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 1-8 of 8 results.

A135911 Number of 4-tuples (x,y,z,t) of nonnegative integers such that x^2+y^3+z^4+t^5 = n.

Original entry on oeis.org

1, 4, 6, 4, 2, 3, 3, 1, 1, 4, 6, 4, 2, 2, 1, 0, 2, 7, 8, 3, 1, 2, 1, 0, 2, 6, 7, 5, 5, 4, 1, 1, 4, 8, 7, 2, 3, 7, 5, 1, 2, 5, 5, 4, 6, 5, 1, 1, 3, 7, 7, 3, 5, 6, 2, 0, 2, 5, 5, 4, 5, 2, 0, 2, 6, 11, 9, 3, 4, 6, 2, 0, 2, 5, 5, 3, 4, 3, 1, 2, 4, 10, 11, 7, 6, 4, 2, 1, 1, 6, 9, 7, 5, 3, 1, 1, 4, 8, 8, 3, 4
Offset: 0

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Author

N. J. A. Sloane, Mar 07 2008

Keywords

Crossrefs

Cf. A111151 (n such that a(n)=0).

A297970 Numbers that are not the sum of 3 squares and a nonnegative 7th power.

Original entry on oeis.org

112, 240, 368, 496, 624, 752, 880, 1008, 1136, 1264, 1392, 1520, 1648, 1776, 1904, 2032, 2160
Offset: 1

Views

Author

XU Pingya, Jan 10 2018

Keywords

Comments

The last term in this sequence is 2160. The reasons are as follows (let b, c, d, i, j, k, m, r, s, t, w, x, y and z be nonnegative integers).
For the Diophantine equation x^2 + y^2 + z^2 + w^7 = m:
(1) If m is not of the form 4^c * (8b + 7), then it follows from Legendre's three-square theorem that the equation has a solution with w = 0.
(2) 8b + 7 - 1^7 = 8b + 6. Then m = 8b + 7, the equation has a solution with w = 1.
(3) 4 * (8b + 7) - 1^7 = (8 * (4b + 3) + 3) = 8d + 3. Then m = 4 * (8b + 7), the equation has a solution with w = 1.
(4) For b >= 17, 16 * (8b + 7) - 3^7 = 8 * (16 * (b - 17) + 12) + 5 = 8i + 5. Then m = 16 * (8b + 7) and b >= 17, the equation has a solution with w = 3.
(5) 4^3 * (8b + 7) - 2^7 = 4^3 * (8b + 5). Then m = 4^3 * (8b + 7), the equation has a solution with w = 2. And 4^3 * (8b + 7) - 3^7 = 8 * (4^3 * (b - 4) + 38) + 5 = 8j + 5. Then m = 4^3 * (8b + 7) and b >= 4, the equation has a solution with w = 3.
(6) 4^4 * (8b + 7) - 2^7 = 4^3 * (8 * (4b + 3) + 3) = 4^3 * (8k + 3). 4^4 * (8b + 7) - 3^7 = 8 * (256b - 217) + 3 = 8r + 3. Then m = 4^4 * (8b + 7), the equation has a solution with w = 2 and when b > 0, the equation has a solution with w = 3.
(7) When c >= 5, 4^c * (8b + 7) - 2^7 = 4^3 * (8 * (b * 4^(c - 3) + 14 * 4^(c - 5) + 5) = 4^3 * (8s + 5). 4^c * (8b + 7) - 3^7 = 8 * (b * 4^(c - 3) + 14 * 4^(c - 3) - 273) + 3 = 8t + 3. Then n = 4^c * (8b + 7), the equation has solutions with w = 2 and 3.
In short, except for the 17 numbers in the sequence, every nonnegative integer can be represented as the sum of 3 squares and a nonnegative 7th power.

Crossrefs

Finite subsequence of A004215 and A296185.

Programs

  • Mathematica
    t1={};
    Do[Do[If[x^2+y^2+z^2+w^7==n, AppendTo[t1,n]&&Break[]], {x,0,n^(1/2)}, {y,x,(n-x^2)^(1/2)}, {z,y,(n-x^2-y^2)^(1/2)}, {w,0,(n-x^2-y^2-z^2)^(1/7)}], {n,0,3000}];
    t2={};
    Do[If[FreeQ[t1,k]==True, AppendTo[t2,k]], {k,0,3000}];
    t2

Formula

a(n) = 128n - 16 = 16 * A004771(n - 1), 1 <= n <= 17.

A351064 Minimal number of positive perfect powers, with different exponents, whose sum is n (considering only minimal possible exponents for bases equal to 1).

