|
| |
|
|
A123317
|
|
Smallest prime power m such that n+m is a prime number.
|
|
1
| |
|
|
1, 1, 2, 1, 2, 1, 4, 3, 2, 1, 2, 1, 4, 3, 2, 1, 2, 1, 4, 3, 2, 1, 8, 5, 4, 3, 2, 1, 2, 1, 16, 5, 4, 3, 2, 1, 4, 3, 2, 1, 2, 1, 4, 3, 2, 1, 32, 5, 4, 3, 2, 1, 8, 5, 4, 3, 2, 1, 2, 1, 256, 5, 4, 3, 2, 1, 4, 3, 2, 1, 2, 1, 16, 5, 4, 3, 2, 1, 4, 3, 2, 1, 128, 5, 4, 3, 2, 1, 8, 7, 16, 5, 4, 3, 2, 1, 4, 3, 2, 1, 2, 1
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,3
|
|
|
COMMENTS
| A123318(n) = n + a(n);
a(A006093(n)) = 1; a(A040976(n)) = 2 for n>2.
|
|
|
EXAMPLE
| n=23: 23+1=3*2^3, 23+2=5^2, 23+3=13*2, 23+2^2=3^3, 23+5=7*2^2, 23+7=5*3*2, but 23+8=31=A000040(11), therefore a(23)=8;
n=24: 24+1=5^2, 24+2=13*2, 24+3=3^3, 24+2^2=7*2^2, but 24+5=29=A000040(10), therefore a(24)=5;
the smallest occurring proper odd prime power is 9=3^2:
n=118: 118+1=17*7, 118+2=5*3*2^3, 118+3=11^2, 118+2^2=61*2, 118+5=41*3, 118+7=5^3, 118+2^3=7*2*3^2, but 118+3^2=127=A000040(31), therefore a(118)=9.
|
|
|
CROSSREFS
| Cf. A013632, A000961.
Sequence in context: A184342 A030767 A135545 * A171453 A164879 A200219
Adjacent sequences: A123314 A123315 A123316 * A123318 A123319 A123320
|
|
|
KEYWORD
| nonn
|
|
|
AUTHOR
| Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Sep 27 2006
|
| |
|
|
