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Equally: even bisection of A246678 halved.

The array in A135764 is identical to the array in A054582 [up to the transposition and different indexing. - Clark Kimberling, Dec 03 2010; comment amended by Antti Karttunen, Feb 03 2015; please see the illustration in Example section].

The array gives a bijection between the natural numbers N and N^2. A more usual bijection is to take the natural numbers A000027 and write them in the usual OEIS square array format. However this bijection has the advantage that it can be formed by iterating the usual bijection between N and 2N. - Joshua Zucker, Nov 04 2011

The array can be used to determine the configurations of k-th Towers of Hanoi moves, by labeling odd row terms C,B,A,C,B,A,... and even row terms B,C,A,B,C,A,.... Then given k equal to or greater than term "a" in each n-th row, but less than the next row term, record the label A, B, or C for term "a". This denotes the peg position for the disc corresponding to the n-th row. For example, with k = 25, five discs are in motion since the binary for 25 = 11001, five bits. We find that 25 in row 5 is greater than 16 labeled C, but less than 48. Thus, disc 5 is on peg C. In the 4th row, 25 is greater than 24 (a C), but less than 40, so goes onto the C peg. Similarly, disc 3 is on A, 2 is on A, and disc 1 is on A. Thus, discs 2 and 3 are on peg A, while 1, 4, and 5 are on peg C. - Gary W. Adamson, Jun 22 2012

Shares with arrays A253551 and A254053 the property that A001511(n) = k for all terms n on row k and when going downward in each column, terms grow by doubling. - Antti Karttunen, Feb 03 2015

Let P be the infinite palindromic word having initial word 0 and midword sequence (1,2,3,4,...) = A000027. Row n of the array A135764 gives the positions of n-1 in S. ("Infinite palindromic word" is defined at A260390.) - Clark Kimberling, Aug 13 2015

The probability distribution series 1 = 2/3 + 4/15 + 16/255 + 256/65535 + ... + A001146(n-1)/A051179(n) governs the proportions of terms in A001511 from row n of the array. In A001511(1..15) there are ((2/3) * 15) = ten terms from row one of the array, ((4/15) * 15) = four terms from row two, and ((16/255) * 15) = one (rounded), giving one term from row three (a 4). - Gary W. Adamson, Dec 16 2021

From Gary W. Adamson, Dec 30 2021: (Start)

Subarrays representing the number of divisors of an integer can be mapped on the table. For 60, write the odd divisors on the top row: 1, 3, 5, 15. Since 60 has 12 divisors, let the left column equal 1, 2, 4, where 4 is the highest power of 2 dividing 60. Multiplying top row terms by left column terms, we get the result:

1 3 5 15

2 6 10 30

4 12 20 60. (End)

Each row i is a multiplicative function, being in essence "the i-th power" of A003961, i.e., A(i,j) = A003961^i (j). Zeroth power gives an identity function, A001477, which occurs as the row zero.

The terms in the same column have the same prime signature.

The array is read by antidiagonals: A(0,0), A(0,1), A(1,0), A(0,2), A(1,1), A(2,0), ... .

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.

See the comments in A246676. This is otherwise similar permutation, except that after having reached an odd number 2m-1 when we have shifted the binary representation of n right k steps, here, in contrary to A246676, we don't shift the primes in the prime factorization of the even number 2m, but instead of an even number (2*a(m)), shifting it the same number (k) of positions towards larger primes, whose product is then decremented by one to get the final result.

From Antti Karttunen, Jan 18 2015: (Start)

This can be viewed as an entanglement or encoding permutation where the complementary pairs of sequences to be interwoven together are even and odd numbers (A005843/A005408) which are entangled with another complementary pair: even numbers in the order they appear in A253885 and odd numbers in their usual order: (A253885/A005408).

From the above follows also that this sequence can be represented as a binary tree. Each child to the left is obtained by doubling the parent and subtracting one, and each child to the right is obtained by applying A253885 to the parent:

1

|

...................2...................

3 4

5......../ \........8 7......../ \........6

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

9 14 15 24 13 26 11 10

17 20 27 34 29 80 47 48 25 32 51 124 21 44 19 12

(End)

In the essence, a(n) tells which number in square array A249741 (the sieve of Eratosthenes minus 1) is at the same position where n is in array A135764, which is formed from odd numbers whose binary expansions are shifted successively leftwards on the successive rows. As the topmost row in both arrays is A005408 (odd numbers), they are fixed, i.e., a(2n+1) = 2n+1 for all n.

Equally: a(n) tells which number in array A114881 is at the same position where n is in the array A054582, as they are the transposes of above two arrays.

See the comments in A246676. This is a similar permutation, except for odd numbers, which are here recursively permuted by the emerging permutation itself. The even bisection halved gives A246680, the odd bisection from a(3) onward with one subtracted and then halved gives this sequence back.