Original entry on oeis.org

1, 2, 3, 1, 2, 3, 4, 1, 1, 2, 3, 2, 3, 4, 5, 1, 2, 3, 4, 2, 3, 4, 5, 2, 1, 2, 1, 2, 3, 4, 2, 1, 2, 3, 4, 1, 2, 3, 4, 2, 2, 3, 2, 2, 3, 4, 3, 2, 1, 2, 3, 2, 3, 4, 5, 3, 2, 3, 2, 3, 4, 5, 2, 1, 2, 3, 4, 2, 3, 4, 5, 2, 2, 3, 3, 2, 3, 4, 3, 2, 1, 2, 3, 3, 2, 3, 4, 3, 2, 2, 2, 3, 3, 4, 3, 2, 2, 3, 4, 1, 2, 3, 4, 3, 3, 2
Offset: 1

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Author

Alberto Zanoni, Feb 22 2022

Keywords

Comments

Conjecture: the only numbers for which 5 addends are needed are 15, 23, 55, 62, 71.
The numbers mentioned in the conjecture are also the first five terms of A111151. - Omar E. Pol, Mar 01 2022

Examples

			a(1) = 1 because 1 can be represented with a single positive perfect power: 1 = 1^2.
a(2) = 2 because 2 can be represented with two (and not fewer) positive perfect powers with different exponents: 2 = 1^2 + 1^3.
a(6) = 3 because 6 can be represented with three (and not fewer) positive perfect powers with different exponents: 6 = 2^2 + 1^3 + 1^4.
a(7) = 4 because 7 can be represented with four (and not fewer) positive perfect powers with different exponents: 7 = 2^2 + 1^3 + 1^4 + 1^5.
a(15) = 5 because 15 can be represented with five (and not fewer) positive perfect powers with different exponents: 15 = 2^2 + 2^3 + 1^4 + 1^5 + 1^6.
		

Crossrefs

A296185 Numbers that are not the sum of 3 squares and an 8th power.

Original entry on oeis.org

112, 240, 368, 496, 624, 752, 880, 1008, 1136, 1264, 1392, 1520, 1648, 1776, 1904, 2032, 2160, 2288, 2416, 2544, 2672, 2800, 2928, 3056, 3184, 3312, 3440, 3568, 3696, 3824, 3952, 4080, 4208, 4336, 4464, 4592, 4720, 4848, 4976, 5104, 5232, 5360, 5488, 5616
Offset: 1

Views

Author

XU Pingya, Jan 13 2018

Keywords

Comments

When m is in this sequence, 9m and m^9 are also in this sequence.
For nonnegative integers a, b, k, n, x, y, z and w, n = x^2 + y^2 + z^2 + w^8 if and only if n is not of the form 4^(4k + 2) * (8b + 7).
1. If n is not of the form 4^a * (8b + 7), then it follows from Legendre's three-square theorem that the equation x^2 + y^2 + z^2 + w^8 = n has a solution with w = 0.
2. If n = 4^a * (8b + 7), then (c, d and j are nonnegative integers):
(1) If a = 4k, then n - (2^k)^8 = 4^(4k) * (8b + 6), and the equation has a solution with w = 2^k.
(2) If a = 4k + 1, then n - (2^k)^8 = 4^(4k) * (32b + 27) is of the form 4^c * (8d + 3), and the equation has a solution with w = 2^k.
(3) If a = 4k + 3, then n - (2^(k + 1))^8 = 4^a * (8b + 3), and the equation has a solution with w = 2^(k + 1).
(4) If a = 4k + 2 and w = 2j + 1, then n == 0 (mod 8), w^8 == 1 (mod 8), and n - w^8 is number of the form 8c + 7. I.e., the equation does not have a solution with w odd.
If a = 4k + 2 and w = 4j + 2, then n - w^8 = 16 * (4^(4k) * (8b + 7) - 16 * (2j + 1)^8) = 4^4 * (4^(4k - 4) * (8b + 7) - (2j + 1)^8). When k = 0, n - w^8 is a number of the form 16 * (8c + 7); when k > 0, n - w^8 is a number of the form 4^4 * (8d + 7). Therefore, the equation does not have a solution with w = 4j + 2. Similarly, it can be proved that there is no solution with w = 4j.

Crossrefs

Programs

  • Mathematica
    t1={};
    Do[Do[If[x^2+y^2+z^2+w^8==n, AppendTo[t1,n]&&Break[]], {x,0,n^(1/2)}, {y,x,(n-x^2)^(1/2)}, {z,y,(n-x^2-y^2)^(1/2)}, {w,0,(n-x^2-y^2-z^2)^(1/8)}], {n,0,5700}];
    t2={};
    Do[If[FreeQ[t1,k]==True, AppendTo[t2,k]], {k,0,5700}];
    t2
  • Python
    from itertools import count, islice
    def A296185_gen(): # generator of terms
        for k in count(0):
            r = 1<<((k<<1)+1<<2)
            yield from range(7*r,r*((r<<8)+7),r<<3)
    A296185_list = list(islice(A296185_gen(),44)) # Chai Wah Wu, May 21 2025

A296579 Numbers that are not the sum of 3 squares and a nonnegative 9th power.

