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.

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A277700 a(n) = A000120(A264977(n)); number of odd terms on row n of A125184.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 2, 3, 1, 2, 1, 3, 2, 1, 3, 4, 1, 3, 2, 3, 1, 2, 3, 3, 2, 3, 1, 4, 3, 1, 4, 5, 1, 4, 3, 3, 2, 3, 3, 2, 1, 1, 2, 3, 3, 2, 3, 3, 2, 3, 3, 4, 1, 3, 4, 3, 3, 4, 1, 5, 4, 1, 5, 6, 1, 5, 4, 3, 3, 4, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 1, 2, 1, 3, 2, 1, 3, 4, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 4, 3, 3, 4, 3, 1, 2, 3, 3, 4, 3, 3, 4, 3, 3, 4, 5, 1, 4, 5, 3, 4
Offset: 0

Views

Author

Antti Karttunen, Oct 27 2016

Keywords

Comments

Positions of even and odd terms are given by A008585 and A001651, which means that parity-wise the terms match with the Fibonacci numbers, A000045.

Links

Crossrefs

Cf. A000120, A001221, A001222, A125184, A264977, A277330.
Cf. also A000045, A001651, A008585, A011655.

Programs

Formula

a(n) = A000120(A264977(n)).
a(n) = A001221(A277330(n)) = A001222(A277330(n)).
Other identities. For all n >= 0:
a(2n) = a(n).
A000035(a(n)) = A011655(n).

A000027 The positive integers. Also called the natural numbers, the whole numbers or the counting numbers, but these terms are ambiguous.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

For some authors, the terms "natural numbers" and "counting numbers" include 0, i.e., refer to the nonnegative integers A001477; the term "whole numbers" frequently also designates the whole set of (signed) integers A001057.
a(n) is smallest positive integer which is consistent with sequence being monotonically increasing and satisfying a(a(n)) = n (cf. A007378).
Inverse Euler transform of A000219.
The rectangular array having A000027 as antidiagonals is the dispersion of the complement of the triangular numbers, A000217 (which triangularly form column 1 of this array). The array is also the transpose of A038722. - Clark Kimberling, Apr 05 2003
For nonzero x, define f(n) = floor(nx) - floor(n/x). Then f=A000027 if and only if x=tau or x=-tau. - Clark Kimberling, Jan 09 2005
Numbers of form (2^i)*k for odd k (i.e., n = A006519(n)*A000265(n)); thus n corresponds uniquely to an ordered pair (i,k) where i=A007814, k=A000265 (with A007814(2n)=A001511(n), A007814(2n+1)=0). - Lekraj Beedassy, Apr 22 2006
If the offset were changed to 0, we would have the following pattern: a(n)=binomial(n,0) + binomial(n,1) for the present sequence (number of regions in 1-space defined by n points), A000124 (number of regions in 2-space defined by n straight lines), A000125 (number of regions in 3-space defined by n planes), A000127 (number of regions in 4-space defined by n hyperplanes), A006261, A008859, A008860, A008861, A008862 and A008863, where the last six sequences are interpreted analogously and in each "... by n ..." clause an offset of 0 has been assumed, resulting in a(0)=1 for all of them, which corresponds to the case of not cutting with a hyperplane at all and therefore having one region. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
Define a number of points on a straight line to be in general arrangement when no two points coincide. Then these are the numbers of regions defined by n points in general arrangement on a straight line, when an offset of 0 is assumed. For instance, a(0)=1, since using no point at all leaves one region. The sequence satisfies the recursion a(n) = a(n-1) + 1. This has the following geometrical interpretation: Suppose there are already n-1 points in general arrangement, thus defining the maximal number of regions on a straight line obtainable by n-1 points, and now one more point is added in general arrangement. Then it will coincide with no other point and act as a dividing wall thereby creating one new region in addition to the a(n-1)=(n-1)+1=n regions already there, hence a(n)=a(n-1)+1. Cf. the comments on A000124 for an analogous interpretation. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
The sequence a(n)=n (for n=1,2,3) and a(n)=n+1 (for n=4,5,...) gives to the rank (minimal cardinality of a generating set) for the semigroup I_n\S_n, where I_n and S_n denote the symmetric inverse semigroup and symmetric group on [n]. - James East, May 03 2007
The sequence a(n)=n (for n=1,2), a(n)=n+1 (for n=3) and a(n)=n+2 (for n=4,5,...) gives the rank (minimal cardinality of a generating set) for the semigroup PT_n\T_n, where PT_n and T_n denote the partial transformation semigroup and transformation semigroup on [n]. - James East, May 03 2007
"God made the integers; all else is the work of man." This famous quotation is a translation of "Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Menschenwerk," spoken by Leopold Kronecker in a lecture at the Berliner Naturforscher-Versammlung in 1886. Possibly the first publication of the statement is in Heinrich Weber's "Leopold Kronecker," Jahresberichte D.M.V. 2 (1893) 5-31. - Clark Kimberling, Jul 07 2007
Binomial transform of A019590, inverse binomial transform of A001792. - Philippe Deléham, Oct 24 2007
Writing A000027 as N, perhaps the simplest one-to-one correspondence between N X N and N is this: f(m,n) = ((m+n)^2 - m - 3n + 2)/2. Its inverse is given by I(k)=(g,h), where g = k - J(J-1)/2, h = J + 1 - g, J = floor((1 + sqrt(8k - 7))/2). Thus I(1)=(1,1), I(2)=(1,2), I(3)=(2,1) and so on; the mapping I fills the first-quadrant lattice by successive antidiagonals. - Clark Kimberling, Sep 11 2008
a(n) is also the mean of the first n odd integers. - Ian Kent, Dec 23 2008
Equals INVERTi transform of A001906, the even-indexed Fibonacci numbers starting (1, 3, 8, 21, 55, ...). - Gary W. Adamson, Jun 05 2009
These are also the 2-rough numbers: positive integers that have no prime factors less than 2. - Michael B. Porter, Oct 08 2009
Totally multiplicative sequence with a(p) = p for prime p. Totally multiplicative sequence with a(p) = a(p-1) + 1 for prime p. - Jaroslav Krizek, Oct 18 2009
Triangle T(k,j) of natural numbers, read by rows, with T(k,j) = binomial(k,2) + j = (k^2-k)/2 + j where 1 <= j <= k. In other words, a(n) = n = binomial(k,2) + j where k is the largest integer such that binomial(k,2) < n and j = n - binomial(k,2). For example, T(4,1)=7, T(4,2)=8, T(4,3)=9, and T(4,4)=10. Note that T(n,n)=A000217(n), the n-th triangular number. - Dennis P. Walsh, Nov 19 2009
Hofstadter-Conway-like sequence (see A004001): a(n) = a(a(n-1)) + a(n-a(n-1)) with a(1) = 1, a(2) = 2. - Jaroslav Krizek, Dec 11 2009
a(n) is also the dimension of the irreducible representations of the Lie algebra sl(2). - Leonid Bedratyuk, Jan 04 2010
Floyd's triangle read by rows. - Paul Muljadi, Jan 25 2010
Number of numbers between k and 2k where k is an integer. - Giovanni Teofilatto, Mar 26 2010
Generated from a(2n) = r*a(n), a(2n+1) = a(n) + a(n+1), r = 2; in an infinite set, row 2 of the array shown in A178568. - Gary W. Adamson, May 29 2010
1/n = continued fraction [n]. Let barover[n] = [n,n,n,...] = 1/k. Then k - 1/k = n. Example: [2,2,2,...] = (sqrt(2) - 1) = 1/k, with k = (sqrt(2) + 1). Then 2 = k - 1/k. - Gary W. Adamson, Jul 15 2010
Number of n-digit numbers the binary expansion of which contains one run of 1's. - Vladimir Shevelev, Jul 30 2010
From Clark Kimberling, Jan 29 2011: (Start)
Let T denote the "natural number array A000027":
1 2 4 7 ...
3 5 8 12 ...
6 9 13 18 ...
10 14 19 25 ...
T(n,k) = n+(n+k-2)*(n+k-1)/2. See A185787 for a list of sequences based on T, such as rows, columns, diagonals, and sub-arrays. (End)
The Stern polynomial B(n,x) evaluated at x=2. See A125184. - T. D. Noe, Feb 28 2011
The denominator in the Maclaurin series of log(2), which is 1 - 1/2 + 1/3 - 1/4 + .... - Mohammad K. Azarian, Oct 13 2011
As a function of Bernoulli numbers B_n (cf. A027641: (1, -1/2, 1/6, 0, -1/30, 0, 1/42, ...)): let V = a variant of B_n changing the (-1/2) to (1/2). Then triangle A074909 (the beheaded Pascal's triangle) * [1, 1/2, 1/6, 0, -1/30, ...] = the vector [1, 2, 3, 4, 5, ...]. - Gary W. Adamson, Mar 05 2012
Number of partitions of 2n+1 into exactly two parts. - Wesley Ivan Hurt, Jul 15 2013
Integers n dividing u(n) = 2u(n-1) - u(n-2); u(0)=0, u(1)=1 (Lucas sequence A001477). - Thomas M. Bridge, Nov 03 2013
For this sequence, the generalized continued fraction a(1)+a(1)/(a(2)+a(2)/(a(3)+a(3)/(a(4)+...))), evaluates to 1/(e-2) = A194807. - Stanislav Sykora, Jan 20 2014
Engel expansion of e-1 (A091131 = 1.71828...). - Jaroslav Krizek, Jan 23 2014
a(n) is the number of permutations of length n simultaneously avoiding 213, 231 and 321 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014
a(n) is also the number of permutations simultaneously avoiding 213, 231 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014
a(n) = least k such that 2*Pi - Sum_{h=1..k} 1/(h^2 - h + 3/16) < 1/n. - Clark Kimberling, Sep 28 2014
a(n) = least k such that Pi^2/6 - Sum_{h=1..k} 1/h^2 < 1/n. - Clark Kimberling, Oct 02 2014
Determinants of the spiral knots S(2,k,(1)). a(k) = det(S(2,k,(1))). These knots are also the torus knots T(2,k). - Ryan Stees, Dec 15 2014
As a function, the restriction of the identity map on the nonnegative integers {0,1,2,3...}, A001477, to the positive integers {1,2,3,...}. - M. F. Hasler, Jan 18 2015
See also A131685(k) = smallest positive number m such that c(i) = m (i^1 + 1) (i^2 + 2) ... (i^k+ k) / k! takes integral values for all i>=0: For k=1, A131685(k)=1, which implies that this is a well defined integer sequence. - Alexander R. Povolotsky, Apr 24 2015
a(n) is the number of compositions of n+2 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016
Does not satisfy Benford's law [Berger-Hill, 2017] - N. J. A. Sloane, Feb 07 2017
Parametrization for the finite multisubsets of the positive integers, where, for p_j the j-th prime, n = Product_{j} p_j^(e_j) corresponds to the multiset containing e_j copies of j ('Heinz encoding' -- see A056239, A003963, A289506, A289507, A289508, A289509). - Christopher J. Smyth, Jul 31 2017
The arithmetic function v_1(n,1) as defined in A289197. - Robert Price, Aug 22 2017
For n >= 3, a(n)=n is the least area that can be obtained for an irregular octagon drawn in a square of n units side, whose sides are parallel to the axes, with 4 vertices that coincide with the 4 vertices of the square, and the 4 remaining vertices having integer coordinates. See Affaire de Logique link. - Michel Marcus, Apr 28 2018
a(n+1) is the order of rowmotion on a poset defined by a disjoint union of chains of length n. - Nick Mayers, Jun 08 2018
Number of 1's in n-th generation of 1-D Cellular Automata using Rules 50, 58, 114, 122, 178, 186, 206, 220, 238, 242, 250 or 252 in the Wolfram numbering scheme, started with a single 1. - Frank Hollstein, Mar 25 2019
(1, 2, 3, 4, 5, ...) is the fourth INVERT transform of (1, -2, 3, -4, 5, ...). - Gary W. Adamson, Jul 15 2019

