Ponder This Challenge - September 2026 - Loeschian Arithmetic Progressions

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.

Solvers

• *Paul Lupascu (1/9/2026 9:45 AM IDT)
• *Alper Halbutogullari (1/9/2026 11:08 AM IDT)
• *Daniel Chong Jyh Tar (1/9/2026 11:30 AM IDT)
• *Lazar Ilic (1/9/2026 1:01 PM IDT)
• *Bertram Felgenhauer (1/9/2026 2:24 PM IDT)
• *George Jiri Spitalsky (1/9/2026 3:40 PM IDT)
• *King Pig (1/9/2026 9:33 PM IDT)
• Aayaam Panigrahi (1/9/2026 10:11 PM IDT)
• *Daniel Bitin (1/9/2026 11:01 PM IDT)
• *Jean-François Hermant (2/9/2026 12:24 AM IDT)
• Alex Fleischer (2/9/2026 12:57 PM IDT)
• Dan Ismailescu (2/9/2026 2:32 PM IDT)
• *Lawrence Hon (3/9/2026 4:35 AM IDT)
• Hoang Vu (3/9/2026 4:45 AM IDT)
• Rethna Pulikkoonattu (3/9/2026 7:22 AM IDT)
• Guangxi Liu (3/9/2026 9:11 AM IDT)
• Michael Vahle (3/9/2026 2:17 PM IDT)
• *Kang Jin Cho (3/9/2026 4:13 PM IDT)
• Florian Fischer (4/9/2026 12:35 AM IDT)
• *Nickita Khylkouski (4/9/2026 4:03 AM IDT)
• Franciraldo Cavalcante (4/9/2026 9:26 AM IDT)
• **Henk Waßmann (4/9/2026 11:59 AM IDT)
• *Vladimir Volevich (4/9/2026 2:53 PM IDT)
• *Tamir Ganor & Shouky Dan (4/9/2026 7:18 PM IDT)
• Ahmet Yüksel (5/9/2026 12:18 PM IDT)
• *Prashant Wankhede (6/9/2026 5:23 AM IDT)
• *Peyo (6/9/2026 9:14 AM IDT)
• Juergen Koehl (7/9/2026 4:44 PM IDT)
• *Daniel Cahill (7/9/2026 6:26 PM IDT)
• Puvichakravarthy Ramachandran (8/9/2026 5:50 AM IDT)
• Siamak (8/9/2026 3:38 PM IDT)
• Paul Revenant (8/9/2026 5:21 PM IDT)
• Michele Missiroli (8/9/2026 9:38 PM IDT)
• *Rahid Zaman (9/9/2026 2:08 AM IDT)
• Björn Richard Nerbe (9/9/2026 8:21 AM IDT)
• *Liubing Yu (10/9/2026 8:00 PM IDT)
• Ewoud Storm van's Gravesande (10/9/2026 11:15 PM IDT)
• *Emin-Ali CALYAKA (10/9/2026 11:54 PM IDT)
• Dominik Reichl (12/9/2026 5:25 PM IDT)
• *Micah Day-O'Connell (12/9/2026 7:38 PM IDT)
• Naren Swaminathan (12/9/2026 8:17 PM IDT)
• Nicolas Lopez (12/9/2026 9:52 PM IDT)
• Peter Ji (13/9/2026 9:14 PM IDT)
• Karl D’Souza (13/9/2026 11:06 PM IDT)
• Resnina Tathyana (14/9/2026 2:28 AM IDT)
• John Tromp (14/9/2026 9:44 PM IDT)
• Devom Brahmbhatt (15/9/2026 5:37 AM IDT)
• Sanandan Swaminathan (15/9/2026 9:18 AM IDT)

The ** holders were, in reverse order of appearance with their original, **-worthy, record:

1. Henk Waßmann (49)
2. Bertram Felgenhauer (45)
3. Paul Lupascu (44)

• n = 52: Henk Waßmann
• n = 51: Daniel Chong Jyh Tar
• n = 50: Bertram Felgenhauer
• n = 48: Peyo, Daniel Bitin, Paul Lupascu, Alper Halbutogullari
• n = 47: Lawrence Hon, George Jiri Spitalsky, Paul Lupascu, Prashant Wankhede, Tamir Ganor & Shouky Dan
• n = 46: King Pig, Daniel Cahill, Jean-François Hermant, Emin-Ali CALYAKA
• n = 45: Kang Jin Cho, Lazar Ilic, Vladimir Volevich