cp's OEIS Frontend

Also (1) number of ways to write n as difference of two triangular numbers (A000217), see A136107; (2) number of ways to arrange n identical objects in a trapezoid. - Tom Verhoeff

Also number of partitions of n into consecutive positive integers including the trivial partition of length 1 (e.g., 9 = 2+3+4 or 4+5 or 9 so a(9)=3). (Useful for cribbage players.) See A069283. - Henry Bottomley, Apr 13 2000

This has been described as Sylvester's theorem, but to reduce ambiguity I suggest calling it Sylvester's enumeration. - Gus Wiseman, Oct 04 2022

a(n) is also the number of factors in the factorization of the Chebyshev polynomial of the first kind T_n(x). - Yuval Dekel (dekelyuval(AT)hotmail.com), Aug 28 2003

Number of factors in the factorization of the polynomial x^n+1 over the integers. See also A000005. - T. D. Noe, Apr 16 2003

a(n) = 1 if and only if n is a power of 2 (see A000079). - Lekraj Beedassy, Apr 12 2005

Number of occurrences of n in A049777. - Philippe Deléham, Jun 19 2005

For n odd, n is prime if and only if a(n) = 2. - George J. Schaeffer (gschaeff(AT)andrew.cmu.edu), Sep 10 2005

Also number of partitions of n such that if k is the largest part, then each of the parts 1,2,...,k-1 occurs exactly once. Example: a(9)=3 because we have [3,3,2,1],[2,2,2,2,1] and [1,1,1,1,1,1,1,1,1]. - Emeric Deutsch, Mar 07 2006

Also the number of factors of the n-th Lucas polynomial. - T. D. Noe, Mar 09 2006

Lengths of rows of triangle A182469;

Denoted by Delta_0(n) in Glaisher 1907. - Michael Somos, May 17 2013

Also the number of partitions p of n into distinct parts such that max(p) - min(p) < length(p). - Clark Kimberling, Apr 18 2014

Row sums of triangle A247795. - Reinhard Zumkeller, Sep 28 2014

Row sums of triangle A237048. - Omar E. Pol, Oct 24 2014

A069288(n) <= a(n). - Reinhard Zumkeller, Apr 05 2015

A000203, A000593 and this sequence have the same parity: A053866. - Omar E. Pol, May 14 2016

a(n) is equal to the number of ways to write 2*n-1 as (4*x + 2)*y + 4*x + 1 where x and y are nonnegative integers. Also a(n) is equal to the number of distinct values of k such that k/(2*n-1) + k divides (k/(2*n-1))^(k/(2*n-1)) + k, (k/(2*n-1))^k + k/(2*n-1) and k^(k/(2*n-1)) + k/(2*n-1). - Juri-Stepan Gerasimov, May 23 2016, Jul 15 2016

Also the number of odd divisors of n*2^m for m >= 0. - Juri-Stepan Gerasimov, Jul 15 2016

a(n) is odd if and only if n is a square or twice a square. - Juri-Stepan Gerasimov, Jul 17 2016

a(n) is also the number of subparts in the symmetric representation of sigma(n). For more information see A279387 and A237593. - Omar E. Pol, Nov 05 2016

a(n) is also the number of partitions of n into an odd number of equal parts. - Omar E. Pol, May 14 2017 [This follows from the g.f. Sum_{k >= 1} x^k/(1-x^(2*k)). - N. J. A. Sloane, Dec 03 2020]

This is a triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists successive blocks of k consecutive terms, where the m-th block starts with m, m>=1, and the first element of column k is in row k*(k+1)/2.

The partitions of n into consecutive parts are represented from the row n up to row A288529(n) as maximum, but in increasing order, exclusively in the columns where the blocks begin.

More precisely, the partition of n into exactly k consecutive parts (if such partition exists) is represented in the column k from the row n up to row n + k - 1 (see examples).

A288772(n) is the minimum number of rows that are required to represent in this table the partitions of all positive integers <= n into consecutive parts.