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 1.
  • T. M. Apostol, Modular Functions and Dirichlet Series in Number Theory, Springer-Verlag, 1990, page 25.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 22.
  • W. Fulton and J. Harris, Representation theory: a first course, (1991), page 149. [From Leonid Bedratyuk, Jan 04 2010]
  • I. S. Gradstein and I. M. Ryshik, Tables of series, products, and integrals, Volume 1, Verlag Harri Deutsch, 1981.
  • R. E. Schwartz, You Can Count on Monsters: The First 100 numbers and Their Characters, A. K. Peters and MAA, 2010.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

A001477 = nonnegative numbers.
Partial sums of A000012.
Cf. A001478, A001906, A007931, A007932, A027641, A074909, A089353 (multisets), A178568, A194807.
Cf. A026081 = integers in reverse alphabetical order in U.S. English, A107322 = English name for number and its reverse have the same number of letters, A119796 = zero through ten in alphabetical order of English reverse spelling, A005589, etc. Cf. A185787 (includes a list of sequences based on the natural number array A000027).
Cf. Boustrophedon transforms: A000737, A231179;
Cf. A038722 (mirrored when seen as triangle), A056011 (boustrophedon).
Cf. A048993, A048994, A000110 (see the Feb 03 2015 formula).
Cf. A289187, A000010, A008683, A000203, A049310.

Programs

Formula

a(2k+1) = A005408(k), k >= 0, a(2k) = A005843(k), k >= 1.
Multiplicative with a(p^e) = p^e. - David W. Wilson, Aug 01 2001
Another g.f.: Sum_{n>0} phi(n)*x^n/(1-x^n) (Apostol).
When seen as an array: T(k, n) = n+1 + (k+n)*(k+n+1)/2. Main diagonal is 2n*(n+1)+1 (A001844), antidiagonal sums are n*(n^2+1)/2 (A006003). - Ralf Stephan, Oct 17 2004
Dirichlet generating function: zeta(s-1). - Franklin T. Adams-Watters, Sep 11 2005
G.f.: x/(1-x)^2. E.g.f.: x*exp(x). a(n)=n. a(-n)=-a(n).
Series reversion of g.f. A(x) is x*C(-x)^2 where C(x) is the g.f. of A000108. - Michael Somos, Sep 04 2006
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = u^2 - v - 4*u*v. - Michael Somos, Oct 03 2006
Convolution of A000012 (the all-ones sequence) with itself. - Tanya Khovanova, Jun 22 2007
a(n) = 2*a(n-1)-a(n-2); a(1)=1, a(2)=2. a(n) = 1+a(n-1). - Philippe Deléham, Nov 03 2008
a(n) = A000720(A000040(n)). - Juri-Stepan Gerasimov, Nov 29 2009
a(n+1) = Sum_{k=0..n} A101950(n,k). - Philippe Deléham, Feb 10 2012
a(n) = Sum_{d | n} phi(d) = Sum_{d | n} A000010(d). - Jaroslav Krizek, Apr 20 2012
G.f.: x * Product_{j>=0} (1+x^(2^j))^2 = x * (1+2*x+x^2) * (1+2*x^2+x^4) * (1+2*x^4+x^8) * ... = x + 2x^2 + 3x^3 + ... . - Gary W. Adamson, Jun 26 2012
a(n) = det(binomial(i+1,j), 1 <= i,j <= n). - Mircea Merca, Apr 06 2013
E.g.f.: x*E(0), where E(k) = 1 + 1/(x - x^3/(x^2 + (k+1)/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 03 2013
From Wolfdieter Lang, Oct 09 2013: (Start)
a(n) = Product_{k=1..n-1} 2*sin(Pi*k/n), n > 1.
a(n) = Product_{k=1..n-1} (2*sin(Pi*k/(2*n)))^2, n > 1.
These identities are used in the calculation of products of ratios of lengths of certain lines in a regular n-gon. For the first identity see the Gradstein-Ryshik reference, p. 62, 1.392 1., bringing the first factor there to the left hand side and taking the limit x -> 0 (L'Hôpital). The second line follows from the first one. Thanks to Seppo Mustonen who led me to consider n-gon lengths products. (End)
a(n) = Sum_{j=0..k} (-1)^(j-1)*j*binomial(n,j)*binomial(n-1+k-j,k-j), k>=0. - Mircea Merca, Jan 25 2014
a(n) = A052410(n)^A052409(n). - Reinhard Zumkeller, Apr 06 2014
a(n) = Sum_{k=1..n^2+2*n} 1/(sqrt(k)+sqrt(k+1)). - Pierre CAMI, Apr 25 2014
a(n) = floor(1/sin(1/n)) = floor(cot(1/(n+1))) = ceiling(cot(1/n)). - Clark Kimberling, Oct 08 2014
a(n) = floor(1/(log(n+1)-log(n))). - Thomas Ordowski, Oct 10 2014
a(k) = det(S(2,k,1)). - Ryan Stees, Dec 15 2014
a(n) = 1/(1/(n+1) + 1/(n+1)^2 + 1/(n+1)^3 + ...). - Pierre CAMI, Jan 22 2015
a(n) = Sum_{m=0..n-1} Stirling1(n-1,m)*Bell(m+1), for n >= 1. This corresponds to Bell(m+1) = Sum_{k=0..m} Stirling2(m, k)*(k+1), for m >= 0, from the fact that Stirling2*Stirling1 = identity matrix. See A048993, A048994 and A000110. - Wolfdieter Lang, Feb 03 2015
a(n) = Sum_{k=1..2n-1}(-1)^(k+1)*k*(2n-k). In addition, surprisingly, a(n) = Sum_{k=1..2n-1}(-1)^(k+1)*k^2*(2n-k)^2. - Charlie Marion, Jan 05 2016
G.f.: x/(1-x)^2 = (x * r(x) *r(x^3) * r(x^9) * r(x^27) * ...), where r(x) = (1 + x + x^2)^2 = (1 + 2x + 3x^2 + 2x^3 + x^4). - Gary W. Adamson, Jan 11 2017
a(n) = floor(1/(Pi/2-arctan(n))). - Clark Kimberling, Mar 11 2020
a(n) = Sum_{d|n} mu(n/d)*sigma(d). - Ridouane Oudra, Oct 03 2020
a(n) = Sum_{k=1..n} phi(gcd(n,k))/phi(n/gcd(n,k)). - Richard L. Ollerton, May 09 2021
a(n) = S(n-1, 2), with the Chebyshev S-polynomials A049310. - Wolfdieter Lang, Mar 09 2023
From Peter Bala, Nov 02 2024: (Start)
For positive integer m, a(n) = (1/m)* Sum_{k = 1..2*m*n-1} (-1)^(k+1) * k * (2*m*n - k) = (1/m) * Sum_{k = 1..2*m*n-1} (-1)^(k+1) * k^2 * (2*m*n - k)^2 (the case m = 1 is given above).
a(n) = Sum_{k = 0..3*n} (-1)^(n+k+1) * k * binomial(3*n+k, 2*k). (End)

Extensions

Links edited by Daniel Forgues, Oct 07 2009.

A007814 Exponent of highest power of 2 dividing n, a.k.a. the binary carry sequence, the ruler sequence, or the 2-adic valuation of n.

Original entry on oeis.org

0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 6, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0
Offset: 1

