login
Self-inverse permutation of natural numbers: a(n) = A263273(4*n) / 4.
15

%I #17 May 25 2017 03:16:44

%S 0,1,2,3,4,5,6,7,16,9,10,19,12,13,14,15,8,17,18,11,20,21,34,43,48,25,

%T 52,27,28,55,30,37,46,57,22,49,36,31,58,39,40,41,42,23,50,45,32,59,24,

%U 35,44,51,26,53,54,29,56,33,38,47,60,61,142,63,88,169,102,115,124,129,70,151,144,97,178,75,106,133,156,79,160,81

%N Self-inverse permutation of natural numbers: a(n) = A263273(4*n) / 4.

%H Antti Karttunen, <a href="/A264974/b264974.txt">Table of n, a(n) for n = 0..4921</a>

%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>

%F a(n) = A263273(4*n) / 4.

%F a(n) = A264986(n) / 2 = A263272(2*n) / 2.

%F As a composition of related permutations:

%F a(n) = A264975(A263272(n)) = A263272(A264976(n)).

%F Other identities. For all n >= 0:

%F a(3*n) = 3*a(n).

%F A000035(a(n)) = A000035(n). [This permutation preserves the parity of n.]

%F A264978(n) = a(2n)/2. [Thus the restriction onto even numbers induces yet another permutation.]

%o (Scheme, three variants)

%o (define (A264974 n) (/ (A263273 (* 4 n)) 4))

%o (define (A264974 n) (/ (A264986 n) 2))

%o (define (A264974 n) (/ (A263272 (* 2 n)) 2))

%o (Python)

%o from sympy import factorint

%o from sympy.ntheory.factor_ import digits

%o from operator import mul

%o def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)

%o def a038502(n):

%o f=factorint(n)

%o return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])

%o def a038500(n): return n/a038502(n)

%o def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)

%o def a(n): return a263273(4*n)/4 # _Indranil Ghosh_, May 25 2017

%Y Terms of A264986 halved.

%Y Cf. A263272, A263273, A264978.

%Y Cf. also A264975, A264976.

%K nonn,base

%O 0,3

%A _Antti Karttunen_, Dec 05 2015