A015910 -id:A015910 - cp's OEIS frontend

A213859 a(n) = 2^n mod (n+2).

1, 2, 0, 3, 4, 4, 0, 2, 6, 6, 4, 7, 8, 2, 0, 9, 16, 10, 4, 2, 12, 12, 16, 8, 14, 20, 4, 15, 16, 16, 0, 2, 18, 22, 16, 19, 20, 2, 24, 21, 16, 22, 4, 38, 24, 24, 16, 32, 6, 2, 4, 27, 34, 52, 8, 2, 30, 30, 4, 31, 32, 2, 0, 8, 16, 34, 4, 2, 46, 36, 16, 37, 38, 17

Offset: 0

Alex Ratushnyak, Jun 22 2012

Conjectures:
Indices of zeros: 2^(x+2)-2, x >= 0.
There are infinitely many n's such that a(n)=n.
Every integer k >= 0 appears in a(n) at least once.
Every k >= 0 appears in a(n) infinitely many times.

G. C. Greubel, Table of n, a(n) for n = 0..10000

Cf. A015910, A066606, A062173, A106262.

[Modexp(2,n,n+2): n in [0..120]]; // G. C. Greubel, Jan 11 2023

Table[PowerMod[2, n, n+2], {n, 0, 100}] (* T. D. Noe, Jun 26 2012 *)

print([2**n % (n+2) for n in range(222)])

[power_mod(2,n,n+2) for n in range(121)] # G. C. Greubel, Jan 11 2023

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

a(n) = A106262(2*n, n). - G. C. Greubel, Jan 11 2023

A060154 Table T(n,k) by antidiagonals of n^k mod k [n,k >= 1].

Original entry on oeis.org

0, 1, 0, 1, 0, 0, 1, 2, 1, 0, 1, 0, 0, 0, 0, 1, 2, 1, 1, 1, 0, 1, 4, 3, 0, 2, 0, 0, 1, 2, 3, 4, 1, 0, 1, 0, 1, 0, 3, 4, 0, 0, 1, 0, 0, 1, 8, 1, 4, 1, 1, 1, 2, 1, 0, 1, 4, 0, 0, 5, 0, 2, 0, 0, 0, 0, 1, 2, 9, 1, 1, 6, 1, 3, 1, 1, 1, 0, 1, 4, 3, 6, 8, 0, 0, 4, 4, 0, 2, 0, 0, 1, 2, 9, 4, 5, 0, 1, 1, 3, 0, 1, 0, 1, 0

Offset: 1

Henry Bottomley, Mar 12 2001

			T(5,3) = 5^3 mod 3 = 125 mod 3 = 2.
Rows start:
  0, 1, 1, 1, 1, ...
  0, 0, 2, 0, 2, ...
  0, 1, 0, 1, 3, ...
  0, 0, 1, 0, 4, ...
  0, 1, 2, 1, 0, ...

Seiichi Manyama, Antidiagonals n = 1..140, flattened

Rows include A057427, A015910, A056969.
Columns include A000004, A000035 (several times), A010872, A010874, A010876, A021559 and other periodic sequences.
Diagonals include A000004 and A057427.
Cf. A114448.

T(n, k) = A051129(n, k)-n*A060155(n, k).

A073797 a(n) = 2^n mod pi(n).

Original entry on oeis.org

0, 0, 0, 2, 1, 0, 0, 0, 0, 3, 1, 2, 4, 2, 4, 4, 1, 0, 0, 0, 0, 5, 1, 2, 4, 8, 7, 2, 4, 2, 4, 8, 5, 10, 9, 8, 4, 8, 4, 6, 12, 2, 4, 8, 2, 8, 1, 2, 4, 8, 1, 0, 0, 0, 0, 0, 0, 8, 16, 2, 4, 8, 16, 14, 10, 3, 6, 12, 5, 8, 16, 2, 4, 8, 16, 11, 1, 6, 12, 2, 4, 18, 13, 3, 6, 12, 1, 8, 16, 8, 16, 8, 16, 8, 16

Offset: 2

Labos Elemer, Aug 12 2002

nonn

			From _Michael De Vlieger_, Dec 09 2018: (Start)
a(2) = 0 since 2^2 mod PrimePi(2) = 4 mod 1 = 0.
a(5) = 2 since 2^5 mod PrimePi(5) = 32 mod 3 = 2. (End)

Michael De Vlieger, Table of n, a(n) for n = 2..10000

Cf. A000079, A000720, A015910, A062173, A064367.