A288773(n) is the largest of all positive integers whose partitions into consecutive parts can be totally represented in the first n rows of this table.

A288774(n) is the largest positive integers whose partitions into consecutive parts can be totally represented in the first n rows of this table.

For a theorem related to this table see A286000.

This is a triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists successive blocks of k consecutive terms in decreasing order, where the m-th block starts with k + m - 1, m>=1, and the first element of column k is in the row k*(k+1)/2.

The partitions of n into consecutive parts are represented from the row n up to row A288529(n) as maximum, exclusively in the columns where the blocks begin.

More precisely, the partition of n into exactly k consecutive parts (if such partition exists) is represented in the column k from the row n up to row n + k - 1 (see examples).

A288772(n) is the minimum number of rows that are required to represent in this table the partitions of all positive integers <= n into consecutive parts.

A288773(n) is the largest of all positive integers whose partitions into consecutive parts can be totally represented in the first n rows of this table.

A288774(n) is the largest positive integers whose partitions into consecutive parts can be totally represented in the first n rows of this table.

Theorem: the smallest part of the partition of n into exactly k consecutive parts (if such partition exists) equals the number of positive integers <= n having a partition into exactly k consecutive parts.

The number of positive terms in row n is A001227(n).

If n = 2^j then the only positive integer in row n is T(n,1) = n

If n is an odd prime then the only two positive integers in row n are T(n,1) = n and T(n,2) = (n - 1)/2.

From Omar E. Pol, Apr 30 2017: (Start)

Conjecture 1: T(n,k) is the smallest part of the partition of n into k consecutive parts, if T(n,k) > 0.

Conjecture 2: the last positive integer in the row n is in the column A109814(n). (End)

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)

Conjecture 1: T(n,k) is the largest part of the partition of n into k consecutive parts, if T(n,k) > 0.

Conjecture 2: row sums give A286015.

Trapezoidal interpretation from Peter Munn, Jun 18 2017: (Start)

There is one to one correspondence between nonzero T(n,k) and trapezoidal area patterns of n dots on a triangular grid, if we include the limiting cases of triangular patterns, straight lines (k=1) or a single dot (k=n=1). The corresponding pattern has T(n,k) dots in its longest side, k dots in the two adjacent sides and T(n,k)-k+1 dots in the fourth side (where a count of 1 dot may be understood as signifying that side's absence).

Reason: From the definition, for k >= 1, m >= 0, T(A000217(k)+km,k) = k+m, where A000217(k) = k(k+1)/2, the k-th triangular number. First element of column k is T(A000217(k),k) = k: this matches a triangular pattern of A000217(k) dots with 3 sides of k dots. Looking at this pattern as k rows of 1..k dots, extend each row by m dots to create a trapezoidal pattern of A000217(k)+km dots with a longest side of k+m dots and adjacent sides of k dots: this matches T(A000217(k)+km,k) = k+m. As nonzero elements in column k occur at intervals of k, every nonzero T(n,k) has a match. Every trapezoidal pattern can be produced by extending a triangular pattern as described, so they all have a match.

The truth of conjecture 1 follows, since each nonzero T(n,k) = k+m corresponds to a trapezoidal pattern of n dots having k rows with lengths (1+m)..(k+m).

The A270877 sieve is related to this sequence because it eliminates n if it is the sum of consecutive numbers whose largest term has survived the sifting (which may likewise be seen in terms of a trapezoidal dot pattern and its longest side). So the sieve eliminates n if any lesser numbers in A270877 are in row n of this sequence.

(End)

Right border of A299765. - Omar E. Pol, Jul 24 2018

In other words: a(n) is smallest part of the partitions of n into consecutive parts. - Omar E. Pol, Mar 12 2019

a(n) has the same definition related to the table A286001 which is another version of the table A286000.

First differs from A288772 at a(11), which shares infinitely many terms.

a(n) has the same definition related to the table A286001 which is another version of the table A286000.

First differs from A288529 at a(11), which shares infinitely many terms.

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