A006521 -id:A006521 - cp's OEIS frontend

A015969 Numbers k that divide 16^k + 1.

1, 17, 289, 4913, 83521, 1419857, 6029713, 12027313, 24137569, 85525793, 102505121, 204464321, 410338673, 1453938481, 1742587057, 3475893457, 6975757441, 24716954177, 29623979969, 59090188769, 111612202577, 118587876497, 420188221009, 500540685121, 503607659473

Offset: 1

Robert G. Wilson v

nonn

Giovanni Resta, Table of n, a(n) for n = 1..34 (terms < 10^13)

Cf. A006521, A015949, A015950, A015951, A015953, A015954, A015955, A015957, A015958, A015960, A015961, A015963, A015965, A015968, A014957.

Column k=16 of A333429.

More terms from Max Alekseyev, Oct 02 2010

Missing terms a(10), a(14), a(18), and a(23) from Giovanni Resta, Mar 23 2020

A093546 Numbers n such that n divides 2^n^2 + 1.

Original entry on oeis.org

1, 3, 9, 27, 57, 81, 171, 243, 513, 729, 1083, 1467, 1539, 2187, 3249, 4401, 4617, 6561, 9747, 13203, 13851, 19683, 20577, 27873, 29241, 32547, 39393, 39609, 41553, 59049, 61731, 83619, 87723, 97641, 118179, 118827, 124659, 177147, 185193, 239121

Offset: 1

Farideh Firoozbakht, Mar 31 2004

nonn

This sequence is closed under multiplication. A006521 is a subsequence of this sequence. A006521 is also closed under multiplication. In fact if m is even and k is a natural number then the sequence "n divides m^n^k + 1" is a subsequence of the sequence "n divides m^n^(k+1)+ 1" and both are closed under multiplication.

"Closed under multiplication" means that if x and y are terms then so is x*y.

Charles R Greathouse IV, Table of n, a(n) for n = 1..415

Cf. A006521, A067945, A093547.

Select[ Range[250857], PowerMod[2, #^2, # ] == # - 1 &] (* Robert G. Wilson v, Apr 02 2004 *)

is(n)=Mod(2,n)^n^2==-1 \\ Charles R Greathouse IV, Aug 01 2016

Corrected and extended by Robert G. Wilson v, Apr 02 2004

A096196 a(n) = (1 + 2^n) mod n.

Original entry on oeis.org

0, 1, 0, 1, 3, 5, 3, 1, 0, 5, 3, 5, 3, 5, 9, 1, 3, 11, 3, 17, 9, 5, 3, 17, 8, 5, 0, 17, 3, 5, 3, 1, 9, 5, 19, 29, 3, 5, 9, 17, 3, 23, 3, 17, 18, 5, 3, 17, 31, 25, 9, 17, 3, 29, 44, 33, 9, 5, 3, 17, 3, 5, 9, 1, 33, 65, 3, 17, 9, 45, 3, 65, 3, 5, 69, 17, 19, 65, 3, 17, 0, 5, 3, 65, 33, 5, 9, 81, 3

Offset: 1

Labos Elemer, Jul 26 2004

Michael De Vlieger, Table of n, a(n) for n = 1..10000
OEIS Wiki, 2^n mod n

Cf. A006521, A015910, A082495.

[(1+2^n) mod (n): n in [1..100]]; // Vincenzo Librandi, Dec 11 2015

seq(1 + 2 &^ n mod n, n = 1 .. 250); # Robert Israel, Dec 10 2015

Table[Mod[1 + Mod[2, n]^n, n], {n, 89}] (* Michael De Vlieger, Dec 10 2015 *)

a(n)=(1+2^n)%n \\ Anders Hellström, Dec 10 2015

a(n)=lift(1+Mod(2,n)^n); \\ Michel Marcus, Dec 12 2015

A327943 Numbers m that divide 6^m + 5.

