Search: a001481 -id:a001481
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A128106
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Sizes of possible gaps around a Gaussian prime: 1 and the even numbers in A001481.
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+20
5
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1, 2, 4, 8, 10, 16, 18, 20, 26, 32, 34, 36, 40, 50, 52, 58, 64, 68, 72, 74, 80, 82, 90, 98, 100, 104, 106, 116, 122, 128, 130, 136, 144, 146, 148, 160, 162, 164, 170, 178, 180, 194, 196, 200, 202, 208, 212, 218, 226, 232, 234, 242, 244, 250, 256, 260, 272, 274
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OFFSET
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1,2
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COMMENTS
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For a given Gaussian prime u, the size of its gap is the minimum of norm(u-v) as v varies over all other Gaussian primes, where norm(a+b*i)=a^2+b^2. Only the small Gaussian primes 1+i and 2+i (and their associates and reflections) have gaps of diameter 1.
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LINKS
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MATHEMATICA
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q=12; imax=2*q^2; lst=Select[Union[Flatten[Table[2*x^2+2*y^2, {x, 0, q}, {y, 0, x}]]], #<=imax&]; Join[{1}, Drop[lst, 1]] (* Vladimir Joseph Stephan Orlovsky, Apr 20 2011 *)
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PROG
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(Sage)
R = []; s = 1; sq = 1
for n in (0..max//2):
if n == s:
sq += 1;
s = sq*sq;
for k in range(sq):
if is_square(n-k*k):
R.append(2*n)
break
R[0] = 1
return R
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A070176
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Let s(n) be smallest number >= n which is a sum of two squares (A001481); sequence gives s(n) - n.
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+20
4
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0, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 2, 1, 0, 2, 1, 0, 0, 0, 1, 0, 4, 3, 2, 1, 0, 0, 2, 1, 0, 2, 1, 0, 1, 0, 1, 0, 0, 2, 1, 0, 0, 3, 2, 1, 0, 3, 2, 1, 0, 0, 1, 0, 0, 4, 3, 2, 1, 0, 2, 1, 0, 2, 1, 0, 0, 2, 1, 0, 3, 2, 1, 0, 0, 0, 5, 4, 3, 2, 1, 0, 0, 0, 2, 1, 0, 3, 2, 1, 0, 0, 6, 5, 4, 3, 2, 1, 0, 0, 1, 0, 0, 2, 1, 0
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OFFSET
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0,7
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COMMENTS
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It is an unsolved problem to determine the rate of growth of this sequence.
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REFERENCES
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H. L. Montgomery, Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis, Amer. Math. Soc., 1996, p. 208.
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LINKS
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MATHEMATICA
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sumOfTwoSquaresQ[n_] := With[{r = Ceiling[Sqrt[n]]}, Do[ Which[n == x^2 + y^2, Return[True], x == r && y == r, Return[False]], {x, 0, r}, {y, x, r}]]; a[n_] := For[s = n, True, s++, If[sumOfTwoSquaresQ[s], Return[s - n]]]; Table[a[n], {n, 0, 104}](* Jean-François Alcover, May 23 2012 *)
s2s[n_]:=Module[{i=0}, While[SquaresR[2, n+i]==0, i++]; i]; Array[s2s, 110, 0] (* Harvey P. Dale, Jun 16 2012 *)
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PROG
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(Haskell)
a070176 n = (head $ dropWhile (< n) a001481_list) - n
a070176_list = map a070176 [0..]
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A328803
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The minimum value of j + k where j and k are positive integers with j^2 + k^2 = A001481(n).
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+20
3
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0, 1, 2, 2, 3, 4, 3, 4, 5, 4, 5, 6, 6, 5, 6, 7, 8, 8, 6, 7, 8, 9, 9, 7, 8, 10, 9, 10, 11, 8, 9, 10, 12, 11, 12, 12, 9, 10, 11, 13, 12, 13, 14, 10, 11, 12, 14, 13, 15, 14, 15, 11, 12, 13, 16, 14, 16, 15, 12, 13, 16, 14, 17, 15, 17, 16, 18, 18, 13, 14, 15, 16
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OFFSET
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1,3
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LINKS
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EXAMPLE
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For n = 14, A001481(14) = 25 = 0^2 + 5^2 = 3^2 + 4^2, so a(14) = min{0+5, 3+4} = 5.
