A006517 -id:A006517 - cp's OEIS frontend

A015898 Numbers k such that k | 9^k + 9.

1, 2, 3, 6, 9, 10, 18, 30, 90, 123, 730, 7086, 172426, 300243, 372390, 449674, 1812298, 5027914, 5307130, 17214354, 20788426, 84929451, 101992426, 102177370, 103544442, 106494770, 161388338, 181327546, 193710249, 207083626

Offset: 1

Robert G. Wilson v

nonn

Giovanni Resta, Table of n, a(n) for n = 1..100

Cf. A006517, A015888, A015889, A015891, A015892, A015893, A015897, A015902, A015903, A015904, A015905, A015906.

Select[Range[10^5], Mod[PowerMod[9, #, #] + 9, #] == 0 &] (* Giovanni Resta, Oct 23 2018 *)

Missing a(2)-a(4) from Giovanni Resta, Oct 23 2018

A015902 Numbers k such that k | 10^k + 10.

Original entry on oeis.org

1, 2, 5, 10, 22, 70, 110, 130, 154, 190, 286, 365, 685, 910, 2002, 2794, 2926, 6526, 9730, 15862, 17290, 24130, 50005, 59158, 60346, 68926, 72010, 131326, 163306, 278278, 313045, 478126, 525790, 552370, 618046, 642874, 883246

Offset: 1

Robert G. Wilson v

nonn

Giovanni Resta, Table of n, a(n) for n = 1..250

Cf. A006517, A015888, A015889, A015891, A015892, A015893, A015897, A015898, A015903, A015904, A015905, A015906.

Select[Range[10^5], Mod[PowerMod[10, #, #] + 10, #] == 0 &] (* Giovanni Resta, Oct 23 2018 *)

Missing a(2)-a(3) from Giovanni Resta, Oct 23 2018

A015904 Numbers k such that k | 12^k + 12.

Original entry on oeis.org

1, 2, 3, 4, 6, 12, 15, 26, 28, 52, 76, 78, 87, 156, 364, 435, 532, 988, 2356, 3052, 5356, 6916, 7588, 10095, 11476, 13636, 15051, 15964, 16458, 24388, 31996, 36114, 40132, 42484, 46396, 50428, 50726, 72436, 114846, 132652, 148276, 171076, 208468

Offset: 1

Robert G. Wilson v

nonn

Giovanni Resta, Table of n, a(n) for n = 1..700

Cf. A006517, A015888, A015889, A015891, A015892, A015893, A015897, A015898, A015902, A015903, A015905, A015906.

Select[Range[10^5], Mod[PowerMod[12, #, #] + 12, #] == 0 &] (* Giovanni Resta, Oct 23 2018 *)

Missing a(2)-a(5) from Giovanni Resta, Oct 23 2018

A015905 Numbers k such that k | 13^k + 13.

Original entry on oeis.org

1, 2, 13, 14, 26, 182, 286, 2626, 2926, 8866, 10186, 57226, 62686, 118378, 185653, 187838, 284662, 396526, 689546, 792946, 927706, 1110214, 1140686, 1501526, 1943266, 2034406, 2402686, 2427706, 2759926, 3032926, 4450706, 6586426, 6849766, 7319026, 9357518

Offset: 1

Robert G. Wilson v

nonn

Giovanni Resta, Table of n, a(n) for n = 1..400

Cf. A006517, A015888, A015889, A015891, A015892, A015893, A015897, A015898, A015902, A015903, A015904, A015906.

Select[Range[10^6], Mod[13 + PowerMod[13, #, #], #] == 0 &] (* Giovanni Resta, Oct 21 2018 *)

Missing a(2) and a(33)-a(35) from Giovanni Resta, Oct 21 2018

A296369 Numbers m such that 2^m == -1/2 (mod m).

Original entry on oeis.org

1, 5, 65, 377, 1189, 1469, 25805, 58589, 134945, 137345, 170585, 272609, 285389, 420209, 538733, 592409, 618449, 680705, 778805, 1163065, 1520441, 1700945, 2099201, 2831009, 4020029, 4174169, 4516109, 5059889, 5215769

Offset: 1

Max Alekseyev, Dec 10 2017

nonn

Equivalently, 2^(m+1) == -1 (mod m), or m divides 2^(m+1) + 1.

The sequence is infinite, see A055685.

