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A293448
Self-inverse permutation of natural numbers: replace (with multiplicity) each prime factor A000040(k) with A000040(min+(max-k)) in the prime factorization of n, where min = A055396(n) and max = A061395(n).
5
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 18, 13, 14, 15, 16, 17, 12, 19, 50, 21, 22, 23, 54, 25, 26, 27, 98, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 250, 41, 70, 43, 242, 75, 46, 47, 162, 49, 20, 51, 338, 53, 24, 55, 686, 57, 58, 59, 150, 61, 62, 147, 64, 65, 154, 67, 578, 69, 42, 71, 108, 73, 74, 45, 722, 77, 286, 79, 1250, 81, 82, 83
OFFSET
1,2
COMMENTS
Reverse the prime-indices in such a way that the smallest and the greatest prime dividing n (A020639 and A006530) are preserved.
a(n) = n iff n belongs to A236510. - Rémy Sigrist, Nov 22 2017
FORMULA
For all even squarefree numbers coincides with A273258, that is, for all n, a(A039956(n)) = A273258(A039956(n)).
EXAMPLE
For n = 25 = 5^2 = prime(3)^2, thus min = max = 3, and we form a product prime(3+(3-3))^2, thus a(25) = prime(3)^2 = 25.
For n = 42 = 2*3*7 = prime(1)*prime(2)*prime(4), thus min = 1 and max = 4, so we form a product prime(1+(4-1))*prime(1+(4-2))*prime(1+(4-4)), thus a(42) = prime(4)*prime(3)*prime(1) = 7*5*2 = 70.
For n = 126 = 2 * 3^2 * 7 = prime(1) * prime(2)^2 * prime(4), thus min = 1 and max = 4, so we form a product prime(1+(4-1)) * prime(1+(4-2))^2 * prime(1+(4-4)), thus a(126) = prime(4) * prime(3)^2 * prime(1) = 7 * 5^2 * 2 = 350.
PROG
(PARI) A293448(n) = { if(1==n, return(n)); my(f=factor(n), mini = primepi(f[1, 1]), maxi = primepi(f[#f~, 1])); for(i=1, #f~, f[i, 1] = prime((maxi-primepi(f[i, 1]))+mini)); factorback(f); }
CROSSREFS
Cf. A000720, A055396, A057889, A061395, A236510 (fixed points), A273258.
Differs from A069799 (and some other related permutations) for the first time at n=42.
Sequence in context: A069799 A225891 A295417 * A085079 A289234 A033000
KEYWORD
nonn
AUTHOR
Antti Karttunen, Nov 09 2017
STATUS
approved