A204388 -id:A204388 - cp's OEIS frontend

A253772 Numbers k such that 4^k + 13 is prime.

1, 2, 4, 10, 19, 32, 40, 146, 566, 2054, 9967, 62639, 87814, 141092

Offset: 1

Vincenzo Librandi, Jan 12 2015

Numbers of the form 4^n+k (for n>0) are never primes when k is even (obviously) or when k == -1 (mod 6): in the last case, in fact, (3+1)^n + 6*h-1 is divisible by 3. - Bruno Berselli, Oct 06 2015

Cf. A104067.

Cf. Numbers k such that 4^k + d is prime: A089437 (d=3), A217349 (d=7), A217350 (d=9), this sequence (d=13), A253773 (d=15), A253774 (d=19), A262345 (d=21), A204388 (d=25), A262969 (d=27), A262971 (d=31), A262972 (d=33).

Magma
```
[n: n in [0..2000] | IsPrime(4^n+13)];
```
Mathematica
```
Select[Range[4000], PrimeQ[4^# + 13] &]
```

is(n)=ispseudoprime(4^n+13) \\ Charles R Greathouse IV, Feb 17 2017

a(n) = A102634(n)/2. - Elmo R. Oliveira, Nov 12 2023

a(11)-a(14) derived from A102634 by Robert Price, Sep 06 2015

A157006 Numbers k such that 2^k + 25 is prime.

Original entry on oeis.org

2, 4, 6, 8, 10, 20, 22, 34, 70, 92, 112, 118, 236, 250, 378, 438, 570, 654, 800, 1636, 2848, 4948, 5670, 6772, 7494, 8006, 9056, 11038, 16268, 21416, 21738, 33370, 78706, 112130, 126446, 164046, 219250, 236432, 368048, 524154, 530810, 640854, 699740, 746302, 754038, 754376, 931976, 989562

Offset: 1

Edwin Dyke (ed.dyke(AT)btinternet.com), Feb 20 2009

nonn

a(40) > 5*10^5. - Robert Price, Oct 15 2015

Since each term is even (n = 2*k), prime numbers of the form 2^k + 25 (see A104072) also have the form 4^k + 25. Those values of k are given in A204388. - Timothy L. Tiffin, Aug 06 2016

			For k = 2, 2^2 + 25 = 29.
For k = 4, 2^4 + 25 = 41.
For k = 6, 2^6 + 25 = 89.

Henri Lifchitz and Renaud Lifchitz, Search for 2^n+25, PRP Top Records.

Cf. A094076, A104072, A123250, A204388.

Cf. A019434 (primes 2^k+1), A057732 (2^k+3), A059242 (2^k+5), A057195 (2^k+7), A057196 (2^k+9), A102633 (2^k+11), A102634 (2^k+13), A057197 (2^k+15), A057200 (2^k+17), A057221 (2^k+19), A057201 (2^k+21), A057203 (2^k+23), this sequence (2^k+25), A157007 (2^k+27), A156982 (2^k+29), A247952 (2^k+31), A247953 (2^k+33), A220077 (2^k+35).

[n: n in [1..1000] | IsPrime(2^n+25)]; // Vincenzo Librandi, Aug 07 2016

Delete[Union[Table[If[PrimeQ[2^n + 25], n, 0], {n, 1, 1000}]], 1]
Select[Range[0, 10000], PrimeQ[2^# + 25] &] (* Vincenzo Librandi, Aug 07 2016 *)

is(n)=ispseudoprime(2^n+5^2) \\ Charles R Greathouse IV, Feb 20 2017

a(n) = 2*A204388(n). - Timothy L. Tiffin, Aug 09 2016

Extended by Vladimir Joseph Stephan Orlovsky, Feb 27 2011
a(29)-a(39) from Robert Price, Oct 15 2015
a(40)-a(48) found by Stefano Morozzi, added by Elmo R. Oliveira, Nov 25 2023

A275702 Numbers n whose deficiency is 26: 2n - sigma(n) = 26.

