cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A270706 Number of ordered ways to write n as x^2*T(x) + y^2 + T(z), where x, y and z are integers with x nonzero, y positive and z nonnegative, and T(m) denotes the triangular number m*(m+1)/2.

Original entry on oeis.org

1, 2, 1, 2, 4, 2, 2, 4, 2, 3, 5, 2, 3, 4, 3, 6, 4, 2, 6, 5, 2, 4, 6, 2, 3, 7, 3, 5, 6, 4, 8, 5, 2, 5, 3, 5, 9, 7, 3, 5, 8, 3, 6, 5, 2, 8, 4, 2, 9, 6, 4, 7, 7, 4, 5, 7, 5, 9, 5, 3, 7, 4, 5, 12, 9, 4, 5, 8, 4, 6, 11, 3, 9, 5, 3, 10, 3, 4, 9, 6, 5, 11, 8, 5, 7, 9, 3, 5, 4, 1
Offset: 1

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Author

Zhi-Wei Sun, Mar 21 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 90, 438, 480, 7108.
(ii) Let pen(x) = x*(3x+1)/2. Any natural number can be written as a*f(x)*g(x) + f(y) + g(z) with x, y and z integers, whenever (a,f(x),g(x)) is among the following ordered triples: (1,T(x),x^2), (1,T(x),pen(x)), (1,T(x),x*(5x+1)/2), (1,T(x),x*(5x+3)/2), (1,T(x),x*(3x+j)) (j = 1,2), (1,pen(x),3*T(x)), (1,pen(x),x*(7x+j)/2) (j = 1,3,5), (1,pen(x),x*(4x+1)), (2,T(x),x^2), (2,T(x),pen(x)), (2,T(x),x(5x+j)/2) (j = 1,3), (2,T(x),x*(3x+j)) (j = 1,2), (2,2*T(x),pen(x)), (2,pen(x),x(7x+j)/2) (j = 3,5), (k,x^2,pen(x)) (k = 1,2,3,4,5,8,11).
(iii) Each 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)*x(5x+1)/2+T(y)+2*T(z), a*T(x)*pen(x)+pen(y)+pen(z) (a = 1,2,3,4), T(x)*pen(x)+pen(y)+3*pen(z), T(x)*pen(x)+pen(y)+4*pen(z), 2*T(x)*pen(x)+pen(y)+3*pen(z), pen(x)*x(5x+j)/2+pen(y)+3*pen(z) (j = 1,3), x(3x+2)*pen(x)+pen(y)+4*pen(z), pen(x)*x(7x+1)/2+pen(y)+pen(z), pen(x)*x(9x+7)/2+pen(y)+pen(z).
See also A270594 and A270705 for some other similar conjectures.

Examples

			a(1) = 1 since 1 = (-1)^2*T(-1) + 1^2 + T(0).
a(3) = 1 since 3 = 1^2*T(1) + 1^2 + T(1).
a(90) = 1 since 90 = 3^2*T(3) + 6^2 + T(0).
a(438) = 1 since 438 = 4^2*T(4) + 5^2 + T(22).
a(480) = 1 since 480 = 1^2*T(1) + 17^2 + T(19).
a(7108) = 1 since 1^2*T(1) + 69^2 + T(68).
		

Crossrefs

Programs

  • Mathematica
    TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]]
    Do[r=0;Do[If[x!=0&&TQ[n-y^2-x^3*(x+1)/2],r=r+1],{y,1,Sqrt[n]},{x,-1-Floor[(2(n-y^2))^(1/4)],(2(n-y^2))^(1/4)}];Print[n," ",r];Continue,{n,1,90}]
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