cp's OEIS Frontend

Leibniz's series: Pi/4 = Sum_{n>=0} (-1)^n/(2n+1) (cf. A072172).

Beginning of the ordering of the natural numbers used in Sharkovski's theorem - see the Cielsielski-Pogoda paper.

The Sharkovski ordering begins with the odd numbers >= 3, then twice these numbers, then 4 times them, then 8 times them, etc., ending with the powers of 2 in decreasing order, ending with 2^0 = 1.

Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0(6).

Also continued fraction for coth(1) (A073747 is decimal expansion). - Rick L. Shepherd, Aug 07 2002

a(1) = 1; a(n) is the smallest number such that a(n) + a(i) is composite for all i = 1 to n-1. - Amarnath Murthy, Jul 14 2003

Smallest number greater than n, not a multiple of n, but containing it in binary representation. - Reinhard Zumkeller, Oct 06 2003

Numbers n such that phi(2n) = phi(n), where phi is Euler's totient (A000010). - Lekraj Beedassy, Aug 27 2004

Pi*sqrt(2)/4 = Sum_{n>=0} (-1)^floor(n/2)/(2n+1) = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 ... [since periodic f(x)=x over -Pi < x < Pi = 2(sin(x)/1 - sin(2x)/2 + sin(3x)/3 - ...) using x = Pi/4 (Maor)]. - Gerald McGarvey, Feb 04 2005

For n > 1, numbers having 2 as an anti-divisor. - Alexandre Wajnberg, Oct 02 2005

a(n) = shortest side a of all integer-sided triangles with sides a <= b <= c and inradius n >= 1.

First differences of squares (A000290). - Lekraj Beedassy, Jul 15 2006

The odd numbers are the solution to the simplest recursion arising when assuming that the algorithm "merge sort" could merge in constant unit time, i.e., T(1):= 1, T(n):= T(floor(n/2)) + T(ceiling(n/2)) + 1. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 14 2006

2n-5 counts the permutations in S_n which have zero occurrences of the pattern 312 and one occurrence of the pattern 123. - David Hoek (david.hok(AT)telia.com), Feb 28 2007

For n > 0: number of divisors of (n-1)th power of any squarefree semiprime: a(n) = A000005(A001248(k)^(n-1)); a(n) = A000005(A000302(n-1)) = A000005(A001019(n-1)) = A000005(A009969(n-1)) = A000005(A087752(n-1)). - Reinhard Zumkeller, Mar 04 2007

For n > 2, a(n-1) is the least integer not the sum of < n n-gonal numbers (0 allowed). - Jonathan Sondow, Jul 01 2007

A134451(a(n)) = abs(A134452(a(n))) = 1; union of A134453 and A134454. - Reinhard Zumkeller, Oct 27 2007

Numbers n such that sigma(2n) = 3*sigma(n). - Farideh Firoozbakht, Feb 26 2008

a(n) = A139391(A016825(n)) = A006370(A016825(n)). - Reinhard Zumkeller, Apr 17 2008

Number of divisors of 4^(n-1) for n > 0. - J. Lowell, Aug 30 2008

Equals INVERT transform of A078050 (signed - cf. comments); and row sums of triangle A144106. - Gary W. Adamson, Sep 11 2008

Odd numbers(n) = 2*n+1 = square pyramidal number(3*n+1) / triangular number(3*n+1). - Pierre CAMI, Sep 27 2008

A000035(a(n))=1, A059841(a(n))=0. - Reinhard Zumkeller, Sep 29 2008

Multiplicative closure of A065091. - Reinhard Zumkeller, Oct 14 2008

a(n) is also the maximum number of triangles that n+2 points in the same plane can determine. 3 points determine max 1 triangle; 4 points can give 3 triangles; 5 points can give 5; 6 points can give 7 etc. - Carmine Suriano, Jun 08 2009

Binomial transform of A130706, inverse binomial transform of A001787(without the initial 0). - Philippe Deléham, Sep 17 2009

Also the 3-rough numbers: positive integers that have no prime factors less than 3. - Michael B. Porter, Oct 08 2009

Or n without 2 as prime factor. - Juri-Stepan Gerasimov, Nov 19 2009

Given an L(2,1) labeling l of a graph G, let k be the maximum label assigned by l. The minimum k possible over all L(2,1) labelings of G is denoted by lambda(G). For n > 0, this sequence gives lambda(K_{n+1}) where K_{n+1} is the complete graph on n+1 vertices. - K.V.Iyer, Dec 19 2009

