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%I #59 Jan 04 2022 03:01:02
%S 0,0,0,0,1,0,1,0,1,1,1,0,2,0,2,1,2,0,2,1,2,1,1,0,4,1,2,1,2,1,3,0,2,2,
%T 2,1,5,0,1,2,4,0,4,0,3,2,2,1,4,2,2,2,2,0,5,1,4,1,1,2,6,1,2,2,4,1,4,0,
%U 3,3,3,0,6,1,2,3,4,1,4,1,4,2,2,0,7,2,3,2,4,2,5,1,2,2,2,2,8,0,3,4
%N Number of ways to write n as n = x*y*z + x + y + z where 1 <= x <= y <= z <= n.
%C From _Vladimir Shevelev_, Aug 02 2015: (Start)
%C Conjecture: liminf_{n->infinity} (a(n)) = 0.
%C Note that this limit should realize on a subsequence of primes. Indeed, if n>=4 is a composite number, then n = p*q, p>=2, q>=2. If p <= q, then, for x=1, y = p-1, z = q-1, we have x*y*z + x + y + z = 1*(p-1)*(q-1) + 1 + (p-1) + (q-1) = p*q = n; so a(n) >= 1. If p > q, then we set x=1, y = q-1, z = p-1, and again a(n) >= 1.
%C Note also that primes r for which a(r) = 0 should grow fast enough. Indeed, r should not be a prime of the form (2*t+1)*k + t + 2, 2 <= t <= k, (*) where t==0 or 2 (mod 3).
%C Indeed, in this case r = x*y*z + x + y + z for x = 2, y = t, z = k. Since gcd(2*t+1, t+2) = gcd(2*(t+2)-3, t+2)=1, then for every considered t and k>=t, the progression (*) contains infinitely many primes r for which a(r) >= 1.
%C Finally, note that limsup_{n->infinity} (a(n)) = infinity. Indeed, this limit is realized, say, on primorials (A002110), since, when m goes to infinity, the number of representations of A002110(m) of the form p*q tends to infinity. So on primorials >1 we have a strictly monotonic subsequence: 0,1,3,8,25,46,78,164 ... (the terms 46 and 78 were calculated by _Michel Marcus_, 164 - by _David A. Corneth_). (End)
%H David A. Corneth, <a href="/A260803/b260803.txt">Table of n, a(n) for n = 0..9999</a>
%H Brian Conrey and Neil Shah, <a href="https://arxiv.org/abs/2112.15551">Which numbers are not the sum plus the product of three positive integers?</a>, arXiv:2112.15551 [math.NT], 2021.
%H Vladimir Shevelev, <a href="http://arxiv.org/abs/1508.03970">Representation of positive integers by the form x1...xk+x1+...+xk</a>, arXiv:1508.03970 [math.NT], 2015.
%F a(n) = A071693(n) - A008619(n) = A071693(n) - floor(n/2) - 1.
%t a[n_] := Sum[Sum[Boole[Mod[n-x-y, x y + 1] == 0 && n-x >= y(x y + 2)], {y, x, (n - x(1+x^2))/2 // Floor}], {x, 1, n/3 // Floor}];
%t Table[a[n], {n, 0, 99}] (* _Jean-François Alcover_, Sep 20 2018, after _M. F. Hasler_ *)
%o (PARI) a(n)=sum(x=1,n\3,sum(y=x,(n-x*(1+x^2))\2,(n-x-y)%(x*y+1)==0&&n-x>=y*(x*y+2))) \\ _M. F. Hasler_, Jul 31 2015
%Y Cf. A071693, A008619, A260804.
%K nonn
%O 0,13
%A _David A. Corneth_, Jul 31 2015
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