A111925 -id:A111925 - cp's OEIS frontend

A300565 Numbers z such that there is a solution to x^3 + y^4 = z^5 with x, y, z >= 1.

32, 250, 1944, 2744, 3888, 19208, 27648, 55296, 59049, 59582, 81000, 82944, 131072, 135000, 185193, 200000

Offset: 1

M. F. Hasler, Apr 16 2018

Consider a solution (x, y, z), x^3 + y^4 = z^5. For any m, (x*m^20, y*m^15, z*m^12) will also be a solution. If (x/m^20, y/m^15, z/m^12) is a triple of integers, it is also a solution. A solution is called primitive if there is no such m > 1.

If S = a^3 + b^8/4 is a square, for some a,b > 0, then z = b^4/2 + sqrt(S) is in the sequence, with x = a*z and y = b*z. All known terms are of this form, with b in {2, 6, 7, 9, 12}, only for a(2) and a(10) one must consider half-integral b = 5/2 resp. 31/2. Also of this form is z = 81000, 82944, 131072, 135000, 185193, 200000, 243000, 395307, 474552, 574992, 800000, 820125, 862488, 864000, 972000, ... (with integer b), and 444528 (with b = 33/2).

			a(1) = 32 = 2^5 is in the sequence because (2^5)^5 = (2^6)^4 + (2^8)^3, using 1 + 1 = 2.
a(2) = 250 = 2*5^3 is in the sequence because 250^5 = 2^5*5^15 = (5^4)^4 + (3*5^5)^3, using 5 + 3^3 = 2^5. This solution is special because x and y are not multiples of z.
a(3) = 1944 = 2^3*3^5 is in the sequence because 1944^5 = (2^4*3^6)^4 + (2^5*3^8)^3, using 2 + 1 = 3.
a(7) = 27648 = 2^10*3^3 is in the sequence because (2^10*3^3)^5 = (2^12*3^4)^4 + (2^16*3^5)^3, using 3 + 1 = 2^2.
a(10) = 59582 = 2*31^3 is in the sequence because (2*31^3)^5 = (31^4)^4 + (31^5)^3, using 31 + 1 = 2^5. This is the second case where x and y are not multiples of z.

Cf. A300564 (z^4 = x^2 + y^3), A111925 (z^5 = x^2 + y^4), A302174, A300566 (z^6 = x^4 + y^5).

is(z)=for(y=1,sqrtnint(-1+z=z^5,4),ispower(z-y^4,3)&&return(y))

a(11)-a(16) from Giovanni Resta, Jan 23 2020

A303374 Numbers of the form a^4 + b^6, with integers a, b > 0.

Original entry on oeis.org

2, 17, 65, 80, 82, 145, 257, 320, 626, 689, 730, 745, 810, 985, 1297, 1354, 1360, 2025, 2402, 2465, 3130, 4097, 4112, 4160, 4177, 4352, 4721, 4825, 5392, 6497, 6562, 6625, 7290, 8192, 10001, 10064, 10657, 10729, 14096, 14642, 14705, 15370, 15626, 15641, 15706, 15881

Offset: 1

M. F. Hasler, Apr 22 2018

A subsequence of A000404 (a^2 + b^2), A055394 (a^2 + b^3), A111925 (a^4 + b^2), A100291 (a^4 + b^3), A303372 (a^2 + b^6).

Although it is easy to produce many terms of this sequence, it is nontrivial to check whether a very large number is of this form. Maybe the most efficient way is to consider decompositions of n into sums of two positive squares (see sum2sqr in A133388), and check if one of the terms is a third power and the other a fourth power.

Cf. A055394 (a^2 + b^3), A111925 (a^2 + b^4), A100291 (a^4 + b^3), A100292 (a^5 + b^2), A100293 (a^5 + b^3), A100294 (a^5 + b^4).

Cf. A303372 (a^2 + b^6), A303373 (a^3 + b^6), A303375 (a^5 + b^6).

