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 31-34 of 34 results.

A224997 Floor(1/f(x^(1/n))) for x = 17, where f computes the fractional part.

Original entry on oeis.org

8, 1, 32, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 16, 16, 17, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 23, 24, 24, 24, 25
Offset: 2

Views

Author

T. D. Noe, Apr 26 2013

Keywords

Crossrefs

Programs

  • Mathematica
    th = 17; t = Table[Floor[1/FractionalPart[th^(1/n)]], {n, 2, 100}]

Formula

a(n) = floor(n/log(17)-1/2) for n>7. - Andrey Zabolotskiy, Dec 01 2017

A224998 Floor(1/f(x^(1/n))) for x = Pi, where f computes the fractional part.

Original entry on oeis.org

7, 1, 2, 3, 3, 4, 5, 6, 7, 8, 9, 9, 10, 11, 12, 13, 14, 15, 16, 16, 17, 18, 19, 20, 21, 22, 23, 23, 24, 25, 26, 27, 28, 29, 30, 30, 31, 32, 33, 34, 35, 36, 37, 37, 38, 39, 40, 41, 42, 43, 44, 44, 45, 46, 47, 48, 49, 50, 51, 51, 52, 53, 54, 55, 56, 57, 58, 58
Offset: 1

Views

Author

T. D. Noe, Apr 26 2013

Keywords

Crossrefs

Programs

  • Mathematica
    th = Pi; t = Table[Floor[1/FractionalPart[th^(1/n)]], {n, 100}]

Formula

a(n) = floor(n/log(Pi)-1/2) for n>4. - Andrey Zabolotskiy, Dec 01 2017

A305319 Triangle T(n,k) read by rows: coefficients in order of decreasing exponents of characteristic polynomial P_n(t) of the matrix M(i,j) = [(i+j>n) or (i+j)=n-1], 1 <= i,j <= n.

Original entry on oeis.org

1, 1, -1, 1, -1, -1, 1, -3, 1, 1, 1, -2, -4, 1, 1, 1, -4, -1, 6, 1, -1, 1, -3, -8, 3, 9, 1, -1, 1, -5, -4, 15, 5, -11, -1, 1, 1, -4, -13, 8, 27, -3, -14, 1, 1, 1, -6, -8, 29, 15, -42, -6, 18, -1, -1, 1, -5, -19, 17, 60, -19, -63, 9, 21, -1, -1, 1, -7, -13, 49, 35, -110, -29, 93, 6, -25, -1, 1, 1, -6, -26, 31, 114, -58, -189, 45, 129, -10, -30, -1, 1
Offset: 0

Views

Author

Gheorghe Coserea, May 30 2018

Keywords

Comments

Related to conjecture from entry A047211.

Examples

			P(0) = 1;
P(1) = t - 1;
P(2) = t^2 - t - 1;
P(3) = t^3 - 3*t^2 + t + 1;
P(4) = t^4 - 2*t^3 - 4*t^2 + t + 1;
...
Triangle starts:
n\k [0]  [1]  [2]  [3]  [4]   [5]   [6]  [7] [8]  [9] [10] [11] [12]
[0]  1
[1]  1,  -1;
[2]  1,  -1,  -1;
[3]  1,  -3,   1,   1;
[4]  1,  -2,  -4,   1,   1;
[5]  1,  -4,  -1,   6,   1,   -1;
[6]  1,  -3,  -8,   3,   9,    1,   -1;
[7]  1,  -5,  -4,  15,   5,  -11,   -1,   1;
[8]  1,  -4, -13,   8,  27,   -3,  -14,   1,  1;
[9]  1,  -6,  -8,  29,  15,  -42,   -6,  18, -1,  -1;
[10] 1,  -5, -19,  17,  60,  -19,  -63,   9, 21,  -1,  -1;
[11] 1,  -7, -13,  49,  35, -110,  -29,  93,  6, -25,  -1,  1;
[12] 1,  -6, -26,  31, 114,  -58, -189,  45, 29, -10, -30, -1,  1;
...
For n=7 the n X n matrix M (dots for zeros):
  [. . . . 1 . 1]
  [. . . 1 . 1 1]
  [. . 1 . 1 1 1]
  [. 1 . 1 1 1 1]
  [1 . 1 1 1 1 1]
  [. 1 1 1 1 1 1]
  [1 1 1 1 1 1 1]
has characteristic polynomial P(7) = det(t*I-M) = t^7 - 5*t^6 - 4*t^5 + 15*t^4 + 5*t^3 - 11*t^2 - t + 1 (which is irreducible over Q: an elementary check shows that P(7)(25) = 4849680601 is a prime and 25 >= 17 = 2 + max(abs([1,-5,-4,15,5,-11,-1,1]))).
		

