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.

Showing 1-3 of 3 results.

A246675 Permutation of natural numbers: a(n) = A000079(A055396(n+1)-1) * ((2*A246277(n+1))-1).

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 7, 6, 9, 16, 11, 32, 13, 10, 15, 64, 17, 128, 19, 18, 21, 256, 23, 12, 25, 14, 27, 512, 29, 1024, 31, 26, 33, 20, 35, 2048, 37, 42, 39, 4096, 41, 8192, 43, 22, 45, 16384, 47, 24, 49, 50, 51, 32768, 53, 36, 55, 66, 57, 65536, 59, 131072, 61, 38, 63, 52, 65, 262144, 67, 74, 69
Offset: 1

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Author

Antti Karttunen, Sep 01 2014

Keywords

Comments

Consider the square array A246278, and also A246275 which is obtained from the former when one is subtracted from each term.
In A246278 the even numbers occur at the top row, and all the rows below that contain only odd numbers, those subsequent terms in each column having been obtained by shifting all primes present in the prime factorization of number immediately above to one larger indices with A003961.
To compute a(n): we do the same process in reverse, by shifting primes in the prime factorization of n+1 step by step to smaller primes, until after k >= 0 such shifts with A064989, the result is even, with the smallest prime present being 2.
We subtract one from this even number and shift the binary expansion of the resulting odd number k positions left (i.e. multiply it with 2^k), which will be the result of a(n).
In the essence, a(n) tells which number in the array A135764 is at the same position where n is in the array A246275. As the topmost row in both arrays is A005408 (odd numbers), they are fixed, i.e., a(2n+1) = 2n+1 for all n.
A055396(n+1) tells on which row of A246275 n is, which is equal to the row of A246278 on which n+1 is.
A246277(n+1) tells in which column of A246275 n is, which is equal to the column of A246278 in which n+1 is.

Examples

			Consider 54 = 55-1. To find 55's position in array A246278, we start shifting its prime factorization 55 = 5 * 11 = p_3 * p_5, step by step: p_2 * p_4 (= 3 * 7 = 21), until we get an even number: p_1 * p_3 = 2*5 = 10.
This tells us that 55 is on row 3 and column 5 (= 10/2) of array A246278, thus 54 occurs in the same position at array A246275. In array A135764 the same position contains number (2^(3-1)) * (10-1) = 4*9 = 36, thus a(54) = 36.

Links

Crossrefs

Inverse: A246676.
More recursed variants: A246677, A246683.
Even bisection halved: A246679.
Other related permutations: A054582, A135764, A246274, A246275, A246276.
Cf. A000079, A003961, A005408, A055396, A064989, A246277, A246278.
a(n) differs from A156552(n+1) for the first time at n=13, where a(13) = 14, while A156552(14) = 17.

Programs

  • PARI
    A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
A246675(n) = { my(k=0); n++; while((n%2), n = A064989(n); k++); n--; while(k>0, n = 2*n; k--); n; };
for(n=1, 2048, write("b246675.txt", n, " ", A246675(n)));
  • Scheme
    (define (A246675 n) (* (A000079 (- (A055396 (+ 1 n)) 1)) (-1+ (* 2 (A246277 (+ 1 n))))))

Formula

a(n) = A000079(A055396(n+1)-1) * ((2*A246277(n+1))-1).
As a composition of related permutations:
a(n) = A135764(A246276(n)).
a(n) = A054582(A246274(n)-1).
Other identities. For all n >= 0:
a(A005408(n)) = A005408(n). [Fixes the odd numbers.]

A246677 Permutation of natural numbers: a(1) = 1, a(2n) = A000079(A055396(2n+1)-1) * ((2*A246277(2n+1))-1), a(2n+1) = 1 + 2*a(n).

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 7, 6, 9, 16, 11, 32, 17, 10, 15, 64, 13, 128, 19, 18, 33, 256, 23, 12, 65, 14, 35, 512, 21, 1024, 31, 26, 129, 20, 27, 2048, 257, 42, 39, 4096, 37, 8192, 67, 22, 513, 16384, 47, 24, 25, 50, 131, 32768, 29, 36, 71, 66, 1025, 65536, 43, 131072, 2049, 38, 63, 52, 53, 262144, 259, 74, 41
Offset: 1

Views

Author

Antti Karttunen, Sep 01 2014

Keywords

Comments

See the comments in A246675. This is otherwise similar permutation, except for odd numbers, which are here recursively permuted by the emerging permutation itself. The even bisection halved gives A246679, the odd bisection from a(3) onward with one subtracted and then halved gives this sequence back.

Links

Crossrefs

Inverse: A246678. Variants: A246675, A246683.
Even bisection halved: A246679.
Cf. A000079, A055396, A246277.
a(n) differs from A156552(n+1) for the first time at n=32, where a(32) = 26, while A156552(33) = 34.

Formula

a(1) = 1, a(2n) = A000079(A055396(2n+1)-1) * ((2*A246277(2n+1))-1), a(2n+1) = 1 + 2*a(n).

A246680 Permutation of natural numbers, even bisection of A246676 halved: a(n) = A246676(2*n)/2.

Original entry on oeis.org

1, 2, 4, 3, 7, 12, 13, 5, 10, 17, 22, 24, 16, 62, 40, 6, 37, 27, 31, 38, 19, 87, 67, 60, 25, 32, 49, 171, 52, 312, 121, 8, 28, 122, 112, 45, 34, 137, 94, 71, 82, 42, 58, 269, 43, 437, 202, 84, 73, 47, 76, 59, 187, 162, 148, 665, 46, 192, 157, 1200, 55, 1562, 364, 9, 97, 57, 85, 423, 115, 612, 337, 93, 61, 72
Offset: 1

Views

Author

Antti Karttunen, Sep 02 2014

Keywords

Comments

Equally: even bisection of A246678 halved.

Links

Crossrefs

Inverse: A246679.
Cf. A246676, A246678.

Programs

Formula

a(n) = A246676(2*n)/2.
a(n) = A246678(2*n)/2.
Showing 1-3 of 3 results.

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