cp's OEIS Frontend

Old name: Triangle T(n,k) = n-k, n >= 1, 0 <= k < n. Fractal sequence formed by repeatedly appending strings m, m-1, ..., 2, 1.

The PARI functions t1 (this sequence), t2 (A002260) can be used to read a square array T(n,k) (n >= 1, k >= 1) by antidiagonals upwards: n -> T(t1(n), t2(n)). - Michael Somos, Aug 23 2002, edited by M. F. Hasler, Mar 31 2020

A004736 is the mirror of the self-fission of the polynomial sequence (q(n,x)) given by q(n,x) = x^n+ x^(n-1) + ... + x + 1. See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011

Seen as flattened list: a(A000217(n)) = 1; a(A000124(n)) = n and a(m) <> n for m < A000124(n). - Reinhard Zumkeller, Jul 22 2012

Sequence B is called a reverse reluctant sequence of sequence A, if B is triangle array read by rows: row number k lists first k elements of the sequence A in reverse order. Sequence A004736 is the reverse reluctant sequence of sequence 1,2,3,... (A000027). - Boris Putievskiy, Dec 13 2012

The row sums equal A000217(n). The alternating row sums equal A004526(n+1). The antidiagonal sums equal A002620(n+1) respectively A008805(n-1). - Johannes W. Meijer, Sep 28 2013

From Peter Bala, Jul 29 2014: (Start)

Riordan array (1/(1-x)^2,x). Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array

/I_k 0\

\ 0 M/

having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the infinite matrix product M(0)*M(1)*M(2)*... is equal to A078812. (End)

T(n, k) gives the number of subsets of [n] := {1, 2, ..., n} with k consecutive numbers (consecutive k-subsets of [n]). - Wolfdieter Lang, May 30 2018

a(n) gives the distance from (n-1) to the smallest triangular number > (n-1). - Ctibor O. Zizka, Apr 09 2020

To construct the sequence, start from 1,2,,3,,,4,,,,5,,,,,6... where there are n commas after each "n". Then fill the empty places by the sequence itself. - Benoit Cloitre, Aug 17 2021

T(n,k) is the number of cycles of length 2*(k+1) in the (n+1)-ladder graph. There are no cycles of odd length. - Mohammed Yaseen, Jan 14 2023

The first 77 entries are also the signature sequence of log(3)=A002391. Then the two sequences start to differ. - R. J. Mathar, May 27 2024

Warning: contributions from Kimberling refer to an alternate version indexed by 1 instead of 0. Other contributors (Adamson in A125026/A130301/A130295) refer to this considering the upper right triangle set to zero, T[i,j]=0 for j>i. - M. F. Hasler, Aug 15 2015

From Clark Kimberling, Feb 05 2011: (Start)

A member of the accumulation chain:

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

(See A144112 for the definition of accumulation array.)

In the m-th accumulation array of A051340,

row_1 = C(m,1) and column_1 = C(1,m+1), for m>=0. (End)

n is the sum of at most a(n) consecutive positive integers. As suggested by David W. Wilson, Aug 15 2005: Suppose n is to be written as sum of k consecutive integers starting with m, then 2n = k(2m + k - 1). Only one of the factors is odd. For each odd divisor d of n there is a unique corresponding k = min(d,2n/d). a(n) is the largest among those k. - Jaap Spies, Aug 16 2005

The numbers that can be written as a sum of k consecutive positive integers are those in column k of A141419 (as a triangle). - Peter Munn, Mar 01 2019

The numbers that cannot be written as a sum of two or more consecutive positive integers are the powers of 2. So a(n) = 1 iff n = 2^k for k >= 0. - Bernard Schott, Mar 03 2019

(Conjecture) Let N=2*n and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n-1} be the n X n unit-primitive matrices (see [Jeffery]) associated with N. 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 B^T denotes the transpose of matrix B. Let S_N=[V_0,V_1,...,V_(n-1)] be the n X n matrix formed by taking the components of vector V_(k-1) as the entries in column k-1 (V_(k-1) gives the ordered spectrum of A_{N,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 also T(n,k)=[X_N](k-1,k-1); that is, row n of the triangle is given by the main diagonal entries of X_N. 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]=[V_(k-1)]^T*V_(k-1). - L. Edson Jeffery, Jan 20 2012

Conjecture that antidiagonal sums are A023855. - L. Edson Jeffery, Jan 20 2012

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 A066680 U {1}; this is the only one that contains its own row numbers only once. - Peter Munn, Dec 04 2019

n = A000217(a(n)) - A000217(a(n) - A109814(n)).

