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

1,2

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.

Formula

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

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 *)

Prog

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

Crossrefs

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

Sequence in context: A057262 A057232 A036145 * A271351 A036143 A248960

Adjacent sequences: A014947 A014948 A014949 * A014951 A014952 A014953

Keyword

nonn

Author

Olivier Gérard

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

Status

approved