login
The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A136475 Irregular triangle read by rows: row n gives prime factors of (2^(3^(n+1))+1)/(2^(3^n)+1). 3
3, 3, 19, 3, 87211, 3, 163, 135433, 272010961, 3, 1459, 139483, 10429407431911334611, 918125051602568899753, 3, 227862073, 3110690934667, 216892513252489863991753, 1102099161075964924744009, 393063301203384521164229656203691748263012766081190297429488962985651210769817 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

1. 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.

2. Note: (2^(3^(k+1)+1)/(2^(3^k)+1) = 2^(2.3^k) - 2^(3^k) + 1.

3. For the primes dividing 2^(3^k)+1 for some k see A136474.

4. Are these numbers always squarefree?

LINKS

Table of n, a(n) for n=0..19.

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

FORMULA

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.

EXAMPLE

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.

MAPLE

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);

CROSSREFS

Cf. A006521, A136473, A136474.

Sequence in context: A189737 A261567 A096935 * A143180 A074249 A234018

Adjacent sequences:  A136472 A136473 A136474 * A136476 A136477 A136478

KEYWORD

nonn,tabf

AUTHOR

Christopher J. Smyth, Feb 16 2008, Feb 19 2008, Feb 22 2008

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified October 23 10:17 EDT 2021. Contains 348211 sequences. (Running on oeis4.)