[2^n mod #PrimesUpTo(n): n in [2..100]]; // G. C. Greubel, Dec 10 2018

Array[Mod[2^#, PrimePi@ #] &, 95, 2] (* Michael De Vlieger, Dec 09 2018 *)
Table[PowerMod[2,n,PrimePi[n]],{n,2,100}] (* Harvey P. Dale, Aug 30 2021 *)

for(n=2, 100, print1(lift(Mod(2^n, primepi(n))), ", ")) \\ G. C. Greubel, Dec 10 2018

[mod(2^n, prime_pi(n)) for n in (2..100)] # G. C. Greubel, Dec 10 2018

a(n) = A000079(n) mod A000720(n).

A082493 a(n) = n*ceiling(2^n/n) - 2^n.

Original entry on oeis.org

0, 0, 1, 0, 3, 2, 5, 0, 1, 6, 9, 8, 11, 10, 7, 0, 15, 8, 17, 4, 13, 18, 21, 8, 18, 22, 1, 12, 27, 26, 29, 0, 25, 30, 17, 8, 35, 34, 31, 24, 39, 20, 41, 28, 28, 42, 45, 32, 19, 26, 43, 36, 51, 26, 12, 24, 49, 54, 57, 44, 59, 58, 55, 0, 33, 2, 65, 52, 61, 26, 69, 8, 71, 70, 7, 60, 59, 14

Offset: 1

Vladeta Jovovic, Apr 28 2003

Least nonnegative k such that (2^n+k)/n is an integer.

If n is a power of 2, a(n) = 0; otherwise a(n) = n - A015910(n). - Robert Israel, Apr 08 2015

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A053638, A015910, A000799, A082494, A215747.

seq(-2&^n mod n, n = 1 .. 100); # Robert Israel, Apr 08 2015

def A082493(n): return (-pow(2,n,n))%n # Chai Wah Wu, Aug 24 2023

a(n) = -(2^n) mod n. - Robert Israel, Apr 08 2015

A155836 2^(2^n) mod n.

Original entry on oeis.org

0, 0, 1, 0, 1, 4, 4, 0, 4, 6, 3, 4, 9, 2, 1, 0, 1, 16, 4, 16, 4, 16, 3, 16, 21, 16, 13, 16, 16, 16, 8, 0, 4, 18, 11, 16, 33, 16, 22, 16, 37, 16, 4, 20, 31, 6, 21, 16, 4, 16, 1, 16, 42, 52, 36, 16, 28, 54, 20, 16, 57, 16, 4, 0, 61, 16, 21, 52, 64, 16, 12, 16, 4, 16, 31, 24, 4, 16, 73, 16, 40

Offset: 1

T. D. Noe, Jan 28 2009

nonn

From the randomness of the graph, it seems likely that every number will eventually occur. a(n)=1 for the n in A094358. When do 5 and 23 occur? The number 14 finally appears at n=34913. a(n) can be computed rapidly using two applications of the powermod function.

			a(1941491)=a(43228711)=a(75548489)=5 and a(100867561)=23. See A155886 for the first occurrence of each number. [From _T. D. Noe_, Jan 31 2009]

T. D. Noe, Table of n, a(n) for n = 1..10000

A015910 (2^n mod n), A036236.

Table[e=IntegerExponent[n,2]; d=n/2^e; k=MultiplicativeOrder[2,d]; r=PowerMod[2,n,k]-e; r=Mod[r,k]; 2^e PowerMod[2,r,d], {n, 100}]
Table[PowerMod[2,2^n,n],{n,100}] (* Harvey P. Dale, Oct 16 2022 *)

a(n)=my(ph=eulerphi(n));lift(Mod(2,n)^(ph+lift(Mod(2,ph)^n))) \\ Charles R Greathouse IV, Feb 24 2012

A179976 a(n) = 2^(2n+1) mod (2n+1).

Original entry on oeis.org

0, 2, 2, 2, 8, 2, 2, 8, 2, 2, 8, 2, 7, 26, 2, 2, 8, 18, 2, 8, 2, 2, 17, 2, 30, 8, 2, 43, 8, 2, 2, 8, 32, 2, 8, 2, 2, 68, 18, 2, 80, 2, 32, 8, 2, 37, 8, 13, 2, 17, 2, 2, 92, 2, 2, 8, 2, 78, 44, 60, 112, 8, 57, 2, 8, 2, 128, 53, 2, 2, 8, 85, 32, 50, 2, 2, 53, 63, 2, 8, 151

Offset: 0

Juri-Stepan Gerasimov, Jan 13 2011

nonn

			a(0) = (2^(2*0+1) mod (2*0+1))=(2 mod 1)=0.

a(n) = A015910(2n+1).

A212844 a(n) = 2^(n+2) mod n.

