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.

User: Vincent Chan

Vincent Chan's wiki page.

Vincent Chan has authored 3 sequences.

A338138 Nested cube root (b(n) + (b(n-1) + ... + (b(1))^(1/3)...)^(1/3))^(1/3), where b(n) = A338137(n).

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 3, 3, 4, 3, 5, 3, 6, 3, 7, 4, 4, 5, 4, 6, 4, 7, 5, 5, 6, 5, 7, 6, 6, 7, 7, 8, 3, 8, 4, 8, 5, 8, 6, 8, 7, 9, 3, 9, 4, 9, 5, 9, 6, 9, 7, 10, 3, 10, 4, 10, 5, 10, 6, 10, 7, 11, 3, 11, 4, 11
Offset: 1

Author

Vincent Chan, Oct 12 2020

Keywords

Comments

Contains every positive integer: letting s_n = (b(n) + (b(n-1) + ... + (b(1))^(1/3)...)^(1/3))^(1/3), we have b_n = s_n^3-s_{n-1}. We claim that if s_1,...s_{n-1} <= k, then s_n <=k+1. Indeed, the given condition implies b_1,...,b_{n-1} <= k^3. Since (k+1)^3-s_{n-1} >= k^3+3k^2+2k+1 > b_j for j < n and b_n is the smallest positive integer not already in the sequence for which b_n+s_{n-1} is a cube, then s_n <= k+1. Then we note that b_n = s_n^3-s_{n-1} cannot repeat, so that s_n cannot be a single constant infinitely often, so {s_n} contains every positive integer.

Crossrefs

Programs

  • Python
    myList = [1]
    roots = [1]
    s = 1
    t = 0
    for n in range(9999):
        b = 2
        while t == 0:
            if(b**3-s > 0 and not b**3-s in myList):
                myList.append(b**3-s)
                s = b
                roots.append(s)
                t = 1
            else:
                b += 1
        t=0
    print("roots: ",roots)

Formula

a(n) = (a(n-1) + A338137(n))^(1/3). - Andrew Howroyd, Oct 15 2020

A338137 Lexicographically earliest sequence of distinct positive integers such that the nested cube root (a(n) + (a(n-1) + ... + (a(1))^(1/3)...)^(1/3))^(1/3) is an integer.

Original entry on oeis.org

1, 7, 6, 25, 5, 62, 4, 123, 3, 214, 2, 341, 20, 24, 61, 23, 122, 22, 213, 21, 340, 57, 60, 121, 59, 212, 58, 339, 118, 120, 211, 119, 338, 209, 210, 337, 336, 505, 19, 509, 56, 508, 117, 507, 208, 506, 335, 722, 18, 726, 55, 725, 116, 724, 207, 723, 334, 993, 17, 997, 54, 996
Offset: 1

Author

Vincent Chan, Oct 12 2020

Keywords

Comments

A permutation of positive integers: letting s_n = (a(n) + (a(n-1) + ... + (a(1))^(1/3)...)^(1/3))^(1/3), we have a_n = s_n^3-s_{n-1}. We claim that if s_1,...s_{n-1} <= k, then s_n <=k+1. Indeed, the given condition implies a_1,...,a_{n-1} <= k^3. Since (k+1)^3-s_{n-1} >= k^3+3k^2+2k+1 > a_j for j < n and a_n is the smallest positive integer not already in the sequence for which a_n+s_{n-1} is a cube, then s_n <= k+1. Then we note that a_n = s_n^3-s_{n-1} cannot repeat, so that s_n cannot be a single constant infinitely often, so {s_n} contains every positive integer. Finally, for an integer k, k appears in the sequence {a_n} no later than the first time s_{n-1} = k^3-k.

Crossrefs

Cf. A323635 (similar definition with square roots).

Programs

  • PARI
    lista(nn) = {my(va = vector(nn), lastcb); va[1] = 1; lastcb = 1; for (n=2, nn, my(k = ceil(sqrtn(sqrtnint(lastcb, 3), 3))); while (#select(x->(x==(k^3-sqrtnint(lastcb, 3))), va), k++); va[n] = k^3-sqrtnint(lastcb, 3); lastcb = k^3;); va; } \\ Michel Marcus, Oct 13 2020
  • Python
    myList = [1]
    s = 1
    t = 0
    for n in range(9999):
        b = 2
        while t == 0:
            if(b**3-s > 0 and not b**3-s in myList):
                myList.append(b**3-s)
                s = b
                t = 1
            else:
                b += 1
        t=0
    print("myList: ",myList)
    

A273935 Number of ways to arrange n women and n men around a circular table so that they can be divided into n nonintersecting pairs of 1 woman and 1 man sitting side-by-side.

Original entry on oeis.org

1, 1, 6, 84, 2160, 89280, 5443200, 460857600, 51819264000, 7476605337600, 1347105779712000, 296508121620480000, 78297264318873600000, 24431729220414996480000, 8893692297263669575680000, 3735464765667589501747200000, 1793050447716486548029440000000
Offset: 0

Author

Vincent Chan, Jun 16 2016

Keywords

Comments

a(n) is the number of monotonic paths from (0,0) to (n,n) which are equivalent to a path which is no more than 1 step from the main diagonal, where two paths are equivalent if they are circular shifts of one another.

Examples

			For n = 2: Represent the women and men as the numbers 1,2,3,4, where 1,3 are women and 2,4 are men. Then all 6 arrangements around the circular table are valid (parentheses included to emphasize a valid pairing, including wraparound):
(1,2),(3,4);
(1,2),(4,3);
1),(3,2),(4;
1),(3,4),(2;
(1,4),(2,3);
(1,4),(3,2).
		

Crossrefs

Programs

  • PARI
    a(n) = if (n==0, 1, n!*(n-1)!*(2^n-1)); \\ Michel Marcus, Jun 20 2016

Formula

a(n) = n!*(n-1)!*(2^n-1) for n>0, a(0) = 1.
a(n) = A000142(n)*A000142(n-1)*A000225(n), n >= 1. - Omar E. Pol, Jun 16 2016