A176271 = odd numbers seen as a triangle read by rows: a(n) = A176271(A002024(n+1), A002260(n+1)). - Reinhard Zumkeller, Apr 13 2010

For n >= 1, a(n-1) = numbers k such that arithmetic mean of the first k positive integers is an integer. A040001(a(n-1)) = 1. See A145051 and A040001. - Jaroslav Krizek, May 28 2010

Union of A179084 and A179085. - Reinhard Zumkeller, Jun 28 2010

For n>0, continued fraction [1,1,n] = (n+1)/a(n); e.g., [1,1,7] = 8/15. - Gary W. Adamson, Jul 15 2010

Numbers that are the sum of two sequential integers. - Dominick Cancilla, Aug 09 2010

Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n + (h-4)*(-1)^n - h)/4 (h and n in A000027), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 4). Also a(n)^2 - 1 == 0 (mod 8). - Bruno Berselli, Nov 17 2010

A004767 = a(a(n)). - Reinhard Zumkeller, Jun 27 2011

A001227(a(n)) = A000005(a(n)); A048272(a(n)) < 0. - Reinhard Zumkeller, Jan 21 2012

a(n) is the minimum number of tosses of a fair coin needed so that the probability of more than n heads is at least 1/2. In fact, Sum_{k=n+1..2n+1} Pr(k heads|2n+1 tosses) = 1/2. - Dennis P. Walsh, Apr 04 2012

A007814(a(n)) = 0; A037227(a(n)) = 1. - Reinhard Zumkeller, Jun 30 2012

1/N (i.e., 1/1, 1/2, 1/3, ...) = Sum_{j=1,3,5,...,infinity} k^j, where k is the infinite set of constants 1/exp.ArcSinh(N/2) = convergents to barover(N). The convergent to barover(1) or [1,1,1,...] = 1/phi = 0.6180339..., whereas c.f. barover(2) converges to 0.414213..., and so on. Thus, with k = 1/phi we obtain 1 = k^1 + k^3 + k^5 + ..., and with k = 0.414213... = (sqrt(2) - 1) we get 1/2 = k^1 + k^3 + k^5 + .... Likewise, with the convergent to barover(3) = 0.302775... = k, we get 1/3 = k^1 + k^3 + k^5 + ..., etc. - Gary W. Adamson, Jul 01 2012

Conjecture on primes with one coach (A216371) relating to the odd integers: iff an integer is in A216371 (primes with one coach either of the form 4q-1 or 4q+1, (q > 0)); the top row of its coach is composed of a permutation of the first q odd integers. Example: prime 19 (q = 5), has 5 terms in each row of its coach: 19: [1, 9, 5, 7, 3] ... [1, 1, 1, 2, 4]. This is interpreted: (19 - 1) = (2^1 * 9), (19 - 9) = (2^1 * 5), (19 - 5) = (2^1 - 7), (19 - 7) = (2^2 * 3), (19 - 3) = (2^4 * 1). - Gary W. Adamson, Sep 09 2012

A005408 is the numerator 2n-1 of the term (1/m^2 - 1/n^2) = (2n-1)/(mn)^2, n = m+1, m > 0 in the Rydberg formula, while A035287 is the denominator (mn)^2. So the quotient a(A005408)/a(A035287) simulates the Hydrogen spectral series of all hydrogen-like elements. - Freimut Marschner, Aug 10 2013

This sequence has unique factorization. The primitive elements are the odd primes (A065091). (Each term of the sequence can be expressed as a product of terms of the sequence. Primitive elements have only the trivial factorization. If the products of terms of the sequence are always in the sequence, and there is a unique factorization of each element into primitive elements, we say that the sequence has unique factorization. So, e.g., the composite numbers do not have unique factorization, because for example 36 = 4*9 = 6*6 has two distinct factorizations.) - Franklin T. Adams-Watters, Sep 28 2013

These are also numbers k such that (k^k+1)/(k+1) is an integer. - Derek Orr, May 22 2014

a(n-1) gives the number of distinct sums in the direct sum {1,2,3,..,n} + {1,2,3,..,n}. For example, {1} + {1} has only one possible sum so a(0) = 1. {1,2} + {1,2} has three distinct possible sums {2,3,4} so a(1) = 3. {1,2,3} + {1,2,3} has 5 distinct possible sums {2,3,4,5,6} so a(2) = 5. - Derek Orr, Nov 22 2014

