OFFSET
0,2
COMMENTS
a(2^n-1) = A002110(n) for any n >= 0.
a(2^(n-1)) = A000961(n+1) for any n > 0.
From Antti Karttunen, Jan 01 2019: (Start)
This sequence can be regarded also as an irregular triangle with rows of lengths 1, 1, 2, 4, 8, 16, ..., that is, it can be represented as a binary tree, where each left hand child contains A322991(k), and each right hand child contains A322992(k), when their parent contains k:
1
|
...................2...................
3 6
4......../ \........10 12......../ \........30
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
5 14 15 42 20 70 60 210
7 18 21 66 28 90 84 330 35 126 105 462 140 630 420 2310
etc.
The leftmost edge is A000961, the next lefmost is A278568 (after 2: 6, 10, 14, 18, ...), the righmost edge is A002110, the next rightmost A088860 but with 3 instead of 4.
(End)
LINKS
EXAMPLE
A289271(1) = 0, hence a(0) = 1.
A289271(2) = 1, hence a(1) = 2.
A289271(3) = 2, hence a(2) = 3.
A289271(4) = 4, hence a(4) = 4.
A289271(5) = 8, hence a(8) = 5.
A289271(6) = 3, hence a(3) = 6.
A289271(7) = 16, hence a(16) = 7.
A289271(8) = 32, hence a(32) = 8.
A289271(9) = 64, hence a(64) = 9.
A289271(10) = 5, hence a(5) = 10.
PROG
(PARI) See Links section.
(PARI) A289272(n) = { my(m=1, pp=1); while(n>0, pp++; while(!isprimepower(pp)||(gcd(pp, m)>1), pp++); if(n%2, m *= pp); n >>=1); (m); }; \\ Antti Karttunen, Jan 01 2019
CROSSREFS
KEYWORD
AUTHOR
Rémy Sigrist, Jun 30 2017
STATUS
approved