cp's OEIS Frontend

Permutation of the natural numbers.

a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers.

From Philippe Deléham, Feb 19 2014: (Start)

A(0,k) = 2*k = A005843(k),

A(1,k) = 4*k + 1 = A016813(k),

A(2,k) = 8*k + 3 = A017101(k),

A(n,0) = A000225(n),

A(n,1) = A153893(n),

A(n,2) = A153894(n),

A(n,3) = A086224(n),

A(n,4) = A052996(n+2),

A(n,5) = A086225(n),

A(n,6) = A198274(n),

A(n,7) = A238087(n),

A(n,8) = A198275(n),

A(n,9) = A198276(n),

A(n,10) = A171389(n). (End)

A permutation of the nonnegative integers. - Alzhekeyev Ascar M, Jun 05 2016

The values in array row n, when expressed in binary, have n trailing 1-bits. - Ruud H.G. van Tol, Mar 18 2025

Essentially same as A091257 but computed starting from offset 0 instead of 1.

Each polynomial in GF(2)[X] is encoded as the number whose binary representation is given by the coefficients of the polynomial, e.g., 13 = 2^3 + 2^2 + 2^0 = 1101_2 encodes 1*X^3 + 1*X^2 + 0*X^1 + 1*X^0 = X^3 + X^2 + X^0. - Antti Karttunen and Peter Munn, Jan 22 2021

To listen to this sequence, I find instrument 99 (crystal) works well with the other parameters defaulted. - Peter Munn, Nov 01 2022

Fractal sequence obtained from powers of 2.

k occurs at (2*k-1)*A000079(m), m >= 0. - Robert G. Wilson v, May 23 2006

Sequence is T^(oo)(1) where T is acting on a word w = w(1)w(2)..w(m) as follows: T(w) = "1"w(1)"2"w(2)"3"(...)"m"w(m)"m+1". For instance T(ab) = 1a2b3. Thus T(1) = 112, T(T(1)) = 1121324, T(T(T(1))) = 112132415362748. - Benoit Cloitre, Mar 02 2009

Note that iterating the post-numbering operator U(w) = w(1) 1 w(2) 2 w(3) 3... produces the same limit sequence except with an additional "1" prepended, i.e., 1,1,1,2,1,3,2,4,... - Glen Whitney, Aug 30 2023

In the binary expansion of n, first swallow all zeros from the right, then add 1, and swallow the now-appearing 0 bit as well. - Ralf Stephan, Aug 22 2013

Although A264646 and this sequence initially agree in their digit-streams, they differ after 48 digits. - N. J. A. Sloane, Nov 20 2015

"[This is a] fractal because we get the same sequence after we delete from it the first appearance of all positive integers" - see Cobeli and Zaharescu link. - Robert G. Wilson v, Jun 03 2018

From Peter Munn, Jun 16 2022: (Start)

The sequence is the list of positive integers interleaved with the sequence itself. Provided the offset is suitable (which is the case here) a term of such a self-interleaved sequence is determined by the odd part of its index. Putting some of the formulas given here into words, a(n) is the position of the odd part of n in the list of odd numbers.

Applying the interleaving transform again, we get A110963.

(End)

Omitting all 1's leaves A131987 + 1. - David James Sycamore, Jul 26 2022

a(n) is also the smallest positive number not among the terms between a(a(n-1)) and a(n-1) inclusive (with a(0)=1 prepended). - Neal Gersh Tolunsky, Mar 07 2023

Number of binary trees of size n and height n-1, computed from size n=3 onward; i.e. A014480(n) = A073345(n+3,n+2). (For sizes n=0 through 2 there are no such trees.)

Also determinant of the n X n matrix M(i,j)=binomial(2i+2j,i+j). - Benoit Cloitre, Mar 27 2004

Subdiagonal in triangle displayed in A128196. - Peter Luschny, Feb 26 2007

From Jaume Oliver Lafont, Nov 08 2009: (Start)

From two BBP-type formulas by Knuth, (page 6 of the reference)

Sum_{n>=0} 1/a(n) = 2^(1/2)*log(1+2^(1/2))

Sum_{n>=0} (-1)^n/a(n) = 2^(1/2)*atan(1/2^(1/2))

(End)

Create a triangle with first column T(n,1)=1+4*n for n=0 1 2... The remaining terms T(r,c)=T(r,c-1)+T(r-1,c-1). T(n,n+1)=a(n). - J. M. Bergot, Dec 18 2012

Row 1: A084639.

For a background discussion of dispersions, see A191426.

