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

%I A277780 #41 Feb 17 2024 04:04:40
%S A277780 8,16,24,32,40,48,56,27,72,80,88,96,104,112,120,54,136,144,152,160,
%T A277780 168,176,184,81,200,208,64,224,232,240,248,108,264,272,280,288,296,
%U A277780 304,312,135,328,336,344,352,360,368,376,162,392,400,408,416,424,128,440
%N A277780 a(n) is the least k > n such that n*k^2 is a cube.
%C A277780 a(n) is bounded above by 8*n (A008590) because n*(8*n)^2 = (4*n)^3.
%C A277780 If and only if n is cubefree, a(n) = 8n. - _David A. Corneth_, Nov 01 2016
%C A277780 Theorem: If n = q*m^3 with q cubefree then k = q*(m+1)^3. - _Hartmut F. W. Hoft_, Nov 02 2016
%C A277780 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
%H A277780 Antti Karttunen, <a href="/A277780/b277780.txt">Table of n, a(n) for n = 1..10000</a>
%F A277780 a(n) = A050985(n) * A000578(1+A048766(A008834(n))). [Formula given in comments expressed with A-numbers] - _Antti Karttunen_, Nov 02 2016.
%F A277780 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
%e A277780 a(24) = 81  because 24 *  81^2 =  54^3;
%e A277780 a(25) = 200 because 25 * 200^2 = 100^3;
%e A277780 a(26) = 208 because 26 * 208^2 = 104^3;
%e A277780 a(27) = 64  because 27 *  64^2 =  48^3.
%e A277780 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
%t A277780 Table[k = n + 1; While[! IntegerQ[(n k^2)^(1/3)], k++]; k, {n, 55}] (* _Michael De Vlieger_, Nov 04 2016 *)
%o A277780 (PARI) a(n) = {my(k = n+1); while (!ispower(n*k^2, 3), k++); k;} \\ _Michel Marcus_, Oct 31 2016
%o A277780 (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
%Y A277780 Cf. A000578, A008590, A008834, A048766, A050985, A072905, A254767, A277781.
%Y A277780 Cf. A002117, A013662, A013663, A013664.
%K A277780 nonn,easy
%O A277780 1,1
%A A277780 _Peter Kagey_, Oct 30 2016