A232395 -id:A232395 - cp's OEIS frontend

A232397 a(n) = ceiling(sqrt(n^4 + n^3 + n^2 + n + 1))^2 - (n^4 + n^3 + n^2 + n + 1).

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

Offset: 0

Vladimir Shevelev, Nov 23 2013

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).

Robert Israel, Table of n, a(n) for n = 0..10001
Max Alekseyev, Re: oddness of iterations of sigma(), SeqFan Mailing List.

Cf. A053699, A068527, A232395, A232423.

[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

0, 4, seq(op([5*k^2, k^2-1]),k=1..100); # Robert Israel, Feb 02 2016

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 *)

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

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

More terms from Peter J. C. Moses

A232423 a(n) = ceiling(sqrt(n^4 - n^3 - n^2 + n + 1))^2 - (n^4 - n^3 - n^2 + n + 1).

Original entry on oeis.org

0, 0, 2, 0, 15, 3, 38, 8, 71, 15, 114, 24, 167, 35, 230, 48, 303, 63, 386, 80, 479, 99, 582, 120, 695, 143, 818, 168, 951, 195, 1094, 224, 1247, 255, 1410, 288, 1583, 323, 1766, 360, 1959, 399, 2162, 440, 2375, 483, 2598, 528, 2831, 575, 3074, 624, 3327, 675

Offset: 0

Vladimir Shevelev, Nov 23 2013

nonn

a(n)=0, iff n^4 - n^3 - n^2 + n + 1 is a perfect square.
One can prove that a(n)=0 iff n=0, 1 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 or 1, m=1 and n=3, m=7.
For m>=0, if we also consider negative values of n, we obtain only one more solution: n=-1, m=1.
Indeed, if we consider 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, b(n) = a(n) + 2*n, while for odd n, b(n) = a(n+2).

Cf. A232395, A232397.

from math import isqrt
def A232423(n): return (1+isqrt(m:=n*(n*(n*(n-1)-1)+1)))**2-m-1 # Chai Wah Wu, Jul 29 2022

For odd n>=3, a(n) = A232397(n); for even n>=2, a(n) = 5/4 * n^2 - n - 1.

More terms from Peter J. C. Moses

cp's OEIS Frontend

A232397 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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