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%I #18 Feb 18 2019 14:28:39
%S 1,2,1,1,2,2,3,3,1,2,4,3,2,1,2,6,5,1,3,3,4,6,1,1,4,6,3,3,5,3,6,4,2,3,
%T 5,5,4,6,4,2,5,5,3,5,2,6,7,3,5,5,7,5,4,2,5,8,5,3,2,6,6,4,5,5,6,7,5,5,
%U 4,6,9,6,6,5,1,7,8,3,2,6,7,5,6,5,7,8,5,3,2,6,10,6,8,7,7,5,4,6,5,5
%N Number of ways to write n as x*(2x-1) + y*(y+1)/2 + z*(z+1)*(z+2)/6, where x,y,z are nonnegative integers with x > 0.
%C Conjecture 1: a(n) > 0 for all n > 0. In other words, any positive integer n can be written as the sum of a positive hexagonal number, a triangular number and a tetrahedral number.
%C We have verified a(n) > 0 for all n = 1..10^7.
%C Conjecture 2: Let c be 1 or 3. Then each n = 0,1,... can be written as c*x(x+1) + y*(y+1)/2 + z*(z+1)*(z+2)/6 with x,y,z nonnegative integers.
%C Conjecture 3: Let t(x) = x*(x+1)*(x+2)/6. Then each n = 0,1,... can be written as 2*t(w) + t(x) + t(y) + t(z) with w,x,y,z nonnegative integers.
%C We have verified Conjecture 3 for all n = 0..2*10^5. Clearly, Conjecture 3 implies Pollock's conjecture which states that any natural number is the sum of five tetrahedral numbers.
%H Zhi-Wei Sun, <a href="/A306460/b306460.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TetrahedralNumber.html">Tetrahedral Number</a>
%e a(3) = 1 with 3 = 1*(2*1-1) + 1*2/2 + 1*2*3/6.
%e a(14) = 1 with 14 = 1*(2*1-1) + 2*3/2 + 3*4*5/6.
%e a(75) = 1 with 75 = 5*(2*5-1) + 4*5/2 + 4*5*6/2.
%e a(349) = 1 with 349 = 5*(2*5-1) + 24*25/2 + 2*3*4/6.
%e a(369) = 1 with 369 = 4*(2*4-1) + 10*11/2 + 11*12*13/6.
%e a(495) = 1 with 495 = 8*(2*8-1) + 20*21/2 + 9*10*11/6.
%e a(642) = 1 with 642 = 16*(2*16-1) + 16*17/2 + 3*4*5/6.
%t f[n_]:=f[n]=n(n+1)(n+2)/6;
%t TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
%t tab={};Do[r=0;Do[If[f[z]>=n, Goto[aa]]; Do[If[TQ[n-f[z]-x(2x-1)],r=r+1],{x,1,(Sqrt[8(n-f[z])+1]+1)/4}];Label[aa],{z,0,n}];tab=Append[tab,r],{n,1,100}];Print[tab]
%Y Cf. A000217, A000292, A000384, A000797, A104246, A262813, A306462, A306471.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Feb 17 2019
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