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June 22

Repeating decimals 1/((k-1)k+1) in base k^2+1 don't include multiples of k in expansion?

I noticed that 1/7 in base that 3,6 & 9 are missing from the expansion in base 10, but it appears that there is a pattern.

k = 2, 1/3 in base 5 is .13(rep), 2/3 in base 5 is .31 (no digit which is a multiple of 2 appears in the pentary expansion, neither 2 or 4 occur
k = 3, 1/7 in base 10 is .142857(rep) (no digit which is a multiple of 3 occurs in the decimal expansion, none of 3, 6 or 9 occur
k = 4, 1/13 (1/C) in base 17 is .153FBD(rep) (not only doesn't any digit which is a multiple of 4 occur (4,8,C,G), it appears that digits which are a multiple of 2 occur.
k = 5, 1/26 (1/L) in base 26 is .164OJL(rep) (no digit which is a multiple of 5 occurs (5,A,F,K,P)

Any idea for a proof or extension of this? (I can't find an easy calculation in base 37)Naraht (talk) 02:11, 22 June 2024 (UTC) .[reply]

There seems to be a typo in the k = 5 case; (k-1)k+1=21, not 26, and 1/26 base 26 is just .1. It might be easier to use a different notation for large bases, say by inserting a comma between each digit and leaving the digits in base 10. So 1/21 = .1,6,4,24,19,21(rep). In this notation 1/31 = (base 37) .1,7,5,35,29,31. I think you can probably see what's going on by looking at k=10 and k=100: 1/91 = (base 101) .1,11,9,99,89,91(rep), and 1/9901 = (base 10001) .1,101,99,9999,9899,9901(rep). The pattern is 1/(k²-k+1) = (base k²+1) .1,k+1,k-1,k²-1,k²-k-1,k²-k+1, which shouldn't be too hard to verify. There is almost certainly python code to compute fractions in large bases to do further experimentation. k²-k+1 and k²+1 are both cyclotomic polynomials, and there are probably similar patterns when you look at 1/Phi_m(k) base Phi_n(k) for various m and n. --RDBury (talk) 15:15, 22 June 2024 (UTC)[reply]

PS. The fact that the denominator is a cyclotomic polynomial is the part of what makes this work, the other part isn't that the base is a cyclotomic polynomial in k, but that it's equal to ±k mod the denominator. In this case the denominator is Φ₆(k) and the denominator is Φ₆(k)+k. A generalization is that if P(x) is congruent to x mod Φ_n(x) (taken as polynomials in x) then for a 1/Φ_n(k) has period dividing n base P(k). For example Φ₁₀(x)=x⁴-x³+x²-x+1 and P(x)=x⁸+x⁴+x²+1 is congruent to x mod Φ₁₀(x). So the generalization states that Φ₁₀(k) has period dividing 10 base P(k). For k=10 we get 1/9091 = (base 100010101) .11001,110010,1100100,11001001,9999909,99999099,99900090,98910000,89009099,90010191(rep) and this indeed has period 10. If P(x) is congruent to -x mod Φ_n(x) the rule is that the period divides 2n. For n=3, k=3, 1/13 = .076923(rep) base 10; for k=4, 1/21 = .0,13,12,16,3,4(rep) base 17; for k=10, 1/111=.0,91,90,100,9,10; and in general 1/(k²+k+1) = .0,k²-k+1,k²-k,k²,k-1,k(rep). There are probably additional tweaks and generalizations you can make on this. --RDBury (talk) 16:03, 23 June 2024 (UTC)[reply]

thank you for all of this, my noticing of the missing "k" was a small part of what could actually be found.Naraht (talk) 16:48, 25 June 2024 (UTC)[reply]