Take[Flatten[Table[a^4+b^6,{a,20},{b,20}]]//Union,50] (* Harvey P. Dale, Jul 17 2025 *)

is(n,k=4,m=6)=for(b=1,sqrtnint(n-1,m),ispower(n-b^m,k)&&return(b)) \\ Returns b > 0 if n is in the sequence, else 0.
is(n,L=sum2sqr(n))={for(i=1,#L,L[i][1]&&for(j=1,2,ispower(L[i][j],3)&&issquare(L[i][3-j])&&return(L[i][j])))} \\ See A133388 for sum2sqr(). Much faster than the above for n >> 10^30.
A303374(L=10^5,k=4,m=6,S=[])={for(a=1,sqrtnint(L-1,m),for(b=1,sqrtnint(L-a^m,k),S=setunion(S,[a^m+b^k])));S}

A345645 Numbers whose square can be represented in exactly one way as the sum of a square and a biquadrate (fourth power).

Original entry on oeis.org

5, 15, 20, 34, 39, 41, 45, 60, 80, 85, 111, 125, 135, 136, 150, 156, 164, 175, 180, 194, 219, 240, 245, 255, 265, 306, 313, 320, 325, 340, 351, 353, 369, 371, 375, 405, 410, 444, 445, 455, 500, 505, 514, 540, 544, 600, 605, 609, 624, 629, 656, 671, 674, 689

Offset: 1

Mohammad Tejabwala, Jun 21 2021

nonn

Numbers z such that there is exactly one solution to z^2 = x^2 + y^4.
From Karl-Heinz Hofmann, Oct 21 2021: (Start)
No term can be a square (see the comment from Altug Alkan in A111925).
Terms must have at least one prime factor of the form p == 1 (mod 4), a Pythagorean prime (A002144).
Additionally, if the terms have prime factors of the form p == 3 (mod 4), which are in A002145, then they must appear in the prime divisor sets of x and y too.
The special prime factor 2 has the same behavior, i.e., if the term is even, x and y must be even too. (End)

			3^2 + 2^4 = 9 + 16 = 25 = 5^2, so 5 is a term.
60^2 + 5^4 = 63^2 + 4^4 = 65^2, so 65 is not a term.

Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000

Cf. A000290, A000583, A180241, A271576 (all solutions).
Cf. A345700 (2 solutions), A345968 (3 solutions), A346110 (4 solutions), A348655 (5 solutions), A349324 (6 solutions), A346115 (the least solutions).
Cf. A002144 (p == 1 (mod 4)), A002145 (p == 3 (mod 4)).

Select[Range@100,Length@Solve[x^2+y^4==#^2&&x>0&&y>0,{x,y},Integers]==1&] (* Giorgos Kalogeropoulos, Jun 25 2021 *)

inlist(list, v) = for (i=1, #list, if (list[i]==v, return(1)));
isok(m) = {my(list = List()); for (k=1, sqrtnint(m^2, 4), if (issquare(j=m^2-k^4) && !inlist(vecsort([k^4,j^2])), listput(list, vecsort([k^4,j^2])));); #list == 1;} \\ Michel Marcus, Jun 26 2021

terms = []
for i in range(1, 700):
    occur = 0
    ii = i*i
    for j in range(1, i):
        k = int((ii - j*j) ** 0.25)
        if k*k*k*k + j*j == ii:
            occur += 1
    if occur == 1:
        terms.append(i)
print(terms)

A345700 Numbers whose square can be represented in exactly two ways as the sum of a positive square and a positive fourth power.

Original entry on oeis.org

65, 145, 260, 580, 585, 1025, 1040, 1105, 1305, 2320, 2340, 2465, 3185, 3625, 4100, 4160, 4335, 4420, 4810, 5220, 5265, 6625, 7105, 7585, 7865, 8405, 9225, 9280, 9360, 9860, 9945, 10985, 11745, 12740, 14500, 16400, 16465, 16640, 17340, 17545, 17680, 19240

Offset: 1

Karl-Heinz Hofmann, Jun 24 2021

nonn

Terms are numbers z such that there are exactly two solutions to z^2 = x^2 + y^4, where x, y and z belong to the set of positive integers.
Terms cannot be a square (see the comment from Altug Alkan in A111925).
Terms must have at least one prime factor of the form p == 1 (mod 4), a Pythagorean prime (A002144).
If the terms additionally have prime factors of the form p == 3 (mod 4), which are in A002145, then they must appear in the prime divisor sets of x and y too.
The lower limit of the ratio x/y is sqrt(2).