Conjecture: n appears in row a(n) of A209260.

From Daniel Forgues, Jan 06 2016: (Start)

n = Sum_{i=k+1..M} i = T(M) - T(k) = (M-k)*(M+k+1)/2.

n = 2^m, m >= 0, iff M = n = 2^m and k = n - 1 = 2^m - 1. (Points on line with slope 1.) (Powers of 2 can't be the sum of consecutive numbers.)

n is odd prime iff k = M-2. Thus M = (n+1)/2 when n is odd prime. (Points on line with slope 1/2.) (Odd primes can't be the sum of more than 2 consecutive numbers.) (End)

If n = 2^m*p where p is an odd prime, then a(n) = 2^m + (p-1)/2. - Robert Israel, Jan 14 2016

This also expresses the following geometry: along a circle having (n) points on its circumference, a(n) expresses the minimum number of hops from a start point, in a given direction (CW or CCW), when each hop is increased by one, before returning to a visited point. For example, on a clock (n=12), starting at 12 (same as zero), the hops would lead to the points 1, 3, 6, 10 and then 3, which was already visited: 5 hops altogether, so a(12) = 5. - Joseph Rozhenko, Dec 25 2023

Conjecture: a(n) is the smallest of the largest parts of the partitions of n into consecutive parts. - Omar E. Pol, Jan 07 2025

A member of the accumulation chain: A051340 < A141419 < A185874 < A185875 < A185876 < ... (See A144112 for the definition of accumulation array.)

From Reinhard Zumkeller, Aug 04 2014: (Start)

n-th row = half of Dynkin diagram weights for the Cartan Groups D_n.

n-th row = partial sums of n-th row of A025581. (End)

A member of the accumulation chain A051340 < A141419 < A185874 < A185875 < A185876 < ... (See A144112 for the definition of accumulation array.)

A member of the accumulation chain A051340 < A141419 < A185874 < A185875 < A185876 < ... (See A144112 for the definition of accumulation array.)

One way to derive this sequence row by row is by eliminating from row n of A141419 any integer already present in row m < n of A141419, with the remaining entries comprising row n of this triangle. The resulting irregular triangle then begins as {1}; {2, 3}; {5, 6}; {4, 7, 9, 10}; ... (see the examples below).

If 2*n+1 is a prime, then 2*n+1 appears in row n+1.

Conjecture 1. 2*n+1 is in the set S_n = {union of first n rows of A209260} if and only if 2*n+1 is composite. (Verified for all n <= 10^6 by Charles R Greathouse IV.) This conjecture has been proved (see [Jeffery]).

Conjecture 2. Let N be an integer, N>0, and let D_N be the set of all positive integral divisors of N. Let B_N be the set of all integral solutions of d + (h-1)/2 for which d in D_N, with h=N/d and N=d*h (some solutions will not be integers and so are not in B_N). Then N appears in row r = min(B_N) of A209260. Also, N appears in row b of A141419, for each b in B_N. - L. Edson Jeffery, Feb 13 2013

From Hartmut F. W. Hoft, Apr 14 2016: (Start)

For a proof of Conjecture 2 see the link.

The row-column index pair in A141419 for number v having factorization v = d*h with h odd is: (d+(h-1)/2, h) if h+1 <= 2*d, and (d+(h-1)/2, 2*d) if h+1 > 2*d. The position of number v in this triangle occurs for the minimum value of d+(h-1)/2 among all divisor pairs d,h with v = d*h and h odd. (End)