Views

Author

John Tromp, Dec 11 1996

Keywords

Comments

This sequence is an exception to my usual rule that when every other term of a sequence is 0 then those 0's should be omitted. In this case we would get A001511. - N. J. A. Sloane
To construct the sequence: start with 0,1, concatenate to get 0,1,0,1. Add + 1 to last term gives 0,1,0,2. Concatenate those 4 terms to get 0,1,0,2,0,1,0,2. Add + 1 to last term etc. - Benoit Cloitre, Mar 06 2003
The sequence is invariant under the following two transformations: increment every element by one (1, 2, 1, 3, 1, 2, 1, 4, ...), put a zero in front and between adjacent elements (0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, ...). The intermediate result is A001511. - Ralf Hinze (ralf(AT)informatik.uni-bonn.de), Aug 26 2003
Fixed point of the morphism 0->01, 1->02, 2->03, 3->04, ..., n->0(n+1), ..., starting from a(1) = 0. - Philippe Deléham, Mar 15 2004
Fixed point of the morphism 0->010, 1->2, 2->3, ..., n->(n+1), .... - Joerg Arndt, Apr 29 2014
a(n) is also the number of times to repeat a step on an even number in the hailstone sequence referenced in the Collatz conjecture. - Alex T. Flood (whiteangelsgrace(AT)gmail.com), Sep 22 2006
Let F(n) be the n-th Fermat number (A000215). Then F(a(r-1)) divides F(n)+2^k for r = k mod 2^n and r != 1. - T. D. Noe, Jul 12 2007
The following relation holds: 2^A007814(n)*(2*A025480(n-1)+1) = A001477(n) = n. (See functions hd, tl and cons in [Paul Tarau 2009].)
a(n) is the number of 0's at the end of n when n is written in base 2.
a(n+1) is the number of 1's at the end of n when n is written in base 2. - M. F. Hasler, Aug 25 2012
Shows which bit to flip when creating the binary reflected Gray code (bits are numbered from the right, offset is 0). That is, A003188(n) XOR A003188(n+1) == 2^A007814(n). - Russ Cox, Dec 04 2010
The sequence is squarefree (in the sense of not containing any subsequence of the form XX) [Allouche and Shallit]. Of course it contains individual terms that are squares (such as 4). - Comment expanded by N. J. A. Sloane, Jan 28 2019
a(n) is the number of zero coefficients in the n-th Stern polynomial, A125184. - T. D. Noe, Mar 01 2011
Lemma: For n < m with r = a(n) = a(m) there exists n < k < m with a(k) > r. Proof: We have n=b2^r and m=c2^r with b < c both odd; choose an even i between them; now a(i2^r) > r and n < i2^r < m. QED. Corollary: Every finite run of consecutive integers has a unique maximum 2-adic valuation. - Jason Kimberley, Sep 09 2011
a(n-2) is the 2-adic valuation of A000166(n) for n >= 2. - Joerg Arndt, Sep 06 2014
a(n) = number of 1's in the partition having Heinz number n. We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product_{j=1..r} p_j-th prime (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] we get 2*2*3*7*29 = 2436. Example: a(24)=3; indeed, the partition having Heinz number 24 = 2*2*2*3 is [1,1,1,2]. - Emeric Deutsch, Jun 04 2015
a(n+1) is the difference between the two largest parts in the integer partition having viabin number n (0 is assumed to be a part). Example: a(20) = 2. Indeed, we have 19 = 10011_2, leading to the Ferrers board of the partition [3,1,1]. For the definition of viabin number see the comment in A290253. - Emeric Deutsch, Aug 24 2017
Apart from being squarefree, as noted above, the sequence has the property that every consecutive subsequence contains at least one number an odd number of times. - Jon Richfield, Dec 20 2018
a(n+1) is the 2-adic valuation of Sum_{e=0..n} u^e = (1 + u + u^2 + ... + u^n), for any u of the form 4k+1 (A016813). - Antti Karttunen, Aug 15 2020
{a(n)} represents the "first black hat" strategy for the game of countably infinitely many hats, with a probability of success of 1/3; cf. the Numberphile link below. - Frederic Ruget, Jun 14 2021
a(n) is the least nonnegative integer k for which there does not exist i+j=n and a(i)=a(j)=k (cf. A322523). - Rémy Sigrist and Jianing Song, Aug 23 2022

Examples

			2^3 divides 24, so a(24)=3.
From _Omar E. Pol_, Jun 12 2009: (Start)
Triangle begins:
  0;
  1,0;
  2,0,1,0;
  3,0,1,0,2,0,1,0;
  4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0;
  5,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0;
  6,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,5,0,1,0,2,...
(End)

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 27.
  • K. Atanassov, On the 37th and the 38th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 2, 83-85.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Links

Crossrefs

Cf. A011371 (partial sums), A094267 (first differences), A001511 (bisection), A346070 (mod 4).
Bisection of A050605 and |A088705|. Pairwise sums are A050603 and A136480. Difference of A285406 and A281264.
This is Guy Steele's sequence GS(1, 4) (see A135416). Cf. A053398(1,n). Column/row 1 of table A050602.
Cf. A007949 (3-adic), A235127 (4-adic), A112765 (5-adic), A122841 (6-adic), A214411 (7-adic), A244413 (8-adic), A122840 (10-adic).
Cf. A000079, A002487, A006519, A051064, A055457, A063787, A215366, A220466.
Cf. A086463 (Dgf at s=2).

Programs

  • Haskell
    a007814 n = if m == 0 then 1 + a007814 n' else 0
            where (n', m) = divMod n 2
-- Reinhard Zumkeller, Jul 05 2013, May 14 2011, Apr 08 2011
  • Haskell
    a007814 n | odd n = 0 | otherwise = 1 + a007814 (n `div` 2)
--  Walt Rorie-Baety, Mar 22 2013
  • Magma
    [Valuation(n, 2): n in [1..120]]; // Bruno Berselli, Aug 05 2013
  • Maple
    ord := proc(n) local i,j; if n=0 then return 0; fi; i:=0; j:=n; while j mod 2 <> 1 do i:=i+1; j:=j/2; od: i; end proc: seq(ord(n), n=1..111);
A007814 := n -> padic[ordp](n,2): seq(A007814(n), n=1..111); # Peter Luschny, Nov 26 2010
  • Mathematica
    Table[IntegerExponent[n, 2], {n, 64}] (* Eric W. Weisstein *)
IntegerExponent[Range[64], 2] (* Eric W. Weisstein, Feb 01 2024 *)
p=2; Array[ If[ Mod[ #, p ]==0, Select[ FactorInteger[ # ], Function[ q, q[ [ 1 ] ]==p ], 1 ][ [ 1, 2 ] ], 0 ]&, 96 ]
DigitCount[BitXor[x, x - 1], 2, 1] - 1; a different version based on the same concept: Floor[Log[2, BitXor[x, x - 1]]] (* Jaume Simon Gispert (jaume(AT)nuem.com), Aug 29 2004 *)
Nest[Join[ #, ReplacePart[ #, Length[ # ] -> Last[ # ] + 1]] &, {0, 1}, 5] (* N. J. Gunther, May 23 2009 *)
Nest[ Flatten[# /. a_Integer -> {0, a + 1}] &, {0}, 7] (* Robert G. Wilson v, Jan 17 2011 *)
  • PARI
    A007814(n)=valuation(n,2);
  • Python
    import math
def a(n): return int(math.log(n - (n & n - 1), 2)) # Indranil Ghosh, Apr 18 2017
  • Python
    def A007814(n): return (~n & n-1).bit_length() # Chai Wah Wu, Jul 01 2022
  • R
    sapply(1:100,function(x) sum(gmp::factorize(x)==2)) # Christian N. K. Anderson, Jun 20 2013
  • Scheme
    (define (A007814 n) (let loop ((n n) (e 0)) (if (odd? n) e (loop (/ n 2) (+ 1 e))))) ;; Antti Karttunen, Oct 06 2017

Formula

a(n) = A001511(n) - 1.
a(2*n) = A050603(2*n) = A001511(n).
a(n) = A091090(n-1) + A036987(n-1) - 1.
a(n) = 0 if n is odd, otherwise 1 + a(n/2). - Reinhard Zumkeller, Aug 11 2001
Sum_{k=1..n} a(k) = n - A000120(n). - Benoit Cloitre, Oct 19 2002
G.f.: A(x) = Sum_{k>=1} x^(2^k)/(1-x^(2^k)). - Ralf Stephan, Apr 10 2002
G.f. A(x) satisfies A(x) = A(x^2) + x^2/(1-x^2). A(x) = B(x^2) = B(x) - x/(1-x), where B(x) is the g.f. for A001151. - Franklin T. Adams-Watters, Feb 09 2006
Totally additive with a(p) = 1 if p = 2, 0 otherwise.
Dirichlet g.f.: zeta(s)/(2^s-1). - Ralf Stephan, Jun 17 2007
Define 0 <= k <= 2^n - 1; binary: k = b(0) + 2*b(1) + 4*b(2) + ... + 2^(n-1)*b(n-1); where b(x) are 0 or 1 for 0 <= x <= n - 1; define c(x) = 1 - b(x) for 0 <= x <= n - 1; Then: a(k) = c(0) + c(0)*c(1) + c(0)*c(1)*c(2) + ... + c(0)*c(1)...c(n-1); a(k+1) = b(0) + b(0)*b(1) + b(0)*b(1)*b(2) + ... + b(0)*b(1)...b(n-1). - Arie Werksma (werksma(AT)tiscali.nl), May 10 2008
a(n) = floor(A002487(n - 1) / A002487(n)). - Reikku Kulon, Oct 05 2008
Sum_{k=1..n} (-1)^A000120(n-k)*a(k) = (-1)^(A000120(n)-1)*(A000120(n) - A000035(n)). - Vladimir Shevelev, Mar 17 2009
a(A001147(n) + A057077(n-1)) = a(2*n). - Vladimir Shevelev, Mar 21 2009
For n>=1, a(A004760(n+1)) = a(n). - Vladimir Shevelev, Apr 15 2009
2^(a(n)) = A006519(n). - Philippe Deléham, Apr 22 2009
a(n) = A063787(n) - A000120(n). - Gary W. Adamson, Jun 04 2009
a(C(n,k)) = A000120(k) + A000120(n-k) - A000120(n). - Vladimir Shevelev, Jul 19 2009
a(n!) = n - A000120(n). - Vladimir Shevelev, Jul 20 2009
v_{2}(n) = Sum_{r>=1} (r / 2^(r+1)) Sum_{k=0..2^(r+1)-1} e^(2(k*Pi*i(n+2^r))/(2^(r+1))). - A. Neves, Sep 28 2010, corrected Oct 04 2010
a(n) mod 2 = A096268(n-1). - Robert G. Wilson v, Jan 18 2012
a(A005408(n)) = 1; a(A016825(n)) = 3; A017113(a(n)) = 5; A051062(a(n)) = 7; a(n) = (A037227(n)-1)/2. - Reinhard Zumkeller, Jun 30 2012
a((2*n-1)*2^p) = p, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 04 2013
a(n) = A067255(n,1). - Reinhard Zumkeller, Jun 11 2013
a(n) = log_2(n - (n AND n-1)). - Gary Detlefs, Jun 13 2014
a(n) = 1 + A000120(n-1) - A000120(n), where A000120 is the Hamming weight function. - Stanislav Sykora, Jul 14 2014
A053398(n,k) = a(A003986(n-1,k-1)+1); a(n) = A053398(n,1) = A053398(n,n) = A053398(2*n-1,n) = Min_{k=1..n} A053398(n,k). - Reinhard Zumkeller, Aug 04 2014
a((2*x-1)*2^n) = a((2*y-1)*2^n) for positive n, x and y. - Juri-Stepan Gerasimov, Aug 04 2016
a(n) = A285406(n) - A281264(n). - Ralf Steiner, Apr 18 2017
a(n) = A000005(n)/(A000005(2*n) - A000005(n)) - 1. - conjectured by Velin Yanev, Jun 30 2017, proved by Nicholas Stearns, Sep 11 2017
Equivalently to above formula, a(n) = A183063(n) / A001227(n), i.e., a(n) is the number of even divisors of n divided by number of odd divisors of n. - Franklin T. Adams-Watters, Oct 31 2018
a(n)*(n mod 4) = 2*floor(((n+1) mod 4)/3). - Gary Detlefs, Feb 16 2019
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1. - Amiram Eldar, Jul 11 2020
a(n) = 2*Sum_{j=1..floor(log_2(n))} frac(binomial(n, 2^j)*2^(j-1)/n). - Dario T. de Castro, Jul 08 2022
a(n) = A070939(n) - A070939(A030101(n)). - Andrew T. Porter, Dec 16 2022
a(n) = floor((gcd(n, 2^n)^(n+1) mod (2^(n+1)-1)^2)/(2^(n+1)-1)) (see Lemma 3.4 from Mazzanti's 2002 article). - Lorenzo Sauras Altuzarra, Mar 10 2024
a(n) = 1 - A088705(n). - Chai Wah Wu, Sep 18 2024

Extensions

Formula index adapted to the offset of A025480 by R. J. Mathar, Jul 20 2010
Edited by Ralf Stephan, Feb 08 2014

A002487 Stern's diatomic series (or Stern-Brocot sequence): a(0) = 0, a(1) = 1; for n > 0: a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1).

