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 A270594 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
 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, 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, 2, 5, 12, 2, 2, 9, 6, 3, 4, 4, 5, 6, 6, 6, 5, 5, 6, 11, 2, 4, 8, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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. (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. See also A270566 for a similar conjecture involving four powers. It is known that any positive integer can be written as the sum of a triangular number, a square and an odd square. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 B. K. Oh and Z.-W. Sun, Mixed sums of squares and triangular numbers (III), J. Number Theory 129(2009), 964-969. Z.-W. Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113. Z.-W. Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396. Z.-W. Sun, On universal sums ax^2+by^2+f(z), aT_x+bT_y+f(z) and zT_x+by^2+f(z), preprint, arXiv:1502.03056 [math.NT], 2015. EXAMPLE a(21) = 1 since 21 = 1*2/2 + 4^2 + (1*(3*1+1)/2)^2. a(24) = 1 since 24 = 5*6/2 + 3^2 + (0*(3*0-1)/2)^2. a(468) = 1 since 468 = 0*1/2 + 18^2 + (3*(3*3-1)/2)^2. a(7538) = 1 since 7538 = 64*65/2 + 47^2 + (6*(3*6+1)/2)^2. a(7731) = 1 since 7731 = 82*83/2 + 62^2 + (4*(3*4-1)/2)^2. a(8522) = 1 since 8522 = 127*128/2 + 13^2 + (3*(3*3+1)/2)^2. a(14763) = 1 since 14763 = 164*165/2 + 33^2 + (3*(3*3-1)/2)^2. a(16222) = 1 since 16222 = 168*169/2 + 45^2 + (1*(3*1-1)/2)^2. a(17143) = 1 since 17143 = 182*183/2 + 21^2 + (2*(3*2+1)/2)^2. a(24958) = 1 since 24958 = 216*217/2 + 39^2 + (1*(3*1-1)/2)^2. a(26148) = 1 since 26148 = 10*11/2 + 142^2 + (7*(3*7+1)/2)^2. MATHEMATICA pQ[n_]:=pQ[n]=IntegerQ[n]&&IntegerQ[Sqrt[24n+1]] 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}] CROSSREFS Cf. A000217, A000290, A001318, A001082, A085787, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559, A270566. Sequence in context: A302290 A145173 A361656 * A270706 A082793 A114929 Adjacent sequences: A270591 A270592 A270593 * A270595 A270596 A270597 KEYWORD nonn AUTHOR Zhi-Wei Sun, Mar 19 2016 STATUS approved

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Last modified May 27 02:41 EDT 2024. Contains 372847 sequences. (Running on oeis4.)