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A232397
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a(n) = ceiling(sqrt(n^4 + n^3 + n^2 + n + 1))^2 - (n^4 + n^3 + n^2 + n + 1).
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2
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0, 4, 5, 0, 20, 3, 45, 8, 80, 15, 125, 24, 180, 35, 245, 48, 320, 63, 405, 80, 500, 99, 605, 120, 720, 143, 845, 168, 980, 195, 1125, 224, 1280, 255, 1445, 288, 1620, 323, 1805, 360, 2000, 399, 2205, 440, 2420, 483, 2645, 528, 2880, 575, 3125, 624, 3380, 675
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OFFSET
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0,2
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COMMENTS
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a(n) = 0 if and only if n^4 + n^3 + n^2 + n + 1 is a perfect square.
Using formula below, we immediately prove that a(n)=0 iff n=0 or n=3. This means that all nonnegative solutions of the Diophantine equation n^4 + n^3 + n^2 + n + 1 = m^2 are n=0, m=1 and n=3, m=11.
For m >=0, if we also consider negative values of n, we obtain only one more solution: n=-1, m=1.
Indeed, if one considers sequence b(n) = ceiling(sqrt(n^4 - n^3 + n^2 - n + 1))^2 - (n^4 - n^3 + n^2 - n +1 ), then, for even n, a(n) = b(n), while for odd n>=3, a(n) = b(n-2).
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LINKS
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FORMULA
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a(1) = 4, for other odd n, a(n) = ((n-1)/2)^2 - 1; for even n>=0, a(n) = 5/4 * n^2.
a(n) = A068527(A053699(n)). [Straight from the description: Difference between smallest square >= (n^4 + n^3 + n^2 + n + 1) and (n^4 + n^3 + n^2 + n + 1)]. - Antti Karttunen, Nov 28 2013
a(n) = (6*n^2-2*n-3+(4*n^2+2*n+3)*(-1)^n+20*(1-(-1)^(2^abs(n-1))))/8. - Luce ETIENNE, Jan 30 2016
G.f.: 4*x+x^2*(x^5-3*x^3-5*x^2-5)/(x^2-1)^3. - Robert Israel, Feb 02 2016
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MAPLE
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0, 4, seq(op([5*k^2, k^2-1]), k=1..100); # Robert Israel, Feb 02 2016
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MATHEMATICA
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Table[Ceiling[Sqrt[n^4 + n^3 + n^2 + n + 1]]^2 - (n^4 + n^3 + n^2 + n + 1), {n, 0, 60}] (* Vincenzo Librandi, Jan 31 2016 *)
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PROG
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(Magma) [Ceiling(Sqrt(n^4+n^3+n^2+n+1))^2-(n^4+n^3+n^2+n+1): n in [0..60]]; // Vincenzo Librandi, Jan 31 2016
(Python)
from math import isqrt
def A232397(n): return (1+isqrt(m:=n*(n*(n*(n+1)+1)+1)))**2-m-1 # Chai Wah Wu, Jul 29 2022
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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