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1, 6, 5, 9, 4, 11, 7, 10, 3, 11, 8, 13, 5, 12, 7, 9, 2, 9, 7, 12, 5, 13, 8, 11, 3, 10, 7, 11, 4, 9, 5, 6, 1, 7, 6, 11, 5, 14, 9, 13, 4, 15, 11, 18, 7, 17, 10, 13, 3, 14, 11, 19, 8, 21, 13, 18, 5, 17, 12, 19
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Also called fusc(n) [Dijkstra].
a(n)/a(n+1) runs through all the reduced nonnegative rationals exactly once [Stern; Calkin and Wilf].
If the terms are written as an array:
column 0 1 2 3 4 5 6 7 8 9 ...
row 0: 0
row 1: 1
row 2: 1,2
row 3: 1,3,2,3
row 4: 1,4,3,5,2,5,3,4
row 5: 1,5,4,7,3,8,5,7,2,7,5,8,3,7,4,5
row 6: 1,6,5,9,4,11,7,10,3,11,8,13,5,12,7,9,2,9,7,12,5,13,8,11,3,10,...
...
then (ignoring row 0) the sum of the k-th row is 3^(k-1), each column is an arithmetic progression and the steps are nothing but the original sequence. - Takashi Tokita (butaneko(AT)fa2.so-net.ne.jp), Mar 08 2003
From N. J. A. Sloane, Oct 15 2017: (Start)
The above observation can be made more precise. Let A(n,k), n >= 0, 0 <= k <= 2^(n-1)-1 for k > 0, denote the entry in row n and column k of the left-justified array above.
The equations for columns 0,1,2,3,4,... are successively (ignoring row 0):
1 (n >= 1),
n (n >= 2),
n-1 (n >= 3),
2n-3 (n >= 3),
n-2 (n >= 4),
3n-7 (n >= 4),
...
and in general column k > 0 is given by
A(n,k) = a(k)*n - A156140(k) for n >= ceiling(log_2(k+1))+1, and 0 otherwise.
(End)
a(n) is the number of odd Stirling numbers S_2(n+1, 2r+1) [Carlitz].
Moshe Newman proved that the fraction a(n+1)/a(n+2) can be generated from the previous fraction a(n)/a(n+1) = x by 1/(2*floor(x) + 1 - x). The successor function f(x) = 1/(floor(x) + 1 - frac(x)) can also be used.
a(n+1) = number of alternating bit sets in n [Finch].
If f(x) = 1/(1 + floor(x) - frac(x)) then f(a(n-1)/a(n)) = a(n)/a(n+1) for n >= 1. If T(x) = -1/x and f(x) = y, then f(T(y)) = T(x) for x > 0. - Michael Somos, Sep 03 2006
a(n+1) is the number of ways of writing n as a sum of powers of 2, each power being used at most twice (the number of hyperbinary representations of n) [Carlitz; Lind].
a(n+1) is the number of partitions of the n-th integer expressible as the sum of distinct even-subscripted Fibonacci numbers (= A054204(n)), into sums of distinct Fibonacci numbers [Bicknell-Johnson, theorem 2.1].
a(n+1) is the number of odd binomial(n-k, k), 0 <= 2*k <= n. [Carlitz], corrected by Alessandro De Luca, Jun 11 2014
a(2^k) = 1. a(3*2^k) = a(2^(k+1) + 2^k) = 2. Sequences of terms between a(2^k) = 1 and a(2^(k+1)) = 1 are palindromes of length 2^k-1 with a(2^k + 2^(k-1)) = 2 in the middle. a(2^(k-1) + 1) = a(2^k - 1) = k+1 for k > 1. - Alexander Adamchuk, Oct 10 2006
The coefficients of the inverse of the g.f. of this sequence form A073469 and are related to binary partitions A000123. - Philippe Flajolet, Sep 06 2008
It appears that the terms of this sequence are the number of odd entries in the diagonals of Pascal's triangle at 45 degrees slope. - Javier Torres (adaycalledzero(AT)hotmail.com), Aug 06 2009
Let M be an infinite lower triangular matrix with (1, 1, 1, 0, 0, 0, ...) in every column shifted down twice:
1;
1, 0;
1, 1, 0;
0, 1, 0, 0;
0, 1, 1, 0, 0;
0, 0, 1, 0, 0, 0;
0, 0, 1, 1, 0, 0, 0;
...
Then this sequence A002487 (without initial 0) is the first column of lim_{n->oo} M^n. (Cf. A026741.) - Gary W. Adamson, Dec 11 2009 [Edited by M. F. Hasler, Feb 12 2017]
Member of the infinite family of sequences of the form a(n) = a(2*n); a(2*n+1) = r*a(n) + a(n+1), r = 1 for A002487 = row 1 in the array of A178239. - Gary W. Adamson, May 23 2010
Equals row 1 in an infinite array shown in A178568, sequences of the form
a(2*n) = r*a(n), a(2*n+1) = a(n) + a(n+1); r = 1. - Gary W. Adamson, May 29 2010
Row sums of A125184, the Stern polynomials. Equivalently, B(n,1), the n-th Stern polynomial evaluated at x = 1. - T. D. Noe, Feb 28 2011
The Kn1y and Kn2y triangle sums, see A180662 for their definitions, of A047999 lead to the sequence given above, e.g., Kn11(n) = A002487(n+1) - A000004(n), Kn12(n) = A002487(n+3) - A000012(n), Kn13(n) = A002487(n+5) - A000034(n+1) and Kn14(n) = A002487(n+7) - A157810(n+1). For the general case of the knight triangle sums see the Stern-Sierpiński triangle A191372. This triangle not only leads to Stern's diatomic series but also to snippets of this sequence and, quite surprisingly, their reverse. - Johannes W. Meijer, Jun 05 2011
Maximum of terms between a(2^k) = 1 and a(2^(k+1)) = 1 is the Fibonacci number F(k+2). - Leonid Bedratyuk, Jul 04 2012
Probably the number of different entries per antidiagonal of A223541. That would mean there are exactly a(n+1) numbers that can be expressed as a nim-product 2^x*2^y with x + y = n. - Tilman Piesk, Mar 27 2013
Let f(m,n) be the frequency of the integer n in the interval [a(2^(m-1)), a(2^m-1)]. Let phi(n) be Euler's totient function (A000010). Conjecture: for all integers m,n n<=m f(m,n) = phi(n). - Yosu Yurramendi, Sep 08 2014
Back in May 1995, it was proved that A000360 is the modulo 3 mapping, (+1,-1,+0)/2, of this sequence A002487 (without initial 0). - M. Jeremie Lafitte (Levitas), Apr 24 2017
Define a sequence chf(n) of Christoffel words over an alphabet {-,+}: chf(1) = '-'; chf(2*n+0) = negate(chf(n)); chf(2*n+1) = negate(concatenate(chf(n),chf(n+1))). Then the length of the chf(n) word is fusc(n) = a(n); the number of '-'-signs in the chf(n) word is c-fusc(n) = A287729(n); the number of '+'-signs in the chf(n) word is s-fusc(n) = A287730(n). See examples below. - I. V. Serov, Jun 01 2017
The sequence can be extended so that a(n) = a(-n), a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1) for all n in Z. - Michael Somos, Jun 25 2019
Named after the German mathematician Moritz Abraham Stern (1807-1894), and sometimes also after the French clockmaker and amateur mathematician Achille Brocot (1817-1878). - Amiram Eldar, Jun 06 2021
It appears that a(n) is equal to the multiplicative inverse of A007305(n+1) mod A007306(n+1). For example, a(12) is 2, the multiplicative inverse of A007305(13) mod A007306(13), where A007305(13) is 4 and A007306(13) is 7. - Gary W. Adamson, Dec 18 2023

Examples

			Stern's diatomic array begins:
  1,1,
  1,2,1,
  1,3,2,3,1,
  1,4,3,5,2,5,3,4,1,
  1,5,4,7,3,8,5,7,2,7,5,8,3,7,4,5,1,
  1,6,5,9,4,11,7,10,3,11,8,13,5,12,7,9,2,9,7,12,5,13,8,11,3,10,7,11,4,9,...
  ...
a(91) = 19, because 91_10 = 1011011_2; b_6=b_4=b_3=b_1=b_0=1, b_5=b_2=0;  L=5; m_1=0, m_2=1, m_3=3, m_4=4, m_5=6; c_1=2, c_2=3, c_3=2, c_4=3; f(1)=1, f(2)=2, f(3)=5, f(4)=8, f(5)=19. - _Yosu Yurramendi_, Jul 13 2016
From _I. V. Serov_, Jun 01 2017: (Start)
a(n) is the length of the Christoffel word chf(n):
n  chf(n) A070939(n)   a(n)
1   '-'       1          1
2   '+'       2          1
3   '+-'      2          2
4   '-'       3          1
5   '--+'     3          3
6   '-+'      3          2
... (End)
G.f. = x + x^2 + 2*x^3 + x^4 + 3*x^5 + 2*x^6 + 3*x^7 + x^8 + ... - _Michael Somos_, Jun 25 2019

References

  • M. Aigner and G. M. Ziegler, Proofs from THE BOOK, 3rd ed., Berlin, Heidelberg, New York: Springer-Verlag, 2004, p. 97.
  • Elwyn R. Berlekamp, John H. Conway and Richard K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 114.
  • Krishna Dasaratha, Laure Flapan, Chansoo Lee, Cornelia Mihaila, Nicholas Neumann-Chun, Sarah Peluse and Matthew Stroegeny, A family of multi-dimensional continued fraction Stern sequences, Abtracts Amer. Math. Soc., Vol. 33 (#1, 2012), #1077-05-2543.
  • Edsger W. Dijkstra, Selected Writings on Computing, Springer, 1982, p. 232 (sequence is called fusc).
  • F. G. M. Eisenstein, Eine neue Gattung zahlentheoretischer Funktionen, welche von zwei Elementen abhaengen und durch gewisse lineare Funktional-Gleichungen definirt werden, Verhandlungen der Koenigl. Preuss. Akademie der Wiss. Berlin (1850), pp. 36-42, Feb 18, 1850. Werke, II, pp. 705-711.
  • Graham Everest, Alf van der Poorten, Igor Shparlinski and Thomas Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 2.16.3; pp. 148-149.
  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 117.
  • Thomas Koshy, Fibonacci and Lucas numbers with applications, Wiley, 2001, p. 98.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A000120, A000123, A000360, A001045, A002083, A011655, A020950, A026741, A037227, A046815, A070871, A070872, A071883, A073459, A084091, A101624, A126606, A168081, A174980, A174981, A178239, A178568, A212288, A213369, A260443, A277020, A277325, A287729, A287730, A293160.
Record values are in A212289.
If the 1's are replaced by pairs of 1's we obtain A049456.
Inverse: A020946.
Cf. a(A001045(n)) = A000045(n). a(A062092(n)) = A000032(n+1).
Cf. A064881-A064886 (Stern-Brocot subtrees).
A column of A072170.
Cf. A049455 for the 0,1 version of Stern's diatomic array.
Cf. A000119, A262097 for analogous sequences in other bases and A277189, A277315, A277328 for related sequences with similar graphs.
Cf. A086592 and references therein to other sequences related to Kepler's tree of fractions.