Original entry on oeis.org

1, 11, 341, 186787, 8607491, 9791567, 11703131, 14320387, 50168819, 952168003, 71654478989, 1328490399527

Offset: 1

Juri-Stepan Gerasimov, Sep 30 2019

Conjecture: For k > 1, k^m == 1 - k (mod m) has infinitely many positive solutions.
Also includes 11834972807906571233 = 31*381773316384082943. - Robert Israel, Oct 03 2019
a(13) > 10^15. - Max Alekseyev, Nov 10 2022

Solutions to k^m == 1-k (mod m): A006521 (k = 2), A015973 (k = 3), A327840 (k = 4), A123047 (k = 5), this sequence (k = 6).

Cf. A245318, A277288, A015891, A253209.

[1] cat [n: n in [1..10^8] | Modexp(6, n, n) + 5 eq n];

Join[{1},Select[Range[98*10^5],PowerMod[6,#,#]==#-5&]] (* The program generates the first six terms of the sequence. To generate more, increase the Range constant but the program may take a long time to run. *) (* Harvey P. Dale, Feb 05 2022 *)

a(11) from Giovanni Resta, Oct 02 2019

a(12) from Max Alekseyev, Nov 10 2022

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.

A327840 Numbers m that divide 4^m + 3.

Original entry on oeis.org

1, 7, 16387, 4509253, 24265177, 42673920001, 103949349763, 12939780075073

Offset: 1

Juri-Stepan Gerasimov, Sep 27 2019

Number of solutions < 10^9 to k^n == k-1 (mod n): 1 (if k = 1), 188 (if k = 2, see A006521), 5 (if k = 3, see A015973), 5 (if k = 4, see this sequence), 5 (if k = 5), 10 (if k = 6), 10 (if k = 7), 7 (if k = 8), 5 (if k = 9), 8 (if k = 10), 11 (if k = 11), 8 (if k = 12), 9 (if k = 13), 4 (if k = 14), 3 (if k = 15), 6 (if k = 16), 7 (if k = 17), 7 (if k = 18), ...

a(9) > 10^15. - Max Alekseyev, Nov 10 2022

Solutions to k^n == 1-k (mod n): A006521 (k = 2), A015973 (k = 3), this sequence (k = 4), A123047 (k = 5), A327943 (k = 6).
Solutions to 4^n == k (mod n): A000079 (k = 0), A015950 (k = -1), A014945 (k = 1), A130421 (k = 2), this sequence (k = -3), A130422 (k = 3).
Cf. A015940, A253208.

[1] cat [n: n in [1..10^8] | Modexp(4,n,n) + 3 eq n];

Select[Range[10^7], IntegerQ[(PowerMod[4, #, # ]+3)/# ]&] (* Metin Sariyar, Sep 28 2019 *)

is(n)=Mod(4,n)^n==-3 \\ Charles R Greathouse IV, Sep 29 2019

a(6)-a(7) from Giovanni Resta, Sep 29 2019

a(8) from Max Alekseyev, Nov 10 2022

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

A087965 Numbers k such that 2^k - 1 is divisible by (k-1).

Original entry on oeis.org

2, 4, 16, 36, 120, 256, 456, 1296, 2556, 2704, 3816, 3856, 4356, 5544, 6480, 8008, 9216, 10440, 10620, 11952, 16212, 22896, 23436, 26320, 26796, 27840, 28680, 35208, 43056, 44100, 47520, 47880, 49680, 51120, 57240, 61920, 62568, 63168, 63936

Offset: 1

Labos Elemer, Sep 22 2003

nonn

Solutions to 2^x == 1 (mod x-1).

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..300 from Harvey P. Dale)

Cf. A006521.

Join[{2},Select[Range[2,64000],PowerMod[2,#,#-1]==1&]] (* Harvey P. Dale, Jun 01 2021 *)

is(k) = !(k % 2) && Mod(2, k-1)^k == 1; \\ Amiram Eldar, Jul 10 2024

A136475 Irregular triangle read by rows: row n gives prime factors of (2^(3^(n+1))+1)/(2^(3^n)+1).

