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.

Showing 1-5 of 5 results.

A066648 Cubes of the form a^2 + b^3 with a, b > 0.

Original entry on oeis.org

512, 1000, 2744, 21952, 32768, 35937, 64000, 175616, 185193, 274625, 357911, 373248, 405224, 474552, 729000, 1157625, 1404928, 1481544, 2000376, 2097152, 2197000, 2299968, 2744000, 3241792, 3652264, 3723875, 4096000, 5451776, 7189057, 8000000, 10218313, 10360232
Offset: 1

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Author

Reinhard Zumkeller, Dec 17 2001

Keywords

Examples

			8^3 = a(0) = 512 = 169 + 343 = 13^2 + 7^3;
10^3 = a(1) = 1000 = 784 + 216 = 28^2 + 6^3.
		

Crossrefs

Programs

  • Mathematica
    q[n_] := Length[Reduce[a^2 + b^3 == n && a > 0 && b > 0, {a, b}, Integers]] > 0; Select[Range[220]^3, q] (* Amiram Eldar, Mar 20 2025 *)

Formula

a(n) = A228946(n)^3. - R. J. Mathar, Dec 03 2015

A293283 Numbers n such that n^2 = a^2 + b^5 for positive integers a b and n.

Original entry on oeis.org

6, 9, 18, 40, 42, 68, 75, 90, 99, 105, 122, 126, 130, 174, 192, 196, 225, 251, 257, 288, 315, 325, 330, 350, 405, 490, 499, 504, 516, 528, 546, 550, 576, 614, 651, 665, 684, 726, 735, 744, 849, 882, 900, 920, 936, 974, 1025, 1032, 1036, 1107, 1140, 1183, 1200
Offset: 1

Views

Author

XU Pingya, Oct 04 2017

Keywords

Comments

For n > 0, k = (n + 1)(2n + 1)^2 is a term in this sequence, because k^2 = (n * (2n + 1)^2)^2 + (2n + 1)^5. Examples: 18, 75, 196, 405, 726, 1183.
When z^2 = x^2 + y^2 (i.e., z = A009003(n)), (z * y^4)^2 = (x * y^4)^2 + (y^2)^5. Thus z * y^4 is a term in this sequence. For example, 1200. More generally, for positive integer i, j and k, x^(5i - 5) * y^(5j - 1) * z^(5k - 5) is in this sequence.
When z^2 = x^2 + y^3 (i.e., z = A070745(n)), (z * y)^2 = (x * y)^2 + y^5. Thus z * y is in this sequence. E.g. 6, 18, 40, ... . More generally, for positive integer i, j and k, x^(5i - 5) * y^(5j - 4) * z^(5k - 4) is in this sequence.
When z^2 = x^2 + y^4 (i.e., z = A271576(n)), (z * y^3)^2 = (x * y^3)^2 + (y^2)^5. Thus z * y^3 is also in this sequence. E.g. 40, 405, 1107, ... . More generally, for positive integer i, j and k, x^(5i - 5) * y^(5j - 2) * z^(5k - 4) is in this sequence.

Examples

			6^2 = 2^2 + 2^5.
9^2 = 7^2 + 2^5.
		

Crossrefs

Programs

  • Mathematica
    c[n_]: = Count[n^2 - Range[(n^2 - 1)^(1/5)]^5, _?(IntegerQ[Sqrt[#]] &)] > 0;
    Select[Range[1200], c]
  • PARI
    isok(n) = for (k=1, n-1, if (ispower(n^2-k^2, 5), return (1));); return (0); \\ Michel Marcus, Oct 06 2017

A293692 Numbers z such that x^2 + y^7 = z^2 for positive integers x and y.

Original entry on oeis.org

12, 18, 33, 54, 126, 160, 272, 366, 375, 520, 531, 540, 594, 630, 756, 825, 945, 1028, 1044, 1094, 1350, 1372, 1506, 1536, 1575, 1980, 2050, 2219, 2304, 2619, 2940, 3250, 3500, 3645, 3906, 3925, 4097, 4224, 4390, 4625, 5500, 5844, 5988, 6048, 6192, 6283, 6422
Offset: 1

Views

Author

XU Pingya, Oct 14 2017

Keywords

Comments

Let i, j and k are nonnegative integers, n is positive integer. As [(n^)^(7i+1) * (2n+1)^(7j + 3) * (n + 1)^(7k)]^2 + [n)^(2i) * (2n + 1)^(2j + 1) * (n + 1)^(2k)]^7 = [n^(7i) * (2n + 1)^(7j + 3) * (n+1)^(7k+1)]^2, so that number of form n^(7i) * (2n + 1)^(7j + 3) * (n+1)^(7k+1) is a term in sequence.
When (x, y, z) is solution of x^2 + y^3 = z^2 (i.e., z = A070745(n)), (x^(7i+1) * y^(7j + 2) * z^(7k)]^2, x^(2i) * y^(2j + 1) * z^(2k), x^(7i) * y^(7j + 2) * z^(7k+1) is solution of x^2 + y^7 = z^2.
When (x, y, z) is solution of x^2 + y^5 = z^2, (i.e., z = A293284(n)), x^(7i+1) * y^(7j + 1) * z^(7k), x^(2i) * y^(2j + 1) * z^(2k), x^(7i) * y^(7j + 1) * z^(7k+1) is solution of x^2 + y^7 = z^2.
When (x, y, z) is solution of x^2 + y^7 = z^2, (x^(7i+1) * y^(7j + 2) * z^(7k), x^(7i) * y^(j + 1) * z^(7k), x^(7i) * y^(7j +2) * z^(7k)) is also.
If x^2 + y^7 = z^2 then y^7 = z^2 - x^2 = (z - x)(z + x) and so (z - x, z + x) is a divisor pair of z^7. - David A. Corneth, May 24 2025

Examples

			4^2 + 2^7 = 12^2, 12 is a term.
31^2 + 2^7 = 33^2, 33 is a term.
		

