A225819 - OEIS

%I #27 Oct 23 2017 06:13:29

%S 1,2,3,4,6,7,9,10,12,14,16,17,19,21,23,25,27,29,31,33,35,37,39,42,44,

%T 46,48,51,53,55,58,60,62,65,67,70,72,75,77,80,82,85,88,90,93,96,98,

%U 101,104,106,109,112,115,117,120,123,126,129,132,134,137,140,143,146,149,152,155

%N Consider the set of n-tuples such that the sum of cubes of the elements is equal to square of their sum; sequence gives largest element in all such tuples.

%C Conjecture [Sen]: lim inf log_n a(n) >= 5/4.

%H Balarka Sen, <a href="/A225819/b225819.txt">Table of n, a(n) for n = 1..500</a>

%H John Mason, <a href="http://www.jstor.org/stable/3620469">Generalising 'sums of cubes equal to squares of sums'</a>, The Mathematical Gazette 85:502 (2001), pp. 50-58.

%H W. R. Utz, The Diophantine Equation (x_1 + x_2 + ... + x_n)^2 = x_1^3 + x_2^3 + ... + x_n^3, Fibonacci Quarterly 15:1 (1977), pp. 14, 16. <a href="http://www.fq.math.ca/Scanned/15-1/utz-a.pdf">Part 1</a>, <a href="http://www.fq.math.ca/Scanned/15-1/utz-b.pdf">part 2</a>.

%F n <= a(n) <= n^(4/3), see A158649.

%e Call an n-multiset with the sum of cubes of the elements equal to square of their sum an n-SCESS.

%e a(6) = 7 since the only 6-SCESS with the largest element >= 7 are (2, 4, 4, 5, 5, 7), (3, 3, 3, 3, 5, 7), (3, 4, 5, 5, 6, 7), (3, 5, 5, 5, 6, 7) and (4, 5, 5, 6, 6, 7) and none have an element larger than 7.

%e a(7) = 9 since the only 7-SCESS with the largest element >= 9 are (4, 4, 4, 5, 5, 5, 9), (4, 5, 5, 5, 6, 6, 9) and (6, 6, 6, 6, 6, 6, 9) and none have an element larger than 9.

%e a(8) = 10 since the only 8-SCESS with the largest element >= 10 are (2, 5, 5, 5, 5, 5, 6, 10), (2, 6, 6, 6, 6, 6, 6, 10), (3, 4, 5, 5, 5, 6, 7, 10), (3, 4, 5, 5, 6, 6, 7, 10), (3, 5, 5, 5, 6, 7, 7, 10), (3, 6, 6, 6, 7, 7, 7, 10), (4, 4, 4, 4, 4, 4, 6, 10), (4, 4, 4, 4, 5, 5, 7, 10), (4, 5, 5, 6, 6, 7, 8, 10), (5, 5, 5, 7, 7, 7, 8, 10) and (6, 6, 6, 6, 6, 6, 9, 10) and none have an element larger than 10.

%o (PARI) a(n)=my(v=vector(n, i, 1), N=n^(4/3), m=n); while(v[#v]<N, v[1]++; if(v[1]>N, for(i=2, N, if(v[i]<N, v[i]++; for(j=2, i-1, v[j]=v[i]); v[1]=max(v[i],m); break))); if(sum(i=1, n, v[i]^3)==sum(i=1, n, v[i])^2, m=max(m,v[1])));m

%Y Cf. A158649, A225808.

%K nonn

%O 1,2

%A _Charles R Greathouse IV_, _Jimmy Zotos_, and _Balarka Sen_, Jul 30 2013