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 71-79 of 79 results.

A177730 Expansion of (6*x + 1) / ((x - 1)*(2*x - 1)*(4*x - 1)*(8*x - 1)).

Original entry on oeis.org

1, 21, 245, 2325, 20181, 168021, 1370965, 11075925, 89042261, 714081621, 5719635285, 45785027925, 366392038741, 2931583636821, 23454458533205, 187642826282325, 1501171242849621, 12009484474209621, 96076333921424725, 768612503886583125, 6148907361161794901
Offset: 0

Views

Author

Roger L. Bagula, May 12 2010

Keywords

Crossrefs

Programs

  • GAP
    a := List([0..200],n->((2^(n+1)-1)^2*(2^(n+2)-1))/3); # Muniru A Asiru, Jan 27 2018
    
  • Maple
    a := seq(((2^(n+1)-1)^2 * (2^(n+2)-1))/3, n = 0..200); # Muniru A Asiru, Jan 27 2018
  • Mathematica
    CoefficientList[Series[(6x+1)/((x-1)(2x-1)(4x-1)(8x-1)),{x,0,30}],x] (* or *) LinearRecurrence[{15,-70,120,-64},{1,21,245,2325},30] (* Harvey P. Dale, Jul 16 2018 *)
  • PARI
    Vec((6*x + 1) / ((x - 1)*(2*x - 1)*(4*x - 1)*(8*x - 1)) + O(x^30)) \\ Colin Barker, Jan 27 2018

Formula

From Colin Barker, Jan 27 2018: (Start)
a(n) = ((2^(n+1)-1)^2 * (2^(n+2)-1)) / 3.
a(n) = 15*a(n-1) - 70*a(n-2) + 120*a(n-3) - 64*a(n-4) for n>3.
(End)

Extensions

Heavily edited, with the blessing of Michel Marcus and Joerg Arndt, by Colin Barker, Jan 27 2018

A296306 a(n) = A001157(n)/A050999(n).

Original entry on oeis.org

1, 5, 1, 21, 1, 5, 1, 85, 1, 5, 1, 21, 1, 5, 1, 341, 1, 5, 1, 21, 1, 5, 1, 85, 1, 5, 1, 21, 1, 5, 1, 1365, 1, 5, 1, 21, 1, 5, 1, 85, 1, 5, 1, 21, 1, 5, 1, 341, 1, 5, 1, 21, 1, 5, 1, 85, 1, 5, 1, 21, 1, 5, 1, 5461, 1, 5, 1, 21, 1, 5, 1, 85, 1, 5, 1, 21, 1, 5, 1, 341, 1, 5, 1, 21, 1, 5, 1, 85, 1, 5, 1, 21, 1, 5, 1, 1365
Offset: 1

Views

Author

Ivan N. Ianakiev, Dec 10 2017

Keywords

Comments

a(n) is the sum of the second powers of the divisors of n divided by the sum of the second powers of the odd divisors of n.
Conjecture 1: For any nonnegative integer k and positive integer n, the sum of the k-th powers of the divisors of n is divisible by the sum of the k-th powers of the odd divisors of n.
Conjecture 2: Distinct terms form A002450 without A002450(0). In other words, a(2^(n-1)) = A002450(n), for n > 0.
Conjecture 3: For n > 0, the list of the first 2^n - 1 terms is palindromic.
Conjecture 4: For n > 0, the sum of the first 2^n - 1 terms equals A006095(n+1).
To prove Conjecture 1 all one needs to do is to prove that the sum of the k-th powers of the divisors of n divided by the sum of the k-th powers of the odd divisors of n equals: a) A001511(n), for k = 0, and b) ((2^k)^A001511(n) - 1)/(2^k - 1), for k > 0. - Ivan N. Ianakiev, Jan 29 2020
Conjecture 1 indeed follows from multiplicativity of sigma_k, in particular sigma_k(2^j (2m+1)) = sigma_k(2^j) sigma_k(2m+1). Conjecture 3 follows from Radcliffe's formula, since A007814 has this property. - M. F. Hasler, Jan 31 2020
a(n) is the sum of squares of powers of 2 that divide n. - Amiram Eldar, Nov 12 2020

Examples

			A001157(4) = 21 and A050999(4) = 1, therefore a(4) = A001157(4)/A050999(4) = 21.
		

