cp's OEIS Frontend

Also called Dress's sequence.

This sequence might be better called Glaisher's sequence, since James Glaisher showed that odd binomial coefficients are counted by 2^A000120(n) in 1899. - Eric Rowland, Mar 17 2017 [However, the name "Gould's sequence" is deeply entrenched in the literature. - N. J. A. Sloane, Mar 17 2017] [Named after the American mathematician Henry Wadsworth Gould (b. 1928). - Amiram Eldar, Jun 19 2021]

All terms are powers of 2. The first occurrence of 2^k is at n = 2^k - 1; e.g., the first occurrence of 16 is at n = 15. - Robert G. Wilson v, Dec 06 2000

a(n) is the highest power of 2 dividing binomial(2n,n) = A000984(n). - Benoit Cloitre, Jan 23 2002

Also number of 1's in n-th row of triangle in A070886. - Hans Havermann, May 26 2002. Equivalently, number of live cells in generation n of a one-dimensional cellular automaton, Rule 90, starting with a single live cell. - Ben Branman, Feb 28 2009. Ditto for Rule 18. - N. J. A. Sloane, Aug 09 2014. This is also the odd-rule cellular automaton defined by OddRule 003 (see Ekhad-Sloane-Zeilberger "Odd-Rule Cellular Automata on the Square Grid" link). - N. J. A. Sloane, Feb 25 2015

Also number of numbers k, 0<=k<=n, such that (k OR n) = n (bitwise logical OR): a(n) = #{k : T(n,k)=n, 0<=k<=n}, where T is defined as in A080098. - Reinhard Zumkeller, Jan 28 2003

To construct the sequence, start with 1 and use the rule: If k >= 0 and a(0),a(1),...,a(2^k-1) are the first 2^k terms, then the next 2^k terms are 2*a(0),2*a(1),...,2*a(2^k-1). - Benoit Cloitre, Jan 30 2003

Also, numerator((2^k)/k!). - Mohammed Bouayoun (mohammed.bouayoun(AT)sanef.com), Mar 03 2004

The odd entries in Pascal's triangle form the Sierpiński Gasket (a fractal). - Amarnath Murthy, Nov 20 2004

Row sums of Sierpiński's Gasket A047999. - Johannes W. Meijer, Jun 05 2011

Fixed point of the morphism "1" -> "1,2", "2" -> "2,4", "4" -> "4,8", ..., "2^k" -> "2^k,2^(k+1)", ... starting with a(0) = 1; 1 -> 12 -> 1224 -> = 12242448 -> 122424482448488(16) -> ... . - Philippe Deléham, Jun 18 2005

a(n) = number of 1's of stage n of the one-dimensional cellular automaton with Rule 90. - Andras Erszegi (erszegi.andras(AT)chello.hu), Apr 01 2006

a(33)..a(63) = A117973(1)..A117973(31). - Stephen Crowley, Mar 21 2007

Or the number of solutions of the equation: A000120(x) + A000120(n-x) = A000120(n). - Vladimir Shevelev, Jul 19 2009

For positive n, a(n) equals the denominator of the permanent of the n X n matrix consisting entirely of (1/2)'s. - John M. Campbell, May 26 2011

Companions to A001316 are A048896, A105321, A117973, A151930 and A191488. They all have the same structure. We observe that for all these sequences a((2*n+1)*2^p-1) = C(p)*A001316(n), p >= 0. If C(p) = 2^p then a(n) = A001316(n), if C(p) = 1 then a(n) = A048896(n), if C(p) = 2^p+2 then a(n) = A105321(n+1), if C(p) = 2^(p+1) then a(n) = A117973(n), if C(p) = 2^p-2 then a(n) = (-1)*A151930(n) and if C(p) = 2^(p+1)+2 then a(n) = A191488(n). Furthermore for all a(2^p - 1) = C(p). - Johannes W. Meijer, Jun 05 2011

a(n) = number of zeros in n-th row of A219463 = number of ones in n-th row of A047999. - Reinhard Zumkeller, Nov 30 2012

This is the Run Length Transform of S(n) = {1,2,4,8,16,...} (cf. A000079). The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g., 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product). - N. J. A. Sloane, Sep 05 2014

A105321(n+1) = a(n+1) + a(n). - Reinhard Zumkeller, Nov 14 2014

a(n) = A261363(n,n) = number of distinct terms in row n of A261363 = number of odd terms in row n+1 of A261363. - Reinhard Zumkeller, Aug 16 2015

From Gary W. Adamson, Aug 26 2016: (Start)

A production matrix for the sequence is lim_{k->infinity} M^k, the left-shifted vector of M:

1, 0, 0, 0, 0, ...

