[GCD[w - u, w - v], 3]], {u, (v - 1)}], {v, 2, (w - 1)}], {w, 3, #}] &, 55] (* Michael De Vlieger, Feb 15 2022 *)
  • PARI
    A349216(n) = sum(w=3,n,sum(v=2,(w-1),sum(u=1,(v-1),valuation(w+w-u-v,3) > valuation(gcd(w-u,w-v),3)))); \\ Antti Karttunen, Nov 13 2021
    
  • SageMath
    def a(n):
        t=3^ceil(log(n,3))
        counter=0
        for w in range(n):
            for v in range(w):
                for u in range(v):
                    if min(gcd(w-u,3^t),gcd(w-v,3^t))
    				
Previous Showing 21-28 of 28 results.