cp's OEIS Frontend

This is the generic form of D in the (nontrivially) solvable Pell equation x^2 - D*y^2 = +4. See A077428 and A078355.

Consider all primitive Pythagorean triples (a,b,c) with c-a=8, sequence gives values of a. (Corresponding values for b are A017113(n), while c follows A078370(n).) - Lambert Klasen (Lambert.Klasen(AT)gmx.net), Nov 19 2004

From Vincenzo Librandi, Aug 08 2010: (Start)

The identity (4*n^3 + 18*n^2 + 24*n + 9)^2 - (4*n^2 + 12*n + 5)*(2*n^2 + 6*n + 4)^2 = 1 (see Ramasamy's paper in link) can be written as A141530(n+2)^2 - a(n)*A046092(n+1)^2 = 1.

a(n)^3 + 6*a(n)^2 + 9*a(n) + 4 is a square: in fact, a(n)^3 + 6*a(n)^2 + 9*a(n) + 4 = (a(n) + 1)^2*(a(n) + 4), where a(n) + 4 = (2*n+3)^2. (End)

Products of two positive odd integers with difference 4 (i.e., 1*5, 3*7, 5*9, 7*11, 9*13, ...). - Wesley Ivan Hurt, Nov 19 2013

Starting with stage 1, the number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 675", based on the 5-celled von Neumann neighborhood. - Robert Price, May 21 2016

The continued fraction expansion of (sqrt(a(n))-1)/2 is [n; {1,2*n+1}] with periodic part of length 2: repeat {1,2*n+1}. - Ron Knott, May 11 2017

a(n) is the sum of 2*n+5 consecutive integers starting from n-1. - Bruno Berselli, Jan 16 2018

The continued fraction expansion of sqrt(a(n)) is [2n+2; {1, n, 2, n, 1, 4n+4}]. For n=0, this collapses to [2; {4}]. - Magus K. Chu, Aug 26 2022

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

Union of A004771 and A017113.

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

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)

From Philippe Deléham, Apr 15 2007: (Start)

This triangle can be found in the Laisant reference in the following form:

.......................5...11..

...................4...9...20..

...............3...7..16...36..

...........2...5..12..28.......

.......1...3...8..20..48.......

...0...1...4..12..32..80....... (End)

Triangle A152920 reversed. - Philippe Deléham, Apr 21 2009

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

From Amiram Eldar, Nov 02 2020: (Start)

Numbers k such that A280292(k) = 2.

The asymptotic density of this sequence is 1/Pi^2 (A092742). (End)

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

Numbers of terms in nonnegative integer sequences the sum of which is never a square.

The sum of a sequence of consecutive nonnegative integers starting with k is never a square for any k, if and only if the number of the terms in the sequence can be expressed as (2*m - 1) * 2^(2*n), m and n being any positive integers. (Proved by Alfred Vella, Jun 14 2005.)

Odious and evil terms alternate. - Vladimir Shevelev, Jun 22 2009

Even numbers whose binary representation ends in an even number of zeros. - Amiram Eldar, Jan 12 2021

From Antti Karttunen, Jan 28 2023: (Start)

Numbers k for which the parity of k is equal to that of A048675(k).

A multiplicative semigroup; if m and n are in the sequence then so is m*n. (End)