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%I #26 Jul 21 2023 09:15:15
%S 1,2,2,3,2,1,2,2,4,3,3,4,1,3,2,3,5,3,5,1,1,3,3,4,2,3,3,3,4,6,6,3,2,3,
%T 2,4,6,6,3,2,3,3,4,5,8,2,3,3,5,3,2,5,3,3,3,7,3,4,4,3,3,3,5,8,2,4,3,5,
%U 7,4,7,4,5,3,6,1,4,4,6,7,4,8,5,1,4,7,7,4,4,5,2,3,5,10,6,4,2,1,3,5,7
%N Number of ordered ways to write n as x^3 + y*(y+1)/2 + z*(3*z+1)/2, where x, y and z are nonnegative integers.
%C Conjecture: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 5, 12, 19, 20, 75, 83, 97, 117.
%C Conjecture verified up to 10^11. - _Mauro Fiorentini_, Jul 20 2023
%C See also A262813 and A262816 for similar conjectures.
%C By Theorem 1.7(i) in the linked paper, each natural number can be written as the sum of a triangular number, an even square and a generalized pentagonal number.
%H Zhi-Wei Sun, <a href="/A262815/b262815.txt">Table of n, a(n) for n = 0..10000</a>
%H Zhi-Wei Sun, <a href="https://doi.org/10.1007/s11425-015-4994-4">On universal sums of polygonal numbers</a>, Sci. China Math. 58(2015), no. 7, 1367-1396.
%e a(0) = 1 since 0 = 0^3 + 0*1/2 + 0*(3*0+1)/2.
%e a(5) = 1 since 5 = 0^3 + 2*3/2 + 1*(3*1+1)/2.
%e a(12) = 1 since 12 = 0^3 + 4*5/2 + 1*(3*1+1)/2.
%e a(19) = 1 since 19 = 1^3 + 2*3/2 + 3*(3*3+1)/2.
%e a(20) = 1 since 20 = 2^3 + 4*5/2 + 1*(3*1+1)/2.
%e a(75) = 1 since 75 = 2^3 + 4*5/2 + 6*(3*6+1)/2.
%e a(83) = 1 since 83 = 0^3 + 3*4/2 + 7*(3*7+1)/2.
%e a(97) = 1 since 97 = 3^3 + 10*11/2 + 3*(3*3+1)/2.
%e a(117) = 1 since 117 = 0^3 + 13*14/2 + 4*(3*4+1)/2.
%t TQ[n_]:=IntegerQ[Sqrt[8n+1]]
%t Do[r=0;Do[If[TQ[n-x^3-z(3z+1)/2],r=r+1],{x,0,n^(1/3)},{z,0,(Sqrt[24(n-x^3)+1]-1)/6}];Print[n," ",r];Continue,{n,0,100}]
%Y Cf. A000217, A000578, A005449, A160325, A262813, A262816.
%K nonn
%O 0,2
%A _Zhi-Wei Sun_, Oct 03 2015
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