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%I #36 Jan 24 2023 16:07:07
%S 157,220,227,246,253,260,267,279,283,286,305,316,323,342,344,361,368,
%T 377,379,384,403,410,435,440,442,468,475,487,494,501,523,530,531,549,
%U 562,568,586,592,594,595,599,602,621,625,640,647,657,658,683,703,710
%N Numbers that are the sum of 5 positive cubes in exactly 2 ways.
%C It appears that this sequence has 15416 terms, the last of which is 2243453. - _Donovan Johnson_, Jan 11 2013
%C From a(1) = 157 we see that c(n) = (number of ways n is the sum of 5 cubes) coincides with A010057 = characteristic function of cubes, up to n = 156. This sequence lists the numbers n for which c(n) = 2. See A003328 for c(n) > 0 and A048926 for c(n) = 1. - _M. F. Hasler_, Jan 04 2023
%H Donovan Johnson, <a href="/A048927/b048927.txt">Table of n, a(n) for n = 1..15416</a> (terms < 10^8)
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CubicNumber.html">Cubic Number.</a>
%t Select[ Range[ 1000], (test = Length[ Select[ PowersRepresentations[#, 5, 3], And @@ (Positive /@ #)& ] ] == 2; If[test, Print[#]]; test)& ](* _Jean-François Alcover_, Nov 09 2012 *)
%o (Python)
%o def ways (n, left = 5, last = 1):
%o a = last; a3 = a**3; c = 0
%o while a3 <= n-left+1:
%o if left > 1:
%o c += ways(n-a3, left-1, a)
%o elif a3 == n:
%o c += 1
%o a += 1; a3 = a**3
%o return c
%o for n in range (1,1000): # to print this sequence
%o if ways(n)==2: print(n,end=", ") # in Python2 use, e.g.: print n,
%o # Minor edits by _M. F. Hasler_, Jan 04 2023
%o (PARI) (waycount(n,numcubes,imax)={if(numcubes==0, !n, sum(i=1,imax, waycount(n-i^3,numcubes-1,i)))}); isA048927(n)=(waycount(n,5,floor(n^(1/3)))==2); \\ _Michael B. Porter_, Sep 27 2009
%Y Cf. A003328 (sums of 5 positive cubes), A025404, A048926 (sum of 5 positive cubes in exactly 1 way), A048930, A294736, A343702, A343705, A344237.
%K nonn
%O 1,1
%A _Eric W. Weisstein_
%E More terms from Walter Hofmann (walterh(AT)gmx.de), Jun 01 2000