A064367 -id:A064367 - cp's OEIS frontend

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

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).

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

A073800 Remainder of division 2^n/c(n), where c(n)=A002808(n), the n-th composite.

Original entry on oeis.org

2, 4, 0, 7, 2, 4, 2, 1, 0, 16, 8, 1, 8, 16, 18, 16, 14, 8, 8, 0, 2, 30, 18, 28, 14, 4, 8, 16, 28, 19, 6, 16, 29, 34, 8, 40, 2, 14, 8, 16, 14, 4, 8, 4, 0, 49, 62, 52, 32, 4, 8, 46, 17, 20, 65, 22, 32, 16, 62, 64, 32, 64, 41, 16, 32, 64, 48, 70, 48, 24, 32, 22, 74, 84, 8, 16, 32, 52

Offset: 1

Labos Elemer, Aug 12 2002

nonn

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A000079, A002808, A015910, A062173, A064367, A073797, A073798, A073799.

Table[Mod[2^j, FixedPoint[j+PrimePi[ # ]+1&, j]], {j, 1, 128}]
Module[{c=Select[Range[200],CompositeQ],len},len=Length[c];Table[ PowerMod[ 2,n,c[[n]]],{n,len}]] (* Harvey P. Dale, Mar 03 2018 *)

A297966 a(n) = 3^n mod prime(n).

Original entry on oeis.org

1, 0, 2, 4, 1, 1, 11, 6, 18, 5, 13, 10, 38, 36, 42, 15, 19, 34, 5, 32, 46, 51, 49, 39, 94, 30, 73, 53, 9, 91, 114, 100, 79, 121, 38, 127, 122, 113, 126, 138, 82, 114, 118, 112, 119, 10, 108, 30, 188, 20, 188, 30, 174, 169, 5, 52, 15, 241, 202, 222, 50, 267

Offset: 1

Vincenzo Librandi, Jan 11 2018

Vincenzo Librandi, Table of n, a(n) for n = 1..5000

Cf. A000040, A000244, A064367, A066601.

P:=Filtered([1..1500],IsPrime);;List([1..65],n->PowerModInt(3,n,P[n])); # Muniru A Asiru, Mar 12 2018

[Modexp(3, n, NthPrime(n)): n in [1..80]];

seq(3 &^ n mod ithprime(n), n=1..65); # Muniru A Asiru, Mar 12 2018

Mathematica
```
Array[PowerMod[3, #, Prime@#]&, 80]
```

a(n) = lift(Mod(3, prime(n))^n); \\ Michel Marcus, Jan 11 2018

a(n) = A000244(n) mod A000040(n).

A297967 a(n) = 5^n mod prime(n).

Original entry on oeis.org

1, 1, 0, 2, 1, 12, 10, 4, 11, 20, 25, 10, 39, 36, 41, 13, 36, 58, 52, 1, 17, 26, 19, 64, 13, 5, 94, 14, 25, 36, 40, 74, 117, 81, 6, 123, 24, 155, 134, 117, 20, 42, 69, 36, 185, 111, 121, 16, 206, 159, 42, 220, 47, 123, 130, 61, 57, 83, 6, 79, 270, 14, 91

Offset: 1

Vincenzo Librandi, Jan 15 2018

nonn

Vincenzo Librandi, Table of n, a(n) for n = 1..5000

Cf. A000040, A000351, A064367, A066603, A297966.

[Modexp(5, n, NthPrime(n)): n in [1..80]];

Mathematica
```
Array[PowerMod[5, #, Prime@#]&, 80]
```

a(n) = A000351(n) mod A000040(n).

cp's OEIS Frontend

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

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

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A073800 Remainder of division 2^n/c(n), where c(n)=A002808(n), the n-th composite.

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A297966 a(n) = 3^n mod prime(n).

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A297967 a(n) = 5^n mod prime(n).

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