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MAPLE
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N:= 1000: # for terms where A001481(n)<=N
for s from 0 to isqrt(N) do
for i from 0 to s/2 do
t:= i^2 + (s-i)^2;
if t > N then break fi;
if not assigned(R[t]) then R[t]:= s fi;
od od:
A1481:= sort(map(op, [indices(R)])):
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PROG
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(Python)
from itertools import count, islice
from sympy.solvers.diophantine.diophantine import diop_DN
from sympy import factorint
def A328803_gen(): # generator of terms
return map(lambda n: min((a+b for a, b in diop_DN(-1, n))), filter(lambda n:(lambda m:all(d&3!=3 or m[d]&1==0 for d in m))(factorint(n)), count(0)))
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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1, 4, 2, 6, 5, 5, 6, 0, 6, 3, 5, 1, 2, 5, 9, 2, 8, 7, 8, 6, 9, 6, 8, 0, 9, 3, 1, 6, 1, 5, 5, 0, 8, 1, 6, 3, 6, 1, 2, 7, 6, 6, 9, 3, 6, 3, 6, 7, 7, 0, 3, 9, 0, 2, 8, 8, 7, 9, 9, 2, 2, 3, 0, 4, 4, 1, 2, 9, 6, 0, 4, 5, 2, 8, 6, 1, 5, 1, 9, 0, 1, 9, 1, 4, 6, 7
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OFFSET
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1,2
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COMMENTS
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This constant can be considered as an equivalent of zeta(2) (= Pi^2/6 = A013661), where Euler's zeta(2) is over all positive integers, with prime elements in A000040, while this constant is over all positive integers that can be written as the sum of two squares (A001481) with prime elements given in A055025.
Close to the value of e^(3/2)/Pi.
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LINKS
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FORMULA
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Equals Product_{i > 0} 1/(1-A055025(i)^-2).
Equals 1/(1-prime(1)^(-2)) * Product_{i>1 and prime(i) == 1 (mod 4)} 1/(1-prime(i)^(-2)) * Product_{i>1 and prime(i) == 3 (mod 4)} 1/(1-prime(i)^(-4)), where prime(n) = A000040(n).
Equals zeta_{2,0} (2) * zeta_{4,1} (2) * zeta_{4,3} (4), where zeta_{4,1} (2) = A175647 and zeta_{2,0} (s) = 2^s/(2^s - 1).
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EXAMPLE
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1.4265560635125928786968093161550816361276693636770...
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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0, 1, 2, 4, 8, 9, 16, 17, 18, 25, 32, 34, 36, 41, 49, 50, 64, 68, 72, 73, 81, 82, 89, 97, 98, 100, 113, 121, 128, 136, 137, 144, 146, 153, 162, 164, 169, 178, 193, 194, 196, 200, 225, 226, 233, 241, 242, 256, 257, 272, 274, 281, 288, 289, 292, 306, 313, 324, 328
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OFFSET
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1,3
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COMMENTS
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Contains A155561 as a subsequence (obtained by restricting a,b,c,d to be nonzero). Also contains A000290 (squares) and A001105 (twice the squares) as subsequence.
N is also of the form x^2 - 2y^2.
N = (p^2-q^2-2*r*s)^2+(r^2-s^2-2*p*q)^2
= (p^2+q^2-r^2-s^2)^2+2*(p*r-p*s-q*r-q*s)^2
= (p^2+q^2+r^2+s^2)^2-2*(p*r+p*s+q*r-q*s)^2
for some nonnegative integers p, q, r, s. (End)
Numbers k such that in the prime factorization of k, all odd primes that occur with an odd exponent are congruent to 1 (mod 8). - Robert Israel, Jun 24 2024
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LINKS
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PROG
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(PARI) isA155562(n, /* use optional 2nd arg to get other analogous sequences */c=[2, 1]) = { for(i=1, #c, for(b=0, sqrtint(n\c[i]), issquare(n-c[i]*b^2) & next(2)); return); 1}
for( n=1, 500, isA155562(n) & print1(n", "))
(Python)
from itertools import count, islice
from sympy import factorint
def A155562_gen(): # generator of terms
return filter(lambda n:all((p & 3 != 3 and p & 7 < 5) or e & 1 == 0 for p, e in factorint(n).items()), count(0))
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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1, 1, 5, 4, 5, 3, 8, 3, 3, 0, 4, 7, 6, 3, 8, 8, 9, 4, 3, 9, 2, 2, 1, 0, 6, 5, 9, 4, 5, 5, 5, 5, 1, 6, 8, 2, 9, 8, 9, 8, 7, 7, 5, 1, 9, 7, 4, 4, 8, 7
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OFFSET
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1,3
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COMMENTS
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This constant can be considered as an analog of zeta(3) (= Apéry's constant = A002117), where Euler's zeta(3) is over all positive integers, with prime elements in A000040, while this constant is over all positive integers that can be written as the sum of two squares (A001481) with prime elements given in A055025.
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LINKS
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FORMULA
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Equals Product_{i > 0} 1/(1-A055025(i)^-3).
Equals 1/(1-prime(1)^(-3)) * Product_{i>1 and prime(i) == 1 (mod 4)} 1/(1-prime(i)^(-3)) * Product_{i>1 and prime(i) == 3 (mod 4)} 1/(1-prime(i)^(-6)), where prime(n) = A000040(n).
Equals zeta_{2,0} (3) * zeta_{4,1} (3) * zeta_{4,3} (6), where zeta_{2,0} (s) = 2^s/(2^s - 1).
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EXAMPLE
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1.1545383304763889439221065945555168298987751974487...