Programs

  • Haskell
    a002487 n = a002487_list !! n
a002487_list = 0 : 1 : stern [1] where
   stern fuscs = fuscs' ++ stern fuscs' where
     fuscs' = interleave fuscs $ zipWith (+) fuscs $ (tail fuscs) ++ [1]
   interleave []     ys = ys
   interleave (x:xs) ys = x : interleave ys xs
-- Reinhard Zumkeller, Aug 23 2011
  • Julia
    using Nemo
function A002487List(len)
    a, A = QQ(0), [0,1]
    for n in 1:len
        a = next_calkin_wilf(a)
        push!(A, denominator(a))
    end
A end
A002487List(91) |> println # Peter Luschny, Mar 13 2018
  • Magma
    [&+[(Binomial(k, n-k-1) mod 2): k in [0..n]]: n in [0..100]]; // Vincenzo Librandi, Jun 18 2019
  • Maple
    A002487 := proc(n) option remember; if n <= 1 then n elif n mod 2 = 0 then procname(n/2); else procname((n-1)/2)+procname((n+1)/2); fi; end: seq(A002487(n),n=0..91);
A002487 := proc(m) local a,b,n; a := 1; b := 0; n := m; while n>0 do if type(n,odd) then b := a+b else a := a+b end if; n := floor(n/2); end do; b; end proc: seq(A002487(n),n=0..91); # Program adapted from E. Dijkstra, Selected Writings on Computing, Springer, 1982, p. 232. - Igor Urbiha (urbiha(AT)math.hr), Oct 28 2002. Since A007306(n) = a(2*n+1), this program can be adapted for A007306 by replacing b := 0 by b := 1.
A002487 := proc(n::integer) local k; option remember; if n = 0 then 0 elif n=1 then 1 else add(K(k,n-1-k)*procname(n - k), k = 1 .. n) end if end proc:
K := proc(n::integer, k::integer) local KC; if 0 <= k and k <= n and n-k <= 2 then KC:=1; else KC:= 0; end if; end proc: seq(A002487(n),n=0..91); # Thomas Wieder, Jan 13 2008
# next Maple program:
a:= proc(n) option remember; `if`(n<2, n,
      (q-> a(q)+(n-2*q)*a(n-q))(iquo(n, 2)))
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Feb 11 2021
fusc := proc(n) local a, b, c; a := 1; b := 0;
    for c in convert(n, base, 2) do
        if c = 0 then a := a + b else b := a + b fi od;
    b end:
seq(fusc(n), n = 0..91); # Peter Luschny, Nov 09 2022
Stern := proc(n, u) local k, j, b;
    b := j -> nops({seq(Bits:-Xor(k, j-k), k = 0..j)}):
    ifelse(n=0, 1-u, seq(b(j), j = 2^(n-1)-1..2^n-1-u)) end:
seq(print([n], Stern(n, 1)), n = 0..5); # As shown in the comments.
seq(print([n], Stern(n, 0)), n = 0..5); # As shown in the examples. # Peter Luschny, Sep 29 2024
  • Mathematica
    a[0] = 0; a[1] = 1; a[n_] := If[ EvenQ[n], a[n/2], a[(n-1)/2] + a[(n+1)/2]]; Table[ a[n], {n, 0, 100}] (* end of program *)
Onemore[l_] := Transpose[{l, l + RotateLeft[l]}] // Flatten;
NestList[Onemore, {1}, 5] // Flatten  (*gives [a(1), ...]*) (* Takashi Tokita, Mar 09 2003 *)
ToBi[l_] := Table[2^(n - 1), {n, Length[l]}].Reverse[l]; Map[Length,
Split[Sort[Map[ToBi, Table[IntegerDigits[n - 1, 3], {n, 500}]]]]]  (*give [a(1), ...]*) (* Takashi Tokita, Mar 10 2003 *)
A002487[m_] := Module[{a = 1, b = 0, n = m}, While[n > 0, If[OddQ[n], b = a+b, a = a+b]; n = Floor[n/2]]; b]; Table[A002487[n], {n, 0, 100}] (* Jean-François Alcover, Sep 06 2013, translated from 2nd Maple program *)
a[0] = 0; a[1] = 1;
Flatten[Table[{a[2*n] = a[n], a[2*n + 1] = a[n] + a[n + 1]}, {n, 0, 50}]] (* Horst H. Manninger, Jun 09 2021 *)
nmax = 100; CoefficientList[Series[x*Product[(1 + x^(2^k) + x^(2^(k+1))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Oct 08 2022 *)
  • PARI
    {a(n) = n=abs(n); if( n<2, n>0, a(n\2) + if( n%2, a(n\2 + 1)))};
  • PARI
    fusc(n)=local(a=1,b=0);while(n>0,if(bitand(n,1),b+=a,a+=b);n>>=1);b \\ Charles R Greathouse IV, Oct 05 2008
  • PARI
    A002487(n,a=1,b=0)=for(i=0,logint(n,2),if(bittest(n,i),b+=a,a+=b));b \\ M. F. Hasler, Feb 12 2017, updated Feb 14 2019
  • Python
    from functools import lru_cache
@lru_cache(maxsize=None)
def a(n): return n if n<2 else a(n//2) if n%2==0 else a((n - 1)//2) + a((n + 1)//2) # Indranil Ghosh, Jun 08 2017; corrected by Reza K Ghazi, Dec 27 2021
  • Python
    def a(n):
    a, b = 1, 0
    while n > 0:
        if n & 1:
            b += a
        else:
            a += b
        n >>= 1
    return b
# Reza K Ghazi, Dec 29 2021
  • Python
    def A002487(n): return sum(int(not (n-k-1) & ~k) for k in range(n)) # Chai Wah Wu, Jun 19 2022
  • Python
    # (fast way for big vectors)
from math import log, ceil
import numpy
how_many_terms = 2**20  # (Powers of 2 recommended but other integers are also possible.)
A002487, A002487[1]  = numpy.zeros(2**(ce:=ceil(log(how_many_terms,2))), dtype=object), 1
for exponent in range(1,ce):
    L, L2 = 2**exponent, 2**(exponent+1)
    A002487[L2 - 1] = exponent + 1
    A002487[L:L2][::2] = A002487[L >> 1: L]
    A002487[L + 1:L2 - 2][::2] = A002487[L:L2 - 3][::2]  +  A002487[L + 2:L2 - 1][::2]
print(list(A002487[0:100])) # Karl-Heinz Hofmann, Jul 22 2025
  • R
    N <- 50 # arbitrary
a <- 1
for (n in 1:N)
{
  a[2*n    ] = a[n]
  a[2*n + 1] = a[n] + a[n+1]
  a
}
a
# Yosu Yurramendi, Oct 04 2014
  • R
    # Given n, compute a(n) by taking into account the binary representation of n
a <- function(n){
  b <- as.numeric(intToBits(n))
  l <- sum(b)
  m <- which(b == 1)-1
  d <- 1
  if(l > 1) for(j in 1:(l-1)) d[j] <- m[j+1]-m[j]+1
  f <- c(0,1)
  if(l > 1) for(j in 3:(l+1)) f[j] <- d[j-2]*f[j-1]-f[j-2]
  return(f[l+1])
} # Yosu Yurramendi, Dec 13 2016
  • R
    # computes the sequence as a vector A, rather than function a() as above.
A <- c(1,1)
maxlevel <- 5 # by choice
for(m in 1:maxlevel) {
  A[2^(m+1)] <- 1
  for(k in 1:(2^m-1)) {
    r <- m - floor(log2(k)) - 1
    A[2^r*(2*k+1)] <- A[2^r*(2*k)] + A[2^r*(2*k+2)]
}}
A # Yosu Yurramendi, May 08 2018
  • Sage
    def A002487(n):
    M = [1, 0]
    for b in n.bits():
        M[b] = M[0] + M[1]
    return M[1]
print([A002487(n) for n in (0..91)])
# For a dual see A174980. Peter Luschny, Nov 28 2017
  • Scheme
    ;; An implementation of memoization-macro definec can be found for example in: http://oeis.org/wiki/Memoization
(definec (A002487 n) (cond ((<= n 1) n) ((even? n) (A002487 (/ n 2))) (else (+ (A002487 (/ (- n 1) 2)) (A002487 (/ (+ n 1) 2))))))
;; Antti Karttunen, Nov 05 2016

