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.

A299117 Sorted terms of A277781.

Original entry on oeis.org

4, 8, 9, 16, 18, 24, 25, 27, 32, 36, 40, 48, 49, 50, 54, 56, 64, 72, 75, 80, 81, 88, 96, 98, 100, 104, 108, 112, 120, 121, 125, 128, 135, 136, 144, 147, 152, 160, 162, 168, 169, 176, 180, 184, 189, 192, 196, 200, 208, 216, 224, 225, 232, 240, 242, 243, 245
Offset: 1

Views

Author

Peter Kagey, Feb 02 2018

Keywords

Comments

It appears that all of these numbers are nonsquarefree (i.e., this sequence is a subsequence of A013929).
This sequence is to A277781 what A013929 is to A072905. That is, A277781 is a bijection from the positive integers to this sequence.

Crossrefs

Programs

  • Mathematica
    With[{nn = 57}, Take[#, nn] &@ Sort@ Table[SelectFirst[n + Range[7 + n^2], AnyTrue[Power[#, 1/3] & /@ {n #, n #^2}, IntegerQ] &], {n, 8 nn}]] (* Michael De Vlieger, Feb 03 2018 *)

A277780 a(n) is the least k > n such that n*k^2 is a cube.

Original entry on oeis.org

8, 16, 24, 32, 40, 48, 56, 27, 72, 80, 88, 96, 104, 112, 120, 54, 136, 144, 152, 160, 168, 176, 184, 81, 200, 208, 64, 224, 232, 240, 248, 108, 264, 272, 280, 288, 296, 304, 312, 135, 328, 336, 344, 352, 360, 368, 376, 162, 392, 400, 408, 416, 424, 128, 440
Offset: 1

Views

Author

Peter Kagey, Oct 30 2016

Keywords

Comments

a(n) is bounded above by 8*n (A008590) because n*(8*n)^2 = (4*n)^3.
If and only if n is cubefree, a(n) = 8n. - David A. Corneth, Nov 01 2016
Theorem: If n = q*m^3 with q cubefree then k = q*(m+1)^3. - Hartmut F. W. Hoft, Nov 02 2016
Proof: let q have u distinct prime divisors p_i. Then q = Product_{i=1..u}(p_i^e_i) where e_i > 0 since p_i|q and e_i < 3 since q is cubefree. Therefore, e_i = 1 or e_i = 2. This yields q|k, i.e., q*t = k. Now for n*k^2 = q*m^3*q^2*t^2 = (q*m)^3 * t^2 to be a cube, t must be a cube. Now, k > n, so q*t/(q*m^3) = t/m^3. The least cube > m^3 is (m+1)^3 so k = q*(m+1)^3 which completes the proof. - David A. Corneth, Nov 03 2016

Examples

			a(24) = 81  because 24 *  81^2 =  54^3;
a(25) = 200 because 25 * 200^2 = 100^3;
a(26) = 208 because 26 * 208^2 = 104^3;
a(27) = 64  because 27 *  64^2 =  48^3.
The cubefree part of 144 is 18. The cubefull part of 144 is 8 = 2^3. Therefore, a(144) = 18 * 3^3 = 486. - _David A. Corneth_, Nov 01 2016
		

Crossrefs

Programs

  • Mathematica
    Table[k = n + 1; While[! IntegerQ[(n k^2)^(1/3)], k++]; k, {n, 55}] (* Michael De Vlieger, Nov 04 2016 *)
  • PARI
    a(n) = {my(k = n+1); while (!ispower(n*k^2, 3), k++); k;} \\ Michel Marcus, Oct 31 2016
    
  • PARI
    a(n) = {my(f = factor(n)); f[, 2] = f[, 2]%3; f=factorback(f); n = sqrtnint(n/f,3); (n+1)^3 * f} \\ David A. Corneth, Nov 01 2016

Formula

a(n) = A050985(n) * A000578(1+A048766(A008834(n))). [Formula given in comments expressed with A-numbers] - Antti Karttunen, Nov 02 2016.
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = 1 + (3*zeta(4) + 3*zeta(5) + zeta(6))/zeta(3) = 7.13539675963975495073... . - Amiram Eldar, Feb 17 2024

A343881 Table read by antidiagonals upward: T(n,k) is the least integer m > k such that k^x * m^y = c^n for some positive integers c, x, and y where x < n and y < n; n >= 2, k >= 1.

Original entry on oeis.org

4, 8, 8, 4, 4, 12, 32, 4, 9, 9, 4, 4, 9, 16, 20, 128, 4, 9, 8, 25, 24, 4, 4, 9, 8, 20, 36, 28, 8, 4, 9, 8, 25, 24, 49, 18, 4, 4, 9, 8, 20, 36, 28, 27, 16, 2048, 4, 9, 8, 25, 24, 49, 18, 24, 40, 4, 4, 9, 8, 20, 36, 28, 16, 12, 80, 44, 8192, 4, 9, 8, 25, 24, 49
Offset: 2

Views

Author

Peter Kagey, May 02 2021

Keywords

Comments

For prime p, the p-th row consists of distinct integers.
Conjecture: T(p,k) = A064549(k) for fixed k > 1 and sufficiently large p.

Examples

			Table begins:
  n\k|    1  2   3   4   5   6   7   8   9   10
-----+-----------------------------------------
   2 |    4, 8, 12,  9, 20, 24, 28, 18, 16,  40
   3 |    8, 4,  9, 16, 25, 36, 49, 27, 24,  80
   4 |    4, 4,  9,  8, 20, 24, 28, 18, 12,  40
   5 |   32, 4,  9,  8, 25, 36, 49, 16, 27, 100
   6 |    4, 4,  9,  8, 20, 24, 28,  9, 16,  40
   7 |  128, 4,  9,  8, 25, 36, 49, 16, 27, 100
   8 |    4, 4,  9,  8, 20, 24, 28, 16, 12,  40
   9 |    8, 4,  9,  8, 25, 36, 49, 16, 24,  80
  10 |    4, 4,  9,  8, 20, 24, 28, 16, 16,  40
  11 | 2048, 4,  9,  8, 25, 36, 49, 16, 27, 100
T(2, 3) = 12 with  3   * 12   =  6^2.
T(3,10) = 80 with 10^2 * 80   = 20^3.
T(4, 5) = 20 with  5^2 * 20^2 = 10^4.
T(5, 1) = 32 with  1   * 32   =  2^5.
T(6, 8) =  9 with  8^2 *  9^3 =  6^6.
		

Crossrefs

Rows: A072905 (n=2), A277781 (n=3).

Formula

T(n,1) = 2^A020639(n).
Showing 1-3 of 3 results.