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A270705 Number of ordered ways to write n as x^2*pen(x) + pen(y) + pen(z) with pen(x) = x*(3x+1)/2 and pen(y) <= pen(z), where x, y and z are integers ("pen" stands for "pentagonal"). 2

%I #18 Jul 21 2023 09:15:35

%S 1,2,5,5,6,4,3,4,4,4,3,2,3,3,6,4,4,4,3,3,3,4,6,5,6,5,5,8,8,9,7,5,7,6,

%T 7,9,7,10,5,5,9,6,12,7,8,6,3,10,6,5,7,5,8,7,8,9,5,9,8,7,5,7,7,5,6,6,5,

%U 4,6,4,8,5,9,6,3,7,5,8,8,8,8,6,6,6,6,6,8,3,1,4,6

%N Number of ordered ways to write n as x^2*pen(x) + pen(y) + pen(z) with pen(x) = x*(3x+1)/2 and pen(y) <= pen(z), where x, y and z are integers ("pen" stands for "pentagonal").

%C Conjecture: (i) Any natural number can be written as a*x^2*pen(x) + b*pen(y) + c*pen(z) with x, y and z integers, provided that (a,b,c) is among the following ordered triples: (j,1,k) (j = 1,2; k = 1,2,3,4), (1,2,3), (3,1,4) and (4,1,3).

%C (ii) Every n = 0,1,2,... can be expressed as x^2*pen(x) + T(y) + T(z) with x, y and z integers, where T(m) denotes the triangular number m*(m+1)/2. Also, for each (a,b) = (1,2),(1,4),(2,2), any natural number can be written as a*x^2*T(x) + b*T(y) + T(z) with x, y and z integers.

%C (iii) Each natural number can be written as x^2*P(x) + pen(y) + pen(z) with x, y and z integers, where P(x) is either of the following polynomials: a*T(x) (a = 1,2,3,4,5), x*(5x+3)/2, x*(3x+1), x*(3x+2), x*(7x+1)/2, x*(4x+1), x*(4x+3), x*(9x+5)/2, x*(5x+3), x*(11x+9)/2, x*(13x+5)/2, x*(17x+9)/2, 3x*(3x+2), x*(11x+2).

%C See also A270594 and A270706 for other similar conjectures.

%H Zhi-Wei Sun, <a href="/A270705/b270705.txt">Table of n, a(n) for n = 0..10000</a>

%H 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 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.

%H Zhi-Wei 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(88) = 1 since 88 = 1^2*pen(1) + pen(-5) + pen(-6).

%t pen[x_]:=pen[x]=x(3x+1)/2

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

%t Do[r=0;Do[If[pQ[n-pen[y]-x^2*pen[x]],r=r+1],{y,-Floor[(Sqrt[12n+1]+1)/6],(Sqrt[12n+1]-1)/6},{x,-1-Floor[(2(n-pen[y])/3)^(1/4)],(2(n-pen[y])/3)^(1/4)}];Print[n," ",r];Continue,{n,0,90}]

%Y Cf. A000217, A000290, A001318, A001082, A085787, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559, A270566, A270594, A270616, A270706.

%K nonn

%O 0,2

%A _Zhi-Wei Sun_, Mar 21 2016

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