Chai Wah Wu, Table of n, a(n) for n = 1..1192
OEIS Wiki, 2^n mod n

Solutions to 2^m == k (mod m): A296370 (k=3/2), A187787 (k=1/2), this sequence (k=-1/2), A000079 (k=0), A006521 (k=-1), A015919 (k=2), A006517 (k=-2), A050259 (k=3), A015940 (k=-3), A015921 (k=4), A244673 (k=-4), A128121 (k=5), A245318 (k=-5), A128122 (k=6), A245728 (k=-6), A033981 (k=7), A240941 (k=-7), A015922 (k=8), A245319 (k=-8), A051447 (k=9), A240942 (k=-9), A128123 (k=10), A245594 (k=-10), A033982 (k=11), A128124 (k=12), A051446 (k=13), A128125 (k=14), A033983 (k=15), A015924 (k=16), A124974 (k=17), A128126 (k=18), A125000 (k=19), A015925 (k=2^5), A015926 (k=2^6), A015927 (k=2^7), A015929 (k=2^8), A015931 (k=2^9), A015932 (k=2^10), A015935 (k=2^11), A015937 (k=2^12)

Select[Range[10^5], Divisible[2^(# + 1) + 1, #] &] (* Robert Price, Oct 11 2018 *)

A296369_list = [n for n in range(1,10**6) if pow(2,n+1,n) == n-1] # Chai Wah Wu, Nov 04 2019

a(n) = A055685(n) - 1.

Incorrect term 4285389 removed by Chai Wah Wu, Nov 04 2019

A296370 Numbers m such that 2^m == 3/2 (mod m).

Original entry on oeis.org

1, 111481, 465793, 79036177, 1781269903307, 250369632905747, 708229497085909, 15673900819204067

Offset: 1

Max Alekseyev, Dec 11 2017

Equivalently, 2^(m+1) == 3 (mod m).
Also, numbers m such that 2^(m+1) - 2 is a Fermat pseudoprime base 2, i.e., 2^(m+1) - 2 belongs to A015919 and A006935.
Some larger terms (may be not in order): 2338990834231272653581, 341569682872976768698011746141903924998969680637.

OEIS Wiki, 2^n mod n

Solutions to 2^m == k (mod m): this sequence (k=3/2), A187787 (k=1/2), A296369 (k=-1/2), A000079 (k=0), A006521 (k=-1), A015919 (k=2), A006517 (k=-2), A050259 (k=3), A015940 (k=-3), A015921 (k=4), A244673 (k=-4), A128121 (k=5), A245318 (k=-5), A128122 (k=6), A245728 (k=-6), A033981 (k=7), A240941 (k=-7), A015922 (k=8), A245319 (k=-8), A051447 (k=9), A240942 (k=-9), A128123 (k=10), A245594 (k=-10), A033982 (k=11), A128124 (k=12), A051446 (k=13), A128125 (k=14), A033983 (k=15), A015924 (k=16), A124974 (k=17), A128126 (k=18), A125000 (k=19), A015925 (k=2^5), A015926 (k=2^6), A015927 (k=2^7), A015929 (k=2^8), A015931 (k=2^9), A015932 (k=2^10), A015935 (k=2^11), A015937 (k=2^12)

Select[Range[10^6], Divisible[2^(# + 1) - 3, #] &] (* Robert Price, Oct 11 2018 *)

a(n) = A296104(n) - 1.

A115976 Numbers k that divide 2^(k-2) + 1.

Original entry on oeis.org

1, 3, 49737, 717027, 9723611, 21335267, 32390921, 38999627, 43091897, 86071337, 101848553, 102361457, 228911411, 302948067, 370219467, 393664027, 455781089, 483464027, 1040406177, 1272206987, 2371678553, 2571052241, 2648052857, 3054713937, 3597613307, 3782971499, 3917903851, 4005163577, 5419912241

Offset: 1

Max Alekseyev, Mar 15 2006

nonn

Some larger terms: 4465786944074559659, 1440261542571735083956640176981881665928575750093930787551969

Max Alekseyev, Table of n, a(n) for n = 1..708 (contains all terms below 10^15)

The odd elements of A244673.

Cf. A006521, A006517, A069927, A067945, A067946, A067947, A068382, A068383, A014945, A014946, A014949, A092028.

lst = {}; Do[ If[ PowerMod[2, 2n - 3, 2n - 1] == 2n - 2, AppendTo[lst, 2n - 1]], {n, 10^9}]; lst (* Robert G. Wilson v, Apr 04 2006 *)

More terms from Robert G. Wilson v, Apr 04 2006
Terms a(24) onward from Max Alekseyev, Feb 03 2015
b-file corrected and extended by Max Alekseyev, Oct 27 2018

A219037 Numbers k such that k divides 2^k + 2 and (k-1) divides 2^k + 1.

Original entry on oeis.org

2, 6, 66, 73786976294838206466

Offset: 1

Max Alekseyev, Nov 10 2012

Also, numbers k such that 2^k == k-2 (mod k*(k-1)).
The sequence is infinite: if m is in this sequence, then so is 2^m + 2.
No other terms below 10^20.