Crossrefs

Programs

  • Magma
    [DivisorSigma(2,n)/&+[d^2:d in Divisors(n)|IsOdd(d)]:n in [1..100]]; // Marius A. Burtea, Jan 29 2020
    
  • Mathematica
    f[n_]:=DivisorSigma[2,n]/Total[Select[Divisors[n],OddQ]^2]; f/@Range[100]
    Table[(4^(IntegerExponent[n, 2] + 1) - 1)/3, {n, 1, 100}] (* Amiram Eldar, Nov 12 2020 *)
  • PARI
    a(n) = sigma(n, 2)/sumdiv(n, d, d^2*(d % 2)); \\ Michel Marcus, Dec 11 2017
    
  • Python
    def A296306(n): return ((1<<((n&-n).bit_length()<<1))-1)//3 # Chai Wah Wu, Jul 16 2022

Formula

a(n) = (4^(A007814(n) + 1) - 1)/3. - David Radcliffe, Dec 11 2017
Multiplicative with a(2^e) = (4^(e+1)-1)/3, and a(p^e) = 1 for odd primes p. - Amiram Eldar, Nov 12 2020
G.f.: Sum_{k>=0} 4^k * x^(2^k) / (1 - x^(2^k)). - Ilya Gutkovskiy, Dec 14 2020
Dirichlet g.f.: zeta(s)/(1-4/2^s). - Amiram Eldar, Dec 31 2022

A346465 Numbers k such that (4^k - 2)*(4^k - 1)/Clausen(2*k, 1) is not squarefree, where Clausen(n, m) = A160014(n, m).

Original entry on oeis.org

9, 11, 18, 27, 32, 36, 45, 50, 53, 54, 63, 68, 72, 74, 78, 81, 90, 95, 99, 100, 108, 116, 117, 126, 127, 135, 137, 144, 147, 150, 153, 155, 158, 162, 171, 179, 180, 182, 189, 198, 200, 204, 207, 216, 221, 225, 233, 234, 242, 243, 250, 252, 261, 263, 270, 279
Offset: 1

Views

Author

Peter Luschny, Jul 20 2021

Keywords

Comments

Also numbers k such that 6*GaussBinomial(2*k, 2, 2)/denominator(Bernoulli(2*k, 1)) is not squarefree.

Crossrefs

Programs

  • Maple
    with(NumberTheory): isa := n -> not IsSquareFree(((4^n - 2)*(4^n - 1))/
    mul(i, i = select(isprime, map(i -> i+1, Divisors(2*n))))):
    select(isa, [$(1..100)]);
  • Mathematica
    q[n_] := Product[k, {k, Select[Table[d + 1, {d, Divisors[2 n]}], PrimeQ]}];
    isA[n_] := ! SquareFreeQ[((4^n - 2) (4^n -1)) / q[n]];
    Select[Range[50],  isA]

Formula

The positive multiples of 9 form a subsequence.
k is a term if and only if A346463(k) > A007947(A346463(k)).

Extensions

More terms from Jinyuan Wang, Jul 23 2021

A372230 Triangular array read by rows. T(n,k) is the number of size k circuits in the linear matroid M[A] where A is the n X 2^n-1 matrix whose columns are the nonzero vectors in GF(2)^n, n>=2, 3<=k<=n+1.

Original entry on oeis.org

1, 7, 7, 35, 105, 168, 155, 1085, 5208, 13888, 651, 9765, 109368, 874944, 3999744, 2667, 82677, 1984248, 37039296, 507967488, 4063739904, 10795, 680085, 33732216, 1349288640, 43177236480, 1036253675520, 14737830051840
Offset: 2

Views

Author

Geoffrey Critzer, Apr 28 2024

Keywords

Comments

For n>=2 and 3<=k<=n, to construct a size k circuit of M[A]: Choose a basis b_1,b_2,...,b_{k-1} of a k-1 dimensional subspace of GF(2)^n. Append the vector b_1 + b_2 + ... + b_{k-1}.

