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 11-20 of 20 results.

A115098 a(0) = 2, a(n) = 3*a(n-1) - 3.

Original entry on oeis.org

2, 3, 6, 15, 42, 123, 366, 1095, 3282, 9843, 29526, 88575, 265722, 797163, 2391486, 7174455, 21523362, 64570083, 193710246, 581130735, 1743392202, 5230176603, 15690529806, 47071589415, 141214768242, 423644304723, 1270932914166
Offset: 0

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Author

Miklos Kristof, Mar 02 2006

Keywords

Comments

Also the domination and connected domination number of the (n+2)-Dorogovtsev-Goltsev-Mendes graph, where the convention DGM(0) = P_2 is used. - Eric W. Weisstein, Jan 14 2024 and Mar 13 2025

Examples

			a(4) = (3^4 + 3)/2 = 84/2 = 42 = 3*a(3) - 3 = 3*15 - 3.
		

Crossrefs

Cf. A067771.

Programs

Formula

a(n) = (3^n + 3)/2.
a(n) = A067771(n-1), n > 0. - R. J. Mathar, Aug 11 2008
G.f.: (2-5*x)/((1-x)*(1-3*x)). - Vincenzo Librandi, Sep 13 2014

A132753 a(n) = 2^(n+1) - n + 1.

Original entry on oeis.org

3, 4, 7, 14, 29, 60, 123, 250, 505, 1016, 2039, 4086, 8181, 16372, 32755, 65522, 131057, 262128, 524271, 1048558, 2097133, 4194284, 8388587, 16777194, 33554409, 67108840, 134217703, 268435430, 536870885, 1073741796, 2147483619
Offset: 0

Views

Author

Gary W. Adamson, Aug 28 2007

Keywords

Comments

Apart from a(0): Row sums of triangle A132752 (old name).
Apart from a(0): Binomial transform of [1, 3, 0, 4, 0, 4, 0, 4, ...].

Examples

			a(3) = 14 = sum of row 3 terms of triangle A132752: (3 + 5 + 5 + 1).
a(3) = 14 = (1, 3, 3, 1) dot (1, 3, 0, 4) = (1 + 9 + 0 + 4).
		

Crossrefs

Programs

  • Magma
    [2^(n+1) -n+1: n in [0..40]]; // G. C. Greubel, Feb 16 2021
  • Maple
    A132753:= n-> 2^(n+1) -n+1; seq(A132753(n), n=0..40) # G. C. Greubel, Feb 16 2021
  • Mathematica
    Table[2^(n+1) -n+1, {n, 0, 30}] (* Bruno Berselli, Aug 31 2013 *)
  • PARI
    a(n)=2^(n+1)-n+1
    
  • PARI
    Vec( (3-8*x+6*x^2)/((1-x)^2*(1-2*x)) + O(x^40)) \\ Colin Barker, Mar 14 2014
    
  • Sage
    [2^(n+1) -n+1 for n in (0..40)] # G. C. Greubel, Feb 16 2021
    

Formula

From Colin Barker, Mar 14 2014: (Start)
a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3).
G.f.: (3 - 8*x + 6*x^2)/((1-x)^2 * (1-2*x)). (End)
E.g.f.: (1-x)*exp(x) + 2*exp(2*x). - G. C. Greubel, Feb 16 2021

Extensions

More terms Vladimir Joseph Stephan Orlovsky, Dec 25 2008
Changed first member, and better name from Ralf Stephan, Aug 31 2013

A032125 "BIK" (reversible, indistinct, unlabeled) transform of 3,3,3,3...

Original entry on oeis.org

3, 9, 30, 108, 408, 1584, 6240, 24768, 98688, 393984, 1574400, 6294528, 25171968, 100675584, 402677760, 1610661888, 6442549248, 25770000384, 103079608320, 412317646848, 1649269014528, 6597072912384, 26388285358080, 105553128849408
Offset: 1

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Author

Keywords

Comments

Number of solutions (x,y,z) to x+y+z = 2^n, x>=0, y>=0, z>=0, gcd(x,y,z)=1. - Vladeta Jovovic, Dec 22 2002

Crossrefs

a(n) = A048240(2^n).