...

Each of the sequences (4n, n>2), (4n+1, n>0), (3n+2, n>=0), generates a dispersion. Each complement (beginning with its first term >1) also generates a dispersion. The six sequences and dispersions are listed here:

...

A191663=dispersion of A042948 (0 or 1 mod 4 and >1)

A054582=dispersion of A005843 (0 or 2 mod 4 and >1; evens)

A191664=dispersion of A014601 (0 or 3 mod 4 and >1)

A191665=dispersion of A042963 (1 or 2 mod 4 and >1)

A191448=dispersion of A005408 (1 or 3 mod 4 and >1, odds)

A191666=dispersion of A042964 (2 or 3 mod 4)

...

EXCEPT for at most 2 initial terms (so that column 1 always starts with 1):

A191663 has 1st col A042964, all else A042948

A054582 has 1st col A005408, all else A005843

A191664 has 1st col A042963, all else A014601

A191665 has 1st col A014601, all else A042963

A191448 has 1st col A005843, all else A005408

A191666 has 1st col A042948, all else A042964

...

There is a formula for sequences of the type "(a or b mod m)", (as in the Mathematica program below):

If f(n)=(n mod 2), then (a,b,a,b,a,b,...) is given by

a*f(n+1)+b*f(n), so that "(a or b mod m)" is given by

a*f(n+1)+b*f(n)+m*floor((n-1)/2)), for n>=1.

Number of words of length n+1 where first element is from {0,1,2}, other elements are from {0,1} and sequence does not decrease (for n=2 there are 3*2^2 sequences, but 000, 100, 110, 111, 200, 210, 211 decrease, so a(2) = 12-7 = 5).

Number of subgroups of C_(2^n) X C_(2^n) (see A060724).

Starting with 1 = row sums of triangle A054582. - Gary W. Adamson, Jun 23 2008

Starting with "1" equals the eigensequence of a triangle with integer squares (1, 4, 9, 16, ...) as the left border and the rest 1's. - Gary W. Adamson, Jul 24 2010

(1 + 2x + 2x^2 + 2x^3 + ...)*(1 + 3x + 7x^2 + 15x^3 + ...) = (1 + 5x + 15x^2 + 37x^3 + ...). - Gary W. Adamson, Mar 14 2012

The partial sums of A033484. - J. M. Bergot, Oct 03 2012

Binomial transform is 0, 1, 7, 33, ... (shifted A066810); inverse binomial transform is 0, 1, 3, 3, ... (3 repeated). - R. J. Mathar, Oct 05 2012

Define a triangle by T(n,0) = n*(n+1) + 1, T(n,n) = n + 1, and T(r,c) = T(r-1,c-1) + T(r-1,c) otherwise; then a(n+1) is the sum of the terms of row n. - J. M. Bergot, Mar 30 2013

Starting with "1" are also the antidiagonal sums of the array formed by partial sums of integer squares (1, 4, 9, 16, ...). - Luciano Ancora, Apr 24 2015

Sums of 2 adjacent terms in diagonal k=2 of Eulerian triangle A008292. I.e., T(n,2)+T(n-1,2) for n > 0. Also, 4th NW-SE diagonal of A126277. In other words, a(n) = A000295(n) + A000295(n+1). - Gregory Gerard Wojnar, Sep 30 2018

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)

Row sums give A014477: Sum_{k=1..n} T(n,k) = A014477(n-1);

central terms give A118415; T(2*k-1,k) = A058962(k-1);

T(n,1) = A000079(n-1);

T(n,2) = A007283(n-1) for n > 1;

T(n,3) = A020714(n-1) for n > 2;

T(n,4) = A005009(n-1) for n > 3;

T(n,5) = A005010(n-1) for n > 4;

T(n,n-1) = A118417(n-1) for n > 1;

T(n,n) = A014480(n-1) = A118413(n,n);

A001511(T(n,k)) = A002024(n,k);

A003602(T(n,k)) = A002260(n,k).

The alternating row sums, Sum_{k=1..n} (-1)^(k+1)*T(n,k), are: (a) in odd rows, the central term, T(n,(n+1)/2) = A058962((n-1)/2); (b) in even rows, the negation of the average of the two central terms, -(T(2n,n) + T(2n,+1))/2 = -A018215(m/2). The absolute values of the alternating row sums give the plain row means, Sum_{k=1..n} T(n,k)/n; the alternating sign row means are (-2)^(n-1). - Gregory Gerard Wojnar, Feb 10 2024

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.