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Liljestrom shows that n is in this sequence if and only if 4n^2+1 is composite.

Complement of A001912.

From Hermann Stamm-Wilbrandt, Sep 16 2014: (Start)

For n > 1, A047209 is a subset of this sequence [ 4*n^2+1 is divisible by 5 if n is (1 or 4) mod 5].

A092464 is a subset of this sequence [4*n^2+1 is divisible by 13 if n is (4 or 9) mod 13].

The above are for divisibility by 5, 13; notation (1,4,5), (4,9,13). Divisibility by p for a and p-a; notation (a,p-a,p). These are the next tuples: (2,15,17), (6,23,29), (3,34,37), (16,25,41), ... . The corresponding sequences are a subset of this sequence [ 2,15,19,32,36,49,... for (2,15,17) ]. 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 [solution to 4*a^2+1=0 (mod p)].

For n>1, A000290 (squares) is a subset of this sequence (4,9,16,25,...) [ 4*(m^2)^2+1 is divisible by m^2+(m+1)^2, tuple (m^2, (m+1)^2, m^2+(m+1)^2) ].

(End)

A094551 generalized to squares. Compare to A094552, which has far fewer solutions. For many values of n (5, 13, 25, 41, 61, 85,...), the value of b-a increases by 2 for each successive n. These n are the same as A001844. In other words, when n=i^2+(i+1)^2, then a=n-i-1 and b=n+i-1. The other values of n (35, 39, 51, 111, 143, 160, 856,...), A094523, have comparatively large values of b-a.

A094550 generalized to squares. Note that equality is attained only for very long sums of squares.

a(31) > 4*10^6. [From Donovan Johnson, Apr 20 2010]