Displaying 51-60 of 231 results found.
page
1
2
3
4
5
6
7
8
9
10
... 24
a(n) is the number of positive integers <= 10^n that can be expressed as a sum of two squares.
+0
2
7, 43, 330, 2749, 24028, 216341, 1985459, 18457847, 173229058, 1637624156, 15570512744, 148736628858, 1426306930865, 13722217893214, 132387263219058, 1280309691127436
EXAMPLE
a(1)=7 since 1 = 0^2 + 1^2, 2 = 1^2 + 1^2, 4 = 0^2 + 2^2, 5 = 1^2 + 2^2, 8 = 2^2 + 2^2, 9 = 0^2 + 3^2, 10 = 1^2 + 3^3.
Number of ways to write n as x^2 + y^2 + z*(3*z-1)/2 with x,y,z integers such that x + 2*y is a square.
+0
6
1, 2, 3, 2, 2, 5, 5, 5, 1, 2, 4, 2, 5, 3, 3, 4, 3, 7, 5, 3, 8, 3, 5, 3, 2, 6, 3, 7, 5, 3, 4, 7, 5, 3, 4, 6, 5, 3, 3, 7, 5, 5, 5, 1, 6, 7, 6, 4, 3, 2, 5, 5, 9, 6, 3, 7, 5, 7, 5, 4, 8, 5, 6, 4, 6, 5, 5, 8, 5, 6, 6, 7, 5, 5, 5, 7, 5, 6, 2, 4, 12
COMMENTS
Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 8, 43, 84, 133, 253, 399, 488, 523, 803, 7369.
(ii) Any integer n > 1 can be written as p + x^2 + y^2 with p prime and x + 2*y (or x + 3*y) a square, where x is an integer and y is a nonnegative integer.
Note that those numbers z*(3*z-1)/2 with z integral are called generalized pentagonal numbers ( A001318). By Theorem 1.7(ii) of the linked paper in Sci. China Math., each n = 0,1,2,... can be written as the sum of two squares and a generalized pentagonal number. Ju. V. Linnik proved in 1960 that any sufficiently large integer can be expressed as the sum of a prime and two squares.
EXAMPLE
a(8) = 1 since 8 = 1^2 + 0^2 + (-2)*(3*(-2)-1)/2 with 1 + 2*0 = 1^2.
a(43) = 1 since 43 = 1^2 + 4^2 + (-4)*(3*(-4)-1)/2 with 1 + 2*4 = 3^2.
a(84) = 1 since 84 = 7^2 + (-3)^2 + (-4)*(3*(-4)-1)/2 with 7 + 2*(-3) = 1^2.
a(133) = 1 since 133 = 4^2 + 0^2 + 9*(3*9-1)/2 with 4 + 2*0 = 2^2.
a(253) = 1 since 253 = (-13)^2 + 7^2 + 5*(3*5-1)/2 with (-13) + 2*7 = 1^2.
a(399) = 1 since 399 = 18^2 + (-7)^2 + (-4)*(3*(-4)-1)/2 with 18 + 2*(-7) = 2^2.
a(488) = 1 since 488 = 9^2 + 20^2 + (-2)*(3*(-2)-1)/2 with 9 + 2*20 = 7^2.
a(523) = 1 since 523 = 9^2 + 0^2 + (-17)*(3*(-17)-1)/2 with 9 + 2*0 = 3^2.
a(803) = 1 since 803 = (-17)^2 + 13^2 + (-15)*(3*(-15)-1)/2 with (-17) + 2*13 = 3^2.
a(7369) = 1 since 7369 = 0^2 + 72^2 + (-38)*(3*(-38)-1)/2 with 0 + 2*72 = 12^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
PenQ[n_]:=PenQ[n]=SQ[24n+1];
Do[r=0; Do[If[PenQ[n-x^2-y^2], Do[If[SQ[(-1)^i*x+2(-1)^j*y], r=r+1], {i, 0, Min[x, 1]}, {j, 0, Min[y, 1]}]], {x, 0, Sqrt[n]}, {y, 0, Sqrt[n-x^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]
Numbers k for which A363763(k) = -1.
