A063664 -id:A063664 - cp's OEIS frontend

A063665 Number of ways 1/n can be written as 1/x^2 + 1/y^2 with y >= x >= 1.

0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Henry Bottomley, Jul 25 2001

nonn

Number of ordered pairs (x,y), with n = (x^2)(y^2)/(x^2 + y^2) and y >= x > 0. - Antti Karttunen, Nov 07 2018

			a(90)=1 since 1/90 = 1/10^2 + 1/30^2
a(98)=2 since 1/98 = 1/10^2 + 1/70^2 = 1/14^2 + 1/14^2.
a(14400) = 3 since 1/14400 = 1/130^2 + 1/312^2 = 1/136^2 + 1/255^2 = 1/150^2 + 1/200^2. - _Antti Karttunen_, Nov 07 2018

Antti Karttunen, Table of n, a(n) for n = 1..100000

Cf. A000161, A025426, A018892, A063664.

A063665(n) = { my(s=0); for(x=1,n,for(y=x,n,if((n*(x*x+y*y)) == (x*x*y*y), s++))); (s); }; \\ Antti Karttunen, Nov 07 2018

A063665(n) = { my(s=0,y); for(x=sqrtint(n),n,my(x2=x*x); if((x2>n)&&issquare((n*x2)/(x2-n),&y)&&(1==denominator(y))&&(y>=x),s++)); (s); }; \\ Antti Karttunen, Nov 07 2018

Definition clarified by Antti Karttunen, Nov 07 2018

A063663 Numbers which can be written as b^2c^2(b^2+c^2).

Original entry on oeis.org

0, 2, 20, 90, 128, 272, 468, 650, 1280, 1332, 1458, 2450, 2900, 3600, 4160, 5760, 6642, 7650, 8192, 10100, 10388, 14580, 14762, 16400, 17408, 20880, 25578, 27540, 28730, 29952, 31250, 38612, 41600, 42048, 50850, 50960, 54900, 60500, 65610

Offset: 1

Henry Bottomley, Jul 25 2001

nonn

			468 is in the sequence since 2^2*3^2*(2^2+3^2) = 4*9*13 = 468.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A063664.

upto(n) = { my(res = List(0)); for(i = 1, sqrtint(n), for(j = 1, i, c = i^2 * j^2 * (i^2 + j^2); if(c <= n, listput(res, c); , next(2) ) ) ); Set(res) } \\ David A. Corneth, Nov 01 2020

Offset 1 from Michel Marcus, Nov 01 2020

A063669 Hypotenuses of reciprocal Pythagorean triangles: number of solutions to 1/(12n)^2 = 1/b^2 + 1/c^2 [with b >= c > 0]; also number of values of A020885 (with repetitions) which divide n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 1, 1, 2, 1, 4, 1, 1, 1, 1, 4, 1, 1, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 2, 1, 1, 1, 4, 1, 1, 1, 1, 2, 1, 1, 1, 1, 6, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 3, 2, 1, 1, 1, 1, 4, 2, 1, 1, 1, 2, 1, 1, 2, 1, 3, 1, 1, 1, 1, 5

Offset: 1

Henry Bottomley, Jul 28 2001

nonn

Primitive reciprocal Pythagorean triangles 1/a^2 = 1/b^2 + 1/c^2 have a=fg, b=ef, c=eg where e^2 = f^2 + g^2; i.e., e,f,g represent the sides of primitive Pythagorean triangles. But the product of the two legs of primitive Pythagorean triangles are multiples of 12 and so the reciprocal of hypotenuses of reciprocal Pythagorean triangles are always multiples of 12 (A008594).

			a(1)=1 since 1/(12*1)^2 = 1/12^2 = 1/15^2 + 1/20^2;
a(70)=6 since 1/(12*70)^2 = 1/840^2 = 1/875^2 + 1/3000^2 = 1/888^2 + 1/2590^2 = 1/910^2 + 1/2184^2 = 1/952^2 + 1/1785^2 = 1/1050^2 + 1/1400^2 = 1/1160^2 + 1/1218^2.
Looking at A020885, 1 is divisible by 1, while 70 is divisible by 1, 5, 10, 14, 35 and again 35.

Cf. A046080, A063664, A063665, A063014.

A259263 Numbers of the form (m*k)^2/(m^2-k^2) for distinct integers m and k.

