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%I A266968 #38 Jul 21 2023 12:52:13
%S A266968 0,0,1,2,2,2,1,1,2,2,2,3,4,2,1,2,2,2,3,3,2,1,1,4,4,2,1,2,3,4,7,5,2,2,
%T A266968 4,3,2,5,6,5,2,1,2,4,5,5,6,4,3,4,4,1,2,4,5,5,4,4,2,3,2,4,5,4,6,5,4,3,
%U A266968 5,6,5,4,4,3,4,5,4,3,2,5,7
%N A266968 Number of ordered ways to write n as x^5+y^4+z^3+w*(w+1)/2, where x, y, z and w are nonnegative integers with z > 0 and w > 0.
%C A266968 Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 6, 7, 14, 21, 22, 26, 41, 51, 184, 189, 206, 225, 229, 526, 708.
%C A266968 (ii) Any natural number can be written as 2*x^5 + y^4 + z^3 + w*(w+1)/2 with x,y,z,w nonnegative integers. Also, each natural number can be written as x^5 + 2*y^4 + z^3 + w*(w+1)/2 with x,y,z,w nonnegative integers.
%C A266968 (iii) For each d = 1,2, every natural number can be written as x^5 + y^4 + z^3 + w*(3w+1)/d with x,y,z nonnegative integers and w an integer.
%C A266968 (iv) Any natural number can be written as x^4 + y^4 + z^3 + w*(3w+1)/2 with x,y,z nonnegative integers and w an integer.
%C A266968 Also, for each P(w) = w(3w+1)/2, w(7w+3)/2, we can write any natural number as x^4 + y^3 + z^3 + P(w) with x,y,z nonnegative integers and w an integer.
%C A266968 (v) Any natural number can be written as the sum of a nonnegative cube and three pentagonal numbers. Also, every n = 0,1,2,... can be expressed as the sum of two nonnegative cubes and two pentagonal numbers.
%C A266968 We have verified that a(n) > 1 for all n = 2..3*10^6.
%C A266968 Compare this conjecture with the conjectures in A262813, A262827, A270559 and A271026.
%H A266968 Zhi-Wei Sun, <a href="/A266968/b266968.txt">Table of n, a(n) for n = 0..10000</a>
%H A266968 Zhi-Wei Sun, <a href="http://dx.doi.org/10.4064/aa127-2-1">Mixed sums of squares and triangular numbers</a>, Acta Arith. 127 (2007), 103-113.
%H A266968 Zhi-Wei Sun, <a href="https://www.sciengine.com/SCM/doi/10.1007/s11425-015-4994-4">On universal sums of polygonal numbers</a>, Sci. China Math. 58 (2015), 1367-1396.
%e A266968 a(2) = 1 since 2 = 0^5 + 0^4 + 1^3 + 1*2/2.
%e A266968 a(6) = 1 since 6 = 1^5 + 1^4 + 1^3 + 2*3/2.
%e A266968 a(7) = 1 since 7 = 0^5 + 0^4 + 1^3 + 3*4/2.
%e A266968 a(14) = 1 since 14 = 0^5 + 0^4 + 2^3 + 3*4/2.
%e A266968 a(21) = 1 since 21 = 1^5 + 2^4 + 1^3 + 2*3/2.
%e A266968 a(22) = 1 since 22 = 0^5 + 0^4 + 1^3 + 6*7/2.
%e A266968 a(26) = 1 since 26 = 1^5 + 2^4 + 2^3 + 1*2/2.
%e A266968 a(41) = 1 since 41 = 2^5 + 0^4 + 2^3 + 1*2/2.
%e A266968 a(51) = 1 since 51 = 2^5 + 1^4 + 2^3 + 4*5/2.
%e A266968 a(184) = 1 since 184 = 0^5 + 0^4 + 4^3 + 15*16/2.
%e A266968 a(189) = 1 since 189 = 1^5 + 2^4 + 1^3 + 18*19/2.
%e A266968 a(206) = 1 since 206 = 2^5 + 3^4 + 3^3 + 11*12/2.
%e A266968 a(225) = 1 since 225 = 0^5 + 3^4 + 2^3 + 16*17/2.
%e A266968 a(229) = 1 since 229 = 1^5 + 3^4 + 3^3 + 15*16/2.
%e A266968 a(526) = 1 since 526 = 3^5 + 1^4 + 6^3 + 11*12/2.
%e A266968 a(708) = 1 since 708 = 1^5 + 5^4 + 3^3 + 10*11/2.
%t A266968 TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
%t A266968 Do[r=0;Do[If[TQ[n-x^5-y^4-z^3],r=r+1],{x,0,n^(1/5)},{y,0,(n-x^5)^(1/4)},{z,1,(n-x^5-y^4)^(1/3)}];Print[n," ",r];Continue,{n,0,80}]
%Y A266968 Cf. A000217, A000326, A000578, A000583, A000584, A001318, A262813, A262815, A262816, A262827, A270469, A270488, A270516, A270533, A270559, A270566, A270920, A271026.
%K A266968 nonn
%O A266968 0,4
%A A266968 _Zhi-Wei Sun_, Mar 28 2016

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