cp's OEIS Frontend

A001844(N) = N^2 + (N+1)^2 is divisible by 5 if and only if N=a(n), n>=1. E.g., n=2: 5|(3^2 + 4^2). But 7^2 + 8^2 is not congruent to 0 (mod 5). - Wolfdieter Lang, May 09 2012

The number of partitions of 5*(n-1) into at most 2 parts. - Colin Barker, Mar 31 2015

The maximum possible number of 6-cycles in an outerplanar graph on n+5 vertices. - Stephen Bartell, Jul 10 2025

Theorem (Sierpinski, 1963): n is a term iff n^2+(n+1)^2 is a composite number. - N. J. A. Sloane, Feb 29 2020

For n > 1, A047219 is a subset of this sequence. This is because n^2 + (n+1)^2 is divisible by 5 if n is (1 or 3) mod 5 (also see A027861). - Dmitry Kamenetsky, Sep 02 2008

From Hermann Stamm-Wilbrandt, Sep 10 2014: (Start)

For n > 0, A212160 is a subset of this sequence (n^2 + (n+1)^2 is divisible by 13 if n == (2 or 10) (mod 13)).

For n >= 0, A212161 is a subset of this sequence (n^2 + (n+1)^2 is divisible by 17 if n == (6 or 10) (mod 17)).

The above are for divisibility by 5, 13, 17; notation (1,3,5), (2,10,13), (6,10,17). Divisibility by p for a and p-a-1; notation (a,p-a-1,p). These are the next tuples: (8,20,29), (15,21,37), (4,36,41), (11,41,53), ... . The corresponding sequences are a subset of this sequence (8,20,37,49,66,78,... for (8,20,29)). These sequences have no entries in the OEIS yet. For any prime of the form 4*k+1 there is exactly one of these tuples/sequences.

For n > 1, A000217 (triangular numbers) is a subset of this sequence (3,6,10,15,...); z=A000217(n), y=z-1, x=n.

For n > 0, A001652 is a subset of this sequence; z=A001652(n), x=y=A053141(n).

For n > 1, A001108(=A115598) is a subset of this sequence; z=A001108(n), x=A076708(n), y=x+1.

For n > 0, A124124(2*n+1)(=A098790(2*n)) is a subset of this sequence (6,37,218,...); z=A124124(2*n+1), x=a(n)-1, y=a(n)+1, a(m) = 6*a(m-1) - a(m-2) + 2, a(0)=0, a(1)=4.

(End)

The companion sequence is A212354.

There are at most two incongruent solutions of this congruence due to the degree. The fact that there are precisely two such solutions for each prime of the form 4*k+1 (see A002144) is due to the reduction of this problem to one of quadratic residues, namely to X^2 == -1 (mod 2p), with p a prime (see the Nagell reference, given in A210848, pp. 132-3, especially theorem 77), adapted to the quadratic form f(x) = 2*x^2 + 2*x + 1, with discriminant D=-4. This congruence with composite modulus has exactly two incongruent solutions because X^2 == -1 (mod 2) has only the solution +1 modulo 2 (odd numbers), and X^2 == -1 (mod p) has (at least one) solution if the Legendre symbol (-1/p) = +1 (i.e., if -1 is a quadratic residue modulo p). Now (-1/p) = (-1)^(p-1)/2 (see, e.g., the Niven-Zuckerman-Montgomery reference given in A001844, Theorem 3.2 (1), p. 132). Hence there is a solution modulo p iff p == 1 (mod 4). Call the smallest positive one X0, with 0 < X0 < p-1. Then one also has the incongruent solution X1 := p-X0. This implies that there are precisely two incongruent solution of the original congruence modulo 2*p for each 1 (mod 4) prime (see, e.g., Nagell's book, pp. 83-4, Theorem 46). If u is a solution for p = A002144(n) (the existence of u has just been proved) then also the companion v := p-1-u satisfies this congruence, and v is incongruent to u modulo p.

Note that x^2 + (x+1)^2 = 4*T(x) + 1, with the triangular numbers A000217.

The primes with x^2 +(x+1)^2 = prime (necessarily from A002144) are found under A027862. The corresponding x values are found under A027861. These x values explain the positions n' where a(n') is smaller than a(n'-1) (for n'>=6): determine k with x=A027861(k), and then n' from A027862(k) = A002144(n'). Note that a(n') = x for such values n'. E.g., n'=6 with a(6)=4: x=4=A027861(3), p=41=A027862(3) = A002144(6). These values n' are n' = 1, 2, 6, 8, 14, 19, 30, ...

All positive solutions of this congruence are provided by the two sequences with entries u(n,k) = a(n) + k*A002144(n) and v(n,k) = A212354(n) + k*A002144(n), n >= 1, k >= 0. For the cases p = 5, 13 and 17 see A047219, A212160 and A212161, respectively, where the even-indexed numbers are the u(n,k) and the odd-indexed ones the v(n,k) (bisection).

2*a(n) + 1 = A206549(n), the smallest positive nontrivial solution of X^2 == +1 (Modd A002144(n)). For the next larger solution 2*A212354(n) + 1 >= p, hence it does not belong to the restricted residue system Modd A002144(n).

The companion sequence is A212353.

See the comments on A212353 for the proof of two incongruent solutions of this congruence for each prime A002144(n). One takes the smallest positive representatives in each case as A212353(n) and a(n), with A212353(n) < a(n).

All positive solutions of this congruence are provided by the two sequences with entries u(n,k) = A212353(n) + k*A002144(n) and v(n,k) = a(n) + k*A002144(n), n >= 1, k >= 0. For the cases p = 5, 13 and 17 see A047219, A212160 and A212161, respectively, where the even-indexed numbers are the u(n,k) and the odd-indexed ones the v(n,k) (bisection).

r2(n) := 2*a(n) + 1 >= A002144(n) iff r(n) := 2*A212353(n) + 1 <= A002144(n)- 2. r2(n)^2 == +1 (Modd A002144(n)) but only r(n) belongs to the relevant restricted residue class. See A206549. Note that floor(r2(n)^2/A002144(n)) is odd. The same holds for r2 replaced by r.