Programs

  • Mathematica
    Table[3*2^(n-2)(2^(n-1)+1),{n,30}] (* or *) LinearRecurrence[{6,-8},{3,9},30] (* Harvey P. Dale, Jan 01 2012 *)
    RecurrenceTable[{a[0]== 3, a[1]== 9, a[n]== 6*a[n-1]  - 8*a[n-2]}, a, {n,50}] (* G. C. Greubel, Aug 22 2015 *)

Formula

a(n) = 3*2^(n-2)*(2^(n-1)+1). - Vladeta Jovovic, Dec 22 2002
Binomial transform of A067771 (if the offset is changed to 0). - Carl Najafi, Sep 09 2011
G.f. -3*x*(-1+3*x) / ( (4*x-1)*(2*x-1) ). a(n)=3*A007582(n-1). - R. J. Mathar, Sep 11 2011
a(1)=3, a(2)=9, a(n) = 6*a(n-1)-8*a(n-2). [Harvey P. Dale, Jan 01 2012]
E.g.f.: (3/8)*(exp(4*x) + 2*exp(2*x) - 3). - G. C. Greubel, Aug 22 2015

A289022 Wiener index of the n-Apollonian network.

Original entry on oeis.org

6, 27, 204, 1941, 19572, 198567, 1999056, 19931337, 196939572, 1930784091, 18802964760, 182062831005, 1754100012108, 16826739416271, 160799296563312, 1531421717572401, 14540848734272388, 137690120683444995, 1300613432805623496, 12258142039717884549
Offset: 1

Views

Author

Andrew Howroyd, Sep 02 2017

Keywords

Crossrefs

Cf. A067771 (edges), A192792, A289521, A289722.

Programs

  • Mathematica
    (* Start from Eric W. Weisstein, Sep 07 2017 *)
    Table[(6655 + 31 (-1)^n 2^(n + 2) + 5 3^(1 + 2 n) (24 + 11 n) + 3^(n + 1) (1197 + 55 n) + 5 2^(5 + n/2) Cos[n Pi/2] - 155 2^((3 + n)/2) Sin[n Pi/2])/3630, {n, 20}]
    LinearRecurrence[{23, -174, 448, -29, -1221, 2088, -4050, 2916}, {6, 27, 204, 1941, 19572, 198567, 1999056, 19931337}, 20]
    CoefficientList[Series[(6 - 111 x + 627 x^2 - 741 x^3 - 1497 x^4 + 2862 x^5 - 5670 x^6 + 8748 x^7)/((1 - x) (1 - 3 x)^2 (1 - 9 x)^2 (1 + 2 x) (1 + 2 x^2)), {x, 0, 20}], x]
    (* End *)
  • PARI
    R(dp, peq, p1, p2, x) = {[3*(dp - x + peq^2 + (2+7*x)*p1^2 + (7+2*x)*p2^2 + (4+2*x)*peq*p1 + 6*peq*p2 + 2*(4+5*x)*p1*p2 + x*(peq+3*p1+3*p2)), x*(1+3*p1), 2*(p1+p2), peq+p2]}
    A(n,x) = {my(v=[6*x,x,0,0,x]); for(i=2, n, v=R(v[1],v[2],v[3],v[4],x)); v[1]}
    Wiener(dp)=sum(i=1,poldegree(dp),i*polcoeff(dp,i));
    a(n) = Wiener(A(n,x));

Formula

a(n) = Sum_{k=1..1+floor(2*n/3)} k*A289722(n,k).
a(n) = 23*a(n-1) - 174*a(n-2) + 448*a(n-3) - 29*a(n-4) - 1221*a(n-5) + 2088*a(n-6) - 4050*a(n-7) + 2916*a(n-8).
G.f.: x*(6 - 111*x + 627*x^2 - 741*x^3 - 1497*x^4 + 2862*x^5 - 5670*x^6 + 8748*x^7)/((1 - x)*(1 - 3*x)^2*(1 - 9*x)^2*(1 + 2*x)*(1 + 2*x^2)).

A281699 Sierpinski stellated octahedron numbers: a(n) = 2*(-3*2^(n+1) + 2^(2n+3) + 5).

Original entry on oeis.org

14, 50, 218, 938, 3914, 16010, 64778, 260618, 1045514, 4188170, 16764938, 67084298, 268386314, 1073643530, 4294770698, 17179475978, 68718690314, 274876334090, 1099508482058, 4398040219658, 17592173461514, 70368719011850, 281474926379018, 1125899806179338, 4503599426043914, 18014398106828810
Offset: 0

Views

Author

Steven Beard, Jan 27 2017

Keywords

Comments

Stella octangula with Sierpinski recursion.