Original entry on oeis.org

3, 3, 19, 3, 87211, 3, 163, 135433, 272010961, 3, 1459, 139483, 10429407431911334611, 918125051602568899753, 3, 227862073, 3110690934667, 216892513252489863991753, 1102099161075964924744009, 393063301203384521164229656203691748263012766081190297429488962985651210769817

Offset: 0

Christopher J. Smyth, Feb 16 2008

The motivation for this sequence is to quickly generate integers n such that n divides 2^n+1 (sequences A006521, A136473). From the link, it is known that if 3^k||n with n|2^n+1 and n not a power of 3, then n is divisible by a prime p dividing 2^(3^k)+1. Thus for any fixed k every n with n|2^n+1 not a power of 3 is divisible by one of the following numbers: 3^k or some 3^j*p, where p>3 is a prime in A136475 before the k-th '3' and j is the number of '3's before p in the sequence.
Note: (2^(3^(k+1))+1)/(2^(3^k)+1) = 2^(2*3^k) - 2^(3^k) + 1.
For the primes dividing 2^(3^k)+1 for some k see A136474.
Are these numbers always squarefree?

			1. (2^(3^4)+1)/(2^(3^3)+1) = 3*163*135433*272010961, the factorization starting at the 4th '3' and ending just before the 5th '3'.
2. From Comment 1 below and k=5, we see that every n not a power of 3 satisfying n|2^n+1 (sequences A006521, A136473) is divisible by 3^5 or 3^2*19 or 3^3*87211 or 3^4*163 or 3^4*135433 or 3^4*272010961.

Toby Bailey and Chris Smyth Primitive solutions of n|2^n+1.(This is the same link as at A136473.)
S. S. Wagstaff, Jr., The Cunningham Project

Cf. A006521, A136473, A136474.

S:=[];for k from 0 to 4 do f:=op(2,ifactors((2^(3^(k+1))+1)/(2^(3^k)+1)));T:=[];for j to nops(f) do T:=[op(T),op(1,op(j,f))];od;S:=[op(S),op(sort(T))];od;op(S);

The prime factors of (2^(3^(k+1))+1)/(2^(3^k)+1) are given in ascending order *for each k*. For each new value of k the factorization starts with a '3', thus delimiting the different factorizations.

A215747 a(n) = (-2)^n mod n.

Original entry on oeis.org

0, 0, 1, 0, 3, 4, 5, 0, 1, 4, 9, 4, 11, 4, 7, 0, 15, 10, 17, 16, 13, 4, 21, 16, 18, 4, 1, 16, 27, 4, 29, 0, 25, 4, 17, 28, 35, 4, 31, 16, 39, 22, 41, 16, 28, 4, 45, 16, 19, 24, 43, 16, 51, 28, 12, 32, 49, 4, 57, 16, 59, 4, 55, 0, 33, 64, 65, 16, 61, 44, 69, 64, 71, 4, 7

Offset: 1

Alex Ratushnyak, Aug 23 2012

n^(n+2) mod (n+2) is essentially the same.
Indices of 0's: 2^k - 1, k>=0.
Indices of 1's: A006521 except the first term.
Indices of 3's: A015940.
Indices of 5's: 7, 133, 1517, 11761, ...
a(A000040(n)) = A000040(n)-2 = A040976(n).

			a(5) = (-2)^5 mod 5 = -32 mod 5 = 3.

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A015910, A082493, A082494, A110146, A213381.

a:= n-> (-2)&^n mod n:
seq(a(n), n=1..100);  # Alois P. Heinz, Apr 08 2015

a[n_]:=Mod[(-2)^n ,n]; Array[a,75] (* Stefano Spezia, Aug 25 2025 *)

for n in range(1, 333):
    print((-2)**n % n, end=',')

cp's OEIS Frontend

A015969 Numbers k that divide 16^k + 1.

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A093546 Numbers n such that n divides 2^n^2 + 1.

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A096196 a(n) = (1 + 2^n) mod n.

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Magma

Maple

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PARI

A327943 Numbers m that divide 6^m + 5.

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Magma

Mathematica

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

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Formula

A327840 Numbers m that divide 4^m + 3.

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Magma

Mathematica

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

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Mathematica

Extensions

A087965 Numbers k such that 2^k - 1 is divisible by (k-1).

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