2, 0, 0, 0, 0, ...

0, 1, 0, 0, 0, ...

0, 2, 0, 0, 0, ...

0, 0, 1, 0, 0, ...

0, 0, 2, 0, 0, ...

0, 0, 0, 1, 0, ...

...

The result is equivalent to the g.f. of Apr 06 2003: Product_{k>=0} (1 + 2*z^(2^k)). (End)

Number of binary palindromes of length n for which the first floor(n/2) symbols are themselves a palindrome (Ji and Wilf 2008). - Jeffrey Shallit, Jun 15 2017

This is a permutation of the positive integers.

From Antti Karttunen, Jun 20 2014: (Start)

Note the indexing: the domain starts from 0, while the range excludes zero, thus this is neither a bijection on the set of nonnegative integers nor on the set of positive natural numbers, but a bijection from the former set to the latter.

Apart from that discrepancy, this could be viewed as yet another "entanglement permutation" where the two complementary pairs to be interwoven together are even and odd numbers (A005843/A005408) which are entangled with the complementary pair even numbers (taken straight) and odd numbers in the order they appear in A003961: (A005843/A003961). See also A246375 which has almost the same recurrence.

Note how the even bisection halved gives the same sequence back. (For a(0)=1, take ceiling of 1/2).

(End)

From Antti Karttunen, Dec 30 2017: (Start)

This irregular table can be represented as a binary tree. Each child to the left is obtained by doubling the parent, and each child to the right is obtained by applying A003961 to the parent:

1

|

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

4 3

8......../ \........9 6......../ \........5

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

16 27 18 25 12 15 10 7

32 81 54 125 36 75 50 49 24 45 30 35 20 21 14 11

etc.

Sequence A005940 is obtained by scanning the same tree level by level in mirror image fashion. Also in binary trees A253563 and A253565 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) gives the distance of n from 1 in all these trees, and A252463 gives the parent of the node containing n.

A252737(n) gives the sum and A252738(n) the product of terms on row n (where 1 is on row 0, 1 on row 1, 3 and 4 on row 2, etc.). A252745(n) gives the number of nodes on level n whose left child is smaller than the right child, and A252744(n) is an indicator function for those nodes.

(End)

Note that the idea behind maps like this (and the mirror image A005940) admits also using alternative orderings of primes, not just standard magnitude-wise ordering (A000040). For example, A332214 is a similar sequence but with primes rearranged as in A332211, and A332817 is obtained when primes are rearranged as in A108546. - Antti Karttunen, Mar 11 2020

From Lorenzo Sauras Altuzarra, Nov 28 2020: (Start)

This sequence is generated from A228351 by applying the following procedure: 1) eliminate the compositions that end in one unless the first one, 2) subtract one unit from every component, 3) replace every tuple [t_1, ..., t_r] by Product_{k=1..r} A000040(k)^(t_k) (see the examples).

Is it true that a(n) = A337909(n+1) if and only if a(n+1) is not a term of A161992?

Does this permutation have any other cycle apart from (1), (2) and (6, 9, 16, 7)? (End)

From Antti Karttunen, Jul 25 2023: (Start)

(In the above question, it is assumed that the starting offset would be 1 instead of 0).

Questions:

Does a(n) = 1+A054429(n) hold only when n is of the form 2^k times 1, 3 or 7, i.e., one of the terms of A029748?

It seems that A007283 gives all fixed points of map n -> a(n), like A335431 seems to give all fixed points of map n -> A332214(n). Is there a general rule for mappings like these that the fixed points (if they exist) must be of the form 2^k times a certain kind of prime, i.e., that any odd composite (times 2^k) can certainly be excluded? See also note in A029747.