Original entry on oeis.org

0, 0, 2, 0, 3, 4, 1, 0, 5, 6, 8, 4, 8, 2, 2, 0, 8, 4, 8, 4, 11, 16, 8, 16, 3, 16, 23, 8, 8, 16, 8, 0, 32, 16, 2, 4, 8, 16, 32, 24, 8, 4, 8, 20, 23, 16, 8, 16, 22, 46, 32, 12, 8, 4, 7, 16, 32, 16, 8, 4, 8, 16, 32, 0, 63, 58, 8, 64, 32, 36, 8, 40, 8, 16, 47

Offset: 1

Alex Ratushnyak, Jul 22 2012

nonn

Also a(n) = x^x mod (x-2), where x = n+2.
Indices of 0's: 2^k, k>=0.
Indices of 1's: 7, 511, 713, 11023, 15553, 43873, 81079, 95263, 323593, 628153, 2275183, 6520633, 6955513, 7947583, 10817233, 12627943, 14223823, 15346303, 19852423, 27923663, 28529473, ...
Conjecture: every integer k >= 0 appears in a(n) at least once.
Each number below 69 appears at least once. Some large first occurrences: a(39806401) = 25, a(259274569) = 33, a(10571927) = 55, a(18039353) = 81. - Charles R Greathouse IV, Jul 21 2015

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

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A000312, A015910, A062173, A112983, A213381, A213859.

A212844 := proc(n)
    modp( 2&^ (n+2),n) ;
end proc: # R. J. Mathar, Jul 24 2012

Table[PowerMod[2, n+2, n], {n, 79}] (* Alonso del Arte, Jul 22 2012 *)

A212844(n)=lift(Mod(2,n)^(n+2)) \\ M. F. Hasler, Jul 23 2012

for n in range(1,99):
    print(2**(n+2) % n, end=',')

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

A245728 Numbers k that divide 2^k + 6.

Original entry on oeis.org

1, 2, 10, 1030, 10009593662, 13957196317, 55299492770, 3764656723270

Offset: 1

Derek Orr, Jul 30 2014

No other terms below 10^15. Some larger terms: 2962089521722084981, 1376243703434217460265762. - Max Alekseyev, Sep 23 2016

			2^10 + 6 = 1030 is divisible by 10. Thus 10 is a term of this sequence.

OEIS Wiki, 2^n mod n

Cf. A015910, A128122, A153972.

select(n -> 2 &^ n + 6 mod n = 0, [$1..10^6]); # Robert Israel, Jul 30 2014

Select[Range[10^5], Divisible[2^# + 6, #] &] (* Robert Price, Oct 12 2018 *)

for(n=1,10^9,if(Mod(2,n)^n==Mod(-6,n),print1(n,", ")))

a(5) from Jason G. Wurtzel, Sep 25 2014

a(6)-a(8) from Max Alekseyev, Sep 23 2016

A264922 Decimal expansion of constant z = Sum_{n>=1} {2^n/n} * n/2^n, where {x} is the fractional part of x.

Original entry on oeis.org

4, 1, 2, 9, 2, 0, 7, 6, 8, 6, 7, 1, 4, 9, 7, 6, 9, 2, 3, 1, 8, 7, 6, 4, 4, 6, 3, 3, 9, 1, 6, 6, 0, 2, 2, 3, 2, 6, 6, 3, 6, 5, 8, 0, 8, 5, 5, 9, 1, 6, 1, 5, 0, 1, 7, 1, 2, 0, 8, 7, 3, 8, 0, 9, 6, 5, 2, 9, 3, 3, 4, 5, 5, 2, 2, 8, 4, 2, 3, 7, 1, 0, 8, 3, 2, 2, 4, 0, 6, 8, 1, 1, 8, 9, 3, 7, 5, 4, 2, 6, 7, 0, 9, 4, 4, 0, 9, 6, 8, 1, 5, 9, 1, 8, 6, 8, 4, 5, 2, 3, 9, 0, 6, 7, 6, 7, 3, 7, 4, 3, 9, 4, 7, 7, 8, 7, 6, 7, 4, 4, 6, 5, 5, 6, 7, 1, 1, 4, 7, 6, 1, 0, 7, 8, 0, 4, 6, 5, 5, 3, 2, 9

