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A285323 revision #9

A285323
a(n) = A065642(A065642(A019565(n))) / A019565(n).
3
1, 4, 9, 3, 25, 4, 5, 3, 49, 4, 7, 3, 7, 4, 5, 3, 121, 4, 9, 3, 11, 4, 5, 3, 11, 4, 7, 3, 7, 4, 5, 3, 169, 4, 9, 3, 13, 4, 5, 3, 13, 4, 7, 3, 7, 4, 5, 3, 13, 4, 9, 3, 11, 4, 5, 3, 11, 4, 7, 3, 7, 4, 5, 3, 289, 4, 9, 3, 17, 4, 5, 3, 17, 4, 7, 3, 7, 4, 5, 3, 17, 4, 9, 3, 11, 4, 5, 3, 11, 4, 7, 3, 7, 4, 5, 3, 17, 4, 9, 3, 13, 4, 5, 3, 13, 4, 7, 3, 7, 4, 5, 3
OFFSET
0,2
COMMENTS
After the initial a(0)=1, the third row of array A285321 divided by its first row. After 1, all terms are either primes or squares of primes. See A285110.
The sequence is completely determined by the positions of two least significant 1-bits of n: After initial zero, if n is a power of two (only one 1-bit present) or if prime(1+A285099(n)) > prime(1+A007814(n))^2, a(n) = prime(1+A007814(n))^2 = A020639(A019565(n))^2, otherwise a(n) = prime(1+A285099(n)) = A014673(A019565(n)).
LINKS
FORMULA
a(n) = A065642(A065642(A019565(n))) / A019565(n).
PROG
(PARI)
A019565(n) = {my(j, v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ This function from M. F. Hasler
A007947(n) = factorback(factorint(n)[, 1]); \\ From Andrew Lelechenko, May 09 2014
A065642(n) = { my(r=A007947(n)); if(1==n, n, n = n+r; while(A007947(n) <> r, n = n+r); n); };
(Scheme)
(define (A285323 n) (/ (A065642 (A065642 (A019565 n))) (A019565 n)))
(define (A285323 n) (cond ((zero? n) 1) ((or (= 1 (A000120 n)) (> (A000040 (+ 1 (A285099 n))) (A000290 (A000040 (+ 1 (A007814 n)))))) (A000290 (A000040 (+ 1 (A007814 n))))) (else (A000040 (+ 1 (A285099 n))))))
KEYWORD
nonn
AUTHOR
Antti Karttunen, Apr 19 2017
STATUS
editing