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.

Showing 1-3 of 3 results.

A206813 Position of 3^n in joint ranking of {2^i}, {3^j}, {5^k}.

Original entry on oeis.org

2, 6, 9, 12, 15, 19, 22, 25, 29, 31, 35, 39, 41, 45, 48, 51, 54, 58, 61, 64, 68, 71, 74, 78, 81, 84, 87, 91, 93, 97, 101, 103, 107, 110, 113, 117, 120, 123, 126, 130, 132, 136, 140, 143, 146, 149, 153, 156, 159, 163, 165, 169, 173, 175, 179, 182, 185, 188
Offset: 1

Views

Author

Clark Kimberling, Feb 17 2012

Keywords

Comments

The exponents i,j,k range through the set N of positive integers, so that the position sequences (A206812 for 2^n, A206813 for 3^n, A206814 for 5^n) partition N.

Examples

			The joint ranking begins with 2,3,4,5,8,9,16,25,27,32,64,81,125,128,243,256, so that
A205812=(1,3,5,7,10,11,14,...)
A205813=(2,6,9,12,15,...)
A205814=(4,8,13,18,23,...)

Crossrefs

Cf. A206805, A206812, A206814.

Programs

  • Mathematica
    f[1, n_] := 2^n; f[2, n_] := 3^n;
f[3, n_] := 5^n; z = 1000;
d[n_, b_, c_] := Floor[n*Log[b, c]];
t[k_] := Table[f[k, n], {n, 1, z}];
t = Sort[Union[t[1], t[2], t[3]]];
p[k_, n_] := Position[t, f[k, n]];
Flatten[Table[p[1, n], {n, 1, z/8}]] (* A206812 *)
Table[n + d[n, 3, 2] + d[n, 5, 2],
  {n, 1, 50}]                        (* A206812 *)
Flatten[Table[p[2, n], {n, 1, z/8}]] (* A206813 *)
Table[n + d[n, 2, 3] + d[n, 5, 3],
  {n, 1, 50}]                        (* A206813 *)
Flatten[Table[p[3, n], {n, 1, z/8}]] (* A206814 *)
Table[n + d[n, 2, 5] + d[n, 3, 5],
  {n, 1, 50}]                        (* A206814 *)

Formula

A205812(n) = n + [n*log(base 3)(2)] + [n*log(base 5)(2)],
A205813(n) = n + [n*log(base 2)(3)] + [n*log(base 5)(3)],
A205814(n) = n + [n*log(base 2)(5)] + [n*log(base 3)(5)],
where []=floor.

A206814 Position of 5^n in joint ranking of {2^i}, {3^j}, {5^k}.

Original entry on oeis.org

4, 8, 13, 18, 23, 27, 33, 37, 42, 47, 52, 56, 62, 66, 70, 76, 80, 85, 90, 95, 99, 105, 109, 114, 119, 124, 128, 134, 138, 142, 147, 152, 157, 161, 167, 171, 176, 181, 186, 190, 196, 200, 204, 210, 214, 219, 224, 229, 233, 239, 243, 248, 253, 258, 262
Offset: 1

Views

Author

Clark Kimberling, Feb 17 2012

Keywords

Comments

The exponents i,j,k range through the set N of positive integers, so that the position sequences (A206812 for 2^n, A206813 for 3^n, A206814 for 5^n) partition N.

Examples

			The joint ranking begins with 2,3,4,5,8,9,16,25,27,32,64,81,125,128,243,256, so that
A205812=(1,3,5,7,10,11,14,...)
A205813=(2,6,9,12,15,...)
A205814=(4,8,13,18,23,...)

Crossrefs

Cf. A206805, A206812, A206813.

Programs

  • Mathematica
    f[1, n_] := 2^n; f[2, n_] := 3^n;
f[3, n_] := 5^n; z = 1000;
d[n_, b_, c_] := Floor[n*Log[b, c]];
t[k_] := Table[f[k, n], {n, 1, z}];
t = Sort[Union[t[1], t[2], t[3]]];
p[k_, n_] := Position[t, f[k, n]];
Flatten[Table[p[1, n], {n, 1, z/8}]] (* A206812 *)
Table[n + d[n, 3, 2] + d[n, 5, 2],
  {n, 1, 50}]                        (* A206812 *)
Flatten[Table[p[2, n], {n, 1, z/8}]] (* A206813 *)
Table[n + d[n, 2, 3] + d[n, 5, 3],
  {n, 1, 50}]                        (* A206813 *)
Flatten[Table[p[3, n], {n, 1, z/8}]] (* A206814 *)
Table[n + d[n, 2, 5] + d[n, 3, 5],
  {n, 1, 50}]                        (* A206814 *)

Formula

A205812(n) = n + [n*log(base 3)(2)] + [n*log(base 5)(2)],
A205813(n) = n + [n*log(base 2)(3)] + [n*log(base 5)(3)],
A205814(n) = n + [n*log(base 2)(5)] + [n*log(base 3)(5)],
where []=floor.

A206022 Riordan array (1, x*exp(arcsinh(-2*x))).

Original entry on oeis.org

1, 0, 1, 0, -2, 1, 0, 2, -4, 1, 0, 0, 8, -6, 1, 0, -2, -8, 18, -8, 1, 0, 0, 0, -32, 32, -10, 1, 0, 4, 8, 30, -80, 50, -12, 1, 0, 0, 0, 0, 128, -160, 72, -14, 1, 0, -10, -16, -28, -112, 350, -280, 98, -16, 1, 0, 0, 0
Offset: 0

Views

Author

Philippe Deléham, Feb 02 2012

Keywords

Comments

Riordan array (1, x*(sqrt(1+4x^2)-2x)); inverse of Riordan array (1, x/sqrt(1-4x)), see A205813.
The g.f. for row sums (1,1,-1,-1,3,1,-9,1,27,13,-81,67,243,...) is (1+2*x^2+x*sqrt(1+4*x^2))/(1+3*x^2).
Triangle T(n,k), read by rows, given by (0, -2, 1, -1, 1, -1, 1, -1, 1, -1, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Examples

			Triangle begins:
  1
  0,   1
  0,  -2,   1
  0,   2,  -4,   1
  0,   0,   8,  -6,    1,
  0,  -2,  -8,  18,   -8,    1
  0,   0,   0, -32,   32,  -10,     1
  0,   4,   8,  30,  -80,   50,   -12,    1
  0,   0,   0,   0,  128, -160,    72,  -14,    1
  0, -10, -16, -28, -112,  350,  -280,   98,  -16,   1
  0,   0,   0,   0,    0, -512,   768, -448,  128, -18,   1
  0,  28,  40,  54,   96,  420, -1512, 1470, -672, 162, -20, 1

Crossrefs

Cf. A104624 (column k=1).

Formula

T(n,n) = 1, T(n+1,n) = -2n = -A005843(n), T(n+2,n) = 2*n^2 = A001105(n), T(n+3,n) = -A130809(n+1), T(2n,n) = A009117(n), T(2n+3,1) = (-1)^n*2*A000108(n).
T(n,k) = T(n-2,k-2) - 4*T(n-2,k-1), for k >= 2.
Showing 1-3 of 3 results.

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