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.

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A186288 Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) before g(j) when f(i)=g(j), where f and g are the squares and pentagonal numbers. Complement of A186289.

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 12, 14, 16, 18, 20, 21, 23, 25, 27, 29, 31, 32, 34, 36, 38, 40, 41, 43, 45, 47, 49, 51, 52, 54, 56, 58, 60, 61, 63, 65, 67, 69, 71, 72, 74, 76, 78, 80, 81, 83, 85, 87, 89, 90, 92, 94, 96, 98, 100, 101, 103, 105, 107, 109, 110, 112, 114, 116, 118, 120, 121, 123, 125, 127, 129, 130, 132, 134, 136, 138, 140, 141, 143, 145, 147, 149, 150, 152, 154, 156, 158, 160, 161, 163, 165, 167, 169, 170, 172, 174, 176, 178, 179, 181
Offset: 1

Views

Author

Clark Kimberling, Feb 17 2011

Keywords

Comments

See A186219 for a discussion of adjusted rank sequences.

Examples

			First, write
1..4...9....16....25..36..49..... (squares)
1....5...12....22....35......51.. (pentagonal)
Replace each number by its rank, where ties are settled by ranking the square number before the pentagonal:
a=(1,3,5,7,9,11,12,14,....)=A186288.
b=(2,4,6,8,10,13,15,17,...)=A186289.

Crossrefs

Cf. A186219, A186289, A186290, A186291, A000290 (squares), A000326 (pentagonal).

Programs

  • Mathematica
    (* adjusted joint ranking; general formula *)
d=1/2; u=1; v=0; w=0; x=3/2; y=-1/2; z=0;
h[n_]:=-y+(4x(u*n^2+v*n+w-z-d)+y^2)^(1/2);
a[n_]:=n+Floor[h[n]/(2x)];
k[n_]:=-v+(4u(x*n^2+y*n+z-w+d)+v^2)^(1/2);
b[n_]:=n+Floor[k[n]/(2u)];
Table[a[n],{n,1,100}]  (* A186288 *)
Table[b[n],{n,1,100}]  (* A186289 *)

A186326 Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) after g(j) when f(i)=g(j), where f and g are the squares and octagonal numbers. Complement of A186327.

Original entry on oeis.org

2, 3, 5, 6, 8, 9, 11, 12, 14, 16, 17, 19, 20, 22, 24, 25, 27, 28, 30, 31, 33, 35, 36, 38, 39, 41, 42, 44, 46, 47, 49, 50, 52, 53, 55, 57, 58, 60, 61, 63, 65, 66, 68, 69, 71, 72, 74, 76, 77, 79, 80, 82, 83, 85, 87, 88, 90, 91, 93, 94, 96, 98, 99, 101, 102, 104, 106, 107, 109, 110, 112, 113, 115, 117, 118, 120, 121, 123, 124, 126, 128, 129, 131, 132, 134, 135, 137, 139, 140, 142, 143, 145, 147, 148, 150, 151, 153, 154, 156, 158
Offset: 1

Views

Author

Clark Kimberling, Feb 17 2011

Keywords

Examples

			First, write
1..4...9..16....25..36....49..64...  (squares)
1....8.......21........40........65. (octagonal)
Replace each number by its rank, where ties are settled by ranking the square number after the octagonal:
a=(2,3,5,6,8,9,11,12,14,...)=A186326
b=(1,4,7,10,13,15,18,21,...)=A186327.

Links

Crossrefs

Cf. A186219, A186324, A186325, A186327.

Programs

  • Mathematica
    (* adjusted joint ranking; general formula *)
d=-1/2; u=1; v=0; w=0; x=3; y=-2; z=0;
h[n_]:=-y+(4x(u*n^2+v*n+w-z-d)+y^2)^(1/2);
a[n_]:=n+Floor[h[n]/(2x)];
k[n_]:=-v+(4u(x*n^2+y*n+z-w+d)+v^2)^(1/2);
b[n_]:=n+Floor[k[n]/(2u)];
Table[a[n], {n, 1, 100}]  (* A186326 *)
Table[b[n], {n, 1, 100}]  (* A186327 *)

A186540 Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) before g(j) when f(i)=g(j), where f(i)=i^2 and g(j)=-2+3j^2. Complement of A186539.

Original entry on oeis.org

2, 5, 8, 10, 13, 16, 19, 21, 24, 27, 30, 32, 35, 38, 40, 43, 46, 49, 51, 54, 57, 60, 62, 65, 68, 71, 73, 76, 79, 81, 84, 87, 90, 92, 95, 98, 101, 103, 106, 109, 112, 114, 117, 120, 122, 125, 128, 131, 133, 136, 139, 142, 144, 147, 150, 152, 155, 158, 161, 163, 166, 169, 172, 174, 177, 180, 183, 185, 188, 191, 193, 196, 199, 202, 204, 207, 210, 213, 215, 218
Offset: 1

Views

Author

Clark Kimberling, Feb 23 2011

Keywords

Comments

See A186219 for a discussion of adjusted joint rank sequences.
Does this differ from A054088? The first 42000 entries of both sequences at least are the same. - R. J. Mathar, Feb 25 2011

Examples

			First, write
1..4..9..16..25..36..49.. (i^2)
.......10....25....46.. (-2+3j^2)
Then replace each number by its rank, where ties are settled by ranking i^2 before -2+3j^2:
b=(1,3,4,6,7,9,11,12,14,15,17,18,..)=A186539
a=(2,5,8,10,13,16,19,21,24,27,30...).

Crossrefs

Cf. A186219, A186539, A186541, A186542.

Programs

Formula

b(n) = n+floor(sqrt((1/3)n^2+1/24)) = A186539(n).
a(n) = n+floor(sqrt(3n^2-3/2)).

A185050 Least k such that G(k) > 3 - 1/2^n, where G(k) is the sum of the first k terms of the geometric series 1 + 2/3 + (2/3)^2 + ....

Original entry on oeis.org

3, 5, 7, 8, 10, 12, 13, 15, 17, 19, 20, 22, 24, 25, 27, 29, 31, 32, 34, 36, 37, 39, 41, 43, 44, 46, 48, 49, 51, 53, 54, 56, 58, 60, 61, 63, 65, 66, 68, 70, 72, 73, 75, 77, 78, 80, 82, 84, 85, 87, 89, 90, 92, 94, 96, 97, 99, 101, 102, 104, 106, 107, 109, 111, 113
Offset: 0

Views

Author

Arkadiusz Wesolowski, Dec 25 2012

Keywords

Comments

Many of terms in this sequence are that same as A186219(n+2) but not all.

Examples

			a(1) = 5 because 1 + 2/3 + (2/3)^2 + (2/3)^3 + (2/3)^4 > 3 - 1/2.

References

  • Mohammad K. Azarian, Geometric Series, Problem 329, Mathematics and Computer Education, Vol. 30, No. 1, Winter 1996, p. 101. Solution published in Vol. 31, No. 2, Spring 1997, pp. 196-197.

Links

Programs

  • Mathematica
    lst = {}; n = s = 0; Do[s = s + (2/3)^k; If[s > 3 - 1/2^n, AppendTo[lst, k + 1]; n++], {k, 0, 112}]; lst
Previous Showing 31-34 of 34 results.

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