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A000534
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Numbers that are not the sum of 4 nonzero squares.
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12
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0, 1, 2, 3, 5, 6, 8, 9, 11, 14, 17, 24, 29, 32, 41, 56, 96, 128, 224, 384, 512, 896, 1536, 2048, 3584, 6144, 8192, 14336, 24576, 32768, 57344, 98304, 131072, 229376, 393216, 524288, 917504, 1572864, 2097152, 3670016, 6291456, 8388608, 14680064
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OFFSET
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1,3
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COMMENTS
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REFERENCES
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J. H. Conway, The Sensual (Quadratic) Form, M.A.A., 1997, p. 140.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 302.
E. Grosswald, Representations of Integers as Sums of Squares. Springer-Verlag, NY, 1985, Theorem 3, pp. 74-75.
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LINKS
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FORMULA
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Consists of the numbers 0, 1, 3, 5, 9, 11, 17, 29, 41, 2*4^m, 6*4^m and 14*4^m (m >= 0). Compare A123069.
From 224 on, a(n) = 4*a(n-3).
Numbers n such that A025428(n) = 0.
G.f.: x^2*(36*x^16 + 32*x^15 + 60*x^14 + 55*x^13 + 36*x^12 + 27*x^11 + 20*x^10 + 19*x^9 + 18*x^8 + 13*x^7 + 11*x^6 + 4*x^5 + 2*x^4 - x^3 - 3*x^2 - 2*x - 1)/(4*x^3 - 1). - Chai Wah Wu, Jul 09 2022
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MATHEMATICA
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q=22; lst={}; Do[Do[Do[Do[z=a^2+b^2+c^2+d^2; If[z<=q^2+3, AppendTo[lst, z]], {d, q}], {c, q}], {b, q}], {a, q}]; lst1=Union@lst lst={}; Do[AppendTo[lst, n], {n, q^2+3}]; lst2=lst Complement[lst2, lst1] (* Vladimir Joseph Stephan Orlovsky, Feb 07 2010 *)
Join[{0, 1, 2, 3, 5, 6, 8, 9, 11, 14, 17, 24, 29, 32, 41}, LinearRecurrence[{0, 0, 4}, {56, 96, 128}, 30]] (* Jean-François Alcover, Feb 09 2016 *)
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PROG
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(PARI) for(n=1, 224, if(sum(a=1, n, sum(b=1, a, sum(c=1, b, sum(d=1, c, if(a^2+b^2+c^2+d^2-n, 0, 1)))))==0, print1(n, ", ")))
(PARI) {a(n)=if( n<2, 0, n<16, [1, 2, 3, 5, 6, 8, 9, 11, 14, 17, 24, 29, 32, 41][n-1], [4, 7, 12][n%3+1] * 2^(n\3*2-7))}; /* Michael Somos, Apr 23 2006 */
(PARI) is(n)=my(k=if(n, n/4^valuation(n, 4), 2)); k==2 || k==6 || k==14 || setsearch([0, 1, 3, 5, 9, 11, 17, 29, 41], n) \\ Charles R Greathouse IV, Sep 03 2014
(Python)
from itertools import count, islice
def A000534_gen(startvalue=0): # generator of terms >= startvalue
return filter(lambda n:n in {0, 1, 3, 5, 9, 11, 17, 29, 41} or n>>((~n&n-1).bit_length()&-2) in {2, 6, 14}, count(max(startvalue, 0)))
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CROSSREFS
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KEYWORD
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nonn,easy,nice,changed
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AUTHOR
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STATUS
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approved
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