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cfErdosMoser

The Erdős-Moser equation revisited using continued fractions

See Yves Gallot, Pieter Moree, Wadim Zudilin,
The Erdős-Moser equation 1k + 2k + ... + (m−1)k = mk revisited using continued fractions.
Math. Comp. 80 (2011), 1221–1237.

On September 2024, Robert Gahan established the new benchmark m > 101010 with cfErdosMoser. The processor was an Intel® Core™ i9-13980HX with 64GB RAM, running Windows 11 Pro. It took about 61 hours with 8 threads and the peak memory usage was 33 GB to prove, using N = 3 · 19#, that conditions (a), (b), (c) are not satisfied if qj < 7.76 · 10 10 381 635 903. With N = 28 · 35 · 54 · 7, it took about 91 hours and the peak memory usage was 56 GB to find the solution qj = 7.69 · 10 16 771 044 760 that fits conditions (a), (b), (c) and partly (d).

Theorem 3 (2024). If an integer pair (m, k) with k ≥ 2 satisfies the Erdős-Moser equation, then m > k > 101010.

About

This new version of the program is not based on the binary expansion of log 2. Instead, the main diagonal Padé approximant to log 2 at the point one is calculated. It can be computed using a generalized continued fraction but then convergents are not irreducible fractions. However, the canonical form can be obtained by dividing the numerator and denominator by a known function.
If the normalized convergents of the generalized continued fraction and the coefficients of the regular continued fraction are evaluated together then the size of the remaining j th and (j + 1)th fractions is smaller than the size of the binary expansion needed for evaluating j coefficients. Because memory size is the main limitation of this computation, this method is more efficient.
If qj is a n-digit number, memory usage is about 4n.

Results

The following table provides the smallest integers j satisfying conditions (a), (b) and (c) of Theorem 2 and (d) is checked for p ≤ 43 such that 3 is a primitive root modulo p.
n# is the primorial function.

The solutions to 28 · 35 · 54 · 7 = 272160000 and 28 · 35 · 54 · 72 = 1905120000, found by Robert Gahan, are the best known results.
23# = 223092870 is the most wanted candidate.

N jN aj+1 qj (rounded down) qj mod 6 p(qj)
23# > 78 · 10 9 > 10 40 000 000 000
28 · 35 · 54 · 72 32 544 936 570 457 913 312 295 7.699688 · 10 16 771 044 760 -1
28 · 35 · 54 · 7 13 988 361 422 107 695 241 629 1.444542 · 10 7 208 444 672 -1 7
32 544 869 308 65 416 187 470 7.699688 · 10 16 771 044 760 -1
3 · 19# 20 145 968 414 8 176 658 150 7.766616 · 10 10 381 635 903 +1
2 · 19# 13 078 292 184 5 890 259 269 6.066802 · 10 6 739 544 561 -1
19# 13 050 765 960 4 245 878 986 1.202797 · 10 6 725 335 083 -1
11! 5 151 307 662 14 371 269 642 1.086746 · 10 2 654 644 376 -1
28 · 35 · 54 2 015 385 392 21 941 638 090 5.565196 · 10 1 038 523 018 -1 19
3 236 257 942 11 535 576 954 5.427816 · 10 1 667 658 416 +1
28 · 35 · 53 2 015 279 170 4 388 327 617 5.565196 · 10 1 038 523 018 -1 19
3 236 170 820 2 307 115 390 5.427816 · 10 1 667 658 416 +1
10! 1 738 197 230 3 345 301 699 1.594940 · 10 895 721 905 +1 5
2 579 384 226 1 116 890 723 5.498581 · 10 1 329 206 901 +1
28 · 35 · 52 1 738 154 976 1 433 700 727 1.594940 · 10 895 721 905 +1 5
1 977 626 256 853 324 651 1.196828 · 10 1 019 133 881 -1
17# 367 396 948 127 090 830 4.724143 · 10 189 330 916 -1 5
586 036 322 331 038 577 9.914162 · 10 302 014 571 +1 19
639 808 912 675 894 301 2.276299 · 10 329 726 891 +1
28 · 35 · 5 351 661 538 85 898 211 9.729739 · 10 181 214 202 -1
28 · 35 155 830 946 31 664 035 2.257099 · 10 80 303 211 -1

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