Crossrefs

Programs

  • Mathematica
    Table[8 (2^(2 n + 1) + 2) - 6 (2^(n + 1) + 1), {n, 0, 25}] (* or *)
    LinearRecurrence[{7, -14, 8}, {14, 50, 218}, 26] (* or *)
    CoefficientList[Series[2 (7 - 24 x + 32 x^2)/((1 - x) (1 - 2 x) (1 - 4 x)), {x, 0, 25}], x] (* Michael De Vlieger, Jan 28 2017 *)
  • PARI
    Vec(2*(7 - 24*x + 32*x^2) / ((1 - x)*(1 - 2*x)*(1 - 4*x)) + O(x^30)) \\ Colin Barker, Jan 28 2017
    
  • PARI
    a(n) = 16*4^n - 12*2^n + 10 \\ Charles R Greathouse IV, Jan 29 2017

Formula

a(n) = 8*(2^(2*n+1)+2) - 6*(2^(n+1)+1).
From Colin Barker, Jan 28 2017: (Start)
a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3) for n>2.
G.f.: 2*(7 - 24*x + 32*x^2) / ((1 - x)*(1 - 2*x)*(1 - 4*x)).
(End)

A290396 a(n) = 3*2^n + 3*4^n + 6^(n+1) + 1.

Original entry on oeis.org

13, 55, 277, 1513, 8593, 49825, 292417, 1729153, 10275073, 61254145, 365945857, 2189371393, 13111037953, 78565515265, 470990340097, 2824331231233, 16939544543233, 101611496669185, 609565899227137, 3656983075356673, 21940249178406913, 131634897988091905, 789782999624318977
Offset: 0

Views

Author

Steven Beard, Jul 29 2017

Keywords

Comments

One of several types of cuboctahedra with Sierpinski recursion. This Sierpinski cuboctahedron is a truncation of the six corners of the Sierpinski octahedron (A279512). The resulting Sierpinski cuboctahedron contains six square pyramids joined along the edges thus it can also be constructed by joining Sierpinski square pyramids (A281698).
Each face of the Sierpinski octahedron is a Sierpinski sieve with all triangular spaces completely formed. The triangles opening within the sieve, on each face, become tetrahedral excavations (cf. A067771). The third term above, for example, begins with an octahedron of 489 counters from A005900. From each of the eight faces a tetrahedron of four counters is removed. Thus the corresponding Sierpinski octahedron has 457 counters at this stage of recursion. Next, to construct the Sierpinski cuboctahedron, the truncation of the six corners is 30 counters each, thus 457 minus 180 equals 277, the third term of the Sierpinski cuboctahedral sequence shown above. For the next term in the sequence, two different sizes of tetrahedra are removed (and not only seen on the outer eight faces but also on the newly opened internal faces, now excavated in like manner.
To arrive at the next term, 1513, there are eight tetrahedra of 56 counters AND forty-eight tetrahedra of 4 counters each excavated from the octahedron, and then the six corners of 188 counters each are truncated thus: (56*8) + (4*48) = 640 total counters removed from the octahedral number 3281 (in A005900) to yield the Sierpinski octahedral number 2641 (in A279512, note that for subsequent terms the number of different size excavations of the octahedron increases, thus the next term would require three different sized excavations for the associated octahedron: 8 tetrahedra of 560 counters, 48 tetrahedra of 56 counters, and 288 tetrahedra of 4 counters).
Next truncate the six corners of 188 counters each: (6*188) = 1128. Sierpinski octahedron 2641 - 1128 = 1513, the fourth term in the Sierpinski cuboctahedral sequence above. The square pyramid of 188 counters truncated from each corner is found in A281698 by, in this case, taking the fourth term of 269 and subtracting its base of 81 counters.
The base of the square pyramid that is not counted is actually the square face of the remaining cuboctahedron. It is therefore easier to construct this Sierpinski cuboctahedral geometry by taking the six square pyramids of 269 counters each, subtracting 5 for the overlapping apices and then subtracting 96 to account for the edges where the square pyramids are overlapped: 6*269 = 1614 - 5 = 1609, - 96 = 1513, the fourth term shown in the Sierpinski cuboctahedral sequence above.

