A014950 Numbers m such that m divides 10^m - 1.

1, 3, 9, 27, 81, 111, 243, 333, 729, 999, 2187, 2997, 4107, 6561, 8991, 12321, 13203, 19683, 20439, 26973, 36963, 39609, 59049, 61317, 80919, 110889, 118827, 151959, 177147, 183951, 242757, 332667, 356481, 455877, 488511, 531441, 551853, 728271

Offset: 1

Author

Olivier Gérard

Keywords

• nonn

Comments

Also, m such that m | R(m) = A002275(m). - Lekraj Beedassy, Mar 25 2005

For n > 1, 3 divides a(n). If m is in the sequence and d divides m then for each positive integer k, d^k*m is in the sequence. So if m is in the sequence then m^k is in the sequence for each positive integer k. In particular, 3^k is in this sequence for all k. - Farideh Firoozbakht, Apr 14 2010

Numbers m such that m divides s(m), where s(1) = 1, s(k) = s(k-1) + k*10^(k-1).

Number of terms <= 10^k, beginning with k = 0: 1, 3, 5, 10, 15, 25, 41, 68, 108, 178, 291, ... - Robert G. Wilson v, Nov 30 2013

Numbers m such that m divides A033713(m). - Hans Havermann, Jan 25 2014

References

• J. D. E. Konhauser et al., Which Way Did The Bicycle Go? Problem 80 pp. 26; 133, Dolciani Math. Exp., No. 18, MAA, Washington DC, 1996.

Links

• Hans Havermann, Table of n, a(n) for n = 1..1600 (first 800 terms from Robert G. Wilson v)
• C. Cooper and R. E. Kennedy, Niven Repunits and 10^n = 1 (mod n), The Fibonacci Quarterly, pp. 139-143, vol 27, May 02 1989.
• Hans Havermann, A014950 factorized and atomized.

Crossrefs

Cf. A066364, A114207, A122787, A127100, A232769.

Programs

Mathematica

Select[ Range[3, 1000000, 6], PowerMod[10, #, #] == 1 &] (* modified by Robert G. Wilson v, Dec 03 2013 *)
k = 3; A014950 = {1}; While[k < 1000000, If[ PowerMod[ 10, k, k] == 1, AppendTo[ A014950, k]; Print@ k]; k += 6]; A014950 (* Robert G. Wilson v, Nov 29 2013 *)

PARI

is(n)=Mod(10,n)^n==1 \\ Charles R Greathouse IV, Nov 29 2013

Formula

Solutions to 10^m == 1 (mod m). - Vladeta Jovovic

Extensions

More terms from Vladeta Jovovic, Dec 18 2001

More terms from Larry Reeves (larryr(AT)acm.org), Jan 06 2005

Edited by Max Alekseyev, May 20 2011