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Number of ordered ways to write n as the sum of a triangular number, a positive square and the square of a generalized pentagonal number (A001318).
4

%I #17 Mar 20 2016 07:15:13

%S 1,2,1,2,4,2,2,4,2,3,5,2,2,3,3,4,3,2,4,5,1,2,5,1,3,7,3,2,6,5,3,6,2,2,

%T 5,4,6,4,3,5,8,2,2,6,2,5,5,1,4,9,5,3,8,5,4,8,4,3,5,5,5,6,3,6,11,2,3,9,

%U 2,5,12,2,2,9,6,3,4,4,5,6,6,6,5,5,6,11,2,4,8,1

%N Number of ordered ways to write n as the sum of a triangular number, a positive square and the square of a generalized pentagonal number (A001318).

%C Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 21, 24, 48, 90, 138, 213, 283, 462, 468, 567, 573, 1998, 2068, 2488, 2687, 5208, 5547, 5638, 6093, 6492, 6548, 6717, 7538, 7731, 8522, 14763, 16222, 17143, 24958, 26148.

%C (ii) Let T(x) = x(x+1)/2, pen(x) = x(3x+1)/2 and hep(x) = x(5x+3)/2. Then any natural number can be written as P(x,y,z) with x, y and z integers, where P(x,y,z) is either of the following polynomials: T(x)^2+T(y)+z(5z+1)/2, T(x)^2+T(y)+z(3z+j) (j = 1,2), T(x)^2+y^2+pen(z), T(x)^2+pen(y)+hep(z), T(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), T(x)^2+pen(y)+z(4z+j) (j = 1,3), T(x)^2+pen(y)+z(5z+j) (j = 1,3,4), T(x)^2+pen(y)+z(11z+7)/2, T(x)^2+y(5y+1)/2+z(3z+2), T(x)^2+hep(y)+z(3z+2), pen(x)^2+T(y)+pen(z), pen(x)^2+T(y)+2*pen(z), pen(x)^2+T(y)+z(9z+7)/2, pen(x)^2+y^2+pen(z), pen(x)^2+2*T(y)+pen(z), pen(x)^2+pen(y)+3*T(z), pen(x)^2+pen(y)+2z^2, pen(x)^2+pen(y)+2*pen(z), pen(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), pen(x)^2+pen(y)+z(4z+3), pen(x)^2+pen(y)+z(9z+1)/2, pen(x)^2+pen(y)+3*pen(z), pen(x)^2+pen(y)+z(5z+j) (j = 1,2,3,4), pen(x)^2+pen(y)+z(11z+j)/2 (j = 7,9), pen(x)^2+pen(y)+z(7z+1), pen(x)^2+pen(y)+3*hep(z), pen(x)^2+y(5y+j)/2+z(3z+k) (j = 1,3; k = 1,2), pen(x)^2+hep(y)+z(7z+j)/2 (j = 1,3,5), pen(x)^2+hep(y)+z(9z+5)/2, pen(y)^2+2pen(y)+z(3z+2), pen(x)^2+2*pen(y)+3*pen(z), (x(5x+1)/2)^2+2*T(y)+pen(z), (x(5x+1)/2)^2+pen(y)+z(7z+3)/2, (x(5x+1)/2)^2+pen(y)+z(4z+1), (x(5x+1)/2)^2+hep(y)+2*pen(z), hep(x)^2+T(y)+2*pen(z), hep(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), hep(x)^2+pen(y)+z(4z+1), hep(x)^2+pen(y)+z(5z+4), 4*pen(x)^2+T(y)+hep(z), 4*pen(x)^2+T(y)+2*pen(z), 4*pen(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), (x(3x+2))^2+y^2+pen(z), (x(3x+2))^2+pen(y)+z(7z+j)/2 (j = 3,5), 2*T(x)^2+T(y)+z(3z+j) (j = 1,2), 2*T(x)^2+y^2+pen(z), 2*T(x)^2+2*T(y)+pen(z), 2*T(x)^2+pen(y)+z(7z+j)/2 (j = 1,5), 2*T(x)^2+pen(y)+z(5z+1), 2*pen(y)^2+T(y)+z(3z+2), 2*pen(x)^2+y^2+pen(z), 2*pen(x)^2+pen(y)+z(7z+3)/2, 2*pen(x)^2+pen(y)+z(4z+j) (j = 1,3), 2*pen(x)^2+pen(y)+z(5z+4), 2*pen(x)^2+pen(y)+z(7z+1), 2*pen(x)^2+hep(y)+2*pen(z), 2*hep(x)^2+pen(y)+z(7z+5)/2, 3*pen(x)^2+T(y)+z(3z+2), 3*pen(x)^2+y^2+pen(z), 3*pen(x)^2+2*T(y)+pen(z), 3*pen(x)^2+pen(y)+z(7z+j)/2 (j = 1,3,5), 3*pen(x)^2+pen(y)+z(4z+1), 6*pen(x)^2+pen(y)+z(7z+3)/2.

