A351830 Distance from the n-th square pyramidal number (sum of the first n positive squares) to the nearest square.

0, 0, 1, 2, 5, 6, 9, 4, 8, 4, 15, 22, 25, 22, 9, 15, 25, 21, 7, 30, 46, 53, 49, 32, 0, 49, 40, 41, 30, 91, 46, 12, 9, 15, 4, 26, 77, 114, 25, 91, 61, 105, 15, 122, 129, 66, 22, 1, 1, 24, 76, 157, 170, 37, 131, 141, 91, 139, 165, 15, 174, 247, 150, 80, 39, 29

Offset

0,4

Comments

As noted by Conway and Sloane (1999), the only zero terms appear at n = 0, n = 1 and n = 24, and the n = 24 case allows for the Lorentzian construction of the Leech lattice through the A351831 vector.

The zero terms are equivalently the subject of the "pile of cannonballs" problem posed by Lucas and solved by Watson. - Peter Munn, Aug 03 2023

References

W. Ljunggren, New solution of a problem proposed by E. Lucas, Norsk Mat. Tidsskr. 34 (1952), pp 65-72.

David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, entry 24, p 101.

Links

Paolo Xausa, Table of n, a(n) for n = 0..9999

Michael A. Bennett, Lucas' Square Pyramid Problem Revisited.

Richard E. Borcherds, How to construct the Leech lattice, YouTube video, 2022.

J. H. Conway and N. J. A. Sloane, Lorentzian forms for the Leech lattice, Bulletin (New Series) of the American Mathematical Society, Volume 6, Number 2, March 1982.

J. H. Conway and N. J. A. Sloane, Sphere Packings, Lattices and Groups, 3rd edition, Springer, New York, NY, 1999, pp. 524-528.

E. Lucas, Problem 1180, Nouvelles Ann. Math. (2) 14 (1875), p 336.

G. N. Watson, The problem of the square pyramid, Messenger of Mathematics 48 (1918), pp. 1-22.

Wikipedia, Leech lattice.

Formula

From Paolo Xausa, Jul 05 2022: (Start)

a(n) = A053188(A000330(n)).

a(n) = abs(A000330(n) - A353295(n)). (End)

Example

a(4) = 5 because the sum of the first 4 positive squares is 1 + 4 + 9 + 16 = 30, the nearest square is 25 and 30 - 25 = 5. - Paolo Xausa, Jul 05 2022

Mathematica

nterms=66; Array[Abs[(s=#(#+1)(2#+1)/6)-Round[Sqrt[s]]^2]&, nterms, 0]

Prog

(Python)
from math import isqrt
def a(n):
    t = n*(n+1)*(2*n+1)//6
    r = isqrt(t)
    return min(t - r**2, (r+1)**2 - t)
print([a(n) for n in range(66)]) # Michael S. Branicky, Feb 21 2022

Crossrefs

Cf. A000290, A000330, A053188, A351831, A353295, A354330, A363284.

Sequence in context: A136369 A305511 A269472 * A244272 A007573 A285663

Adjacent sequences: A351827 A351828 A351829 * A351831 A351832 A351833

Keyword

nonn,easy

Author

Paolo Xausa, Feb 21 2022

Extensions

Name edited by Peter Munn, Aug 04 2023

Status

approved