Mark Underwood - cp's OEIS frontend

User: Mark Underwood

Mark Underwood has authored 3 sequences.

A260549 Primitive values n such that the square with opposite corners (0,0) and (n,n) contains a point (x,y) with integer coordinates, with 0 < x,y < n, at an integer distance from three of the four corners.

52, 195, 700, 740, 867, 996, 1443, 2145, 3364, 6015, 6240, 7800, 8400, 9165, 9375, 10879, 10952, 11184, 11352, 11484, 11492, 11997, 12675, 13156, 14355, 14739, 20280, 20415, 20625, 20988, 21125, 21320, 21853, 22472, 23069, 26180, 26588, 27189, 28168, 30195

Offset: 1

Mark Underwood and Giovanni Resta, Jul 29 2015

nonn

Primitiveness condition prescribes that, if n is in the sequence because of the interior point (x, y), then GCD(x, y, n) = 1.
Some values admit more than one qualifying interior point. For example, for n = 21125 we have both (4416, 3737) and (8357, 7524).
This sequence is a superset of A215365, which only contains even values.

			700 is in the sequence because the point (304,297) is at distance 425 from (0,0), 495 from (700,0) and 565 from (700,700) and GCD(700,304,297) = 1.

Giovanni Resta, Table of n, a(n) for n = 1..75 (terms < 130000)

Cf. A215365.

A216374 Number of ways to express the square of the n-th prime as the sum of four nonzero squares.

Original entry on oeis.org

1, 0, 1, 2, 3, 5, 7, 9, 13, 20, 23, 32, 38, 42, 50, 63, 77, 83, 99, 111, 117, 137, 150, 172, 204, 221, 230, 247, 257, 275, 347, 368, 402, 414, 475, 488, 527, 567, 595, 638, 682, 698, 776, 792, 825, 842, 945, 1055, 1092, 1112, 1150, 1210, 1230, 1333, 1397, 1463, 1530, 1553, 1622, 1668, 1692, 1813, 1989, 2041

Offset: 1

Mark Underwood, Sep 05 2012

nonn

The simple counting and the conjectured first formula agree for all the primes from 3 to 997. The counting and the conjectured second formula agree for all the primes from 5 to 997. The author of this sequence would like to know whether the formulas are already known and/or how it could be proved.

I suspect Jacobi's theorem will suffice. - Charles R Greathouse IV, Sep 30 2012

			prime(n)'s are 2, 3, 5, 7, 11, 13, 17, ... giving the sequence 1, 0, 1, 2, 3, 5, 7, ...

Sergey Beliy and others, Pythagorean "five"tuples and "six"tuples, digest of 9 messages in Yahoo group "Unsolved Problems in Number Theory, Logic, and Cryptography", Sep 04 2012.

Cf. A025428, A001248.

forprime(p=2,1000, k=0; for(s1=1,sqrt((p^2)/4),for(s2=s1,sqrt((p^2 - s1^2)/3), for(s3=s2,sqrt((p^2-s1^2 - s2^2)/2), if(issquare(p^2-s1^2-s2^2-s3^2),k++)))) ; f = floor((p^2+4*p+24)/48.) ; f2 = (p^2 + 4*p + (19*(5*(p%48)+2)^2)%48 - 24)/48 ;                 print1([p,k,f,f2]" "))
/* code above prints [p, k, f, f2] where p is the prime, k is the number of ways the square of p can be expressed as the sum of four nonzero squares, and f and f2 are the formulas derivations. f and k are observed to be the same for p from 3 to 997; f2 and k are observed to be the same for p from 5 to 997. */

A216374(n)=sum(s1=1,.5*n=prime(n+1),my(t);sum(s2=s1,sqrtint((n^2-s1^2)\3),sum(s3=s2,sqrtint((t=n^2-s1^2-s2^2)\2),issquare(t-s3^2)))) \\ M. F. Hasler, Sep 11 2012

a(n) = floor((prime(n)^2 + 4*prime(n) + 24)/48) (conjectured for n>1).
a(n) = (prime(n)^2 + 4*prime(n) + (19*(5*(prime(n) mod 48)+2)^2) mod 48 - 24)/48 (conjectured for n>2).
a(n) = A025428(A001248(n)), where A001248(n) = A000040(n)^2 = prime(n)^2. - M. F. Hasler, Sep 10 2012

A215365 Primitive integer length of the side of an origin-centered square that contains inside its boundary a point with integer coordinates that is an integer distance from three of the four corners.

Original entry on oeis.org

52, 700, 740, 996, 3364, 6240, 7800, 8400, 10952, 11184, 11352, 11492, 11484, 13156, 20280, 20988, 21320, 22472, 26180, 26588, 28168, 34500, 39988, 40680, 43700, 44944, 45976, 49500, 58956, 70448, 77500, 90168, 103896, 105468, 106200, 115752, 118636, 124620, 129000

Offset: 1

Mark Underwood, Aug 08 2012

nonn

No point with integer distance to all four corners is known.

The sequence only contains even values because an odd-sided square centered at the origin has corners with non-integer coordinates, which cannot be at integer distance from interior lattice points. If the square instead of being centered at the origin has a corner on the origin, then the resulting sequence is A260549. - Giovanni Resta, Jul 29 2015

			With n = side length, we find an a,b such that a^2 + b^2 = d1^2, a^2 + (n-b)^2 = d2^2, b^2 + (n-a)^2 = d3^2, (n-a)^2 + (n-b)^2 = d4^2 is true in integers for three of these four equations. n = 52 is the first, with a=7 and b=24.

Yasushi Ieno, Other special cases of a square problem, arXiv:2111.02888 [math.GM], 2021.
Yang Ji, Several special cases of a square problem, arXiv:2105.05250 [math.GM], 2021.
G. Shute and K. L. Yocom, Problem 966. Seven integral distances, Math. Mag. 50, 166 (1977).
UnsolvedProblems Web Site, Rational Distance

Cf. A260549.

has(n)=for(a=1,n-1,for(b=a,n-1, if(issquare(norml2([a,b])) + issquare(norml2([n-a,b])) + issquare(norml2([a,n-b])) + issquare(norml2([n-a,n-b])) > 2, return(1)))); 0
is(n)=sumdiv(n,d,has(d))==1 \\ Charles R Greathouse IV, Jul 28 2015

Data corrected and name improved by Mark Underwood, Jul 28 2015

a(7)-a(39) from Giovanni Resta, Jul 29 2015

cp's OEIS Frontend

User: Mark Underwood

A260549 Primitive values n such that the square with opposite corners (0,0) and (n,n) contains a point (x,y) with integer coordinates, with 0 < x,y < n, at an integer distance from three of the four corners.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A216374 Number of ways to express the square of the n-th prime as the sum of four nonzero squares.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

A215365 Primitive integer length of the side of an origin-centered square that contains inside its boundary a point with integer coordinates that is an integer distance from three of the four corners.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Extensions