Crossrefs

Programs

  • Mathematica
    z[n_] := Count[n^2 - Range[(n^2 - 1)^(1/7)]^7, _?(IntegerQ[Sqrt[#]] &)] > 0; Select[Range[6550], z]

A293694 Numbers z such that x^2 + y^8 = z^2 for positive integers x and y.

Original entry on oeis.org

20, 34, 65, 135, 320, 369, 544, 1040, 1095, 1305, 1350, 1404, 1620, 1625, 1746, 1971, 2056, 2160, 2379, 2754, 3060, 3281, 3996, 4100, 4470, 5120, 5265, 5904, 6625, 7825, 7830, 8194, 8575, 8704, 8796, 10250, 10935, 11125, 11700, 12500, 13154, 14500, 15579
Offset: 1

Views

Author

XU Pingya, Oct 16 2017

Keywords

Comments

Let i, j and k be nonnegative integers, m > n be positive integers. As ((m^2 - n^2)^(4*i+1) * (2*m*n)^(4*j+3) * (m^2 + n^2)^(4*k))^2 + ((m^2 - n^2)^i * (2*m*n)^(j+1) * (m^2 + n^2)^k)^8 = ((m^2 - n^2)^(4*i) * (2*m*n)^(4*j+3) * (m^2 + n^2)^(4*k+1))^2, so that the number of the form (m^2 - n^2)^(4*i) * (2*m*n)^(4*j+3) * (m^2 + n^2)^(4*k+1) is a term.
When (x, y, z) is a solution of x^2 + y^4 = z^2 (i.e., z = A271576(n)), (x^(4*i+1) * y^(4*j+2) * z^(4*k), x^i * y^(j+1) * z^k, x^(4*i) * y^(4*j+2) * z^(4*k+1)) is a solution of x^2 + y^8 = z^2.
When (x, y, z) is a solution of x^2 + y^6 = z^2 (i.e., z = A293690(n)), (x^(4*i+1) * y^(4*j+1) * z^(4*k), x^i * y^(j+1) * z^k, x^(4*i) * y^(4*j+1) * z^(4*k+1)) is a solution of x^2 + y^8 = z^2.
When (x, y, z) is a solution of x^2 + y^8 = z^2, (x^(4*i+1) * y^(4*j) * z^(4*k), x^i * y^(j+1) * z^k, x^(4*i) * y^(4*j) * z^(4*k+1)) is also a solution of x^2 + y^8 = z^2.

Examples

			12^2 + 2^8 = 20^2, 20 is a term.
63^2 + 2^8 = 65^2, 65 is a term.
		

Crossrefs

Programs

  • Mathematica
    z[n_] := Count[n^2 - Range[(n^2 - 1)^(1/8)]^8, _?(IntegerQ[Sqrt[#]] &)] > 0; Select[Range[16000], z]

A293690 Numbers z such that x^2 + y^6 = z^2 for positive integers x and y.

Original entry on oeis.org

10, 17, 45, 80, 123, 136, 225, 234, 260, 270, 291, 325, 360, 365, 459, 510, 514, 640, 666, 745, 984, 1025, 1088, 1215, 1225, 1250, 1305, 1450, 1466, 1565, 1740, 1753, 1800, 1872, 1950, 1970, 2022, 2080, 2125, 2160, 2328, 2600, 2628, 2880, 2920, 3172, 3185
Offset: 1

Views

Author

XU Pingya, Oct 14 2017

Keywords

Comments

Let i, j and k be nonnegative integers, m > n be positive integers. As ((m^2 - n^2)^(3*i+1) * (2*m*n)^(3*j+2) * (m^2 + n^2)^(3*k))^2 + ((m^2 - n^2)^i * (2*m*n)^(j+1) * (m^2 + n^2)^k)^6 = ((m^2 - n^2)^(3*i) * (2*m*n)^(3*j+2) * (m^2 + n^2)^(3*k+1))^2, so that the number of the form (m^2 - n^2)^(3*i) * (2*m*n)^(3*j+2) * (m^2 + n^2)^(3*k+1) is a term.
When (x, y, z) is a solution of x^2 + y^4 = z^2 (i.e., z = A271576(n)), (x^(3*i+1) * y^(3*j+1) * z^(3*k), x^i * y^(j+1) * z^k, x^(3*i) * y^(3*j+1) * z^(3*k+1)) is a solution of x^2 + y^6 = z^2.
When (x, y, z) is a solution of x^2 + y^6 = z^2, (x^(3*i+1) * y^(3*j) * z^(3*k), x^i * y^(j+1) * z^k, x^(3*i) * y^(3*j) * z^(3*k+1)) is also a solution of x^2 + y^6 = z^2.

Examples

			6^2 + 2^6 = 10^2, 10 is a term.
15^2 + 2^6 = 17^2, 17 is a term.
		

Crossrefs

Programs

  • Mathematica
    z[n_] := Count[n^2 - Range[(n^2 - 1)^(1/6)]^6, _?(IntegerQ[Sqrt[#]] &)] > 0; Select[Range[3200], z]
Showing 1-5 of 5 results.