The number of partitions of 4*n into at most 2 parts. - Colin Barker, Mar 31 2015

a(n) is representable as a sum of two but no fewer consecutive nonnegative integers, e.g., 1 = 0 + 1, 3 = 1 + 2, 5 = 2 + 3, etc. (see A138591). - Martin Renner, Mar 14 2016

Unique solution a( ) of the complementary equation a(n) = a(n-1)^2 - a(n-2)*b(n-1), where a(0) = 1, a(1) = 3, and a( ) and b( ) are increasing complementary sequences. - Clark Kimberling, Nov 21 2017

Also the number of maximal and maximum cliques in the n-centipede graph. - Eric W. Weisstein, Dec 01 2017

Lexicographically earliest sequence of distinct positive integers such that the average of any number of consecutive terms is always an integer. (For opposite property see A042963.) - Ivan Neretin, Dec 21 2017

Maximum number of non-intersecting line segments between vertices of a convex (n+2)-gon. - Christoph B. Kassir, Oct 21 2022

a(n) is the number of parking functions of size n+1 avoiding the patterns 123, 132, and 231. - Lara Pudwell, Apr 10 2023

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

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

Twice the odd numbers, also called singly even numbers.

Numbers having equal numbers of odd and even divisors: A001227(a(n)) = A000005(2*a(n)). - Reinhard Zumkeller, Dec 28 2003

Continued fraction for coth(1/2) = (e+1)/(e-1). The continued fraction for tanh(1/2) = (e-1)/(e+1) would be a(0) = 0, a(n) = A016825(n-1), n >= 1.

No solutions to a(n) = b^2 - c^2. - Henry Bottomley, Jan 13 2001

Sequence gives m such that 8 is the largest power of 2 dividing A003629(k)^m-1 for any k. - Benoit Cloitre, Apr 05 2002

k such that Sum_{d|k} (-1)^d = A048272(k) = 0. - Benoit Cloitre, Apr 15 2002

Also k such that Sum_{d|k} phi(d)*mu(k/d) = A007431(k) = 0. - Benoit Cloitre, Apr 15 2002

Also k such that Sum_{d|k} (d/A000005(d))*mu(k/d) = 0, k such that Sum_{d|k}(A000005(d)/d)*mu(k/d) = 0. - Benoit Cloitre, Apr 19 2002

Solutions to phi(x) = phi(x/2); primorial numbers are here. - Labos Elemer, Dec 16 2002

Together with 1, numbers that are not the leg of a primitive Pythagorean triangle. - Lekraj Beedassy, Nov 25 2003

For n > 0: complement of A107750 and A023416(a(n)-1) = A023416(a(n)) <> A023416(a(n)+1). - Reinhard Zumkeller, May 23 2005

Also the minimal value of Sum_{i=1..n+2} (p(i) - p(i+1))^2, where p(n+3) = p(1), as p ranges over all permutations of {1,2,...,n+2} (see the Mihai reference). Example: a(2)=10 because the values of the sum for the permutations of {1,2,3,4} are 10 (8 times), 12 (8 times) and 18 (8 times). - Emeric Deutsch, Jul 30 2005

Except for a(n)=2, numbers having 4 as an anti-divisor. - Alexandre Wajnberg, Oct 02 2005

A139391(a(n)) = A006370(a(n)) = A005408(n). - Reinhard Zumkeller, Apr 17 2008

Also a(n) = (n-1) + n + (n+1) + (n+2), so a(n) and -a(n) are all the integers that are sums of four consecutive integers. - Rick L. Shepherd, Mar 21 2009

The denominator in Pi/8 = 1/2 - 1/6 + 1/10 - 1/14 + 1/18 - 1/22 + .... - Mohammad K. Azarian, Oct 13 2011

This sequence gives the positive zeros of i^x + 1 = 0, x real, where i^x = exp(i*x*Pi/2). - Ilya Gutkovskiy, Aug 08 2015

Numbers k such that Sum_{j=1..k} j^3 is not a multiple of k. - Chai Wah Wu, Aug 23 2017