Original entry on oeis.org

12, 18, 48, 72, 108, 147, 150, 162, 180, 192, 225, 240, 288, 300, 400, 405, 432, 448, 450, 578, 588, 600, 648, 720, 768, 882, 900, 960, 972, 980, 1008, 1100, 1152, 1200, 1260, 1323, 1350, 1452, 1458, 1600, 1620, 1728, 1792, 1800, 2025, 2028, 2100, 2160, 2178, 2312, 2352, 2400, 2592, 2700, 2880, 3042, 3072, 3150

Offset: 1

Derek Orr, Jun 22 2015

nonn

The odd numbers are much more rare than even numbers: 147, 225, 405, 1323, 2025, 3645, 3675, ... For 1 <= m <= 10^4 and 1 <= k <= m, there are 9217 total solutions. Of these solutions, only 679 are odd. See A259288.
Similarly, the reciprocals of these numbers can be represented as the difference in the reciprocals of two squares (i.e., there exists two distinct integers m and k satisfying 1/a(n) = 1/m^2 - 1/k^2).
If a(n) is a square, its square root is in A111200.

			(3*6)^2/(6^2-3^2) = 18^2/(3*9) = 12. So 12 is a member of this sequence.

Cf. A063664, A111200, A259288.

v=[];for(m=1,7500,for(n=1,m-1,if(type(s=(m*n)^2/(m^2-n^2))=="t_INT",v=concat(v,s))));vecsort(v,,8)

A254307 Least k such that there are n positive integers, all less than or equal to k, such that the sum of the reciprocals of their squares equals 1.

Original entry on oeis.org

6, 4, 6, 3, 4, 6, 6, 4, 6, 6, 4, 6, 6, 6, 6, 6, 6, 6, 6, 5, 6, 6, 6, 8, 6, 6, 8, 6, 8, 8, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 7, 8, 8, 8, 9, 8, 8, 9, 8, 8, 9, 9, 8, 9, 9, 8, 9, 9, 9, 9, 10, 9, 9, 10, 9, 10, 10, 9

Offset: 6

Charles R Greathouse IV, Jan 27 2015

nonn

a(2), a(3), and a(5) are undefined, so this sequence starts at offset 6. Gasarch (2015) shows that a(n) exists for all n >= 6, though this was known (folklore?) previously; he also poses three open questions.

First occurrence of n: 1, 4, 9, 7, 25, 6, 49, 29, 53, 69, 121, 87, 140, 179, 221, ..., . - Robert G. Wilson v, Feb 15 2015

			a(1) = 1: 1 = 1/1.
a(4) = 2: 1 = 1/4 + 1/4 +1/4 + 1/4.
a(6) = 6: 1 = 1/4 + 1/4 + 1/4 + 1/9 + 1/9 + 1/36.
a(7) = 4: 1 = 1/4 + 1/4 + 1/4 + 1/16 + 1/16 + 1/16 + 1/16.
a(8) = 6: 1 = 1/4 + 1/4 + 1/9 + 1/9 + 1/9 + 1/9 + 1/36 + 1/36.
a(9) = 3: 1 = 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9.

Bill Gasarch, The Solution to a problem in a Romanian math problem book (2015)

Cf. A000058, A063664.

/* oo = 10^10; \\ uncomment for earlier pari versions */
ssd(n,total,mn,mx)=my(t,best=oo); if(total<=0,return(0)); if(n==1, return(if(issquare(1/total,&t)&&t>=mn&&t<=mx&&denominator(t)==1,t,0))); for(k=mn, min(sqrtint(n\total),mx), t=ssd(n-1,total-1/k^2,k,mx); if(t,best=min(best,t))); best
a(n)=my(k=sqrtint(n-1),t=oo);while(t==oo,k++;t=ssd(n-1,1-1/k^2,2,k));k

sqrt(n) <= a(n) < 2*sqrt(n) for n > 8. The lower bound is sharp since a(n^2) = n.

cp's OEIS Frontend

A063665 Number of ways 1/n can be written as 1/x^2 + 1/y^2 with y >= x >= 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

Extensions

A063663 Numbers which can be written as b^2*c^2*(b^2+c^2).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

Extensions

A063669 Hypotenuses of reciprocal Pythagorean triangles: number of solutions to 1/(12n)^2 = 1/b^2 + 1/c^2 [with b >= c > 0]; also number of values of A020885 (with repetitions) which divide n.

Views

Author

Keywords

Comments

Examples

Crossrefs

A259263 Numbers of the form (m*k)^2/(m^2-k^2) for distinct integers m and k.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

A254307 Least k such that there are n positive integers, all less than or equal to k, such that the sum of the reciprocals of their squares equals 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A063663 Numbers which can be written as b^2c^2(b^2+c^2).