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It is not (particularly) rare that $$ 2^n\equiv1\pmod{2n-1} $$

This happens 81 times for n less than a million, to wit:

1, 2, 8, 128, 228, 648, 1352, 1908, 3240, 4608, 5220, 5976, 11448, 13160, 13920, 21528, 22050, 23760, 23940, 24840, 30960, 31284, 31584, 31968, 32768, 37224, 46092, 46512, 47268, 60480, 65664, 66528, 78540, 78600, 81728, 82800, 84312, 98406, 102672, 103968, 130416, 133560, 139248, 141840, 154980, 162792, 166716, 215100, 238368, 278520, 280368, 288360, 297252, 326200, 353808, 386400, 439740, 516528, 540260, 581400, 594720, 614052, 633420, 648440, 664668, 688968, 720000, 731808, 734448, 739620, 763425, 763848, 824670, 839808, 847000, 864468, 877380, 919100, 961020, 965808, 994728

Of those only two (1 and 763425) are odd. Lest I deceive by use of small numbers, going to $10^8$ adds two additional odds (10888425, 40068105) and 563 additional evens.

Is there a reason for this (apparent) bias? The only obvious difference is that $3\mid4^n-1$ but $3\not\mid2\cdot4^n-1.$ Is this enough?

All the odd n I've found so far, with their (particularly smooth) factorizations:

1
763425 = 3^4 * 5^2 * 13 * 29
10888425 = 3^4 * 5^2 * 19 * 283
40068105 = 3 * 5 * 7 * 11 * 113 * 307
142086921 = 3 * 19 * 29 * 43 * 1999
191345625 = 3^3 * 5^4 * 17 * 23 * 29
462784725 = 3^3 * 5^2 * 13 * 23 * 2293
468545025 = 3 * 5^2 * 13 * 29 * 73 * 227
552451809 = 3 * 7 * 13 * 19 * 73 * 1459
595018305 = 3^5 * 5 * 7 * 43 * 1627
683993905 = 5 * 7 * 43 * 179 * 2539
956917125 = 3^3 * 5^3 * 37 * 79 * 97
1013987349 = 3^3 * 29 * 1295003
1024992045 = 3^2 * 5 * 7^3 * 11 * 6037
1567781325 = 3^5 * 5^2 * 11 * 29 * 809
1581567885 = 3^2 * 5 * 17 * 19 * 233 * 467
3094868865 = 3 * 5 * 11 * 19 * 987199
3312888345 = 3^2 * 5 * 13 * 17 * 43 * 61 * 127
2'5 9'2
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Charles
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  • For what it's worth, these numbers are tabulated at http://oeis.org/A081856 but only the first 40, so not as far as the first (non-trivial) odd one (and no further information is given at that site). – Gerry Myerson Dec 06 '13 at 08:56
  • $n=1135747173843009$ is another one. – lsr314 Dec 09 '13 at 15:46

1 Answers1

2

This cannot be an exact answer, but might give a clue: for an expression $f(n)=2^n-1$ the set of primes $p$, when seen as candidates for becoming primefactors of that expression, becomes "filtered" by the $\operatorname{ord}_2(p)$- function, which is a divisor of $\varphi(p)$ (Euler's totient) : if $\operatorname{ord}_2(p)$ divides $n$ then the prime $p$ becomes a primefactor of $f(n)$.

But many of that $\operatorname{ord}_2()$'s are even, so for odd $n$ there are few prime-factor candidates . For instance, the primefactors $3,5,11,...$ can only occur in $f(n)$ if $n$ is even, and only the primefactors $7,23,31,47,71,...$ have odd such orders . (A quick numerical check says, that about 29% of the odd primes up to some k have odd $\operatorname{ord}_2(p)$).
(...) It seems, there shall be a little more to say, but I've to look at it with more time.

Gottfried Helms
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