A266408 Permutation of natural numbers: a(n) = (1/2) * (1+A263273(A003961(n))).

1, 2, 4, 5, 3, 11, 10, 14, 9, 8, 7, 32, 13, 29, 37, 41, 6, 26, 12, 23, 15, 20, 28, 95, 22, 38, 115, 86, 19, 110, 16, 122, 30, 17, 36, 77, 34, 35, 55, 68, 25, 44, 31, 59, 60, 83, 40, 284, 61, 65, 100, 113, 33, 344, 46, 257, 70, 56, 24, 329, 21, 47, 289, 365, 88, 89, 39, 50, 49, 107, 18, 230, 27, 101, 244, 104, 112, 164, 82, 203, 174, 74

Offset

1,2

Links

Antti Karttunen, Table of n, a(n) for n = 1..8191

Index entries for sequences that are permutations of the natural numbers

Formula

a(n) = (1/2) * (1+A263273(A003961(n))).

Prog

(PARI)
A030102(n) = { my(r=[n%3]); while(0<n\=3, r=concat(n%3, r)); subst(Polrev(r), x, 3); } \\ After M. F. Hasler's Nov 04 2011 code in A030102.
A263273 = n -> if(!n, n, A030102(n/(3^valuation(n, 3))) * (3^valuation(n, 3))); \\ Taking of the quotient probably unnecessary.
A003961(n) = my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); \\ Using code of Michel Marcus
A266408 = n -> (1+A263273(A003961(n)))/2;
for(n=1, 8191, write("b266408.txt", n, " ", A266408(n)));
(Scheme) (define (A266408 n) (/ (+ 1 (A263273 (A003961 n))) 2))

Crossrefs

Inverse: A266407.

Cf. A003961, A263273.

Cf. also A048673, A266401, A266403.

Sequence in context: A118461 A376733 A332667 * A358339 A230564 A011174

Adjacent sequences: A266405 A266406 A266407 * A266409 A266410 A266411

Keyword

nonn,base

Author

Antti Karttunen, Jan 02 2016

Status

approved