			29679^2 = 29640^2 + 39^4, so 29679 is not a term (only one solution).
17680^2 = 15088^2 + 96^4 = 17472^2 + 52^4, so 17680 is a term.
36900^2 = 36000^2 + 90^4 = 36828^2 + 48^4, so 36900 is a term.
18785^2 = 18207^2 + 68^4 = 17340^2 + 85^4 = 13983^2 + 112^4, so 18785 is not a term (three solutions).

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000
Karl-Heinz Hofmann, All valid {z,x1,y1,x2,y2} sets up to 10^7
Karl-Heinz Hofmann, Python code for multiple solutions

Cf. A111925, A271576, A345645 (in exactly 1 way), A345968 (in exactly 3 ways), A346110 (in exactly 4 ways), A348655 (in exactly 5 ways), A349324 (in exactly 6 ways), A346115 (the least solutions).

Cf. A002144 (p == 1 (mod 4)), A002145 (p == 3 (mod 4)).

inlist(list, v) = for (i=1, #list, if (list[i]==v, return(1)));
isok(m) = {my(list = List()); for (k=1, sqrtnint(m^2, 4), if (issquare(j=m^2-k^4) && !inlist(vecsort([k^4,j^2])), listput(list, vecsort([k^4,j^2])));); #list == 2;} \\ Michel Marcus, Jun 26 2021

is(n)=my(n2=n^2,s); for(y=sqrtnint(2*n-2,4)+1,sqrtint(n-1), if(issquare(n2-y^4) && s++>2, return(0))); s==2; \\ Charles R Greathouse IV, Jul 02 2021

Python
```
# see link above.
```

A345968 Numbers whose square can be represented in exactly three ways as the sum of a positive square and a positive fourth power.

Original entry on oeis.org

1625, 6500, 14625, 18785, 24505, 26000, 40625, 58500, 75140, 79625, 88985, 98020, 104000, 120250, 131625, 162500, 169065, 196625, 220545, 234000, 274625, 296225, 300560, 318500, 355940, 365625, 392080, 416000, 481000, 526500, 547230, 586625, 611585, 612625

Offset: 1

Karl-Heinz Hofmann, Jun 30 2021

nonn

Terms are numbers z such that there are exactly 3 solutions to z^2 = x^2 + y^4, where x, y and z belong to the set of positive integers.
No term can be a square (see the comment from Altug Alkan in A111925).
Terms must have at least one prime factor of the form p == 1 (mod 4), a Pythagorean prime (A002144).
Additionally, if the terms have prime factors of the form p == 3 (mod 4), which are in A002145, then they must appear in the prime divisor sets of x and y too.
The special prime factor 2 has the same behavior, i.e., if the term is even, x and y must be even too.

			29640^2 + 39^4 = 29679^2; 29679 is not a term (only 1 solution).
60^2 + 5^4 = 63^2 + 4^4 = 65^2; 65 is not a term (only 2 solutions).
572^2 + 39^4 = 1500^2 + 25^4 = 1575^2 + 20^4 = 1625^2; 1625 is a term (3 solutions).
165308^2 + 663^4 = 349575^2 + 560^4 = 433500^2 + 425^4 = 455175^2 + 340^4 = 469625^2; 469625 is not a term (4 solutions).

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000
Karl-Heinz Hofmann, All valid {z,x1,y1,x2,y2,x3,y3} sets up to 10^8

Cf. A271576 (1 and more solutions), A345645 (1 solution), A345700 (2 solutions), A346110 (4 solutions), A348655 (5 solutions), A349324 (6 solutions), A346115 (the least solutions).

Cf. A002144 (p == 1 (mod 4)), A002145 (p == 3 (mod 4)).

A346115 Least number k such that k^2 can be expressed in exactly n ways as x^2 + y^4 with {x, y} >= 1.

Original entry on oeis.org

5, 65, 1625, 469625, 642916625, 15697403475, 2474052064291275

Offset: 1

Karl-Heinz Hofmann, Jul 05 2021

a(5) <= 642916625. - Hugo Pfoertner, Jul 07 2021
From Karl-Heinz Hofmann, Sep 02 2021: (Start)
a(6) <= 15697403475.
2 * 10^10 < a(7) <= 2474052064291275.
These two conjectured values arise from the "green group". Up to term a(5) the least solutions are in the "blue group". Follow the links below to get more information about the different colored groups.
Terms cannot be a square (see the comment from Altug Alkan in A111925).
Terms must have at least one prime factor of the form p == 1 (mod 4), a Pythagorean prime (A002144).
If the terms additionally have prime factors of the form p == 3 (mod 4), which are in A002145, then they must appear in the prime divisor sets of x and y too. The special prime factor 2 has the same behavior, i.e., if the term is even, x and y must be even too. (End)

			a(1)=A345645(1); a(2)=A345700(1); a(3)=A345968(1); a(4)=A346110(1).