Numbers k such that Lucas(k) is a multiple of 3. - Bruno Berselli, Oct 17 2017

Also numbers k such that t^k == -1 (mod 5), where t is a term of A047221. - Bruno Berselli, Dec 28 2017

The even numbers form a ring, and these are the primes in that ring. Note that unique factorization into primes does not hold, since 60 = 2*30 = 6*10. - N. J. A. Sloane, Nov 11 2019

Also numbers ending with 10 in base 2. - John Keith, May 09 2022

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0(65).

n such that 16 is the largest power of 2 dividing A003629(k)^n - 1 for any k. - Benoit Cloitre, Mar 23 2002

Continued fraction expansion of tanh(1/4). - Benoit Cloitre, Dec 17 2002

Consider all primitive Pythagorean triples (a,b,c) with c - a = 8, sequence gives values for b. (Corresponding values for a are A078371(n), while c follows A078370(n).) - Lambert Klasen (Lambert.Klasen(AT)gmx.net), Nov 19 2004

Also numbers of the form a^2 + b^2 + c^2 + d^2, where a,b,c,d are odd integers. - Alexander Adamchuk, Dec 01 2006

If X is an n-set and Y_i (i=1,2,3) mutually disjoint 2-subsets of X then a(n-5) is equal to the number of 4-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Aug 26 2007

A007814(a(n)) = 2; A037227(a(n)) = 5. - Reinhard Zumkeller, Jun 30 2012

Numbers k such that 3^k + 1 is divisible by 41. - Bruno Berselli, Aug 22 2018

Lexicographically smallest arithmetic progression of positive integers avoiding Fibonacci numbers. - Paolo Xausa, May 08 2023

From Martin Renner, May 24 2024: (Start)

Also number of points in a grid cross with equally long arms and a width of two points, e.g.:

* *

* * * *

* * * * * *

* * * * * * * * * * * * * * * * * * * *

* * * * * * * * * * * * * * * * * * * *

* * * * * *

* * * *

* *

etc. (End)

The a(n) appeared in the analysis of A220002, a sequence related to the Catalan numbers.

The first Maple program makes use of a program by Peter Luschny for the calculation of the a(n) values. The second Maple program shows that this sequence has a beautiful internal structure, see the first formula, while the third Maple program makes optimal use of this internal structure for the fast calculation of a(n) values for large n.

The cross references lead to sequences that have the same internal structure as this sequence.

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0(97).

n such that 32 is the largest power of 2 dividing A003629(k)^n-1 for any k. - Benoit Cloitre, Mar 23 2002

Continued fraction expansion of tanh(1/8). - Benoit Cloitre, Dec 17 2002

If Y and Z are 2-blocks of a (4n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting both Y and Z. - Milan Janjic, Oct 28 2007

General form: (q*n+x)*q x=+1; q=2=A016825, q=3=A017197, q=4=A119413, ... x=-1; q=3=A017233, q=4=A098502, ... x=+2; q=4=A051062, ... - Vladimir Joseph Stephan Orlovsky, Feb 16 2009

a(n)*n+1 = (4n+1)^2 and a(n)*(n+1)+1 = (4n+3)^2 are both perfect squares. - Carmine Suriano, Jun 01 2014

For all positive integers n, there are infinitely many positive integers k such that k*n + 1 and k*(n+1) + 1 are both perfect squares. Except for 8, all the numbers of this sequence are the smallest integers k which are solutions for getting two perfect squares. Example: a(1) = 24 and 24 * 1 + 1 = 25 = 5^2, then 24 * (1+1) + 1 = 49 = 7^2. [Reference AMM] - Bernard Schott, Sep 24 2017

Numbers k such that 3^k + 1 is divisible by 17*193. - Bruno Berselli, Aug 22 2018

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) = A002487(n), the number of '-'-signs in the chf(n) word is c-fusc(n) = a(n) (the current sequence) and the number of '+'-signs in the chf(n) word is s-fusc(n) = A287730(n). See examples below.

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) = A002487(n), the number of '-'-signs in the chf(n) word is c-fusc(n) = A287729(n) and the number of '+'-signs in the chf(n) word is the current sequence a(n) = s-fusc(n). See examples below.

From Andrew Howroyd, Jul 25 2018: (Start)

Moebius transform of A037227.

Multiplicative because A037227 is. (End)