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A232423 a(n) = ceiling(sqrt(n^4 - n^3 - n^2 + n + 1))^2 - (n^4 - n^3 - n^2 + n + 1). 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

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).

LINKS

Table of n, a(n) for n=0..53.

FORMULA

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

PROG

(Python)

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

CROSSREFS

Cf. A232395, A232397.

Sequence in context: A189425 A266169 A277928 * A074031 A086261 A111978

Adjacent sequences: A232420 A232421 A232422 * A232424 A232425 A232426

KEYWORD

nonn

AUTHOR

Vladimir Shevelev, Nov 23 2013

EXTENSIONS

More terms from Peter J. C. Moses

STATUS

approved

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Last modified December 6 05:22 EST 2022. Contains 358594 sequences. (Running on oeis4.)