W. Sierpinski, 250 Problems in Elementary Number Theory. New York: American Elsevier, 1970. Problem #18.

Kin Y. Li et al., Solution to Problem 323, Mathematical Excalibur 14(2), 2009, p. 3.

Intersection of A006517 and A055685.

Cf. A217468, A216822, A171959.

Conjecture: a(n+1) = 2^a(n) + 2 for all n.

A319222 Numbers k such that k divides 2^(2k+1) + 1.

Original entry on oeis.org

1, 3, 129, 2537, 51889, 101617, 226873, 270427, 653467, 945667, 1740979, 5819937, 6520987, 9828587, 15452867, 24950857, 51377539, 89519449, 108627601, 135776371, 160126609, 296338873, 310026163, 400431289, 641706243, 643359937, 678257563, 803419697, 902661523, 952431331, 1004273987, 1243893697, 1796055907

Offset: 1

Altug Alkan, Sep 13 2018

nonn

Also, numbers k such that 4^k == -1/2 (mod k) (cf. A296369). - Max Alekseyev, Sep 15 2018

If k is in the sequence, and m is another divisor of 2^(2*k+1)+1 and is coprime to k, then m*k is in the sequence. - Robert Israel, Sep 14 2018

Cf. A006517, A006521, A015950.

filter:= n -> 2 &^ (2*n+1)+1 mod n = 0:
select(filter, [$1..10^7]); # Robert Israel, Sep 14 2018

PARI
```
is_A319222(n) = Mod(2, n)^(2*n+1)==-1;
```

A370578 Numbers k such that k + 1 divides 3^k + 1.

Original entry on oeis.org

0, 1, 3, 27, 531, 1035, 4635, 6363, 11475, 19683, 40131, 80955, 266475, 280755, 307395, 356643, 490371, 544347, 557955, 565515, 572715, 808227, 1256355, 1695483, 1959075, 1995075, 2771595, 2837835, 3004155, 3208491, 3337635, 3886443, 4670955, 5619411, 6434595, 6942817

Offset: 1

Akiva Weinberger, Feb 22 2024

nonn

The sequence is infinite. It contains all numbers of the form 3^(3^m).
After 3, the smallest term that is not a multiple of 9 is a(13) = 266475.
After 1, the smallest term that is not a multiple of 3 is a(36) = 6942817.
After 1, the smallest term that is not 3 (mod 8) is, also, a(36) = 6942817.
No term can be 2 (mod 3). Proof: Otherwise, k + 1 would be a multiple of 3 while 3^k + 1 would not.
All terms after 0 are odd. Proof: Suppose k is even, so that k+1 is odd. Let p be a prime factor of k+1. Then (by definition of k) 3^k == -1 (mod p) and 3^(2k) == 1 (mod p), so the order of 3 (mod p) divides 2k but not k. Thus the order of 3 is a multiple of 2^(v_2(k)+1) where v_2(k) = A007814(k) is the exponent of 2 in the prime factorization of k. But 3^(p-1) == 1 (mod p) by Fermat's little theorem, so p == 1 (mod 2^(v_2(k)+1)). Multiplying this for all prime factors p of k+1 gives k+1 == 1 (mod 2^(v_2(k)+1)), or k == 0 (mod 2^(v_2(k)+1)). But this contradicts the definition of v_2.
Apart from 0, the only possible residues mod 72 are 1, 3, 13, 25, 27, 37, 49, 51, and 61. It is conjectured that all appear eventually. (See John Omielan's answer to the author's question on Mathematics Stack Exchange.)
Empirically, approximately 80% of the terms are 27 (mod 72).

			(3^27+1)/(27+1) is an integer, so 27 is in the sequence. This can be shown efficiently using a modular exponentiation algorithm to find 3^27 mod 28.

Zubin Mukerjee, Table of n, a(n) for n = 1..214
Mathematics Stack Exchange, Why do so many solutions to n+1 | 3^n+1 satisfy n = 27 (mod 72)?

Cf. A006521, A006517, A007814, A015973, A055777.

Select[Range[0,7000000],TrueQ[PowerMod[3,#,#+1]==#]&] (* James C. McMahon, Feb 25 2024 *)

for n in range(100_000_000):
  if (pow(3,n,n+1)==n):
    print(n)

cp's OEIS Frontend

A015898 Numbers k such that k | 9^k + 9.

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A015902 Numbers k such that k | 10^k + 10.

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A015904 Numbers k such that k | 12^k + 12.

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A015905 Numbers k such that k | 13^k + 13.

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A296369 Numbers m such that 2^m == -1/2 (mod m).

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A296370 Numbers m such that 2^m == 3/2 (mod m).

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A115976 Numbers k that divide 2^(k-2) + 1.

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A219037 Numbers k such that k divides 2^k + 2 and (k-1) divides 2^k + 1.

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A319222 Numbers k such that k divides 2^(2k+1) + 1.

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