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 1-6 of 6 results.

A268856 a(n) = smallest magic sum of any 3 X 3 magic square which contains exactly n numbers that are triangular or square.

Original entry on oeis.org

39, 36, 30, 21, 21, 15, 27, 675
Offset: 0

Views

Author

Arkadiusz Wesolowski, Feb 14 2016

Keywords

Comments

a(8) and a(9) are greater than 10^10. - Arkadiusz Wesolowski, Oct 27 2018
a(8) <= 3*7215727335550^2 = (A221669(1) + A221669(2) + A221669(3))*16978181966^2. - Arkadiusz Wesolowski, May 21 2022

Crossrefs

A269231 a(n) = smallest magic sum of any 3 X 3 semimagic square which contains exactly n cubes.

Original entry on oeis.org

19, 16, 15, 32, 73, 153, 1520, 5104, 241801435
Offset: 0

Views

Author

Arkadiusz Wesolowski, Feb 20 2016

Keywords

Crossrefs

Extensions

a(8) from Christian Boyer, added by Arkadiusz Wesolowski, Apr 14 2016

A268854 a(n) is the smallest magic sum of any 3 X 3 magic square which contains exactly n triangular numbers.

Original entry on oeis.org

39, 24, 21, 15, 18, 24, 189, 67734
Offset: 0

Views

Author

Arkadiusz Wesolowski, Feb 14 2016

Keywords

Crossrefs

A375459 a(n) = smallest magic sum of any 3 X 3 magic square which contains exactly n squared primes, or 0 if no such magic square exists.

Original entry on oeis.org

21, 18, 15, 51, 195, 435, 76035
Offset: 0

Views

Author

Arkadiusz Wesolowski, Aug 15 2024

Keywords

Comments

a(9) = 0.
a(n) = A268855(n+1) for n = 0, 1, 2, and 5.
a(7) and a(8) - if nonzero, they are greater than 10^10.

Crossrefs

A319589 A 3 X 3 magic square with five square entries, read by rows.

Original entry on oeis.org

34969, 83521, 7585, 14641, 42025, 69409, 76465, 529, 49081
Offset: 1

Views

Author

Arkadiusz Wesolowski, Sep 23 2018

Keywords

Comments

Some magic squares of order 3 with five square entries have this parametric form:
|-----------------------|-----------------------|-----------------------|
| (x*y)^2 | y^4 |sqrt(z)*(3*x^2 - y^2)/2|
|-----------------------|-----------------------|-----------------------|
| x^4 | z | 2*z - x^4 |
|-----------------------|-----------------------|-----------------------|
|sqrt(z)*(3*y^2 - x^2)/2| 2*z - y^4 | 2*z - (x*y)^2 |
|-----------------------|-----------------------|-----------------------|
where z = (x^2 + y^2)^2/4, x and y are integers such that (x^4 - y^4 + 2*(x*y)^2)/2 is a square (e.g., x = 11 and y = 17; x = 5337 and y = 6257).
This sequence presents the magic square belonging to this family and having the smallest possible magic sum (S = 126075).

Examples

			The magic square is
+-------+-------+-------+
| 187^2 |  17^4 |  7585 |
+-------+-------+-------+
|  11^4 | 205^2 | 69409 |
+-------+-------+-------+
| 76465 |  23^2 | 49081 |
+-------+-------+-------+
		

Crossrefs

A375361 Odd numbers with at least two prime divisors of the form 4*k + 1 counted with multiplicity.

Original entry on oeis.org

25, 65, 75, 85, 125, 145, 169, 175, 185, 195, 205, 221, 225, 255, 265, 275, 289, 305, 325, 365, 375, 377, 425, 435, 445, 455, 475, 481, 485, 493, 505, 507, 525, 533, 545, 555, 565, 575, 585, 595, 615, 625, 629, 663, 675, 685, 689, 697, 715, 725, 745, 765, 775
Offset: 1

Views

Author

Arkadiusz Wesolowski, Aug 13 2024

Keywords

Comments

Odd numbers k such that k^2 can be expressed as the arithmetic mean of two distinct perfect squares in more than one way. For example, 25^2 = (5^2 + 35^2)/2 = (17^2 + 31^2)/2.
Let x be a squared integer which is the central element of a 3 X 3 magic square in which seven (or more) of the entries are squared integers. If the greatest common divisor of all nine entries is 1, then the square root of x is a composite number that is divisible only by primes congruent to 1 mod 4. For example, sqrt(A221669(5)) = 425 is both in A004613 and in this sequence.

Examples

			65 is in this sequence because 65 has two prime factors of the form 4*k + 1, namely 5 = 4*1 + 1 and 13 = 4*3 + 1.
		

Crossrefs

Programs

  • Magma
    f:=func; nopf:=func; sum:=func; [n: n in [1..775 by 2] | sum(n) gt 1];
    
  • PARI
    isok(n) = my(v=Vec(factor(n))); n%2&&sum(t=1, omega(n), if((v[1]%4)[t]==1, v[2][t]))>1;
    
  • PARI
    isok(n) = my(t); if(n%2, for(k=sqrtint(n^2-1)+2, sqrtint(2*n^2-1), if(issquare(2*n^2-k^2)&&t++>1, return(1)))); 0;
Showing 1-6 of 6 results.