Karl-Heinz Hofmann, All valid {z,x1,y1,x2,y2,x3,y3,x4,y4} sets up to 10^9
Karl-Heinz Hofmann, A 3D Animation of the 4 solutions and the least 5 solution up to 10^9

Cf. A000290, A000583, A111925.
Cf. A271576 (1 or more solutions), A345645 (1 solution),  A345700 (2 solutions), A345968 (3 solutions), A346110 (4 solutions), A348655 (5 solutions), A349324 (6 solutions).
Cf. A002144 (p == 1 (mod 4)), A002145 (p == 3 (mod 4)).

a(5) confirmed by Martin Ehrenstein, Jul 09 2021
a(6) confirmed by Karl-Heinz Hofmann, Oct 15 2021
a(7) confirmed by Jon E. Schoenfield, Nov 15 2021

A348655 Numbers whose square can be represented in exactly five ways as the sum of a positive square and a positive fourth power.

Original entry on oeis.org

642916625, 2571666500, 4418701625, 5786249625, 10286666000, 16072915625, 17674806500, 20931496625, 23144998500, 31502914625, 39768314625, 41146664000, 52076246625, 57801168750, 64291662500, 70699226000, 77792911625, 83725986500, 92579994000, 108652909625

Offset: 1

Karl-Heinz Hofmann, Oct 27 2021

nonn

Numbers z such that there are exactly 5 solutions to z^2 = x^2 + y^4.
Terms cannot be a square (see the comment from Altug Alkan in A111925).
Terms must have at least one prime factor of the form p == 1 (mod 4), a Pythagorean prime (A002144).
If the terms additionally have prime factors of the form p == 3 (mod 4), which are in A002145, then they must appear in the prime divisor sets of x and y too.
Some other terms of the sequence: 20931496625, 23144998500, 31502914625, 41146664000, 52076246625, 64291662500, 77792911625, 83725986500, 92579994000, 108652909625, 126011658500, 144656240625, 164586656000. - Chai Wah Wu, Oct 29 2021

			5786249625^2 = 5785404300^2 +  9945^4
5786249625^2 = 5608211175^2 + 37740^4
5786249625^2 = 5341153500^2 + 47175^4
5786249625^2 = 4307113575^2 + 62160^4
5786249625^2 = 2036759868^2 + 73593^4

Jon E. Schoenfield, Table of n, a(n) for n = 1..3989 (all terms < A346115(7) = 2474052064291275)

Cf. A111925, A271576, A345645 (in exactly 1 way), A345700 (in exactly 2 ways), A345968 (in exactly 3 ways), A346110 (in exactly 4 ways), A349324 (in exactly 6 ways), A346115 (the least solutions).

Cf. A002144 (p == 1 (mod 4)), A002145 (p == 3 (mod 4)).

a(8) and beyond from Jon E. Schoenfield, Nov 14 2021

A349324 Numbers whose square can be represented in exactly six ways as the sum of a positive square and a positive fourth power.

Original entry on oeis.org

15697403475, 62789613900, 141276631275, 251158455600, 392435086875, 565106525100, 769172770275, 1004633822400, 1271489681475, 1569740347500, 1899385820475, 2260426100400, 2652861187275, 3076691081100, 3531915781875, 4018535289600, 4536549604275, 5085958725900

