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A332817
a(n) = A108548(A163511(n)).
12
1, 2, 4, 3, 8, 9, 6, 5, 16, 27, 18, 25, 12, 15, 10, 7, 32, 81, 54, 125, 36, 75, 50, 49, 24, 45, 30, 35, 20, 21, 14, 13, 64, 243, 162, 625, 108, 375, 250, 343, 72, 225, 150, 245, 100, 147, 98, 169, 48, 135, 90, 175, 60, 105, 70, 91, 40, 63, 42, 65, 28, 39, 26, 11, 128, 729, 486, 3125, 324, 1875, 1250, 2401, 216, 1125, 750, 1715, 500
OFFSET
0,2
COMMENTS
This irregular table can be represented as a binary tree. Each child to the left is obtained by doubling the parent, and each child to the right is obtained by applying A332818 to the parent:
1
|
...................2...................
4 3
8......../ \........9 6......../ \........5
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
16 27 18 25 12 15 10 7
32 81 54 125 36 75 50 49 24 45 30 35 20 21 14 13
etc.
This is the mirror image of the tree in A332815.
FORMULA
a(n) = A108548(A163511(n)).
For n >= 1, a(n) = A332815(A054429(n)).
PROG
(PARI)
up_to = 26927;
A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t }; \\ From A005940
A054429(n) = ((3<<#binary(n\2))-n-1); \\ From A054429
A163511(n) = if(!n, 1, A005940(1+A054429(n)));
A108546list(up_to) = { my(v=vector(up_to), p, q); v[1] = 2; v[2] = 3; v[3] = 5; for(n=4, up_to, p = v[n-2]; q = nextprime(1+p); while(q%4 != p%4, q=nextprime(1+q)); v[n] = q); (v); };
v108546 = A108546list(up_to);
A108546(n) = v108546[n]; \\ Antti Karttunen, Mar 05 2020
A108548(n) = { my(f=factor(n)); f[, 1] = apply(A108546, apply(primepi, f[, 1])); factorback(f); };
CROSSREFS
Cf. A332811 (inverse permutation).
Cf. A054429, A108548, A163511, A332815 (mirror image).
Cf. A108546 (the right edge of the tree from 2 downward).
Cf. also A332214.
Sequence in context: A252755 A366275 A163511 * A332214 A285322 A129593
KEYWORD
nonn,tabf
AUTHOR
Antti Karttunen, Mar 05 2020
STATUS
approved