Crossrefs

Cf. A047211.

Programs

  • PARI
    P(n, t='t) = charpoly(matrix(n, n, i, j, (i+j > n) || (i+j)==n-1), t);
    seq(N, t='t) = {
      my(a=vector(N)); for (n=1, 4, a[n] = subst(P(n,'t), 't, t));
      for (n=5, N,
         a[n] +=  (2*t + 3*(-1)^(n%2))*a[n-1] - (t^2-4)*a[n-2];
         a[n] += -(2*t + 3*(-1)^(n%2))*a[n-3] - a[n-4]);
      a;
    };
    concat(1, concat(apply(p->Vec(p), seq(12))))
    \\ test: N=100; vector(N, n, P(n)) == seq(N)

Formula

P(n) = det(t*I - M), where M(i,j) = [(i+j>n) or (i+j)=n-1], 1 <= i,j <= n.
P(n) = (2*t + 3*(-1)^n)*P(n-1) - (t^2 - 4)*P(n-2) - (2*t + 3*(-1)^n)*P(n-3) - P(n-4).
G.f.: A(x;t) = Sum_{n>=0} P(n)*x^n = (t*x^8 + (-t^2 + t - 1)*x^7 + (-t^3 + t^2 + 2*t + 1)*x^6 + (t^4 - 2*t^3 + t^2 + 2*t)*x^5 - t^2*x^4 + (-t^3 - t^2 + 2*t)*x^3 + (-t^2 - t)*x^2 + (t - 1)*x + 1)/(x^8 + (-2*t^2 + 1)*x^6 + t^4*x^4 + (-2*t^2 + 1)*x^2 + 1).

A063093 Dimension of the space of weight 2n cusp forms for Gamma_0( 25 ).

Original entry on oeis.org

0, 5, 9, 15, 19, 25, 29, 35, 39, 45, 49, 55, 59, 65, 69, 75, 79, 85, 89, 95, 99, 105, 109, 115, 119, 125, 129, 135, 139, 145, 149, 155, 159, 165, 169, 175, 179, 185, 189, 195, 199, 205, 209, 215, 219, 225, 229, 235, 239, 245
Offset: 1

Views

Author

N. J. A. Sloane, Jul 08 2001

Keywords

Comments

If b(n) is the sequence of integers congruent to {0,3} (mod 5) and c(n) is the sequence of integers congruent to {2,4}(mod 5). Then a(n) = b(n) + c(n). Equivalently a(n) = A047218(n+1) + A047211(n). - Anthony Hernandez, Aug 16 2016

Crossrefs

Programs

  • Mathematica
    Rest@ CoefficientList[Series[x^2*(5 + 4 x + x^2)/((1 - x)^2*(1 + x)), {x, 0, 50}], x] (* Michael De Vlieger, Aug 26 2016 *)
    LinearRecurrence[{1,1,-1},{0,5,9,15},50] (* Harvey P. Dale, Apr 09 2019 *)

Formula

a(n) = 10*n - a(n-1) - 16 for n>2, with a(1)=0, a(2)=5. - Vincenzo Librandi, Aug 07 2010
From Colin Barker, Sep 26 2012: (Start)
a(n) = ((-1)^n + 10*n - 11)/2 for n>1.
a(n) = a(n-1) + a(n-2) - a(n-3) for n>3.
G.f.: x^2*(5+4*x+x^2)/((1-x)^2*(1+x)). (End)
Sum_{n>=2} (-1)^n/a(n) = sqrt(5+2*sqrt(5))*Pi/20 - 3*sqrt(5)*log(phi)/20 - log(5)/8, where phi is the golden ratio (A001622). - Amiram Eldar, Apr 15 2023
Previous Showing 31-34 of 34 results.