cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Thomas Andrews

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Thomas Andrews has authored 2 sequences.

A186699 Numbers n such that there are n numbers in arithmetic progression whose squares sum to a perfect square.

Original entry on oeis.org

1, 2, 4, 9, 11, 16, 23, 24, 25, 26, 33, 36, 47, 49, 50, 52, 59, 64, 73, 74, 81, 88, 96, 97, 100, 107, 121, 122, 144, 146, 148, 169, 177, 184, 191, 193, 194, 196, 218, 225, 239, 241, 242, 244, 249, 256, 276, 289, 292, 297, 299, 311, 312, 313, 324, 337, 338
Offset: 1

Author

Thomas Andrews, Feb 25 2011, Mar 12 2011

Keywords

Comments

A positive integer n is in this sequence if and only if there is a solution to the Pell-like equation x^2-ny^2=d^2(n-1)n(n+1)/3 for some x,y,d integers.
A positive integer n is in this sequence if and only if it can be written in the form: (u^2-3w^2)/(v^2+3w^2), with u,v,w integers and gcd(v,w)=1. This can also be written as a n(v^2) + 3(n+1)(w^2) = z^2.
If n is in this sequence, then we can find an arithmetic progression of *positive* integers which satisfy this equation. (The description above does not require the sequence to be positive.)
By using the method of Legendre to find whether there exists rational numbers r,s on the curve nr^2 + 3(n+1)s^2 = 1, we get the following necessary and sufficient conditions on n:
A. Factor n=a^2b, with b squarefree, then
1. If 3 does not divide b(n+1), then b ≅ 1 (mod 3)
2. If b is divisible by 3, then b ≅ 6 (mod 9)
3. 3 is a square (mod b.)
B.
1. If n+1 is divisible by 3, then (n+1)/3 is the sum of two perfect squares
2. If n+1 is not divisible by 3, then n+1 is the sum of two perfect squares
When n is a perfect square, we can use the arithmetic sequence starting at m=(3n+2)(sqrt(n)-1)/2 + 6 and common difference 6.

Examples

			For n=4, (13,19,25,31) is an arithmetic progression of length 4, and 13^2+19^2+25^2+31^2 = 46^2, so 4 is in the sequence.
		

Crossrefs

Cf. A134419 is a subsequence.

A185545 Numbers n such that there exists a sequence of n consecutive perfect squares that add up to a perfect square.

Original entry on oeis.org

1, 2, 11, 23, 24, 25, 26, 33, 47, 49, 50, 59, 73, 74, 88, 96, 97, 107, 121, 122, 146, 169, 177, 184, 191, 193, 194, 218, 239, 241, 242, 249, 289, 297, 299, 311, 312, 313, 337, 338, 347, 352, 361, 362, 376, 383, 393, 407, 409, 431, 443, 457, 458, 479, 481, 491, 506
Offset: 1

Author

Thomas Andrews, Feb 13 2011

Keywords

Comments

This is A001032 with the addition of 25, because in this sequence the perfect squares may include 0. Subsequence of A134419.