cp's OEIS Frontend

A306382 Positive integers not representable as Pen(x) + Pen(y) + 2*Pen(z), where x, y, z are nonnegative integers, and Pen(k) denotes the pentagonal number k*(3k-1)/2.

%I A306382 #30 Feb 21 2019 14:09:55
%S A306382 9,18,21,28,31,39,43,55,69,74,89,90,98,109,111,113,134,135,144,193,
%T A306382 202,214,230,243,260,265,273,275,310,510,553,698,699,749,773,780,865,
%U A306382 878,945,965,1219,1398,1413,2153,2168,2335,2828,3178,3793
%N A306382 Positive integers not representable as Pen(x) + Pen(y) + 2*Pen(z), where x, y, z are nonnegative integers, and Pen(k) denotes the pentagonal number k*(3k-1)/2.
%C A306382 Conjecture 1: This sequence only has 49 terms as listed.
%C A306382 Conjecture 2: Any integer n > 6471 can be written as Pen(x) + Pen(y) + 3*Pen(z) with x,y,z nonnegative integers. Any integer n > 7727 can be written as Pen(x) + 2*Pen(y) + 2*Pen(z) with x,y,z nonnegative integers. Also, any integer n > 4451 can be written as Pen(x) + 2*Pen(y) + 4*Pen(z) with x,y,z nonnegative integers.
%C A306382 Conjecture 3: Let N(1) = 5928, N(4) = 2761, N(8) = 8868 and N(11) = 9929. For each c among 1, 4, 8, 11, any integer n > N(c) can be written as x*(3x+1)/2 + y*(3y+1) + c*z*(3z+1)/2 with x,y,z nonnegative integers.
%C A306382 Conjecture 4: Any integer n > 5544 can be written as x*(3x+1)/2 + y*(3y+1)/2 + 3z*(3z+1)/2 with x,y,z nonnegative integers. Any integer n > 7093 can be written as x*(3x+1)/2 + 3*y*(3y+1)/2 + 2z*(3z+1) with x,y,z nonnegative integers. Also, any integer n > 8181 can be written as x*(3x+1)/2 + 2y*(3y+1) + 3z*(3z+1) with x,y,z nonnegative integers.
%C A306382 Conjecture 5: For each positive integer m, there are only finitely many positive integers not representable as x*(x+1)/2 + y*(y+1)/2 + z*(z+1)/2 with x,y,z in the set {m, m+1, ...}.
%C A306382 See also A306383 for similar conjectures.
%H A306382 Georg Fischer, <a href="/A306382/b306382.txt">Table of n, a(n) for n = 1..49</a>
%H A306382 Zhi-Wei Sun, <a href="https://doi.org/10.1007/s11425-017-9354-4">Universal sums of three quadratic polynomials</a>, Sci. China Math., in press.
%e A306382 a(1) = 9 since the set {Pen(x) + Pen(y) + 2*Pen(z): x,y,z = 0,1,2,...} contains 1..8 but it does not contain 9.
%t A306382 PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]]&&(n==0||Mod[Sqrt[24n+1]+1,6]==0);
%t A306382 tab={};Do[Do[If[PenQ[m-x(3x-1)-y(3y-1)/2],Goto[aa]],{x,0,(Sqrt[12m+1]+1)/6},{y,0,(Sqrt[12(m-x(3x-1))+1]+1)/6}];tab=Append[tab,m];Label[aa],{m,1,5000}];Print[tab]
%Y A306382 Cf. A000217, A000326, A005449, A008443, A306383.
%K A306382 nonn
%O A306382 1,1
%A A306382 _Zhi-Wei Sun_, Feb 11 2019