(End)

If the conjecture given in A364297 holds, then it implies the above conjecture about A007283. See also A364963. - Antti Karttunen, Sep 06 2023

Conjecture: a(n^k) is never of the form x^k, for any integers n > 0, k > 1, x >= 1. This holds at least for squares, cubes, seventh and eleventh powers (see A365808, A365801, A366287 and A366391). - Antti Karttunen, Sep 24 2023, Oct 10 2023.

See A365805 for why the above holds for any n^k, with k > 1. - Antti Karttunen, Nov 23 2023

If instead of base 2 we take base 10, then we have the so-called Harshad or Niven numbers (i.e., positive integers divisible by the sum of their digits; A005349). - Emeric Deutsch, Apr 11 2007

A199238(a(n)) = 0. - Reinhard Zumkeller, Nov 04 2011

Or, a(n)=number of 1's ("live" cells) at stage n of a 2-dimensional cellular automata evolving by the rule: 1 if NE+NW+S=1, else 0.

This is the odd-rule cellular automaton defined by OddRule 013 (see Ekhad-Sloane-Zeilberger "Odd-Rule Cellular Automata on the Square Grid" link). - N. J. A. Sloane, Feb 25 2015

Or, start with S=[1]; replace S by [S, 3*S]; repeat ad infinitum.

Fixed point of the morphism 1 -> 13, 3 -> 39, 9 -> 9(27), ... = 3^k -> 3^k 3^(k+1), ... starting from a(0) = 1; 1 -> 13 -> 1339 -> = 1339399(27) -> 1339399(27)399(27)9(27)(27)(81) -> ..., . - Robert G. Wilson v, Jan 24 2006

Equals row sums of triangle A166453 (the square of Sierpiński's gasket, A047999). - Gary W. Adamson, Oct 13 2009

First bisection of A169697=1,5,3,19,3,. a(2n+2)+a(2n+3)=12,12,36,=12*A147610 ? Distribution of terms (in A000244): A011782=1,A000079 for first array, A000079 for second. - Paul Curtz, Apr 20 2010

a(A000225(n)) = A000244(n) and a(m) != A000244(n) for m < A000225(n). - Reinhard Zumkeller, Nov 14 2011

This sequence pertains to phenotype Punnett square mathematics. Start with X=1. Each hybrid cross involves the equation X:3X. Therefore, the ratio in the first (mono) hybrid cross is X=1:3X=3(1) or 3; or 3:1. When you move up to the next hybridization level, replace the previous cross ratio with X. X now represents 2 numbers-1:3. Therefore, the ratio in the second (di) hybrid cross is X=(1:3):3X=[3(1):3(3)] or (3:9). Put it together and you get 1:3:3:9. Each time you move up a hybridization level, replace the previous ratio with X, and use the same equation-X:3X to get its ratio. - John Michael Feuk, Dec 10 2011

Number of odd values in the n-th layer of Pascal's tetrahedron (see A268240). - Caden Le, Mar 03 2025

a(x*y) <= a(x)^A000120(y). - Joe Amos, Mar 28 2025

a(n) = 2^A048881 = 2^{maximal power of 2 dividing the n-th Catalan number (A000108)}. [Comment corrected by N. J. A. Sloane, Apr 30 2018]

Row sums of triangle A128937. - Philippe Deléham, May 02 2007

a(n) = sum of (n+1)-th row terms of triangle A167364. - Gary W. Adamson, Nov 01 2009

a(n), n >= 1: Numerators of Maclaurin series for 1 - ((sin x)/x)^2, A117972(n), n >= 2: Denominators of Maclaurin series for 1 - ((sin x)/x)^2, the correlation function in Montgomery's pair correlation conjecture. - Daniel Forgues, Oct 16 2011

For n > 0: a(n) = A007954(A007931(n)). - Reinhard Zumkeller, Oct 26 2012

a(n) = A261363(2*(n+1), n+1). - Reinhard Zumkeller, Aug 16 2015

From Gus Wiseman, Oct 30 2022: (Start)

Also the number of coarsenings of the (n+1)-th composition in standard order. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. See link for sequences related to standard compositions. For example, the a(10) = 4 coarsenings of (2,1,1) are: (2,1,1), (2,2), (3,1), (4).

Also the number of times n+1 appears in A357134. For example, 11 appears at positions 11, 20, 33, and 1024, so a(10) = 4.

