|
| |
|
|
A116018
|
|
Numbers n such that n + phi(n) is a repdigit.
|
|
4
| |
|
|
1, 2, 3, 4, 5, 6, 17, 21, 63, 167, 201, 389, 603, 1667, 3795, 3889, 4465, 5926, 50394, 166667, 510042, 2000001, 3888889, 5185194, 5798663, 5925926, 6000003, 32050435, 200000001, 335447667, 365110755, 444766346, 600000003, 1558138862
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,2
|
|
|
COMMENTS
| (I). If p=(2*10^(3n+1)+7)/27 is prime then m=2p is in the sequence because m+phi(m)=3p-1=2*(10^(3n+1)-1)/9 is a repdigit number. m=2*(2*10^811+7)/27 (a 811-digit number) is the smallest such terms and the next such terms has 4219 digits. - Farideh Firoozbakht (mymontain(AT)yahoo.com), Aug 24 2006
(II). If p=(8*10^(3n+1)+1)/27 is prime then m=2p is in the sequence because m+phi(m)=8*(10^(3n+1)-1)/9 is a repdigit number. 5926 is the smallest such terms. - Farideh Firoozbakht (mymontain(AT)yahoo.com), Aug 24 2006
(III). If p=(2*10^n+1)/3 then both numbers 3p & 9p are in the sequence because 3p+phi(3p)=5p-2=3*(10^(n+1)-1)/9 & 9p+ phi(9p)=9*(10^(n+1)-1)/9 are repdigit numbers. 21 & 63 are the smallest such terms. - Farideh Firoozbakht (mymontain(AT)yahoo.com), Aug 24 2006
(IV). All primes p of the form (35*10^n+1)/9 are in the sequence because p+phi(p)=7*(10^n-1)/9 is a repdigit number. 389 is the smallest such terms. - Farideh Firoozbakht (mymontain(AT)yahoo.com), Aug 24 2006
(V). All primes p of the form (10^n+2)/6 are in the sequence because p+phi(p)=2p-1=3*(10^n-1)/9 is a repdigit number. 2, 17 & 167 are such terms. - Farideh Firoozbakht (mymontain(AT)yahoo.com), Aug 24 2006, Dec 19 2007
|
|
|
EXAMPLE
| 5185194+phi(5185194)=6666666.
|
|
|
CROSSREFS
| Cf. A116017.
Sequence in context: A166098 A124365 A115896 * A048095 A031015 A024639
Adjacent sequences: A116015 A116016 A116017 * A116019 A116020 A116021
|
|
|
KEYWORD
| nonn,base
|
|
|
AUTHOR
| Giovanni Resta (g.resta(AT)iit.cnr.it), Feb 13 2006
|
|
|
EXTENSIONS
| More terms from Farideh Firoozbakht (mymontain(AT)yahoo.com), Aug 24 2006
|
| |
|
|