Examples

			Triangle begins ...
    1;
    7,    7;
   35,   105,     168;
  155,  1085,    5208,    13888;
  651,  9765,  109368,   874944,   3999744;
 2667, 82677, 1984248, 37039296, 507967488, 4063739904;
...
		

References

  • J. Oxley, Matroid Theory, Oxford Graduate Texts in Mathematics, 1992, page 8.

Crossrefs

Cf. A022166, A053601, A006095, A372350 (row sums).

Programs

  • Mathematica
    nn = 8; Map[Select[#, # > 0 &] &, Table[Table[PadRight[Table[Product[(2^n - 2^i)/(2^k - 2^i), {i, 0, k - 1}], {k, 2, n}], nn], {n, 2, nn}][[All, j]]*    Table[Product[2^n - 2^i, {i, 0, n - 1}]/(n + 1)!, {n, 2, nn}][[j]], {j, 1, nn - 1}] // Transpose] // Grid

Formula

T(n,k) = A022166(n,k-1)*A053601(k-1)/k.
T(n,3) = A006095.
T(n,n+1) = A053601(n)/(n+1).

A374981 a(n) = (n^2 - 1)/6 - Sum_{t=0..n-1} [ A000217(t)/n ], where [ x ] means the fractional part of x here.

Original entry on oeis.org

0, 0, 1, 1, 3, 4, 6, 7, 11, 13, 16, 19, 24, 27, 33, 35, 42, 47, 53, 58, 67, 71, 78, 85, 95, 102, 112, 118, 128, 138, 147, 155, 170, 178, 191, 200, 213, 224, 238, 248, 263, 277, 290, 302, 322, 331, 347, 361, 380, 395, 413, 427, 445, 463, 482, 496, 519, 534, 554, 573, 594, 612, 637, 651, 678, 698
Offset: 1

Views

Author

Thomas Scheuerle, Jul 26 2024

Keywords

Comments

a(c^n) for some constant c can be expressed as a linear recurrence with constant coefficients.

Crossrefs

Programs

  • PARI
    a(n) = (n^2-1)/6-sum(k=1,n,(k*(k+1)/2)%n)/n+((n+1)%2)/2

Formula

a(n) = (n^2 - 1 - A374968(n))/6.
a(2^n) = A006095(n).
a(3^n) has the ordinary generating function: x*(1 - 2*x + 5*x^2 + 12*x^3)/(1 - 13*x + 36*x^2 + 12*x^3 - 117*x^4 + 81*x^5).

A006108 Gaussian binomial coefficient [ 2n,n ] for q=4.

Original entry on oeis.org

1, 5, 357, 376805, 6221613541, 1634141006295525, 6857430062381149327845, 460250514083576206796548772325, 494205307747746503853075131001823990245
Offset: 0

Views

Author

Keywords

References

  • J. Goldman and G.-C. Rota, The number of subspaces of a vector space, pp. 75-83 of W. T. Tutte, editor, Recent Progress in Combinatorics. Academic Press, NY, 1969.
  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration. Wiley, NY, 1983, p. 99.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • M. Sved, Gaussians and binomials, Ars. Combinatoria, 17A (1984), 325-351.

A270218 Number of active (ON, black) cells at stage 2^n-1 of the two-dimensional cellular automaton defined by "Rule 129", based on the 5-celled von Neumann neighborhood.