(End)

Consider a simple cellular automaton, a grid of binary cells c(i,j), where the next state of the grid is calculated by applying the following rule to each cell: c(i,j) = ( c(i+1,j-1) + c(i+1,j+1) + c(i-1,j-1) + c(i-1,j+1) ) mod 2 If we start with a single cell having the value 1 and all the others 0, then the aggregate values of the subsequent states of the grid will be the terms in this sequence. - Andras Erszegi (erszegi.andras(AT)chello.hu), Mar 31 2006. See link for initial states. - N. J. A. Sloane, Feb 12 2015

This is the odd-rule cellular automaton defined by OddRule 033 (see Ekhad-Sloane-Zeilberger "Odd-Rule Cellular Automata on the Square Grid" link). - N. J. A. Sloane, Feb 25 2015

First differences of A116520. - Omar E. Pol, May 05 2010

Highest power of 2 dividing n-th Catalan number (A000108).

a(n) = 0 iff n = 2^k - 1, k=0,1,...

Appears to be number of binary left-rotations (iterations of A006257) to reach fixed point of form 2^k-1. Right-rotation analog is A063250. This would imply that for n >= 0, a(n)=f(n), recursively defined to be 0 for n=0, otherwise as f( ( (1-n)(1-p)(1-s) - (1-n-p-s) ) / 2) + p (s+1) / 2, where p = n mod 2 and s = - signum(n) (f(n<0) is A000120(-n)). - Marc LeBrun, Jul 11 2001

Let f(0) = 01, f(1) = 12, f(2) = 23, f(3) = 34, f(4) = 45, etc. Sequence gives concatenation of 0, f(0), f(f(0)), f(f(f(0))), ... Also f(f(...f(0)...)) converges to A000120. - Philippe Deléham, Aug 14 2003

C(n, k) is the number of occurrence of k in the n-th group of terms in this sequence read by rows: {0}; {0, 1}; {0, 1, 1, 2}; {0, 1, 1, 2, 1, 2, 2, 3}; {0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4 }; ... - Philippe Deléham, Jan 01 2004

Highest power of 2 dividing binomial(n,floor(n/2)). - Benoit Cloitre, Oct 20 2003

2^a(n) are numerators in the Maclaurin series for (sin x)^2. - Jacob A. Siehler, Nov 11 2009

Conjecture: a(n) is the sum of digits of the n-th word in A076478, for n >= 1; has been confirmed for n up to 20000. - Clark Kimberling, Jul 14 2021

Essentially the same sequence as A001316, which has much more information, and also A159913. - N. J. A. Sloane, Jun 05 2009

Smallest number with same number of 1's in its binary expansion as n.

Fixed point of the morphism 0 -> 01, 1 -> 13, 3 -> 37, ... = k -> k, 2k+1, ... starting from a(0) = 0; 1 -> 01 -> 0113 -> 01131337 -> 011313371337377(15) -> ..., . - Robert G. Wilson v, Jan 24 2006

From Gary W. Adamson, Jun 04 2009: (Start)

As an infinite string, 2^n terms per row starting with "1": (1; 1,3; 1,3,3,7; 1,3,3,7,3,7,7,15; 1,3,3,7,3,7,7,15,3,7,7,15,7,15,15,31;...)

Row sums of that triangle = A027649: (1, 4, 14, 46, 454, ...); where the next row sum = current term of A027649 + next term in finite difference row of A027649, i.e., (1, 3, 10, 32, 100, 308, ...) = A053581. (End)

From Omar E. Pol, Jan 24 2016: (Start)

Partial sums give A267700.

a(n) is also the number of cells turned ON at n-th generation of the cellular automaton of A267700 in a 90-degree sector on the square grid.

a(n) is also the number of Y-toothpicks added at n-th generation of the structure of A267700 in a 120-degree sector on the triangular grid. (End)

Row sums of A090971. - Nikolaos Pantelidis, Nov 23 2022

"Least deficient numbers" or "almost perfect numbers" are those k for which A033879(k) = 1, or equally, for which a(k) = -A048881(k-1). The only known solutions are powers of 2 (A000079), all present also in A295296. See also A235796 and A378988. - Antti Karttunen, Dec 16 2024