Crossrefs

Programs

  • Mathematica
    Table[3*2^n + 3*4^n + 6^(n + 1) + 1, {n, 0, 22}] (* Michael De Vlieger, Jul 29 2017 *)
  • PARI
    Vec((13 - 114*x + 290*x^2 - 204*x^3) / ((1 - x)*(1 - 2*x)*(1 - 4*x)*(1 - 6*x)) + O(x^30)) \\ Colin Barker, Jul 29 2017
    
  • PARI
    a(n) = 3*2^n + 3*4^n + 6^(n+1) + 1 \\ Charles R Greathouse IV, Nov 03 2017

Formula

a(n) = 3*2^n + 3*4^n + 6^(n+1) + 1.
From Colin Barker, Jul 29 2017: (Start)
G.f.: (13 - 114*x + 290*x^2 - 204*x^3) / ((1 - x)*(1 - 2*x)*(1 - 4*x)*(1 - 6*x)).
a(n) = 13*a(n-1) - 56*a(n-2) + 92*a(n-3) - 48*a(n-4) for n>3.
(End)

A370933 Number of pairs of antipodal vertices in the level n>1 Sierpiński triangle graph.

Original entry on oeis.org

6, 15, 42, 132, 456, 1680, 6432, 25152, 99456, 395520, 1577472, 6300672, 25184256, 100700160, 402726912, 1610760192, 6442745856, 25770393600, 103080394752, 412319219712, 1649272160256, 6597079203840, 26388297940992, 105553154015232, 422212540563456, 1688850011258880, 6755399743045632
Offset: 2

Views

Author

Allan Bickle, Aug 07 2024

Keywords

Comments

A level 1 Sierpiński triangle graph is a triangle. Level n+1 is formed from three copies of level n by identifying pairs of corner vertices of each pair of triangles.
Antipodal vertices are a pair of vertices at maximum distance in a graph. The diameter of the level n Sierpiński triangle graph is 2^(n-1).

Examples

			3 example graphs:                        o
                                        / \
                                       o---o
                                      / \ / \
                        o            o---o---o
                       / \          / \     / \
             o        o---o        o---o   o---o
            / \      / \ / \      / \ / \ / \ / \
           o---o    o---o---o    o---o---o---o---o
Graph:      S_1        S_2              S_3
For S_2, there are 3 pairs of corners and 3 pairs of a corner and a middle vertex, so a(2) = 6.
		

Crossrefs

Cf. A007283, A029858, A067771, A193277, A233774, A233775, A246959, A298202 (Sierpiński triangle graphs).
Cf. A375256 (antipodal pairs in Hanoi graphs).

Programs

  • Mathematica
    A370933[n_] := 3*2^(n - 3)*(2^(n - 2) + 3);
    Array[A370933, 30, 2] (* or *)
    LinearRecurrence[{6, -8}, {6, 15}, 30] (* Paolo Xausa, Sep 23 2024 *)
  • PARI
    a(n) = 3*2^(n-3)*(2^(n-2)+3); \\ Michel Marcus, Aug 08 2024

Formula

a(n) = 3*2^(n-3)*(2^(n-2)+3).
a(n) = 3*A257273(n-2).
a(n) = A375256(n-1) + 3.

Extensions

More terms from Michel Marcus, Aug 08 2024

A112027 a(1)=1; then successively add 1, divide by 2, add 2 and then total up the last 4 terms.

Original entry on oeis.org

1, 2, 1, 3, 7, 8, 4, 6, 25, 26, 13, 15, 79, 80, 40, 42, 241, 242, 121, 123, 727, 728, 364, 366, 2185, 2186, 1093, 1095, 6559, 6560, 3280, 3282, 19681, 19682, 9841, 9843, 59047, 59048, 29524, 29526, 177145, 177146, 88573, 88575, 531439, 531440, 265720, 265722, 1594321
Offset: 1

Views

Author

N. J. A. Sloane, Nov 24 2005

Keywords

References

  • Joshua Zucker, Posting to Seq Fan mailing list, Nov 24 2005

Crossrefs

Quadrusections: A058481, A024023, A003462, A067771.