Original entry on oeis.org

1, 4, 28, 140, 620, 2604, 10668, 43180, 173740, 697004, 2792108, 11176620, 44722860, 178924204, 715762348, 2863180460
Offset: 0

Views

Author

Robert Price, Mar 13 2016

Keywords

Comments

Initialized with a single black (ON) cell at stage zero.
It appears Rules 385, 425, 465 and 553 also generate this sequence. - Lars Blomberg, Apr 30 2016 (It would be nice to have a proof! - N. J. A. Sloane, May 09 2016)

References

  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Crossrefs

Cf. A270217.

Programs

  • Mathematica
    CAStep[rule_,a_]:=Map[rule[[10-#]]&,ListConvolve[{{0,2,0},{2,1,2},{0,2,0}},a,2],{2}];
    code=129; stages=128;
    rule=IntegerDigits[code,2,10];
    g=2*stages+1; (* Maximum size of grid *)
    a=PadLeft[{{1}},{g,g},0,Floor[{g,g}/2]]; (* Initial ON cell on grid *)
    ca=a;
    ca=Table[ca=CAStep[rule,ca],{n,1,stages+1}];
    PrependTo[ca,a];
    (* Trim full grid to reflect growth by one cell at each stage *)
    k=(Length[ca[[1]]]+1)/2;
    ca=Table[Table[Part[ca[[n]][[j]],Range[k+1-n,k-1+n]],{j,k+1-n,k-1+n}],{n,1,k}];
    on=Map[Function[Apply[Plus,Flatten[#1]]],ca] (* Count ON cells at each stage *)
    Part[on,2^Range[0,Log[2,stages]]] (* Extract relevant terms *)

Formula

Conjectures from Colin Barker, Mar 13 2016: (Start)
a(n) = 4*(1-3*2^n+2^(1+2*n))/3.
a(n) = 7*a(n-1)-14*a(n-2)+8*a(n-3) for n>3.
G.f.: (1-3*x+14*x^2-8*x^3) / ((1-x)*(1-2*x)*(1-4*x)).
(End)
a(n) = 4*A006095(n+1) (conjectured). - Michal Stajszczak, May 20 2020

Extensions

a(8)-a(15) from Lars Blomberg, Apr 30 2016

A374839 Triangle read by rows: T(n,k) is the number of strong subtrees of height k in the complete binary tree of height n.

Original entry on oeis.org

1, 3, 1, 7, 7, 1, 15, 35, 23, 1, 31, 155, 403, 279, 1, 63, 651, 6603, 71827, 65815, 1, 127, 2667, 106299, 18394315, 4313323667, 4295033111, 1, 255, 10795, 1703451, 4709050939, 282677998234827, 18447026751295461523, 18446744078004584727, 1
Offset: 1

Views

Author

John V Siratt, Jul 21 2024

Keywords

Comments

Informally, a strong subtree is one that preserves meets, the relative level of vertices, and the number of immediate successors to non-terminal vertices.
The collection of strong subtrees with height k of some tree T is often denoted by S_k(T). T(n,k) is the cardinality of S_k(2^(
There is a general expression for T(n,k) given in terms of an auxiliary function that can be eliminated for the diagonals, with the following examples:
T(m+1,m) = 3 + Sum_{j=1..m-1} 2^(2^i).
T(m+2,m) = 7 + 3*(Sum_{j=1..m-1} 2^(2^i)) + Sum_{1<=i,j<=m-1} 2^(2^i + 2^j).
T(m+3,m) = 15 + 7*(Sum_{j=1..m-1} 2^(2^i)) + 3*(Sum_{1<=i,j<=m-1} 2^(2^i + 2^j)) + Sum_{1<=i,j,k<=m-1} 2^(2^i + 2^j + 2^k).