Offset: 0

Paul D. Hanna, Dec 03 2015

			z = 0.41292076867149769231876446339166022326636580855916\
15017120873809652933455228423710832240681189375426\
70944096815918684523906767374394778767446556711476\
10780465532930416417809262367754600548943347721936\
55335089998952017672435611201014919700656911176350\
62372182725523627777491225313970963752168821911399\
67310841379582079241875027376200157722800032983503\
52300500273468914504274753388182612540758874330051\
88409791519634550380640194311077029592977832839103\
92762052659306868595889500273010680885518723259637...
INFINITE SERIES.
z = 0/2 + 0/2^2 + 2/2^3 + 0/2^4 + 2/2^5 + 4/2^6 + 2/2^7 + 0/2^8 + 8/2^9 + 4/2^10 + 2/2^11 + 4/2^12 + 2/2^13 + 4/2^14 + 8/2^15 + 0/2^16 + 2/2^17 + 10/2^18 + 2/2^19 + 16/2^20 + 8/2^21 + 4/2^22 + 2/2^23 + 16/2^24 + 7/2^25 + 4/2^26 + 26/2^27 + 16/2^28 + 2/2^29 + 4/2^30 + 2/2^31 + 0/2^32 + 8/2^33 +...+ A015910(n)/2^n +...

Cf. A015910 (2^n mod n), A264918, A264919, A264920, A264921.

z = Sum_{n>=1} (2^n mod n) / 2^n = Sum_{n>=1} A015910(n) / 2^n.

A073799 Numbers that begin a run of consecutive integers k such that PrimePi(k) divides 2^k.

Original entry on oeis.org

2, 7, 19, 53, 131, 311, 719, 1619, 3671, 8161, 17863, 38873, 84017, 180503, 386093, 821641, 1742537, 3681131, 7754077, 16290047, 34136029, 71378569, 148948139, 310248241, 645155197, 1339484197, 2777105129, 5750079047, 11891268401, 24563311309, 50685770167, 104484802057, 215187847711

Offset: 1

Labos Elemer, Aug 12 2002

nonn

It seems that each term is a bit larger than twice the previous one.
Runs have lengths 3, 4, 4, 6, 6, 2, 8, 2, 2, 6, 18, 18, 30, 8, 24, 6, 2, 18, ..., respectively.
From Chai Wah Wu, Jan 27 2020: (Start)
Theorem: a(1) = 2 and a(n) = A033844(n) for n > 1. For n > 1, the length of the n-th run is prime(2^n+1)-prime(2^n) = A051439(n)-A033844(n) = A074325(n).
Proof: Let r > 1. If p = prime(2^r), then primepi(p) = 2^r.
primepi(p-1) = 2^r - 1. Since r > 1, 2^r - 1 > 2 and odd and thus does not divide any power of 2.
In addition 2^r < p and thus divides 2^p. This means that p is a term. Let q be such that p < q < prime(2^r+1). Then primepi(q) = 2^r and divides 2^q. Since primepi(q-1) = 2^r and divides 2^(q-1), this means that q does not start a run and thus is not a term.
Let w be such that prime(2^r+1) <= w < prime(2^(r+1)). Then 2^r + 1 <= primepi(w) < 2^(r+1) and does not divide any power of 2. This means that w is not a term.
(End)

Chai Wah Wu, Table of n, a(n) for n = 1..57

Cf. A000079, A000720, A015910, A062173, A064367, A073797, A073798.
Cf. A033844. - R. J. Mathar, Sep 23 2008
Cf. A051439, A074325.

aQ[k_] := Divisible[2^k, PrimePi[k]]; s = {}; len = {}; n = 2; While[Length[s] < 10, While[! aQ[n], n++]; n1 = n; While[aQ[n], n++]; If[n > n1, AppendTo[s, n1]; AppendTo[len, n - n1]]; n++]; s (* Amiram Eldar, Dec 11 2018 *)

a(n) = if(n==1, 2, prime(2^n)); \\ Jinyuan Wang, Mar 01 2020

from sympy import prime
def A073799(n):
    return 2 if n == 1 else prime(2**n) # Chai Wah Wu, Jan 27 2020

Solutions to 2^(x-1) mod PrimePi(x-1) > 0 but 2^x mod PrimePi(x) = 0.

a(n) = A033844(n) for n > 1. - Chai Wah Wu, Jan 27 2020

Edited by Jon E. Schoenfield, Dec 10 2018
a(15)-a(18) from Amiram Eldar, Dec 11 2018
a(19)-a(33) from Chai Wah Wu, Jan 27 2020

cp's OEIS Frontend

A213859 a(n) = 2^n mod (n+2).

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A060154 Table T(n,k) by antidiagonals of n^k mod k [n,k >= 1].

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A073797 a(n) = 2^n mod pi(n).

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A082493 a(n) = n*ceiling(2^n/n) - 2^n.

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A155836 2^(2^n) mod n.

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A179976 a(n) = 2^(2n+1) mod (2n+1).

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A212844 a(n) = 2^(n+2) mod n.

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A245728 Numbers k that divide 2^k + 6.

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