cp's OEIS Frontend

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

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)

Central terms give A118415; row sums give A118414;

T(n,1) = A005408(n-1);

T(n,2) = A016825(n-1) for n>1;

T(n,3) = A017113(n-1) for n>2;

T(n,4) = A051062(n-1) for n>3;

T(n,n-2) = A052951(n-1) for n>2;

T(n,n) = A014480(n-1) = A118416(n,n);

A001511(T(n,k)) = A002260(n,k);

A003602(T(n,k)) = A002024(n,k).

G.f.: x*y*(1 + x + 2*x*y - 6*x^2*y)/((1 - x)^2*(1 - 2*x*y)^2). - Stefano Spezia, Dec 22 2024

As a rectangle, the accumulation array of A051340.

From Clark Kimberling, Feb 05 2011: (Start)

Here all the weights are divided by two where they aren't in Cahn.

As a rectangle, A141419 is in the accumulation chain

... < A051340 < A141419 < A185874 < A185875 < A185876 < ...

(See A144112 for the definition of accumulation array.)

row 1: A000027

col 1: A000217

diag (1,5,...): A000326 (pentagonal numbers)

diag (2,7,...): A005449 (second pentagonal numbers)

diag (3,9,...): A045943 (triangular matchstick numbers)

diag (4,11,...): A115067

diag (5,13,...): A140090

diag (6,15,...): A140091

diag (7,17,...): A059845

diag (8,19,...): A140672

(End)

Let N=2*n+1 and k=1,2,...,n. Let A_{N,n-1} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1], an n X n unit-primitive matrix (see [Jeffery]). Let M_n=[A_{N,n-1}]^4. Then t(n,k)=[M_n](1,k), that is, the n-th row of the triangle is given by the first row of M_n. - _L. Edson Jeffery, Nov 20 2011

Conjecture. Let N=2*n+1 and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n-1} be the n X n unit-primitive matrices (again see [Jeffery]) associated with N, and define the Chebyshev polynomials of the second kind by the recurrence U_0(x) = 1, U_1(x) = 2*x and U_r(x) = 2*x*U_(r-1)(x) - U_(r-2)(x) (r>1). Define the column vectors V_(k-1) = (U_(k-1)(cos(Pi/N)), U_(k-1)(cos(3*Pi/N)), ..., U_(k-1)(cos((2*n-1)*Pi/N)))^T, where T denotes matrix transpose. Let S_N = [V_0, V_1, ..., V_(n-1)] be the n X n matrix formed by taking V_(k-1) as column k-1. Let X_N = [S_N]^T*S_N, and let [X_N](i,j) denote the entry in row i and column j of X_N, i,j in {0,...,n-1}. Then t(n,k) = [X_N](k-1,k-1), and row n of the triangle is given by the main diagonal entries of X_N. Remarks: Hence t(n,k) is the sum of squares t(n,k) = sum[m=1,...,n (U_(k-1)(cos((2*m-1)*Pi/N)))^2]. Finally, this sequence is related to A057059, since X_N = [sum_{m=1,...,n} A057059(n,m)*A_{N,m-1}] is also an integral linear combination of unit-primitive matrices from the N-th set. - L. Edson Jeffery, Jan 20 2012

Row sums: n*(n+1)*(2*n+1)/6. - L. Edson Jeffery, Jan 25 2013

n-th row = partial sums of n-th row of A004736. - Reinhard Zumkeller, Aug 04 2014

T(n,k) is the number of distinct sums made by at most k elements in {1, 2, ... n}, for 1 <= k <= n, e.g., T(6,2) = the number of distinct sums made by at most 2 elements in {1,2,3,4,5,6}. The sums range from 1, to 5+6=11. So there are 11 distinct sums. - Derek Orr, Nov 26 2014

A number n occurs in this sequence A001227(n) times, the number of odd divisors of n, see A209260. - Hartmut F. W. Hoft, Apr 14 2016

Conjecture: 2*n + 1 is composite if and only if gcd(t(n,m),m) != 1, for some m. - L. Edson Jeffery, Jan 30 2018

From Peter Munn, Aug 21 2019 in respect of the sequence read as a triangle: (Start)

A number m can be found in column k if and only if A286013(m, k) is nonzero, in which case m occurs in column k on row A286013(m, k).

The first occurrence of m is in row A212652(m) column A109814(m), which is the rightmost column in which m occurs. This occurrence determines where m appears in A209260. The last occurrence of m is in row m column 1.

Viewed as a sequence of rows, consider the subsequences (of rows) that contain every positive integer. The lexicographically latest of these subsequences consists of the rows with row numbers in A270877; this is the only one that contains its own row numbers only once.

(End)

Take the number of rightmost zeros in the binary expansion of n, double it, and increment it by 1. - Ralf Stephan, Aug 22 2013

Gives the maximum possible number of n X n complex Hermitian matrices with the property that all of their nonzero real linear combinations are nonsingular (see Adams et al. reference). - Nathaniel Johnston, Dec 11 2013

This sequence and A006519 (greatest power of 2 dividing n) are very similar, the difference being all zeros except for every 16th term (see A101119 for nonzero differences). - Simon Plouffe, Dec 02 2004

For all n congruent to 2^k (mod 2^(k+1)), a(n) is the same. Therefore, for any natural number m, the list of the first 2^m - 1 terms is palindromic. - Ivan N. Ianakiev, Jul 21 2019

Named after the Austrian mathematician Johann Radon (1887-1956) and the German mathematician Adolf Hurwitz (1859-1919). - Amiram Eldar, Jun 15 2021

All terms are nonnegative because for all n, x = A276086(A276085(n)) <= n, as any factor prime(i)^k || n (with k >= prime(i)) will propagate carries (in the image of fully additive A276085) towards more significant digit positions, which A276086 will convert back to the exponents of larger primes, but for each new instance of such larger prime present in x, enough instances of smaller primes in n have been eliminated (by the carry process) so that the net change of magnitude is negative, unless there are no such factors present at all in n (i.e., when n is a term of A048103), then A276086(A276085(n)) = n, and a(n) = 0.

This implies also that the least k for which A276085(k) = n is k = A276086(n).

There are several conspicuous patterns among the terms. For example, for n = 392, 3992, 39992, 399992, 3999992, ..., a(n) = 98, 998, 9998, 99998, 999998, ..., = n/4, but this holds only if n/8 is not in A100716, as generally, for all terms x that are in the intersection of A051062 and A168183 and x/8 is in A048103, it follows that a(x) = x/4. There are many other similar identities.

Differs from similar A376417 for the first time at n=625, 1250, 1875, 2500, 3125, 3375, 3750, 4375, 4500, 5000, 5625, ...