Offset: 1

Jon E. Schoenfield, Nov 14 2021

nonn

Numbers z such that there are exactly 6 solutions to z^2 = x^2 + y^4 with x, y, z positive integers.
See also A348655.
Note that y^4 = z^2 - x^2 = (z - x)*(z + x), so each solution corresponds to a positive integer y and a factorization of y^4 into two positive integers f = z - x and g = z + x. We need f < g so that x > 0, and we need f == g (mod 2) so that (f + g)/2 = (z - x + z - x)/2 = z will be an integer.
Note also that it follows from y^4 = z^2 - x^2 that y < sqrt(z).
Thus, given some value zMax, we can generate the value of z for each solution to z^2 = x^2 + y^4 over the positive integers with z <= zMax by simply computing z = (f + y^4/f)/2 for (1) every divisor f (< y^2) of y^4 for each odd y < sqrt(zMax) and (2) every even divisor f (< y^2) of y^4 such that y^4/f is also even for each even y < sqrt(zMax), and discarding those results that exceed zMax. For any given y, the smaller the value of f, the larger the resulting z. So, for each y, we can test the divisors of y^4 that are less than y^2 in descending order, and move on to the next value of y as soon as we either exhaust all the allowed divisors or get a value of z that exceeds zMax.
For example, to get the z value for every solution with z <= 100, we can factor y^4 for each y in 1..9 as follows:
.
  y   y^4   f  g = y^4/f     z = (f+g)/2 (or comments)
  -  ----  --  ---------  -------------------------------
  1     1   -       -                        -
  2    16   2       8     ( 2 +   8)/2  =    5 (solution)
  3    81   3      27     ( 3 +  27)/2  =   15 (solution)
  "    "    1      81     ( 1 +  81)/2  =   41 (solution)
  4   256   8      32     ( 8 +  32)/2  =   20 (solution)
  "    "    4      64     ( 4 +  64)/2  =   34 (solution)
  "    "    2     128     ( 2 + 128)/2  =   65 (solution)
  5   625   5     125     ( 5 + 125)/2  =   65 (solution)
  "    "    1     625     ( 1 + 625)/2  =  313 > 100
  6  1296  24      54     (24 +  54)/2  =   39 (solution)
  "    "   18      72     (18 +  72)/2  =   45 (solution)
  "    "   12     108     (12 + 108)/2  =   60 (solution)
  "    "    8     162     ( 8 + 162)/2  =   85 (solution)
  "    "    6     216     ( 6 + 216)/2  =  111 > 100
  7  2401   7     343     ( 7 + 343)/2  =  175 > 100
  8  4096  32     128     (32 + 128)/2  =   80 (solution)
  "    "   16     256     (16 + 256)/2  =  136 > 100
  9  6561  27     243     (27 + 243)/2  =  135 > 100
.
so, sorted in ascending order, the z values <= 100 that occur are 5, 15, 20, 34, 39, 41, 45, 65, 65, 60, 80, 85. (The smallest z value, 5, occurs only once, so it's A345645(1); 65, the smallest value that occurs twice, is A345700(1).)
From Karl-Heinz Hofmann, Nov 15 2021: (Start)
Terms cannot be a square (see the comment from Altug Alkan in A111925).
Terms must have at least one prime factor of the form p == 1 (mod 4), a Pythagorean prime (A002144).
If the terms additionally have prime factors of the form p == 3 (mod 4), which are in A002145, then they must appear in the prime divisor sets of x and y too.
The special prime factor 2 has the same behavior, i.e., if the term is even, x and y must be even too. (End)

			15697403475^2 = 13780596525^2 +  86700^4
              = 13195420077^2 +  92208^4
              = 11468350875^2 + 103530^4
              = 10710751443^2 + 107124^4
              =  9221086875^2 + 112710^4
              =  4878327597^2 + 122148^4.

Jon E. Schoenfield, Table of n, a(n) for n = 1..424 (all terms < A346115(7) = 2474052064291275)
Karl-Heinz Hofmann, A 3D animation of the terms and the least one with 7 solutions

Cf. A111925, A271576, A345645 (in exactly 1 way), A345700 (in exactly 2 ways), A345968 (in exactly 3 ways), A346110 (in exactly 4 ways), A348655 (in exactly 5 ways), A346115 (the least solutions).

Cf. A002144 (p == 1 (mod 4)), A002145 (p == 3 (mod 4)).

A303372 Numbers of the form a^2 + b^6, with integers a, b > 0.

Original entry on oeis.org

2, 5, 10, 17, 26, 37, 50, 65, 68, 73, 80, 82, 89, 100, 101, 113, 122, 128, 145, 164, 170, 185, 197, 208, 226, 233, 257, 260, 289, 290, 320, 325, 353, 362, 388, 401, 425, 442, 464, 485, 505, 530, 548, 577, 593, 626, 640, 677, 689, 730, 733, 738, 740, 745, 754, 765, 778

Offset: 1

M. F. Hasler, Apr 22 2018

A subsequence of A055394, the numbers of the form a^2 + b^3.

