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A293445 A multiplicative encoding (base-2 compressed) for the exponents of 3 obtained when using Shevelev's algorithm for computing A053446.

%I A293445 #15 Oct 09 2017 21:53:19
%S A293445 2,2,3,12,36,3,12,24,6,48,12,20736,82944,12,18,864,248832,6,20,
%T A293445 19906560,59719680,80,8640,720,25920,34560,5,80,103195607040,240,480,
%U A293445 622080,137594142720,138240,20,59440669655040,138240,20,14929920,29859840,240,59719680,8640,720,414720,8640,540,447897600,960,46080,34560,59719680,295814814232058265600,5,80
%N A293445 A multiplicative encoding (base-2 compressed) for the exponents of 3 obtained when using Shevelev's algorithm for computing A053446.
%H A293445 Antti Karttunen, <a href="/A293445/b293445.txt">Table of n, a(n) for n = 1..1458</a>
%F A293445 A048675(a(n)) = A053446(n).
%e A293445 A001651(5) = 7 as 7 is the fifth number not divisible by 3. According to the algorithm described in the comment of A053446 we have in the form of a "finite continued fraction"
%e A293445     1 + 14
%e A293445     ------ + 7
%e A293445        3^1
%e A293445     ---------- + 14
%e A293445           3^1
%e A293445     ----------------- + 7
%e A293445             3^2
%e A293445     ---------------------- = 1
%e A293445                3^2
%e A293445 Cumulatively multiplying (with A019565) together the prime-numbers corresponding to 1-bits in the binary expansions of the exponents of 3 in the denominators (that are 1, 1, 2, 2, in binary 1, 1, 10, 10, with 1's in bit-positions 0 and 1), yields prime(0+1) * prime(0+1) * prime(1+1) * prime(1+1) = 2^2 * 3^2 = 36, thus a(5) = 36.
%e A293445 (Adapted from _Vladimir Shevelev_'s explanation in A053446.)
%e A293445 Another example: A001651(19) = 28 as 28 is the 19th number not divisible by 3. (1 + 28) is not a multiple of 3, so we start with (1 + 2*28) = 1+56 = 57 and proceed as:
%e A293445     1 + 56
%e A293445     ------ + 56                     [that is, (57/3) + 56 = 75]
%e A293445        3^1
%e A293445     ---------- + 56                 [that is, (75/3) + 56 = 81]
%e A293445           3^1
%e A293445     -----------------  = 1          [that is, (81/81) = 1]
%e A293445             3^4
%e A293445 So we obtained exponents 1, 1, 4 (in binary "1", "1" and "100") where the 1-bits are in positions 0, 0 and 2. We form a product prime(0+1) * prime(0+1) * prime(2+1) = 2*2*5, thus a(19) = 20.
%o A293445 (Scheme)
%o A293445 (define (A293445 n) (define (next_one k m) (if (zero? (modulo (+ k m) 3)) (+ k m) (+ k m m))) (let* ((u (A001651 n)) (x_init (next_one 1 u))) (let loop ((x x_init) (z (A019565 (A007949 x_init)))) (let ((r (A038502 x))) (if (= 1 r) z (let ((x_next (next_one r u))) (loop x_next (* z (A019565 (A007949 x_next))))))))))
%o A293445 (define (A001651 n) (let ((x (- n 1))) (if (even? x) (+ 1 (* 3 (/ x 2))) (- (* 3 (/ (+ x 1) 2)) 1))))
%o A293445 (define (A038500 n) (A000244 (A007949 n)))
%o A293445 (define (A038502 n) (/ n (A038500 n)))
%Y A293445 Cf. A001651, A007949, A019565, A038502, A053446, A293220.
%Y A293445 Cf. A293446 (restricted growth transform of this sequence).
%Y A293445 Cf. also A292265.
%K A293445 nonn
%O A293445 1,1
%A A293445 _Antti Karttunen_, Oct 09 2017