Formula

a(n+1) = (2*k+1)*a(n) - a(n-1) where k = floor(a(n-1)/a(n)). - David S. Newman, Mar 04 2001
Let e(n) = A007814(n) = exponent of highest power of 2 dividing n. Then a(n+1) = (2k+1)*a(n)-a(n-1), n > 0, where k = e(n). Moreover, floor(a(n-1)/a(n)) = e(n), in agreement with D. Newman's formula. - Dragutin Svrtan (dsvrtan(AT)math.hr) and Igor Urbiha (urbiha(AT)math.hr), Jan 10 2002
Calkin and Wilf showed 0.9588 <= limsup a(n)/n^(log(phi)/log(2)) <= 1.1709 where phi is the golden mean. Does this supremum limit = 1? - Benoit Cloitre, Jan 18 2004. Coons and Tyler show the limit is A246765 = 0.9588... - Kevin Ryde, Jan 09 2021
a(n) = Sum_{k=0..floor((n-1)/2)} (binomial(n-k-1, k) mod 2). - Paul Barry, Sep 13 2004
a(n) = Sum_{k=0..n-1} (binomial(k, n-k-1) mod 2). - Paul Barry, Mar 26 2005
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = v^3 + 2*u*v*w - u^2*w. - Michael Somos, May 02 2005
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^3), A(x^6)) where f(u1, u2, u3, u6) = u1^3*u6 - 3*u1^2*u2*u6 + 3*u2^3*u6 - u2^3*u3. - Michael Somos, May 02 2005
G.f.: x * Product_{k>=0} (1 + x^(2^k) + x^(2^(k+1))) [Carlitz].
a(n) = a(n-2) + a(n-1) - 2*(a(n-2) mod a(n-1)). - Mike Stay, Nov 06 2006
A079978(n) = (1 + e^(i*Pi*A002487(n)))/2, i=sqrt(-1). - Paul Barry, Jan 14 2005
a(n) = Sum_{k=1..n} K(k, n-k)*a(n - k), where K(n,k) = 1 if 0 <= k AND k <= n AND n-k <= 2 and K(n,k) = 0 else. (When using such a K-coefficient, several different arguments to K or several different definitions of K may lead to the same integer sequence. For example, if we drop the condition k <= n in the above definition, then we arrive at A002083 = Narayana-Zidek-Capell numbers.) - Thomas Wieder, Jan 13 2008
a(k+1)*a(2^n - k) - a(k)*a(2^n - (k+1)) = 1; a(2^n - k) + a(k) = a(2^(n+1) + k). Both formulas hold for 0 <= k <= 2^n - 1. G.f.: G(z) = a(1) + a(2)*z + a(3)*z^2 + ... + a(k+1)*z^k + ... Define f(z) = (1 + z + z^2), then G(z) = lim f(z)*f(z^2)*f(z^4)* ... *f(z^(2^n))*... = (1 + z + z^2)*G(z^2). - Arie Werksma (werksma(AT)tiscali.nl), Apr 11 2008
a(k+1)*a(2^n - k) - a(k)*a(2^n - (k+1)) = 1 (0 <= k <= 2^n - 1). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
a(2^n + k) = a(2^n - k) + a(k) (0 <= k <= 2^n). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
Let g(z) = a(1) + a(2)*z + a(3)*z^2 + ... + a(k+1)*z^k + ..., f(z) = 1 + z + z^2. Then g(z) = lim_{n->infinity} f(z)*f(z^2)*f(z^4)*...*f(z^(2^n)), g(z) = f(z)*g(z^2). - Arie Werksma (werksma(AT)tiscali.nl), Apr 18 2008
For 0 <= k <= 2^n - 1, write k = b(0) + 2*b(1) + 4*b(2) + ... + 2^(n-1)*b(n-1) where b(0), b(1), etc. are 0 or 1. Define a 2 X 2 matrix X(m) with entries x(1,1) = x(2,2) = 1, x(1,2) = 1 - b(m), x(2,1) = b(m). Let P(n)= X(0)*X(1)* ... *X(n-1). The entries of the matrix P are members of the sequence: p(1,1) = a(k+1), p(1,2) = a(2^n - (k+1)), p(2,1) = a(k), p(2,2) = a(2^n - k). - Arie Werksma (werksma(AT)tiscali.nl), Apr 20 2008
Let f(x) = A030101(x); if 2^n + 1 <= x <= 2^(n + 1) and y = 2^(n + 1) - f(x - 1) then a(x) = a(y). - Arie Werksma (Werksma(AT)Tiscali.nl), Jul 11 2008
a(n) = A126606(n + 1) / 2. - Reikku Kulon, Oct 05 2008
Equals infinite convolution product of [1,1,1,0,0,0,0,0,0] aerated A000079 - 1 times, i.e., [1,1,1,0,0,0,0,0,0] * [1,0,1,0,1,0,0,0,0] * [1,0,0,0,1,0,0,0,1]. - Mats Granvik and Gary W. Adamson, Oct 02 2009; corrected by Mats Granvik, Oct 10 2009
a(2^(p+2)*n+2^(p+1)-1) - a(2^(p+1)*n+2^p-1) = A007306(n+1), p >= 0 and n >= 0. - Johannes W. Meijer, Feb 07 2013
a(2*n-1) = A007306(n), n > 0. - Yosu Yurramendi, Jun 23 2014
a(n*2^m) = a(n), m>0, n > 0. - Yosu Yurramendi, Jul 03 2014
a(k+1)*a(2^m+k) - a(k)*a(2^m+(k+1)) = 1 for m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Nov 07 2014
a(2^(m+1)+(k+1))*a(2^m+k) - a(2^(m+1)+k)*a(2^m+(k+1)) = 1 for m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Nov 07 2014
a(5*2^k) = 3. a(7*2^k) = 3. a(9*2^k) = 4. a(11*2^k) = 5. a(13*2^k) = 5. a(15*2^k) = 4. In general: a((2j-1)*2^k) = A007306(j), j > 0, k >= 0 (see Adamchuk's comment). - Yosu Yurramendi, Mar 05 2016
a(2^m+2^m'+k') = a(2^m'+k')*(m-m'+1) - a(k'), m >= 0, m' <= m-1, 0 <= k' < 2^m'. - Yosu Yurramendi, Jul 13 2016
From Yosu Yurramendi, Jul 13 2016: (Start)
Let n be a natural number and [b_m b_(m-1) ... b_1 b_0] its binary expansion with b_m=1.
Let L = Sum_{i=0..m} b_i be the number of binary digits equal to 1 (L >= 1).
Let {m_j: j=1..L} be the set of subindices such that b_m_j = 1, j=1..L, and 0 <= m_1 <= m_2 <= ... <= m_L = m.
If L = 1 then c_1 = 1, otherwise let {c_j: j=1..(L-1)} be the set of coefficients such that c_(j) = m_(j+1) - m_j + 1, 1 <= j <= L-1.
Let f be a function defined on {1..L+1} such that f(1) = 0, f(2) = 1, f(j) = c_(j-2)*f(j-1) - f(j-2), 3 <= j <= L+1.
Then a(n) = f(L+1) (see example). (End)
a(n) = A001222(A260443(n)) = A000120(A277020(n)). Also a(n) = A000120(A101624(n-1)) for n >= 1. - Antti Karttunen, Nov 05 2016
(a(n-1) + a(n+1))/a(n) = A037227(n) for n >= 1. - Peter Bala, Feb 07 2017
a(0) = 0; a(3n) = 2*A000360(3n-1); a(3n+1) = 2*A000360(3n) - 1; a(3n+2) = 2*A000360(3n+1) + 1. - M. Jeremie Lafitte (Levitas), Apr 24 2017
From I. V. Serov, Jun 14 2017: (Start)
a(n) = A287896(n-1) - 1*A288002(n-1) for n > 1;
a(n) = A007306(n-1) - 2*A288002(n-1) for n > 1. (End)
From Yosu Yurramendi, Feb 14 2018: (Start)
a(2^(m+2) + 2^(m+1) + k) - a(2^(m+1) + 2^m + k) = 2*a(k), m >= 0, 0 <= k < 2^m.
a(2^(m+2) + 2^(m+1) + k) - a(2^(m+1) + k) = a(2^m + k), m >= 0, 0 <= k < 2^m.
a(2^m + k) = a(k)*(m - floor(log_2(k)) - 1) + a(2^(floor(log_2(k))+1) + k), m >= 0, 0 < k < 2^m, a(2^m) = 1, a(0) = 0. (End)
From Yosu Yurramendi, May 08 2018: (Start)
a(2^m) = 1, m >= 0.
a(2^r*(2*k+1)) = a(2^r*(2*k)) + a(2^r*(2*k+2)), r < - m - floor(log_2(k)) - 1, m > 0, 1 <= k < 2^m. (End)
Trow(n) = [card({k XOR (j-k): k=0..j}) for j = 2^(n-1)-1..2^n-2] when regarded as an irregular table (n >= 1). - Peter Luschny, Sep 29 2024
a(n) = A000120(A168081(n)). - Karl-Heinz Hofmann, Jun 16 2025

Extensions

Additional references and comments from Len Smiley, Joshua Zucker, Rick L. Shepherd and Herbert S. Wilf
Typo in definition corrected by Reinhard Zumkeller, Aug 23 2011
Incorrect formula deleted and text edited by Johannes W. Meijer, Feb 07 2013

A260443 Prime factorization representation of Stern polynomials: a(0) = 1, a(1) = 2, a(2n) = A003961(a(n)), a(2n+1) = a(n)*a(n+1).

Original entry on oeis.org

1, 2, 3, 6, 5, 18, 15, 30, 7, 90, 75, 270, 35, 450, 105, 210, 11, 630, 525, 6750, 245, 20250, 2625, 9450, 77, 15750, 3675, 47250, 385, 22050, 1155, 2310, 13, 6930, 5775, 330750, 2695, 3543750, 128625, 1653750, 847, 4961250, 643125, 53156250, 18865, 24806250, 202125, 727650, 143, 1212750, 282975, 57881250, 29645, 173643750, 1414875, 18191250, 1001
Offset: 0

Views

Author

Antti Karttunen, Jul 28 2015

Keywords

Comments

The exponents in the prime factorization of term a(n) give the coefficients of the n-th Stern polynomial. See A125184 and the examples.
None of the terms have prime gaps in their factorization, i.e., all can be found in A073491.
Contains neither perfect squares nor prime powers with exponent > 1. A277701 gives the positions of the terms that are 2*square. - Antti Karttunen, Oct 27 2016
Many of the derived sequences (like A002487) have similar "Fir forest" or "Gaudian cathedrals" style scatter plot. - Antti Karttunen, Mar 21 2017

Examples

			n    a(n)   prime factorization    Stern polynomial
------------------------------------------------------------
0       1   (empty)                B_0(x) = 0
1       2   p_1                    B_1(x) = 1
2       3   p_2                    B_2(x) = x
3       6   p_2 * p_1              B_3(x) = x + 1
4       5   p_3                    B_4(x) = x^2
5      18   p_2^2 * p_1            B_5(x) = 2x + 1
6      15   p_3 * p_2              B_6(x) = x^2 + x
7      30   p_3 * p_2 * p_1        B_7(x) = x^2 + x + 1
8       7   p_4                    B_8(x) = x^3
9      90   p_3 * p_2^2 * p_1      B_9(x) = x^2 + 2x + 1

Links

Crossrefs

Same sequence sorted into ascending order: A260442.
Cf. A000040, A000079, A000225, A001222, A002487, A003415, A003961, A005811, A007949, A046523, A056239, A073491, A090880, A097249, A101979, A125184, A178590, A186891, A206284, A277314, A277315, A277325, A277326, A277329, A277330, A277701, A277705, A277899, A278243, A278530, A278544, A284010, A284011.
Cf. also A048675, A277333 (left inverses).
Cf. A277323, A277324 (bisections), A277200 (even terms sorted), A277197 (first differences), A277198.
Cf. A277316 (values at primes), A277318.
Cf. A023758 (positions of squarefree terms), A101082 (of terms not squarefree), A277702 (positions of records), A277703 (their values).
Cf. A283992, A283993 (number of irreducible, reducible polynomials in range 1 .. n).
Cf. also A206296 (Fibonacci polynomials similarly represented).