Although it is easy to produce many terms of this sequence, it is nontrivial to check whether a very large number is of this form.

			The first terms are 1^2 + 1^6 = 2, 2^2 + 1^6 = 5, 3^2 + 1^6 = 10, 4^2 + 1^6 = 17, 5^2 + 1^6 = 26, ..., 8^2 + 1^6 = 1^2 + 2^6 = 65, 2^2 + 2^6 = 68, 3^2 + 2^6 = 73, ...

Cf. A055394 (a^2 + b^3), A111925 (a^2 + b^4), A100291 (a^4 + b^3), A100292 (a^5 + b^2), A100293 (a^5 + b^3), A100294 (a^5 + b^4).

Cf. A303373 (a^3 + b^6), A303374 (a^4 + b^6), A303375 (a^5 + b^6).

is(n,k=2,m=6)=for(b=1,sqrtnint(n-1,m),ispower(n-b^m,k)&&return(b)) \\ Returns b > 0 if n is in the sequence, else 0.
A303372_vec(L=10^5,k=2,m=6,S=List())={for(a=1,sqrtnint(L-1,m),for(b=1,sqrtnint(L-a^m,k),listput(S,a^m+b^k)));Set(S)} \\ List of all terms up to limit L

A303373 Numbers of the form a^3 + b^6, with integers a, b > 0.

Original entry on oeis.org

2, 9, 28, 65, 72, 91, 126, 128, 189, 217, 280, 344, 407, 513, 576, 730, 737, 756, 793, 854, 945, 1001, 1064, 1072, 1241, 1332, 1395, 1458, 1729, 1792, 2060, 2198, 2261, 2457, 2745, 2808, 2926, 3376, 3439, 3473, 4097, 4104, 4123, 4160, 4221, 4312, 4439, 4608, 4825, 4914

Offset: 1

M. F. Hasler, Apr 22 2018

A subsequence of the numbers of the form a^3 + b^2, A055394.

Although it is easy to produce many terms of this sequence, it is nontrivial to check whether a very large number is of this form.

			The first terms are 1^3 + 1^6 = 2, 2^3 + 1^6 = 9, 3^3 + 1^6 = 28, 4^3 + 1^6 = 65, 2^3 + 2^6 = 72, 3^3 + 2^6 = 91, 5^3 + 1^6 = 126, 4^3 + 2^6 = 128, ...

Cf. A055394 (a^2 + b^3), A111925 (a^2 + b^4), A100291 (a^4 + b^3), A100292 (a^5 + b^2), A100293 (a^5 + b^3), A100294 (a^5 + b^4).

Cf. A303372 (a^2 + b^6), A303374 (a^4 + b^6), A303375 (a^5 + b^6).

is(n,k=3,m=6)=for(b=1,sqrtnint(n-1,m),ispower(n-b^m,k)&&return(b)) \\ Returns b > 0 if n is in the sequence, else 0.
A303373_vec(L=10^5,k=3,m=6,S=List())={for(a=1,sqrtnint(L-1,m),for(b=1,sqrtnint(L-a^m,k),listput(S,a^m+b^k)));Set(S)} \\ List of all terms up to limit L

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A300565 Numbers z such that there is a solution to x^3 + y^4 = z^5 with x, y, z >= 1.

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A303374 Numbers of the form a^4 + b^6, with integers a, b > 0.

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A345645 Numbers whose square can be represented in exactly one way as the sum of a square and a biquadrate (fourth power).

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A345700 Numbers whose square can be represented in exactly two ways as the sum of a positive square and a positive fourth power.

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PARI

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A345968 Numbers whose square can be represented in exactly three ways as the sum of a positive square and a positive fourth power.

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A346115 Least number k such that k^2 can be expressed in exactly n ways as x^2 + y^4 with {x, y} >= 1.

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A348655 Numbers whose square can be represented in exactly five ways as the sum of a positive square and a positive fourth power.

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A349324 Numbers whose square can be represented in exactly six ways as the sum of a positive square and a positive fourth power.

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A303372 Numbers of the form a^2 + b^6, with integers a, b > 0.

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