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A277780
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a(n) is the least k > n such that n*k^2 is a cube.
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2
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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
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OFFSET
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1,1
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COMMENTS
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a(n) is bounded above by 8*n (A008590) because n*(8*n)^2 = (4*n)^3.
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
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LINKS
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FORMULA
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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
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EXAMPLE
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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
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MATHEMATICA
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Table[k = n + 1; While[! IntegerQ[(n k^2)^(1/3)], k++]; k, {n, 55}] (* Michael De Vlieger, Nov 04 2016 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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