%C See also A270566 for a similar conjecture involving four powers.

%C It is known that any positive integer can be written as the sum of a triangular number, a square and an odd square.

%H Zhi-Wei Sun, <a href="/A270594/b270594.txt">Table of n, a(n) for n = 1..10000</a>

%H B. K. Oh and Z.-W. Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2008.10.002">Mixed sums of squares and triangular numbers (III)</a>, J. Number Theory 129(2009), 964-969.

%H Z.-W. 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 Z.-W. Sun, <a href="http://math.scichina.com:8081/sciAe/EN/abstract/abstract517007.shtml">On universal sums of polygonal numbers</a>, Sci. China Math. 58(2015), no. 7, 1367-1396.

%H Z.-W. Sun, <a href="http://arxiv.org/abs/1502.03056">On universal sums ax^2+by^2+f(z), aT_x+bT_y+f(z) and zT_x+by^2+f(z)</a>, preprint, arXiv:1502.03056 [math.NT], 2015.

%e a(21) = 1 since 21 = 1*2/2 + 4^2 + (1*(3*1+1)/2)^2.

%e a(24) = 1 since 24 = 5*6/2 + 3^2 + (0*(3*0-1)/2)^2.

%e a(468) = 1 since 468 = 0*1/2 + 18^2 + (3*(3*3-1)/2)^2.

%e a(7538) = 1 since 7538 = 64*65/2 + 47^2 + (6*(3*6+1)/2)^2.

%e a(7731) = 1 since 7731 = 82*83/2 + 62^2 + (4*(3*4-1)/2)^2.

%e a(8522) = 1 since 8522 = 127*128/2 + 13^2 + (3*(3*3+1)/2)^2.

%e a(14763) = 1 since 14763 = 164*165/2 + 33^2 + (3*(3*3-1)/2)^2.

%e a(16222) = 1 since 16222 = 168*169/2 + 45^2 + (1*(3*1-1)/2)^2.

%e a(17143) = 1 since 17143 = 182*183/2 + 21^2 + (2*(3*2+1)/2)^2.

%e a(24958) = 1 since 24958 = 216*217/2 + 39^2 + (1*(3*1-1)/2)^2.

%e a(26148) = 1 since 26148 = 10*11/2 + 142^2 + (7*(3*7+1)/2)^2.

%t pQ[n_]:=pQ[n]=IntegerQ[n]&&IntegerQ[Sqrt[24n+1]]

%t Do[r=0;Do[If[pQ[Sqrt[n-x^2-y(y+1)/2]],r=r+1],{x,1,Sqrt[n]},{y,0,(Sqrt[8(n-x^2)+1]-1)/2}];Print[n," ",r];Continue,{n,1,90}]

%Y Cf. A000217, A000290, A001318, A001082, A085787, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559, A270566.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Mar 19 2016