Search: a346115 -id:a346115
|
| |
|
|
A271576
|
|
Integers whose square is of the form x^2 + y^4, with x,y > 0 (see A111925).
|
|
+10
16
|
|
|
|
5, 15, 20, 34, 39, 41, 45, 60, 65, 80, 85, 111, 125, 135, 136, 145, 150, 156, 164, 175, 180, 194, 219, 240, 245, 255, 260, 265, 306, 313, 320, 325, 340, 351, 353, 369, 371, 375, 405, 410, 444, 445, 455, 500, 505, 514, 540, 544, 580, 585, 600, 605, 609, 624, 629, 656, 671, 674, 689
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
If z is in this sequence, z*k^2 is, for k > 0. Also note that since there are no fourth powers in A111925, there are no squares in this sequence. - Altug Alkan, Apr 10 2016
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. (End)
Apparently, the vast majority of the terms of this sequence can be expressed as x^2 + y^4 with x,y > 0 in only one way (A345645 lists those terms), but some can be so expressed in exactly two, three, four, five, six, or more ways. Among the first 3976926961 terms of this sequence,
3948648229 are in A345645 (exactly 1 way),
25415062 are in A345700 (exactly 2 ways),
2697713 are in A345968 (exactly 3 ways),
161543 are in A346110 (exactly 4 ways),
3989 are in A348655 (exactly 5 ways),
424 are in A349324 (exactly 6 ways),
and just 1 -- a(3976926961) = 2474052064291275 = A346115(7) -- is a number whose square can be written as x^2 + y^4 with x,y > 0 in exactly 7 ways. The ratios of successive counts above, i.e., 3948648229/25415062, 25415062/2697713, 2697713/161543, 161543/3989, 3989/424, and 424/1, are approximately 155.4, 9.4, 16.7, 40.5, 9.4, and 424.0. What is it that (over the interval [1, 2474052064291275], at least) makes numbers whose squares can be written as x^2 + y^4 with x,y > 0 in more than 6 ways so much rarer than those that can be written thus in exactly 6 ways? (End)
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
5^2 = 25 = 9 + 16 = 3^2 + 2^4, so 5 is a term.
|
|
|
MATHEMATICA
|
Select[Range@ 200, Resolve[Exists[{x, y}, Reduce[#^2 == x^2 + y^4, {x, y}, Integers], And[x > 0, y > 0]]] &] (* Michael De Vlieger, Apr 10 2016 *)
|
|
|
PROG
|
(PARI) isok(n) = n = n^2; for (k=1, n-1, if (issquare(k) && (p=ispower(n-k, 4)), return (1)))
(PARI) is(n)=my(n2=n^2); for(b=sqrtnint(2*n-2, 4)+1, sqrtint(n-1), if(issquare(n2-b^4), return(1))); 0 \\ Charles R Greathouse IV, Nov 16 2021
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A346110
|
|
Numbers whose square can be represented in exactly four ways as the sum of a positive square and a positive fourth power.
|
|
+10
10
|
|
|
|
469625, 1878500, 2224625, 4226625, 7514000, 8898500, 11740625, 15289625, 16906500, 20021625, 23011625, 25716665, 30056000, 35594000, 38039625, 46962500, 54316275, 55615625, 56824625, 61158500, 67626000, 79366625, 80086500, 92046500, 92481870
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Terms are numbers z such that there are exactly four 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 special prime factor 2 has the same behavior. Means: If the term is even, x and y must be even too.
Apparently, all terms are divisible by 65. The divided terms are in A346594. Are there exceptions for n > 25? - Hugo Pfoertner, Jul 14 2021, Jul 29 2021
Yes, there are exceptions: a(44,46,53,95,97) are not divisible by 65 (5*13) but they have in common: They are divisible by 145 (5*29). - Karl-Heinz Hofmann, Aug 28 2021
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
29679^2 = 29640^2 + 39^4, so 29679 is not a term (only one solution).
60^2 + 5^4 = 63^2 + 4^4 = 65^2, so 65 is not a term (only two solutions).
572^2 + 39^4 = 1500^2 + 25^4 = 1575^2 + 20^4 = 1625^2, so 1625 is not a term (only three solutions).
165308^2 + 663^4 = 349575^2 + 560^4 = 433500^2 + 425^4 = 455175^2 + 340^4 = 469625^2, so 469625 is a term (four solutions).
|
|
|
PROG
|
(Python) see link above
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A345645
|
|
Numbers whose square can be represented in exactly one way as the sum of a square and a biquadrate (fourth power).
|
|
+10
9
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Numbers z such that there is exactly one solution to z^2 = x^2 + y^4.
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)
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
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.
|
|
|
MATHEMATICA
|
Select[Range@100, Length@Solve[x^2+y^4==#^2&&x>0&&y>0, {x, y}, Integers]==1&] (* Giorgos Kalogeropoulos, Jun 25 2021 *)
|
|
|
PROG
|
(Python)
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)
(PARI) 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
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A345700
|
|
Numbers whose square can be represented in exactly two ways as the sum of a positive square and a positive fourth power.
|
|
+10
9
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
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).
