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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A261241 One half of numbers representable in at least two different ways as sums of four nonvanishing cubes. See A259060 for these numbers and their representations.

Original entry on oeis.org

3213, 3950, 4807, 5796, 6929, 8218, 9675, 11312, 13141, 15174, 17423, 19900, 22617, 25586, 28819, 32328, 36125, 40222, 44631, 49364, 54433, 59850, 65627, 71776, 78309, 85238, 92575, 100332, 108521, 117154, 126243, 135800, 145837, 156366
Offset: 0

Views

Author

Wolfdieter Lang, Aug 12 2015

Keywords

Comments

See A259060. There may be other numbers with this property.

References

  • W. Sierpiński, 250 Problems in Elementary Number Theory, American Elsevier Publ. Comp., New York, PWN-Polish Scientific Publishers, Warszawa, 1970, Problem 227, p. 20 and p. 110.

Links

Crossrefs

Cf. A259060.

Programs

  • Magma
    [(n+9)*(2*n^2 + 36*n + 357): n in [0..50]]; // Vincenzo Librandi, Aug 13 2015
  • Magma
    I:=[3213,3950,4807,5796]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Aug 13 2015
  • Mathematica
    CoefficientList[Series[(3213 - 8902 x + 8285 x^2 - 2584 x^3)/(1 - x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, Aug 13 2015 *)

Formula

a(n) = (n+9)*(2*n^2 + 36*n + 357), n >= 0.
O.g.f.: (3213 - 8902*x + 8285*x^2 - 2584*x^3)/(1-x)^4.
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Vincenzo Librandi, Aug 13 2015

A259058 Numbers that are representable in at least two ways as sums of four distinct nonvanishing squares.

Original entry on oeis.org

454, 530, 614, 706, 806, 914, 1030, 1154, 1286, 1426, 1574, 1730, 1894, 2066, 2246, 2434, 2630, 2834, 3046, 3266, 3494, 3730, 3974, 4226, 4486, 4754, 5030, 5314, 5606, 5906, 6214, 6530, 6854, 7186, 7526, 7874, 8230, 8594, 8966, 9346, 9734
Offset: 0

Views

Author

Wolfdieter Lang, Aug 12 2015

Keywords

Comments

This is part one of Exercise 229 in Sierpiński's problem book. See p. 20 and p. 110 for the solution. He uses the identity (n-8)^2 + (n-1)^2 + (n+1)^2 + (n+8)^2 = 4*n^2 + 130 = (n-7)^2 + (n-4)^2 + (n+4)^2 + (n+7)^2, for n >= 9.
Here n was replaced by n + 9: (n+1)^2 + (n+8)^2 +(n+10)^2 + (n+17)^2 = 4*n^2 + 72*n + 454 = (n+2)^2 + (n+5)^2 + (n+13)^2 + (n+16)^2, for n >= 0.
There may be other numbers having this property.
Because the summands have no common factor > 1 each of these two representations is called primitive. Therefore, this is a proper subsequence of A223727, hence of A004433. - Wolfdieter Lang, Aug 20 2015

Examples

			n=0: 454 = 1^2 + 8^2 + 10^2 + 17^2 = 2^2 + 5^2 + 13^2 + 16^2.
n=2: 614 = 3^2 + 10^2 + 12^2 + 19^2 = 4^2 + 7^2 + 15^2 + 18^2.

References

  • W. Sierpiński, 250 Problems in Elementary Number Theory, American Elsevier Publ. Comp., New York, PWN-Polish Scientific Publishers, Warszawa, 1970.

Links

Crossrefs

Cf. A259059, A223727, A004433, A259060 (four cubes).

Programs

  • Magma
    [4*n^2 + 72*n + 454: n in [0..50]]; // Vincenzo Librandi, Aug 13 2015
  • Magma
    I:=[454, 530, 614]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Aug 13 2015
  • Mathematica
    CoefficientList[Series[2 (227 - 416 x + 193 x^2)/(1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Aug 13 2015 *)
  • PARI
    a(n)=4*n^2+72*n+454 \\ Charles R Greathouse IV, Jun 17 2017

Formula

a(n) = 4*n^2 + 72*n + 454 = 2*A259059(n). See the comment for the sum of four squares in two ways.
O.g.f.: 2*(227 - 416*x + 193*x^2)/(1-x)^3.
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). - Vincenzo Librandi, Aug 13 2015
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