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 1-4 of 4 results.

A259912 Discriminant of the field of the number having constant continued fraction [n,n,n,...].

Original entry on oeis.org

5, 8, 13, 5, 29, 40, 53, 17, 85, 104, 5, 37, 173, 8, 229, 65, 293, 328, 365, 101, 445, 488, 533, 145, 629, 680, 733, 197, 5, 904, 965, 257, 1093, 1160, 1229, 13, 1373, 1448, 61, 401, 1685, 1768, 1853, 485, 2029, 2120, 2213, 577, 2405, 2504, 2605, 677, 2813
Offset: 1

Views

Author

Clark Kimberling, Jul 20 2015

Keywords

Comments

Central numbers of the triangle at A259911.
It appears that a(n) = 5 for n in A002878 = (1,4,11,29,...), a bisection of the Lucas sequence.

Examples

			[3,3,3,...] = (1/2)(3 + sqrt(13)), so that a(3) = 13.
		

Crossrefs

Programs

  • Mathematica
    t = Table[FromContinuedFraction[{n, {n}}], {n, 1, 100}];
    Flatten[NumberFieldDiscriminant[t]]

A259913 Discriminant of the number field containing the number with periodic continued fraction [1,n,1,n,1,n,...].

Original entry on oeis.org

5, 12, 21, 8, 5, 60, 77, 24, 13, 140, 165, 12, 221, 28, 285, 5, 357, 44, 437, 120, 21, 572, 69, 168, 29, 780, 93, 56, 957, 1020, 1085, 8, 1221, 1292, 1365, 40, 1517, 1596, 1677, 440, 205, 1932, 2021, 33, 5, 92, 2397, 156, 53, 12, 2805, 728, 3021, 348, 3245
Offset: 1

Views

Author

Clark Kimberling, Jul 20 2015

Keywords

Comments

a(n) is the first term in row n of the triangle at A259911.
It appears that a(n) = 5 if n is a nonzero term of A004146.

Examples

			[1,3,1,3,1,3,...] = (1/6)(3 + sqrt(21)), so that a(3) = 21.
		

Crossrefs

Cf. A259911.

Programs

  • Mathematica
    v = Table[FromContinuedFraction[{1, {n, 1}}], {n, 1, 60}];
    Flatten[NumberFieldDiscriminant[v]]

A378873 Squarefree part of A378872(n) (the discriminant of the minimal polynomial of a number whose continued fraction expansion has periodic part given by the n-th composition (in standard order)).

Original entry on oeis.org

5, 2, 5, 13, 3, 3, 5, 5, 21, 2, 10, 21, 10, 10, 5, 29, 2, 15, 17, 15, 85, 85, 6, 2, 17, 85, 6, 17, 6, 6, 5, 10, 5, 6, 26, 13, 37, 37, 165, 6, 37, 2, 221, 37, 3, 221, 65, 5, 26, 37, 165, 37, 221, 3, 65, 26, 165, 221, 65, 165, 65, 65, 5, 53, 15, 35, 37, 3, 229
Offset: 1

Views

Author

Pontus von Brömssen, Dec 10 2024

Keywords

Comments

Any number x whose continued fraction expansion is eventually periodic can be written uniquely as x = (c+f*sqrt(d))/b, where b, c, f, d are integers, b > 0, d > 0 is squarefree, and GCD(b,c,f) = 1. a(n) is equal to d when the periodic part of the continued fraction of x is given by the n-th composition. If two numbers have eventually periodic continued fraction expansions with the same periodic part, their respective values of d are the same.

Examples

			For n = 6, the 5th composition is (1,2). The value of the continued fraction 1+1/(2+1/(1+1/(2+...))) is (1+sqrt(3))/2, so a(6) = 3.
		

Crossrefs

Cf. A007913, A066099 (compositions in standard order), A246904, A246922, A259911, A259912, A305311, A378872, A378874.

Formula

a(n) = A007913(A378872(n)) = A378872(n)/A378874(n)^2.
a(2^n) = A259912(n+1) if a(2^n) == 1 (mod 4), a(2^n) = A259912(n+1)/4 otherwise.
For n > k >= 0, a(2^n+2^k) = A259911(n,k+1) if a(2^n+2^k) == 1 (mod 4), a(2^n+2^k) = A259911(n,k+1)/4 otherwise.

A259532 Triangular array: row k shows the discriminant of sqrt(j) + sqrt(k) for j=1..k.

Original entry on oeis.org

1, 8, 8, 12, 8, 12, 1, 2304, 2304, 1, 5, 8, 12, 8, 5, 24, 1600, 12, 12, 1600, 24, 28, 2304, 3600, 1, 3600, 2304, 28, 8, 12544, 2304, 5, 5, 2304, 12544, 8, 1, 8, 7056, 24, 5, 24, 7056, 8, 1, 40, 8, 2304, 28, 14400, 14400, 28, 2304, 8, 40, 44, 1600, 12, 8
Offset: 1

Views

Author

Clark Kimberling, Jul 20 2015

Keywords

Examples

			First seven rows:
1
8   8
12  8     12
1   2304  2304   1
5   8     12     8   5
24  1600  12     12  1600  24
28  2304  3600   1   3600  2304  28
		

Crossrefs

Programs

  • Mathematica
    t = Table[NumberFieldDiscriminant[Sqrt[j] + Sqrt[k + 1 - j]], {k, 1, 16}, {j, 1, k}]
    TableForm[t] (* A259532 array *)
    Flatten[t]   (* A259532 sequence *)
Showing 1-4 of 4 results.