This puzzle was suggested by Hugo Pfoertner - thanks Hugo!

In a now-famous 2004 article, Ben Green and Terence Tao proved that arbitrarily long arithmetic progressions exist in the primes. This holds true analogously for other sets of numbers if their density is sufficiently high, for example, numbers that are the sum of two squares. There are also enough numbers of the form x² + y² + x*y (with x,y being integers), called "Loesch numbers," that arbitrarily long arithmetic progressions can be found among them as well.

Example: The first Loesch numbers are 0, 1, 3, 4, 7, 9, 12, 13, 16, 19, 21, ..., and the numbers 1, 7, 13, 19 form an arithmetic progression with 4 terms and a difference of 6. This progression can be described by its initial value of 1 and the step size of 6.

Your goal: Find an arithmetic progression of 35 terms from the Loesch numbers such that the end of the progression becomes as small as possible, specifying the starting value and the step size.

A bonus "*" will be given for finding a progression with at least 42 terms, again specified by its starting value and step size (the end of the progression needs not be as small as possible).

An extra bonus "**" will be awarded to the participant(s) who find the progression with the most (>42) terms. The allocation of (**) to participants may change during the submission process.