Programs

  • Maple
    b:= n-> mul(nextprime(i[1])^i[2], i=ifactors(n)[2]):
a:= proc(n) option remember; `if`(n<2, n+1,
      `if`(irem(n, 2, 'h')=0, b(a(h)), a(h)*a(n-h)))
    end:
seq(a(n), n=0..56);  # Alois P. Heinz, Jul 04 2024
  • Mathematica
    a[n_] := a[n] = Which[n < 2, n + 1, EvenQ@ n, Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[# == 1] &@ a[n/2], True, a[#] a[# + 1] &[(n - 1)/2]]; Table[a@ n, {n, 0, 56}] (* Michael De Vlieger, Apr 05 2017 *)
  • PARI
    A003961(n) = my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); \\ From Michel Marcus
A260443(n) = if(n<2, n+1, if(n%2, A260443(n\2)*A260443(n\2+1), A003961(A260443(n\2)))); \\ After Charles R Greathouse IV's code for "ps" in A186891.
\\ Antti Karttunen, Oct 11 2016
  • Python
    from sympy import factorint, prime, primepi
from functools import reduce
from operator import mul
def a003961(n):
    F = factorint(n)
    return 1 if n==1 else reduce(mul, (prime(primepi(i) + 1)**F[i] for i in F))
def a(n): return n + 1 if n<2 else a003961(a(n//2)) if n%2==0 else a((n - 1)//2)*a((n + 1)//2)
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 21 2017
  • Scheme
    ;; Uses memoization-macro definec:
(definec (A260443 n) (cond ((<= n 1) (+ 1 n)) ((even? n) (A003961 (A260443 (/ n 2)))) (else (* (A260443 (/ (- n 1) 2)) (A260443 (/ (+ n 1) 2))))))
;; A more standalone version added Oct 10 2016, requiring only an implementation of A000040 and the memoization-macro definec:
(define (A260443 n) (product_primes_to_kth_powers (A260443as_coeff_list n)))
(define (product_primes_to_kth_powers nums) (let loop ((p 1) (nums nums) (i 1)) (cond ((null? nums) p) (else (loop (* p (expt (A000040 i) (car nums))) (cdr nums) (+ 1 i))))))
(definec (A260443as_coeff_list n) (cond ((zero? n) (list)) ((= 1 n) (list 1)) ((even? n) (cons 0 (A260443as_coeff_list (/ n 2)))) (else (add_two_lists (A260443as_coeff_list (/ (- n 1) 2)) (A260443as_coeff_list (/ (+ n 1) 2))))))
(define (add_two_lists nums1 nums2) (let ((len1 (length nums1)) (len2 (length nums2))) (cond ((< len1 len2) (add_two_lists nums2 nums1)) (else (map + nums1 (append nums2 (make-list (- len1 len2) 0)))))))

Formula

a(0) = 1, a(1) = 2, a(2n) = A003961(a(n)), a(2n+1) = a(n)*a(n+1).
Other identities. For all n >= 0:
A001221(a(n)) = A277314(n). [#nonzero coefficients in each polynomial.]
A001222(a(n)) = A002487(n). [When each polynomial is evaluated at x=1.]
A048675(a(n)) = n. [at x=2.]
A090880(a(n)) = A178590(n). [at x=3.]
A248663(a(n)) = A264977(n). [at x=2 over the field GF(2).]
A276075(a(n)) = A276081(n). ["at factorials".]
A156552(a(n)) = A277020(n). [Converted to "unary-binary" encoding.]
A051903(a(n)) = A277315(n). [Maximal coefficient.]
A277322(a(n)) = A277013(n). [Number of irreducible polynomial factors.]
A005361(a(n)) = A277325(n). [Product of nonzero coefficients.]
A072411(a(n)) = A277326(n). [And their LCM.]
A007913(a(n)) = A277330(n). [The squarefree part.]
A000005(a(n)) = A277705(n). [Number of divisors.]
A046523(a(n)) = A278243(n). [Filter-sequence.]
A284010(a(n)) = A284011(n). [True for n > 1. Another filter-sequence.]
A003415(a(n)) = A278544(n). [Arithmetic derivative.]
A056239(a(n)) = A278530(n). [Weighted sum of coefficients.]
A097249(a(n)) = A277899(n).
a(A000079(n)) = A000040(n+1).
a(A000225(n)) = A002110(n).
a(A000051(n)) = 3*A002110(n).
For n >= 1, a(A000918(n)) = A070826(n).
A007949(a(n)) is the interleaving of A000035 and A005811, probably A101979.
A061395(a(n)) = A277329(n).
Also, for all n >= 1:
A055396(a(n)) = A001511(n).
A252735(a(n)) = A061395(a(n)) - 1 = A057526(n).
a(A000040(n)) = A277316(n).
a(A186891(1+n)) = A277318(n). [Subsequence for irreducible polynomials].

Extensions

More linking formulas added by Antti Karttunen, Mar 21 2017

A186891 Numbers n such that the Stern polynomial B(n,x) is irreducible.

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 37, 41, 43, 47, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 121, 125, 127, 131, 133, 137, 139, 143, 145, 149, 151, 157, 161, 163, 167, 169, 173, 175, 179, 181, 185, 191, 193, 197, 199
Offset: 1

Views

Author

T. D. Noe, Feb 28 2011

Keywords

Comments

Ulas and Ulas conjecture that all primes are here. The nonprime n are in A186892. See A186886 for the least number having n prime factors.

Links

Crossrefs

Cf. A057526 (degree of Stern polynomials), A125184, A260443 (Stern polynomials).
Cf. A186892 (subsequence of nonprime terms).
Cf. A186893 (subsequence for self-reciprocal polynomials).
Positions of 0 and 1's in A277013, Positions of 1 and 2's in A284011.
Cf. A283991 (characteristic function for terms > 1).
Cf. also A186886, A277190, A277318, A283992.

Programs

  • Mathematica
    ps[n_] := ps[n] = If[n<2, n, If[OddQ[n], ps[Quotient[n, 2]] + ps[Quotient[n, 2] + 1], x ps[Quotient[n, 2]]]];
selQ[n_] := IrreduciblePolynomialQ[ps[n]];
Join[{1}, Select[Range[200], selQ]] (* Jean-François Alcover, Nov 02 2018, translated from PARI *)
  • PARI
    ps(n)=if(n<2, n, if(n%2, ps(n\2)+ps(n\2+1), 'x*ps(n\2)))
is(n)=polisirreducible(ps(n)) \\ Charles R Greathouse IV, Apr 07 2015

Formula

From Antti Karttunen, Mar 21 2017: (Start)
A283992(a(1+n)) = n.
A260443(a(1+n)) = A277318(n).
(End)

A264977 a(0) = 0, a(1) = 1, a(2*n) = 2*a(n), a(2*n+1) = a(n) XOR a(n+1).

Original entry on oeis.org

0, 1, 2, 3, 4, 1, 6, 7, 8, 5, 2, 7, 12, 1, 14, 15, 16, 13, 10, 7, 4, 5, 14, 11, 24, 13, 2, 15, 28, 1, 30, 31, 32, 29, 26, 7, 20, 13, 14, 3, 8, 1, 10, 11, 28, 5, 22, 19, 48, 21, 26, 15, 4, 13, 30, 19, 56, 29, 2, 31, 60, 1, 62, 63, 64, 61, 58, 7, 52, 29, 14, 19, 40, 25, 26, 3, 28, 13, 6, 11, 16, 9, 2, 11, 20, 1, 22
Offset: 0

Views

Author

Antti Karttunen, Dec 10 2015

Keywords

Comments

a(n) is the n-th Stern polynomial (cf. A260443, A125184) evaluated at X = 2 over the field GF(2).
For n >= 1, a(n) gives the index of the row where n occurs in array A277710.

Examples

			In this example, binary numbers are given zero-padded to four bits.
a(2) = 2a(1) = 2 * 1 = 2.
a(3) = a(1) XOR a(2) = 1 XOR 2 = 0001 XOR 0010 = 0011 = 3.
a(4) = 2a(2) = 2 * 2 = 4.
a(5) = a(2) XOR a(3) = 2 XOR 3 = 0010 XOR 0011 = 0001 = 1.
a(6) = 2a(3) = 2 * 3 = 6.
a(7) = a(3) XOR a(4) = 3 XOR 4 = 0011 XOR 0100 = 0111 = 7.

Links

Crossrefs

Cf. A000225, A001511, A002487, A003987, A010060, A011655, A048675, A055396, A125184, A248663, A260443, A265397, A277330.
Cf. A023758 (the fixed points).
Cf. also A068156, A168081, A265407.
Cf. A277700 (binary weight of terms).
Cf. A277701, A277712, A277713 (positions of 1's, 2's and 3's in this sequence).
Cf. A277711 (position of the first occurrence of each n in this sequence).
Cf. also arrays A277710, A099884.