+0
7
46, 55, 62, 71, 80, 86, 107, 130, 172, 187, 195, 208, 222, 247, 259, 263, 268, 272, 280, 297, 314, 330, 358, 363, 370, 372, 379, 394, 400, 405, 429, 449, 450, 462, 489, 500, 529, 534, 587, 629, 641, 646, 652, 667, 668, 672, 704, 715, 733, 736, 749, 769, 775, 776, 778, 785, 793, 799
PROG
(Python)
from itertools import count, islice
from sympy import factorint
def A363762_gen(startvalue=1): # generator of terms >= startvalue for n in count(max(startvalue, 1)):
for k in range(n>>1, ((n+1)**2<<1)+1):
c = 0
for m in range(k**2+1, (k+1)**2):
if all(p==2 or p&3==1 or e&1^1 for p, e in factorint(m).items()):
c += 1
if c>n:
break
if c==n:
break
else:
yield n
CROSSREFS
Numbers not occurring as terms of A077773.
Minimum numbers of squares needed to write n! as a sum of nonzero squares.
+0
1
1, 2, 3, 3, 3, 2, 3, 3, 3, 4, 3, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3
EXAMPLE
a(2)=2 because 2!=1^2+1^2; a(3)=3: 3!=6=2^2+1+1; a(6)=2: 6!=720=24^2+12^2
Integers k that are the sum of two nonzero squares while k*(k+1) is not.
+0
1
2, 5, 10, 13, 18, 20, 26, 29, 32, 34, 37, 41, 45, 50, 53, 58, 61, 65, 68, 74, 82, 85, 90, 98, 101, 104, 106, 109, 113, 117, 122, 125, 128, 130, 137, 146, 149, 153, 157, 160, 162, 164, 170, 173, 178, 181, 185, 194, 197, 200, 202, 205, 208, 212, 218, 221, 226, 229, 234, 242, 245, 250, 257, 261
COMMENTS
Values of a^2 + b^2 such that (a^2 + b^2)^2 + a^2 + b^2 is not of the form x^2 + y^2 where a, b, x, y are nonzero integers.
EXAMPLE
5 is a term because 5 = 1^2 + 2^2 and 5^2 + 5 = 30 is not a term of A000404.
MATHEMATICA
Select[Range@ 270, Length@ First@ # >= 1 && Last@ # == {} &[PowersRepresentations[#, 2, 2] /. {0, _} -> Nothing & /@ {#, # (# + 1)} &@ #] &] (* Michael De Vlieger, Apr 14 2016 *)
PROG
(PARI) isA000404(n) = {for( i=1, #n=factor(n)~%4, n[1, i]==3 && n[2, i]%2 && return); n && ( vecmin(n[1, ])==1 || (n[1, 1]==2 && n[2, 1]%2))}
for(n=1, 1e3, if(!isA000404(n*(n+1)) && isA000404(n), print1(n, ", ")));
a(n) is the least k < 3*n such that there are exactly n distinct numbers j that can be expressed as sum of two squares with k^2 < j < (k+1)^2, or -1 if such a k does not exist.
+0
5
0, 1, 2, 4, 5, 7, 8, 10, 13, 12, 15, 17, 19, 23, 21, 24, 25, 28, 32, 31, 34, 37, 39, 44, 41, 43, 45, 50, 51, 48, 57, 55, 56, 59, 64, 63, 68, 69, 74, 77, 78, 75, 72, 80, 88, 84, -1, 94, 89, 96, 93, 99, 97, 102, 108, -1, 106, 111, 110, 113, 117, 120, -1, 123, 133, 127, 130, 137, 142, 138, 139, -1, 135
FORMULA
If a(n) != -1, then a(n) >= n/2. - Chai Wah Wu, Jun 22 2023 a(n) = -1 for n > 15898.
