A129614 -id:A129614 - cp's OEIS frontend

A050474 Solutions to 2*phi(x) = x+1.

1, 3, 15, 255, 65535, 83623935, 4294967295, 6992962672132095

Offset: 1

Jud McCranie, Dec 24 1999

If n is in the sequence and n+2 is prime then m=n*(n+2) is in the sequence because 2*phi(m) = 2*phi(n*(n+2)) = 2*phi(n)*(n+1) = (n+1)^2 = m+1. We can obtain the terms 3, 15, 255, 65535 & 4294967295 from 1 (the first term) in this way. Also since 83623935 is a term and 83623935+2 is prime 83623935*(83623935+2)=6992962672132095 is in the sequence. So 1 and 83623935 are the only known independent terms and next term of this sequence if it exists is the third such term. - Farideh Firoozbakht, May 01 2007
The next term, if it exists, has at least 7 distinct prime factors (see Beiler, p. 92). - Jud McCranie, Dec 13 2012
From Chris Boyd, Mar 22 2015: (Start)
Solutions to k*phi(x) = x + 1, including a(1) - a(8), were published in 1932 by D. H. Lehmer. In the paper's summing up, "3*5*353*929" (= 4919055) was printed in error; it should have read "3*5*17*353*929" (= 83623935), i.e., a(6). This error has been propagated in several subsequent texts, including Wong's thesis.
Lehmer identified solutions where x has fewer than 7 distinct prime factors. Wong showed that no additional solutions exist unless x has at least 8 distinct prime factors. It appears not to be excluded by either author that an unidentified solution < a(8) with 8 or more distinct prime factors may exist. (End)
There are no other terms below 10^25. - Max Alekseyev, May 04 2024

			2*phi(15) = 2*8 = 15 + 1, so 15 is a member of the sequence.

A. H. Beiler, Recreations in the Theory of Numbers, page 92.

D. H. Lehmer, On Euler's totient function, Bulletin of the American Mathematical Society, 38 (1932), 745-751.
E. Wong, Computations on normal families of primes, Simon Fraser University, 1997, MSc thesis.

Subsequence of A203966.

Cf. A000010, A129613, A129614, A129615, A202855.

[n: n in [1..2*10^6] | 2*EulerPhi(n) eq (n+1)]; // Vincenzo Librandi, Mar 22 2015

Select[Range[700000], (# + 1)== 2 EulerPhi[#] &] (* Vincenzo Librandi, Mar 22 2015 *)

is_A050474(n)=if(2*eulerphi(n)==n+1,1,0) \\ Chris Boyd, Mar 22 2015

A number n is in the sequence iff phi(n^2)=1+2+3+...+n because n is in the sequence <=> 2*phi(n)=n+1 <=> n*phi(n)=n*(n+1)/2 <=> phi(n^2)=1+2+3++...+n. For n=1,2,...,5, a(n)=2^2^(n-1)-1. - Farideh Firoozbakht, Jan 26 2006

A129615 a(n) is the smallest natural number m such that 4^4^k + m is prime for k=0,1,...,n.

Original entry on oeis.org

1, 1, 15, 385, 137593, 106908087

Offset: 0

Farideh Firoozbakht, May 13 2007

Next term is greater than 3*10^8.

			For k=0,1,...,5 4^4^k + 106908087 are prime and 106908087 is the smallest number with this property so a(5)=106908087.

Cf. A129613, A129614.

A233464 a(n) is the smallest natural number m such that 10^10^k + m is prime for k = 0, 1, ...., n.

Original entry on oeis.org

1, 19, 5641, 1289743, 2578966671

Offset: 0

Farideh Firoozbakht, Mar 13 2014

			29 (=10^1+19) and 10000000019 (=10^10+19) are primes so a(1)=19.

Cf. A129613, A129614, A129615.

okm(m, n) = {for (k=0, n, if (!isprime(10^10^k + m), return (0)););return (1);}
a(n) = {m = 0; while (!okm(m, n), m++); m;} \\ Michel Marcus, Mar 16 2014

a(4) from Giovanni Resta, Mar 14 2014

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A129613 a(n) is the smallest natural number m such that 2^(2^k) + m is prime for k=0,1,...,n.

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A050474 Solutions to 2*phi(x) = x+1.

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A129615 a(n) is the smallest natural number m such that 4^4^k + m is prime for k=0,1,...,n.

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A233464 a(n) is the smallest natural number m such that 10^10^k + m is prime for k = 0, 1, ...., n.

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