Examples

			Triangle begins:
    1;
    3,    1;
    7,    7,      1;
   15,   35,     23,        1;
   31,  155,    403,      279,          1;
   63,  651,   6603,    71827,      65815,          1;
  127, 2667, 106299, 18394315, 4313323667, 4295033111, 1;
  ...
Formatted as a transposed array:
T(n,k) | n=1  2  3   4    5      6           7                     8
--------------------------------------------------------------------
k=1    |   1  3  7  15   31     63         127                   255
2      |   0  1  7  35  155    651        2667                 10795
3      |   0  0  1  23  403   6603      106299               1703451
4      |   0  0  0   1  279  71827    18394315            4709050939
5      |   0  0  0   0    1  65815  4313323667       282677998234827
6      |   0  0  0   0    0      1  4295033111  18447026751295461523
7      |   0  0  0   0    0      0           1  18446744078004584727
8      |   0  0  0   0    0      0           0                     1
		

Crossrefs

Column 1 is A000225.
Column 2 appears to be A006095.

Programs

  • PARI
    T(n,k)={my(s=0); forvec(x=vector(k,i,[1,n]),s+=prod(i=1, k,  (2^(x[i] - if(i>1, x[i-1]) - 1))^(2^(i - 1))), 2); s}
    { for(n=1, 8, print(vector(n,k,T(n,k)))) } \\ Andrew Howroyd, Jul 23 2024

Formula

T(n,k) = Sum_{1 <= x_1 < ... < x_k <= n} Product_{i=1..k} (2^(x_i - x_{i-1} - 1))^(2^(i - 1)), where x_0 = 0.

A377278 Denominators in a harmonic triangle; q-analog of A126615, here q = 2.

Original entry on oeis.org

1, 3, 3, 3, 21, 7, 3, 21, 105, 15, 3, 21, 105, 465, 31, 3, 21, 105, 465, 1953, 63, 3, 21, 105, 465, 1953, 8001, 127, 3, 21, 105, 465, 1953, 8001, 32385, 255, 3, 21, 105, 465, 1953, 8001, 32385, 130305, 511, 3, 21, 105, 465, 1953, 8001, 32385, 130305, 522753, 1023
Offset: 1

Author

Werner Schulte, Oct 22 2024

Keywords

Comments

The harmonic triangle uses the terms of this sequence as denominators, numerators = 1. The inverse of the harmonic triangle has entries -2^(n-k-1) for 1<=k
Conjecture: Row sums of the harmonic triangle are A204243(n) / A005329(n).

Examples

			Triangle T(n, k) for 1 <= k <= n starts:
n\ k :  1   2    3    4     5     6      7       8       9    10
================================================================
   1 :  1
   2 :  3   3
   3 :  3  21    7
   4 :  3  21  105   15
   5 :  3  21  105  465    31
   6 :  3  21  105  465  1953    63
   7 :  3  21  105  465  1953  8001    127
   8 :  3  21  105  465  1953  8001  32385     255
   9 :  3  21  105  465  1953  8001  32385  130305     511
  10 :  3  21  105  465  1953  8001  32385  130305  522753  1023
  etc.
The harmonic triangle starts:
  [1]  1/1
  [2]  1/3   1/3
  [3]  1/3  1/21    1/7
  [4]  1/3  1/21  1/105   1/15
  [5]  1/3  1/21  1/105  1/465    1/31
  [6]  1/3  1/21  1/105  1/465  1/1953  1/63
  etc.
The inverse of the harmonic triangle starts:
  [1]    1
  [2]   -1   3
  [3]   -2  -1   7
  [4]   -4  -2  -1  15
  [5]   -8  -4  -2  -1  31
  [6]  -16  -8  -4  -2  -1  63
  etc.
		

Crossrefs

Programs

  • PARI
    T(n,k)=if(k
    				

Formula

T(n, k) = (2^k - 1) * (2^(k+1) - 1) for 1 <= k < n; T(n, n) = 2^n - 1.
Sum_{k=1..n} 2^(k-1) / T(n, k) = 1.
Product_{k=1..n} T(n, k)^((-1)^k) = 1.
Row sums are n + 4 * (2^n - 1) * (2^(n-1) - 1) / 3 = n + 4 * A006095(n).
G.f.: x*y*(1 + 2*x - 4*x*y + 4*x^2*y)/((1 - x)*(1 - x*y)(1 - 2*x*y)*(1 - 4*x*y)). - Stefano Spezia, Oct 23 2024
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