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
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).
|
|
|
PROG
|
(Python) # see link above.
(PARI) 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
(PARI) 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
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A345968
|
|
Numbers whose square can be represented in exactly three ways as the sum of a positive square and a positive fourth power.
|
|
+10
9
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
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.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
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).
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A348655
|
|
Numbers whose square can be represented in exactly five ways as the sum of a positive square and a positive fourth power.
|
|
+10
9
|
|
|
|
642916625, 2571666500, 4418701625, 5786249625, 10286666000, 16072915625, 17674806500, 20931496625, 23144998500, 31502914625, 39768314625, 41146664000, 52076246625, 57801168750, 64291662500, 70699226000, 77792911625, 83725986500, 92579994000, 108652909625
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
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
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
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
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A349324
|
|
Numbers whose square can be represented in exactly six ways as the sum of a positive square and a positive fourth power.
|
|
+10
9
|
|
|
|
15697403475, 62789613900, 141276631275, 251158455600, 392435086875, 565106525100, 769172770275, 1004633822400, 1271489681475, 1569740347500, 1899385820475, 2260426100400, 2652861187275, 3076691081100, 3531915781875, 4018535289600, 4536549604275, 5085958725900
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Numbers z such that there are exactly 6 solutions to z^2 = x^2 + y^4 with x, y, z positive integers.
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).)
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)
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
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.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A349664
|
|
a(n) is the number of solutions for n^4 = z^2 - x^2 with {z,x} >= 1.
|
|
+10
2
|
|
|
|
0, 1, 2, 3, 2, 7, 2, 5, 4, 7, 2, 17, 2, 7, 12, 7, 2, 13, 2, 17, 12, 7, 2, 27, 4, 7, 6, 17, 2, 37, 2, 9, 12, 7, 12, 31, 2, 7, 12, 27, 2, 37, 2, 17, 22, 7, 2, 37, 4, 13, 12, 17, 2, 19, 12, 27, 12, 7, 2, 87, 2, 7, 22, 11, 12, 37, 2, 17, 12, 37, 2, 49, 2, 7, 22
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,3
|
|
|
COMMENTS
|
If n is an odd prime^i, the number of solutions is 2*i.
If n = 2^i, the number of solutions is 2*i-1.
These two facts are not generally valid in reverse for terms > 6.
If a(n) = 2, n is an odd prime. This is generally valid in reverse.
For more information about these facts see the link.
Conditions to be satisfied for a valid, countable solution:
- z cannot be a square.
- z must have at least one prime factor of the form p == 1 (mod 4), a Pythagorean prime (A002144).
- If z has 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 n too.
- If z is even, x and n must be even too.
- The lower limit of the ratio x/n is sqrt(2).
- high limits of z and x:
| n is odd | n is even
---------+------------------+------------------
z limit | (n^4 + 1)/2 | (n^4 + 4)/4
x limit | (n^4 + 1)/2 - 1 | (n^4 + 4)/4 - 2
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
a(6) = 7 (solutions): 6^4 = 1296 = 325^2 - 323^2 = 164^2 - 160^2 = 111^2 - 105^2 = 85^2 - 77^2 = 60^2 - 48^2 = 45^2 - 27^2 = 39^2 - 15^2.
|
|
|
MATHEMATICA
|
a[n_] := Length[Solve[n^4 == z^2 - x^2 && x >= 1 && z >= 1, {x, z}, Integers]]; Array[a, 75] (* Amiram Eldar, Dec 14 2021 *)
|
|
|
PROG
|
(PARI) a(n) = numdiv(if(n%2, n^4, n^4/4))\2; \\ Jinyuan Wang, Dec 19 2021
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
| |
|
|
|
0, 1, 2, 3, 7, 17, 27, 37, 87, 137, 157, 187, 247, 437, 687, 787, 937, 1237, 2187, 3437, 3937, 4687, 6187, 8437, 10937, 17187, 19687, 23437, 30937, 42187, 54687, 55687, 85937, 98437, 117187, 154687, 210937, 223437, 273437, 278437, 304687, 429687, 492187, 585937
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,3
|
|
|
COMMENTS
|
Terms are the record numbers of solutions for the equation: y^4 = z^2 - x^2.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
Number of | y | Factorization
solutions | | of y
----------+---+--------------
0 | 1 | -
1 | 2 | [2]
2 | 3 | [3]
3 | 4 | [2, 2]
7 | 6 | [2, 3]
: : :
For more terms with y and factorization of y see link.
|
|
|
PROG
|
(PARI) lista(nn) = my(f, r); print1("0, 1, 2"); forstep(n=4, nn, 2, f=factor(n)[, 2]; if(r<f=prod(k=2, #f, 4*f[k]+1)*(4*f[1]-1), print1(", ", (r=f)\2))); \\ Jinyuan Wang, Dec 19 2021
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
Search completed in 0.012 seconds
|