Programs

  • Maple
    a[1]:=1; k:=1; for n from 1 to 16 do k:=k+1; a[k]:=a[k-1]+1; k:=k+1; a[k]:=a[k-1]/2; k:=k+1; a[k]:=a[k-1]+2; k:=k+1; a[k]:=a[k-1]+a[k-2]+a[k-3]+a[k-4]; od;
  • Mathematica
    LinearRecurrence[{0, 0, 0, 4, 0, 0, 0, -3}, {1, 2, 1, 3, 7, 8, 4, 6}, 50] (* Paolo Xausa, May 20 2024 *)

Formula

G.f.: -x*(6*x^7-3*x^4-3*x^3-x^2-2*x-1) / ((x-1)*(x+1)*(x^2+1)*(3*x^4-1)). - Colin Barker, Jul 28 2013

Extensions

Definition found by Franklin T. Adams-Watters, Feb 01 2006
More terms from N. J. A. Sloane, Feb 22 2006

A205248 Number of (n+1) X 2 0..1 arrays with the number of clockwise edge increases in every 2 X 2 subblock the same.

Original entry on oeis.org

16, 40, 112, 328, 976, 2920, 8752, 26248, 78736, 236200, 708592, 2125768, 6377296, 19131880, 57395632, 172186888, 516560656, 1549681960, 4649045872, 13947137608, 41841412816, 125524238440, 376572715312, 1129718145928, 3389154437776
Offset: 1

Views

Author

R. H. Hardin, Jan 24 2012

Keywords

Comments

Also, the number of cliques in the n-Apollonian network. Cliques in this graph have a maximum size of 4. - Andrew Howroyd, Sep 02 2017

Examples

			Some solutions for n=4:
  1  0    0  1    1  1    0  1    1  1    1  1    1  0    1  0    1  1    1  1
  0  1    0  0    1  1    0  1    0  1    0  1    0  1    0  0    1  1    1  1
  1  0    1  1    1  1    0  1    0  1    0  0    1  0    0  1    1  1    1  1
  0  1    1  0    1  1    0  0    0  1    1  0    0  1    1  1    1  1    1  1
  1  0    0  0    1  1    0  1    1  1    1  1    1  0    0  1    1  1    1  1
		

Crossrefs

Column 1 of A205255.

Programs

Formula

a(n) = 4*a(n-1) - 3*a(n-2).
From Andrew Howroyd, Sep 02 2017: (Start)
a(n) = 4*(3^n + 1).
G.f.: 8*x*(2 - 3*x)/((1 - x)*(1 - 3*x)).
a(n) = 8*A007051(n).
a(n) = 1 + A289521(n) + A067771(n) + A003462(n+1) + A003462(n).
(End)

A289999 Sierpinski cuboctahedral numbers: a(n) = 16*4^n - 12*2^n + 9.

Original entry on oeis.org

13, 49, 217, 937, 3913, 16009, 64777, 260617, 1045513, 4188169, 16764937, 67084297, 268386313, 1073643529, 4294770697, 17179475977, 68718690313, 274876334089, 1099508482057, 4398040219657, 17592173461513, 70368719011849, 281474926379017, 1125899806179337, 4503599426043913, 18014398106828809
Offset: 0

Views

Author

Steven Beard, Sep 03 2017

Keywords

Comments

Sierpinski cuboctahedron constructed by joining eight Sierpinski tetrahedra of sequence 4, 10, 34, 130, 514, 2050, 8194... (4^n*2)+2 (the double of A052539). This sequence is also Sierpinski recursion for the octahemioctahedron A274974.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(13 - 42 x + 56 x^2)/((1 - x) (1 - 2 x) (1 - 4 x)), {x, 0, 25}], x] (* Michael De Vlieger, Sep 03 2017 *)
    Table[16*4^n-12*2^n+9,{n,0,30}] (* or *) LinearRecurrence[{7,-14,8},{13,49,217},30] (* Harvey P. Dale, Dec 31 2018 *)
  • PARI
    Vec((13 - 42*x + 56*x^2) / ((1 - x)*(1 - 2*x)*(1 - 4*x)) + O(x^30)) \\ Colin Barker, Sep 03 2017
    
  • PARI
    a(n) = 16*4^n - 12*2^n + 9 \\ Charles R Greathouse IV, Nov 03 2017

Formula

a(n) = -3*2^(n + 2) + 2^(2n + 4) + 9.
From Colin Barker, Sep 03 2017: (Start)
G.f.: (13 - 42*x + 56*x^2) / ((1 - x)*(1 - 2*x)*(1 - 4*x)).
a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3) for n>2.
(End)
Previous Showing 11-20 of 20 results.