cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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).

Original entry on oeis.org

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

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Author

Antti Karttunen, Nov 09 2017

Keywords

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

Examples

			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.
		

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.

Programs

  • 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); }

Formula

For all even squarefree numbers coincides with A273258, that is, for all n, a(A039956(n)) = A273258(A039956(n)).