Original entry on oeis.org

58, 75, 328, 850, 1210, 2848, 35968, 537088, 549768921088, 8796145451008

Offset: 1

Timothy L. Tiffin, Aug 05 2016

Any term x = a(m) can be combined with any term y = A275701(n) to satisfy the property (sigma(x)+sigma(y))/(x+y) = 2.  Although this property is a necessary condition for two numbers to be amicable, it is not a sufficient one.  So far, these two sequences have produced only one amicable pair: (1210,1184) = (a(5),A275701(2)) = (A063990(4),A063990(3)).  If more are ever found, then they will also exhibit x-y = 26.
Notice that:
a(1) =     58 =   29*  2 = (4^1+25)*(4^1)/2
a(3) =    328 =   41*  8 = (4^2+25)*(4^2)/2
a(6) =   2848 =   89* 32 = (4^3+25)*(4^3)/2
a(7) =  35968 =  281*128 = (4^4+25)*(4^4)/2
a(8) = 537088 = 1049*512 = (4^5+25)*(4^5)/2.
If p = 4^k+25 is prime and n = p*(p-25)/2, then it is not hard to show that 2*n - sigma(n) = 26. The values of k in A204388 will guarantee that p is prime (A104072). Similarly, if q = 2*4^k+25 is prime and n = q*(q-25)/2, then 2*n - sigma(n) = 26. However, q will never be prime since it will always be divisible by 3: 2*4^k+25 == (2*1^k+25) mod 3 == 27 mod 3 == 0 mod 3. So, the following values will be in this sequence and provide upper bounds for the next seven terms:
(4^10+25)*(4^10)/2 = 549768921088 >= a(9)
(4^11+25)*(4^11)/2 = 8796145451008 >= a(10)
(4^17+25)*(4^17)/2 = 147573952804424777728 >= a(11)
(4^35+25)*(4^35)/2 = 696898287454081973187748591279228938354688 >= a(12)
(4^46+25)*(4^46)/2 = 12259964326927110866866776279099475433218926722425028608 >= a(13)
(4^56+25)*(4^56)/2 = 13479973333575319897333507543509880240529303896615642871755920375808 >= a(14)
(4^59+25)*(4^59)/2 = 55213970774324510299478046898216207773446358605225195265697257166471168 >= a(15).
The rightmost digit of n = p*(p-25)/2 will always be 8. [Proof: If k is odd, then 4^k+25 == 9 mod 10 and (4^k)/2 == 2 mod 10, which implies that p*(p-25)/2 == 8 mod 10. If k is even, then 4^k+25 == 1 mod 10 and (4^k)/2 == 8 mod 10, which implies that p*(p-25)/2 == 8 mod 10.]
a(10) > 2.3*10^12. - Giovanni Resta, Aug 07 2016
a(11) > 10^18. - Hiroaki Yamanouchi, Aug 21 2018

			a(1) = 58, since 2*58-sigma(58) = 116-90 = 26.
a(2) = 75, since 2*75-sigma(75) = 150-124 = 26.
a(3) = 328, since 2*328-sigma(328) = 656-630 = 26.

D. Alpern, Factorization using the Elliptic Curve Method.

Cf. A033879, A063990, A104072, A204388, A275701 (abundance 26).

[n: n in [1..2*10^6] | (2*n-SumOfDivisors(n)) eq 26]; // Vincenzo Librandi, Aug 06 2016

Select[Range[10^6], 2 # - (DivisorSigma[1, #]) == 26 &] (* Vincenzo Librandi, Aug 06 2016 *)

is(n) = 2*n-sigma(n)==26 \\ Felix Fröhlich, Aug 06 2016

a(9) from Giovanni Resta, Aug 07 2016

a(10) from Hiroaki Yamanouchi, Aug 21 2018

A104072 Primes of the form 2^n + 5^2.

Original entry on oeis.org

29, 41, 89, 281, 1049, 1048601, 4194329, 17179869209, 1180591620717411303449, 4951760157141521099596496921, 5192296858534827628530496329220121, 332306998946228968225951765070086169

Offset: 1

Roger L. Bagula, Mar 02 2005

nonn

Primes of the form 4^n + 4! + 1. - Vincenzo Librandi, Nov 13 2010
Indeed, calculating mod 3 we have 2^n + 5^2 = (-1)^n + 1 = 0 if n is odd, so n must be even to yield a prime. - M. F. Hasler, Nov 13 2010
Those even values of n are given in A157006.  Since n = 2k, these prime numbers also have the form 4^k + 25, where k is given in A204388. - Timothy L. Tiffin, Aug 06 2016
These primes a(m) can be used to generate numbers having deficiency 26. The formula a(m)*(a(m)-25)/2 produces those terms in A275702 having rightmost digit 8. - Timothy L. Tiffin, Aug 09 2016

			From _Timothy L. Tiffin_, Aug 07 2016: (Start)
a(1) = 2^2  + 5^2 =       4 + 25 =      29.
a(2) = 2^4  + 5^2 =      16 + 25 =      41.
a(3) = 2^6  + 5^2 =      64 + 25 =      89.
a(4) = 2^8  + 5^2 =     256 + 25 =     281.
a(5) = 2^10 + 5^2 =    1024 + 25 =    1049.
a(6) = 2^20 + 5^2 = 1048576 + 25 = 1048601. (End)

Cf. A157006, A204388, A275702.

a = Delete[Union[Flatten[Table[If [PrimeQ[2^n + 25] == True, 2^n + 25, 0], {n, 1, 400}]]], 1]
Select[2^Range[0,120]+25,PrimeQ] (* Harvey P. Dale, Jun 20 2017 *)

a(m) = 2^(A157006(m)) + 5^2 = 4^(A204388(m)) + 25. - Timothy L. Tiffin, Aug 07 2016

If n == 0 mod 4, then a(m) == 1 mod 10. If n == 2 mod 4, then a(m) == 9 mod 10. - Timothy L. Tiffin, Aug 09 2016

cp's OEIS Frontend

A253772 Numbers k such that 4^k + 13 is prime.

Views

Author

Keywords

Comments

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A157006 Numbers k such that 2^k + 25 is prime.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A275702 Numbers n whose deficiency is 26: 2n - sigma(n) = 26.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Extensions

A104072 Primes of the form 2^n + 5^2.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula