Εμφάνιση αναρτήσεων με ετικέτα SEQUENCES. Εμφάνιση όλων των αναρτήσεων
Εμφάνιση αναρτήσεων με ετικέτα SEQUENCES. Εμφάνιση όλων των αναρτήσεων

Τετάρτη 20 Σεπτεμβρίου 2017

PERIODIC SEQUENCE

Let fk(n) be the periodic sequence 1,2,3,...k,1,2,3...k,1,2,3...1,2,3,...k,...

fk(2^t) is periodic, t=0,1,2,3....

Example: k =7

1*,2*,3,4*,5,6,7,1*,2,3,4,5,6,7,1,2*,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4*,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1*,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2,3,4,5,6,7,1,2*,3,4,5,6,7,.....

The numbers with astetisks is the sequence f7(2^t)

Let k be odd

k = 1 : period : 1

k = 3 : period : 1,2

k = 5 : period : 1,2,4,3

k = 7 : period: 1,2,4

k= 9 : period: 1,2,4,8,7,5

k = 11: period: 1,2,4,8,5,10,9,7,3,6

k = 13: period: 1,2,4,8,3,6,12,11,9,5,10,7

k = 15: period: 1,2,4,8

k = 17: period: 1,2,4,8,16,15,13,9

k = 19: period: 1,2,4,8,16,13,7,14,9,18,17,15,11,3,6,12,5,10

k = 21: period: 1,2,4,8,16,11

Let p(k) be the number of the terms of the period (length of the period)

p(k) is the rank of 2 modulo k ie p(k) is the least number x such that 2^x - 1 = 0 (mod k)

For k = 1, the least x such that 2^x - 1 = 0 (mod 1) is 1: 2^1 - 1 = 1 = 1.1

For k = 3, the least x such that 2^x - 1 = 0 (mod 3) is 2: 2^2 - 1 = 3 = 1.3

For k = 5, the least x such that 2^x - 1 = 0 (mod 5) is 4: 2^4 - 1 = 15 = 3.5

For k = 7, the least x such that 2^x - 1 = 0 (mod 7) is 3: 2^3 - 1 = 7 = 1.7

For k = 9, the least x such that 2^x - 1 = 0 (mod 9) is 6: 2^6 - 1 = 63 = 7.9

For k =11, the least x such that 2^x - 1 = 0 (mod 11) is 10: 2^10 - 1 = 1023 = 93.11

For k =13, the least x such that 2^x - 1 = 0 (mod 13) is 12: 2^12 - 1 = 4095 = 315.13

For k =15, the least x such that 2^x - 1 = 0 (mod 15) is 4: 2^4 - 1 = 15 = 1.15

For k =17, the least x such that 2^x - 1 = 0 (mod 17) is 8: 2^8 - 1 = 255 = 15.17

For k =19, the least x such that 2^x - 1 = 0 (mod 19) is 18: 2^18 - 1 = 262144 = 13797.19

For k =21, the least x such that 2^x - 1 = 0 (mod 21) is 6: 2^6 - 1 = 63 = 3.21

......

For k = 1,3,5,7,..... we have the sequence: 1,3,15,7,63,1023,4095,15,255,262144,3......

Let 2^p(k) -1 = M(k).k

M(k): 1,1,3,1,7,93,315,1,15,13797,3,.... = A165781

p(k) = k - 1 ==> k is prime of the sequence : 3, 5, 11, 13, 19, 29, 37, 53, 59, 61, 67, 83, ...... = A001122

____________FIGURATE_________________

Let 123456...k be a regular k-gon.

We start from 1 and move clockwise and put 1 inside the polygon before 2.

From 2 we move clockwise and put 2 inside the polygon before 4.

From 4 we move clockwise and put 4 inside the poygon before 8.

....

We repeat this counting and moving until we reach a number we have reached already. Then we stop since we have a loop.

Examples for k = 3,5,7,9,11,13,15,17,19,21.


Σάββατο 27 Δεκεμβρίου 2014

ANOPOLIS PRIMES

Pn = n + Qk, n is a natural number (including 0) ie n belongs to No = {0, 1, 2, 3, ....} with n > 0

where Qk is the smallest number in the set No - {Q1, Q2,.... Qk-1} such that Pn is prime

n = 1

Q1 = the smallest number in the set No - {Q0} = No - {0} such that 1 + Q1 is prime = 1

P1 = 1 + 1 = 2

n = 2

Q2 = the smallest number in the set No - {Q0, Q1} = No - {0, 1}such that 2 + Q2 is prime = 3

P2 = 2 + 3 = 5

n = 3

Q3 = the smallest number in the set No - {Q0, Q1, Q2} = No - {0, 1, 3} such that 3 + Q3 is prime = 2

P3 = 3 + 2 = 5

n = 4

Q4 = the smallest number in the set No - {Q0, Q1, Q2, Q3} = No - {0, 1, 3, 2} such that 4 + Q4 is prime = 7

P4 = 4 + 7 = 11

and so on.

Pn : 2, 5, 5, 11, 11, 11, 11, 17, 17, 23, 23, 23, 23, 29, 29, 37, 37, 37, 37, 37, 37, 47, 47, 47, 47, 53, 53, 59, 59, 59, 59, 59, 67, 67, 67, 67, 73, 73, 79, 79, 83, 83, 83, 83, 89, 89, 97, 97, 97, 97, 97, 97,.....

Anopolis Prime Numbers Sequence:

An = the missing primes from Pn: 3, 7, 13, 19, 31, 41, 43, 61, 71,.............................

Antreas P Hatzipolakis, 27 Dec. 2014

Πέμπτη 5 Σεπτεμβρίου 2013

INTEGER SEQUENCES

In the book:

Marc Barbut, Bernard Monjardet: Ordre et Classification: Algebre et Combinatoire. Tome II. Hachette Universite, Paris, 1970.

are listed some integer sequences.

In pp. 44-46 the following:

1,3,19,219,4231,130023,6129859,.... It is A001035 in OEIS (The On-Line Encyclopedia of Integer Sequences)

1,4,29,355,6942,209527,9535241..... It is A000798 in OEIS

1,2,5,16,63,316,..... NOT FOUND IN OEIS

1,3,9,33,139,.... It is A001930 in OEIS 1,3,13,75,541,4683,47293,... It is A000670 in OEIS

In p. 101 the following:

1,2,5,,15,52,203,876,... It is A056273 in OEIS

0,1,3,7,15,31,63,.... It is A000225 in OEIS

0,0,1,6,25,90,301,... It is A000392 in OEIS

0,0,0,1,20,65,350,... NOT FOUND IN OEIS

0,0,0,0,1,15,140,... NOT FOUND IN OEIS

In pp. 165-166 the following:

1,4,18,166,7579,7828352,2414682040996,... It is A007153 in OEIS

1,3,8,28,208,.... NOT FOUND IN OEIS

1,2,4,12,81,... It is A001206 in OEIS

1,1,2,3,7,30,703,... NOT FOUND IN OEIS

BIBLIOGRAPHY:

For pages 1-78:

For pages 83-166:


LOCUS PROBLEM

Problem by Antreas Hatzipolakis Solution by Francisco Javier García Capitán ETC LISTING OF Q X(72803)