Programs

  • Mathematica
    recurXOR[0] = 0; recurXOR[1] = 1; recurXOR[n_] := recurXOR[n] = If[EvenQ[n], 2recurXOR[n/2], BitXor[recurXOR[(n - 1)/2 + 1], recurXOR[(n - 1)/2]]]; Table[recurXOR[n], {n, 0, 100}] (* Jean-François Alcover, Oct 23 2016 *)
  • Python
    class Memoize:
    def _init_(self, func):
        self.func=func
        self.cache={}
    def _call_(self, arg):
        if arg not in self.cache:
            self.cache[arg] = self.func(arg)
        return self.cache[arg]
@Memoize
def a(n): return n if n<2 else 2*a(n//2) if n%2==0 else a((n - 1)//2)^a((n + 1)//2)
print([a(n) for n in range(51)]) # Indranil Ghosh, Jul 27 2017

Formula

a(0) = 0, a(1) = 1, a(2*n) = 2*a(n), a(2*n+1) = a(n) XOR a(n+1).
a(n) = A248663(A260443(n)).
a(n) = A048675(A277330(n)). - Antti Karttunen, Oct 27 2016
Other identities. For all n >= 0:
a(n) = n - A265397(n).
From Antti Karttunen, Oct 28 2016: (Start)
A000035(a(n)) = A000035(n). [Preserves the parity of n.]
A010873(a(n)) = A010873(n). [a(n) mod 4 = n mod 4.]
A001511(a(n)) = A001511(n) = A055396(A277330(n)). [In general, the 2-adic valuation of n is preserved.]
A010060(a(n)) = A011655(n).
a(n) <= n.
For n >= 2, a((2^n)+1) = (2^n) - 3.
The following two identities are so far unproved:
For n >= 2, a(3*A000225(n)) = a(A068156(n)) = 5.
For n >= 2, a(A068156(n)-2) = A062709(n) = 2^n + 3.
(End)

A277324 Odd bisection of A260443 (the even terms): a(n) = A260443((2*n)+1).

Original entry on oeis.org

2, 6, 18, 30, 90, 270, 450, 210, 630, 6750, 20250, 9450, 15750, 47250, 22050, 2310, 6930, 330750, 3543750, 1653750, 4961250, 53156250, 24806250, 727650, 1212750, 57881250, 173643750, 18191250, 8489250, 25467750, 2668050, 30030, 90090, 40020750, 1910081250, 891371250, 9550406250, 455814843750, 212713593750
Offset: 0

Views

Author

Antti Karttunen, Oct 10 2016

Keywords

Comments

From David A. Corneth, Oct 22 2016: (Start)
The exponents of the prime factorization of a(n) are first nondecreasing, then nonincreasing.
The exponent of 2 in the prime factorization of a(n) is 1. (End)

Examples

			A method to find terms of this sequence, explained by an example to find a(7). To find k = a(7), we find k such that A048675(k) = 2*7+1 = 15. 7 has the binary partitions: {[7, 0, 0], [5, 1, 0], [3, 2, 0], [1, 3, 0], [3, 0, 1], [1, 1, 1]}. To each of those, we prepend a 1. This gives the binary partitions of 15 starting with a 1. For example, for the first we get [1, 7, 0, 0]. We see that only [1, 5, 1, 0], [1, 3, 2, 0] and [1, 1, 1, 1] start nondecreasing, then nonincreasing, so we only check those. These numbers will be the exponents in a prime factorization. [1, 5, 1, 0] corresponds to prime(1)^1 * prime(2)^5 * prime(3)^1 * prime(4)^0 = 2430. We find that [1, 1, 1, 1] gives k = 210 for which A048675(k) = 15 so a(7) = 210. - _David A. Corneth_, Oct 22 2016

Links

Crossrefs

Cf. A005811, A005940, A007306, A007949, A156552, A260443, A277189, A277323, A283484, A283975, A284267, A284268, A284563, A284564, A284573.
Cf. A277200 (same sequence sorted into ascending order).

Programs

  • Mathematica
    a[n_] := a[n] = Which[n < 2, n + 1, EvenQ@ n, Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[# == 1] &@ a[n/2], True, a[#] a[# + 1] &[(n - 1)/2]]; Table[a[2 n + 1], {n, 0, 38}] (* Michael De Vlieger, Apr 05 2017 *)
  • Python
    from sympy import factorint, prime, primepi
from operator import mul
def a003961(n):
    F=factorint(n)
    return 1 if n==1 else reduce(mul, [prime(primepi(i) + 1)**F[i] for i in F])
def a260443(n): return n + 1 if n<2 else a003961(a260443(n//2)) if n%2==0 else a260443((n - 1)//2)*a260443((n + 1)//2)
def a(n): return a260443(2*n + 1)
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 21 2017

Formula

a(n) = A260443((2*n)+1).
a(0) = 2; for n >= 1, a(n) = A260443(n) * A260443(n+1).
Other identities. For all n >= 0:
A007949(a(n)) = A005811(n). [See comments in A125184.]
A156552(a(n)) = A277189(n), a(n) = A005940(1+A277189(n)).
A048675(a(n)) = 2n + 1. - David A. Corneth, Oct 22 2016
A001222(a(n)) = A007306(1+n).
A056169(a(n)) = A284267(n).
A275812(a(n)) = A284268(n).
A248663(a(n)) = A283975(n).
A000188(a(n)) = A283484(n).
A247503(a(n)) = A284563(n).
A248101(a(n)) = A284564(n).
A046523(a(n)) = A284573(n).
a(n) = A277198(n) * A284008(n).
a(n) = A284576(n) * A284578(n) = A284577(n) * A000290(A284578(n)).

Extensions

More linking formulas added by Antti Karttunen, Apr 16 2017

A286378 Restricted growth sequence computed for Stern-polynomial related filter-sequence A278243.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 3, 5, 2, 6, 4, 7, 3, 8, 5, 9, 2, 10, 6, 11, 4, 12, 7, 13, 3, 13, 8, 14, 5, 15, 9, 16, 2, 17, 10, 18, 6, 19, 11, 20, 4, 21, 12, 22, 7, 23, 13, 24, 3, 24, 13, 25, 8, 26, 14, 27, 5, 28, 15, 29, 9, 30, 16, 31, 2, 32, 17, 33, 10, 34, 18, 35, 6, 36, 19, 37, 11, 38, 20, 39, 4, 40, 21, 41, 12, 42, 22, 43, 7, 44, 23, 45, 13, 46, 24, 47, 3, 47, 24, 48
Offset: 0

Views

Author

Antti Karttunen, May 09 2017

Keywords

Comments

Construction: we start with a(0)=1 for A278243(0)=1, and then after, for n > 0, we use the least unused natural number k for a(n) if A278243(n) has not been encountered before, otherwise [whenever A278243(n) = A278243(m), for some m < n], we set a(n) = a(m).
When filtering sequences (by equivalence class partitioning), this sequence (with its modestly sized terms) can be used instead of A278243, because for all i, j it holds that: a(i) = a(j) <=> A278243(i) = A278243(j).
For example, for all i, j: a(i) = a(j) => A002487(i) = A002487(j).
For pairs of distinct primes p, q for which a(p) = a(q) see comments in A317945. - Antti Karttunen, Aug 12 2018

Examples

			For n=1, A278243(1) = 2, which has not been encountered before, thus we allot for a(1) the least so far unused number, which is 2, thus a(1) = 2.
For n=2, A278243(2) = 2, which was already encountered as A278243(1), thus we set a(2) = a(1) = 2.
For n=3, A278243(3) = 6, which has not been encountered before, thus we allot for a(3) the least so far unused number, which is 3, thus a(3) = 3.
For n=23, A278243(23) = 2520, which has not been encountered before, thus we allot for a(23) the least so far unused number, which is 13, thus a(23) = 3.
For n=25, A278243(25) = 2520, which was already encountered at n=23, thus we set a(25) = a(23) = 13.

Links

Crossrefs

Cf. A002487, A125184, A260443, A278243, A286377.
Cf. also A101296, A286603, A286605, A286610, A286619, A286621, A286622, A286626 for similarly constructed sequences.
Cf. also A317942, A317943, A317944, A317945.
Differs from A103391(1+n) for the first time at n=25, where a(25)=13, while A103391(26) = 14.

Programs

  • Mathematica
    a[n_] := a[n] = Which[n < 2, n + 1, EvenQ@ n, Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[# == 1] &@ a[n/2], True, a[#] a[# + 1] &[(n - 1)/2]]; With[{nn = 100}, Function[s, Table[Position[Keys@ s, k_ /; MemberQ[k, n]][[1, 1]], {n, nn}]]@ Map[#1 -> #2 & @@ # &, Transpose@ {Values@ #, Keys@ #}] &@ PositionIndex@ Table[Times @@ MapIndexed[Prime[First@#2]^#1 &, Sort[FactorInteger[#][[All, -1]], Greater]] - Boole[# == 1] &@ a@ n, {n, 0, nn}]] (* Michael De Vlieger, May 12 2017 *)
  • PARI
    up_to = 65537;
rgs_transform(invec) = { my(occurrences = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(occurrences,invec[i]), my(pp = mapget(occurrences, invec[i])); outvec[i] = outvec[pp] , mapput(occurrences,invec[i],i); outvec[i] = u; u++ )); outvec; };
write_to_bfile(start_offset,vec,bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)-1, " ", vec[n])); }
A046523(n) = { my(f=vecsort(factor(n)[, 2], , 4), p); prod(i=1, #f, (p=nextprime(p+1))^f[i]); };  \\ From A046523
A003961(n) = my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); \\ From A003961
A260443(n) = if(n<2, n+1, if(n%2, A260443(n\2)*A260443(n\2+1), A003961(A260443(n\2))));
A278243(n) = A046523(A260443(n));
v286378 = rgs_transform(vector(up_to+1,n,A278243(n-1)));
A286378(n) = v286378[1+n];

A277013 a(n) = number of irreducible polynomial factors (counted with multiplicity) in the n-th Stern polynomial B(n,t).

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 3, 1, 2, 2, 4, 1, 3, 1, 3, 2, 2, 1, 4, 1, 2, 3, 3, 1, 3, 1, 5, 2, 2, 2, 4, 1, 2, 2, 4, 1, 3, 1, 3, 3, 2, 1, 5, 2, 2, 2, 3, 1, 4, 1, 4, 2, 2, 1, 4, 1, 2, 3, 6, 1, 3, 1, 3, 2, 3, 1, 5, 1, 2, 3, 3, 1, 3, 1, 5, 2, 2, 1, 4, 2, 2, 2, 4, 1, 4, 1, 3, 2, 2, 1, 6, 1, 3, 3, 3, 1, 3, 1, 4, 3, 2, 1, 5, 1, 2, 2, 5, 1, 3, 1, 3, 2, 2, 2, 5
Offset: 1

Views

Author

Antti Karttunen, Oct 07 2016

Keywords

Examples

			B(11,t) = t^2 + 3t + 1 which is irreducible, so a(11) = 1.
B(12,t) = t^3 + t^2 = t^2(t+1), so a(12) = 3.

Links

Crossrefs

Cf. A186891 (positions of 0 and 1's in this sequence), A277027 (terms squared).
Cf. A000079, A125184, A277322, A260443.
Differs from A001222 for the first time at n=25, where a(25)=1. A277190 gives the positions of differing terms.

Programs

Formula

a(n) = A277322(A260443(n)).
It seems that for all n >= 1, a(2^n) = n.
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