PROG
(PARI) a363761(upto) = {for (n=0, upto, my(kfound=-1);
for (k=0, 3*n, my(k1=k^2+1, k2=k*(k+2), m=0);
for (j=k1, k2, m+= (sumdiv(j, d, (d%4==1)-(d%4==3))>0); if (m>n, break));
if (m==n, kfound=k; break); if (m==n, kfound=k; break)); print1(kfound, ", "))};
a363761(75)
(Python)
from sympy import factorint
for k in range(n>>1, 3*n):
c = 0
for m in range(k**2+1, (k+1)**2):
if all(p==2 or p&3==1 or e&1^1 for p, e in factorint(m).items()):
c += 1
if c>n:
break
if c==n:
return k
CROSSREFS
Identical with A363763 for n <= 11459, but increasingly different afterwards, i.e., a(11460) = -1, whereas A363763(11460) = 34451.
Positive integers which can be expressed as the sum of three or fewer squares, no more than two of which are greater than 1.
+0
2
1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 13, 14, 16, 17, 18, 19, 20, 21, 25, 26, 27, 29, 30, 32, 33, 34, 35, 36, 37, 38, 40, 41, 42, 45, 46, 49, 50, 51, 52, 53, 54, 58, 59, 61, 62, 64, 65, 66, 68, 69, 72, 73, 74, 75, 80, 81, 82, 83, 85, 86, 89, 90, 91, 97, 98, 99, 100
COMMENTS
These are the numbers which can be a clue in a Tasquare puzzle.
Tasquare also known as Tasukuea.
EXAMPLE
A 9 clue can be satisfied in multiple ways:
OOO OO OO
OOO OO9OO
OOO9 O
MAPLE
q:= proc(n) local i; for i to isqrt(n) do if ormap(issqr,
[n-i^2, n-i^2-1]) then return true fi od: false
end:
MATHEMATICA
q[n_] := Module[{i}, For[i = 1, i <= Sqrt[n], i++, If[AnyTrue[ {n-i^2, n-i^2-1}, IntegerQ@Sqrt[#]&], Return[True]]]; False];
a(n) = Sum_{b=0..floor(sqrt(n)), n-b^2 is square} b.
+0
2
1, 1, 0, 2, 3, 0, 0, 2, 3, 4, 0, 0, 5, 0, 0, 4, 5, 3, 0, 6, 0, 0, 0, 0, 12, 6, 0, 0, 7, 0, 0, 4, 0, 8, 0, 6, 7, 0, 0, 8, 9, 0, 0, 0, 9, 0, 0, 0, 7, 13, 0, 10, 9, 0, 0, 0, 0, 10, 0, 0, 11, 0, 0, 8, 20, 0, 0, 10, 0, 0, 0, 6, 11, 12, 0, 0, 0, 0, 0, 12, 9, 10, 0
MATHEMATICA
a[n_]:=Sum[b Boole[IntegerQ[Sqrt[n-b^2]]], {b, 0, Floor[Sqrt[n]]}]; Array[a, 83] (* Stefano Spezia, May 15 2023 *)
PROG
(Python)
from gmpy2 import *
a = lambda n: sum([b for b in range(0, isqrt(n) + 1) if is_square(n - (b*b))])
print([a(n) for n in range(1, 84)])
(Python)
from sympy import divisors
from sympy.solvers.diophantine.diophantine import cornacchia
c = 0
for d in divisors(n):
if (k:=d**2)>n:
break
q, r = divmod(n, k)
if not r:
c += sum(d*(a[0]+(a[1] if a[0]!=a[1] else 0)) for a in cornacchia(1, 1, q) or [])
(PARI) a(n) = sum(b=0, sqrtint(n), if (issquare(n-b^2), b)); \\ Michel Marcus, May 16 2023
Positive integers n such that Fibonacci(n) = a^2 + b^2, where a, b are integers.