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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1, 0, 6, 8, 5, 9, 2, 1, 0, 5, 6, 5, 4, 9, 9, 0, 1, 3, 5, 2, 0, 2, 9, 4, 8, 0, 2, 0, 7, 4, 3, 2, 4, 3, 6, 1, 3, 6, 1, 3, 3, 3, 5, 9, 0, 8, 1, 0, 1, 7
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OFFSET
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1,3
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COMMENTS
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This constant can be considered as an analog of zeta(4) (= Pi^4/90 = A013662), where Euler's zeta(4) is over all positive integers, with prime elements in A000040, while this constant is over all positive integers that can be written as the sum of two squares (A001481) with prime elements given in A055025.
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LINKS
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FORMULA
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Equals Product_{i > 0} 1/(1-A055025(i)^-4).
Equals 1/(1-prime(1)^(-4)) * Product_{i>1 and prime(i) == 1 (mod 4)} 1/(1-prime(i)^(-4)) * Product_{i>1 and prime(i) == 3 (mod 4)} 1/(1-prime(i)^(-8)), where prime(n) = A000040(n).
Equals zeta_{2,0} (4) * zeta_{4,1} (4) * zeta_{4,3} (8), where zeta_{2,0} (s) = 2^s/(2^s - 1).
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EXAMPLE
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1.0685921056549901352029480207432436136133359081017...
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A071011
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Numbers n such that n is a sum of 2 squares (i.e., n is in A001481(k)) and sigma(n) == 0 (mod 4).
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+20
1
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65, 85, 125, 130, 145, 170, 185, 205, 221, 250, 260, 265, 290, 305, 340, 365, 370, 377, 410, 442, 445, 481, 485, 493, 500, 505, 520, 530, 533, 545, 565, 580, 585, 610, 629, 680, 685, 689, 697, 730, 740, 745, 754, 765, 785, 793, 820, 865, 884, 890, 901, 905
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OFFSET
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1,1
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COMMENTS
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It is conjectured that if m is not a sum of 2 squares (i.e., m is in A022544(k)) sigma(m) == 0 (mod 4).
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LINKS
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MATHEMATICA
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Select[Range[10^3], And[SquaresR[2, #] > 0, Divisible[DivisorSigma[1, #], 4]] &] (* Michael De Vlieger, Jul 30 2017 *)
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PROG
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(PARI) for(n=1, 1000, if(1-sign(sum(i=0, n, sum(j=0, i, if(i^2+j^2-n, 0, 1))))+sigma(n)%4==0, print1(n, ", ")))
(Python)
from math import prod
from itertools import count, islice
from sympy import factorint
def A071011_gen(): # generator of terms
return filter(lambda n:(lambda f:all(p & 3 != 3 or e & 1 == 0 for p, e in f) and prod((p**(e+1)-1)//(p-1) & 3 for p, e in f) & 3 == 0)(factorint(n).items()), count(0))
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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0, 1, 4, 9, 13, 16, 25, 36, 37, 49, 52, 61, 64, 73, 81, 97, 100, 109, 117, 121, 144, 148, 157, 169, 181, 193, 196, 208, 225, 229, 241, 244, 256, 277, 289, 292, 313, 324, 325, 333, 337, 349, 361, 373, 388, 397, 400, 409, 421, 433, 436, 441, 457, 468, 481, 484
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OFFSET
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1,3
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COMMENTS
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Contains A155561 as a subsequence (obtained by restricting a,b,c,d to be nonzero). Also contains A000290 (squares) as subsequence.
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LINKS
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PROG
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(PARI) isA155563(n, /* use optional 2nd arg to get other analogous sequences */c=[3, 1]) = { for(i=1, #c, for(b=0, sqrtint(n\c[i]), issquare(n-c[i]*b^2) & next(2)); return); 1}
for( n=0, 500, isA155563(n) & print1(n", "))
(PARI) is(n)=(n==0) || (#bnfisintnorm(bnfinit(z^2+z+1), n) && #bnfisintnorm(bnfinit(z^2+1), n));
select(n->is(n), vector(1500, j, j-1)) \\ Joerg Arndt, Jan 11 2015
(Python)
from itertools import count, islice
from sympy import factorint
def A155563_gen(): # generator of terms
return filter(lambda n: all(e & 1 == 0 or (p & 3 != 3 and p % 3 < 2) for p, e in factorint(n).items()), count(0))
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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0, 1, 4, 5, 9, 16, 20, 25, 29, 36, 41, 45, 49, 61, 64, 80, 81, 89, 100, 101, 109, 116, 121, 125, 144, 145, 149, 164, 169, 180, 181, 196, 205, 225, 229, 241, 244, 245, 256, 261, 269, 281, 289, 305, 320, 324, 349, 356, 361, 369, 389, 400, 401, 404, 405, 409, 421
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OFFSET
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1,3
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COMMENTS
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Contains A155575 as a subsequence (obtained by restricting a,b,c,d to be nonzero). Also contains A000290 (squares) as subsequence.
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LINKS
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PROG
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(PARI) isA155565(n, /* use optional 2nd arg to get other analogous sequences */c=[5, 1]) = { for(i=1, #c, for(b=0, sqrtint(n\c[i]), issquare(n-c[i]*b^2) & next(2)); return); 1}
for( n=1, 500, isA155565(n) & print1(n", "))
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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