Original entry on oeis.org

112, 240, 368, 448, 496, 624, 752, 880, 960, 1008, 1136, 1264, 1392, 1472, 1520, 1648, 1776, 1904, 1984, 2032, 2160, 2288, 2416, 2496, 2544, 2672, 2800, 2928, 3008, 3056, 3184, 3312, 3440, 3520, 3568, 3696, 3824, 3952, 4032, 4080, 4208, 4336, 4464, 4544, 4592
Offset: 1

Views

Author

XU Pingya, Jan 30 2018

Keywords

Comments

a(n) consists of the number of forms 16*(8i + 7) (0 <= i <= 152) and 64*(8j + 7) (0 <= j <= 37).
The last term in this sequence is a(191) = 19568 = 16*(8*152 + 7) (see A297970).

Crossrefs

Finite subsequence of A004215.
A297970 is a subsequence.

Programs

  • Mathematica
    t1=Table[4^2*(8j+7), {j,0,152}];
    t2=Table[4^3*(8j+7), {j,0,37}];
    t=Union[t1, t2]

A297931 Numbers that are not the sum of a square and 3 nonnegative cubes.

Original entry on oeis.org

15, 22, 23, 48, 86, 94, 112, 120, 139, 184, 203, 211, 230, 237, 263, 301, 309, 312, 335, 373, 399, 1014, 1056, 1455, 1644, 2029, 2272, 2658, 3309, 3469, 4019, 6502, 11101
Offset: 1

Views

Author

XU Pingya, Jan 08 2018

Keywords

Comments

After 11101, there are no more terms up to 570000.
No more terms < 10^10; is this sequence finite? - Mauro Fiorentini, Jan 26 2019

Crossrefs

Programs

  • Mathematica
    t1={};
    Do[Do[If[x^2+y^2+z^2+w^3==n, AppendTo[t1,n]&&Break[]], {x,0,n^(1/2)}, {y,x,(n-x^2)^(1/2)}, {z,y,(n-x^2-y^2)^(1/2)}, {w,0,(n-x^2-y^2-z^2)^(1/3)}], {n,0,5.7*10^5}];
    t2={};
    Do[If[FreeQ[t1,k]==True, AppendTo[t2,k]], {k,0,5.7*10^5}];
    t2

A332664 a(n) = number of nonnegative integers that are not the sum of {2 squares, a nonnegative 5th power, and a nonnegative n-th power}.

Original entry on oeis.org

0, 2, 14, 115, 116, 109, 245, 381, 1387, 913, 1234, 1552, 2103, 2838, 3036, 3384, 4693, 5405, 8304, 9088, 11089, 13289, 15815, 18619, 20979, 22755, 24107, 24984, 25548
Offset: 2

Views

Author

XU Pingya, Feb 18 2020

Keywords

Comments

a(2) = 0 by a theorem of Zhi-Wei Sun, see A273915. All terms beyond a(2) are conjectures and have only been checked to 4*10^9.

Examples

			a(2) = 0, since any nonnegative integer k is the sum of 3 squares and a nonnegative 5th power (see A273915).
a(4) = 14. Since any nonnegative integer k (<= 4*10^9) is the sum of {2 squares, a nonnegative 5th power, and a 4th power}, except for 14 numbers: 23, 44, 71, 79, 215, 383, 863, 1439, 1583, 1727, 1759, 1919, 2159, 2543.
		

Crossrefs

Programs

  • Mathematica
    a(5)
    Do[m=1000000 (k-1)+1; n=1000000 k;
      t=Union@Flatten@Table[x^2 + y^2 + z^5 + w^5,
    {x,0,n^(1/2)}, {y,x,(n-x^2)^(1/2)}, {z,0,(n-x^2-y^2)^(1/5)},
    {w, If[x^2 + y^2 + z^5 < m, Floor[(m-1-x^2-y^2-z^5)^(1/5)] + 1, z], (n-x^2-y^2-z^5)^(1/5)}];
      b=Complement[Range[m, n], t];
      Print[Length@b], {k,4000}]

A332665 a(n) = largest number k that is not the sum of {2 squares, a nonnegative 5th power, and a nonnegative n-th power}.

Original entry on oeis.org

71, 2543, 146687, 55871, 63703, 353087, 606239, 21073228, 4606428, 19212172, 16301375, 88507864, 228238087, 126691212, 121909855, 366845415, 448525727, 1818767247, 1319413228, 341968924, 673747279, 188739607, 2366756975, 1770393855, 411629407, 1240489516
Offset: 3

Views

Author

XU Pingya, Feb 19 2020

Keywords

Comments

All terms are conjectural and have only been checked up to 4*10^9.

Examples

			a(4) = 2543, since 2543 is the greatest number such that is not the sum of {2 squares, a nonnegative 5th power, and a 4th power} (at least up to 4*10^9).
		

Crossrefs

Showing 1-8 of 8 results.