+0
3
1, 2, 3, 5, 6, 7, 9, 11, 12, 13, 14, 15, 17, 19, 21, 23, 25, 26, 27, 29, 31, 33, 35, 37, 38, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 62, 63, 65, 67, 69, 71, 73, 74, 75, 77, 79, 81, 83, 85, 86, 87, 89, 91, 93, 95, 97, 98, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 122, 123, 125, 127
COMMENTS
All odd numbers are in this sequence, since the Fibonacci number with index 2m+1 is the sum of the squares of the two Fibonacci numbers with indices m and m+1. Those with even indices ultimately depend on certain Lucas numbers being the sum of two squares (see A124132). Joint work with Kevin O'Bryant and Dennis Eichhorn. Numbers n such that Fibonacci(n) or Fibonacci(n)/2 is a square are only 0, 1, 2, 3, 6, 12. So a and b must be distinct and nonzero for all values of this sequence except 1, 2, 3, 6, 12. - Altug Alkan, May 04 2016
EXAMPLE
14 is in the sequence because F_14=377=11^2+16^2.
16 is not in the sequence because F_16=987 is congruent to 3 (mod 4).
PROG
(PARI) for(n=1, 10^6, t=fibonacci(n); s=sqrtint(t); forstep(i=s, 1, -1, if(issquare(t-i*i), print1(n, ", "); break))) \\ Ralf Stephan, Sep 15 2013 (PARI) is2s(n)={my(f=factor(n)); for(i=1, #f[, 1], if(f[i, 2]%2 && f[i, 1]%4==3, return(0))); 1; } \\ see A001481for(n=1, 10^6, if(is2s(fibonacci(n)), print1(n, ", "))); \\ Joerg Arndt, Sep 15 2013 (Haskell)
a124134 n = a124134_list !! (n-1)
a124134_list = filter ((> 0) . a000161 . a000045) [1..]
(Python)
from itertools import count, islice
from sympy import factorint, fibonacci
def A124134_gen(): # generator of terms return filter(lambda n:n & 1 or all(p & 3 != 3 or e & 1 == 0 for p, e in factorint(fibonacci(n)).items()), count(1))
AUTHOR
Melvin J. Knight (melknightdr(AT)verizon.net), Nov 30 2006
Numbers that are not the sum of two squares and two fourth powers.
+0
4
23, 44, 71, 79, 184, 368, 519, 599, 704, 1136, 1264, 2944, 4024, 5888, 8304, 9584, 11264, 18176, 20224, 47104, 64384, 94208, 132864, 153344, 180224, 290816, 323584, 753664, 1030144, 1507328, 2125824, 2453504, 2883584, 4653056, 5177344, 12058624, 16482304
COMMENTS
When n is a term, 16n is also. This can be proved as follows:
(1) If w is odd, then 16n - w^4 == 7 (mod 8), and it follows from Legendre's three-square theorem that the equation x^2 + y^2 + z^4 + w^4 = 16n has no solution (it is the same when x, y or z are odd numbers).
(2) If x, y, z and w are even numbers (x = 2a, y = 2b, z = 2c, w = 2d) such that x^2 + y^2 + z^4 + w^4 = 16n, then a^2 + b^2 = 4(n - c^4 - d^4). So there are integers u and v satisfying u^2 + v^2 = n - c^4 - d^4. i.e. u^2 + v^2 + c^4 + d^4 = n, which is a contradiction.
(End)
Conjecture: The set {a(n): n > 0} coincides with {16^k*m: k = 0, 1, 2, ... and m = 23, 44, 71, 79, 184, 519, 599, 4024}. - Zhi-Wei Sun, Jan 27 2022
PROG
(PARI)
N=10^6; x='x+O('x^N);
S(e)=sum(j=0, ceil(N^(1/e)), x^(j^e));
v=Vec( S(4)^2 * S(2)^2 );
for(n=1, #v, if(!v[n], print1(n-1, ", ")));
Search completed in 0.122 seconds
| 