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This number triangle results from the array A(n, m) = T(n+m+1) - T(n-1), with T = A000217, for n, m >= 1. For this array see the example by Bob Selcoe, in A111774 (but with rows continued). The present triangle is obtained by reading the array by upwards antidiagonals: T(n, k) = A(n+1-k, k). See also the Jul 09 2019 comment by Ralf Steiner with the formula c_k(n) (rows k >= 1, columns n >= 3), rewritten for A(n, m) = (m+2)*(2*n+m+1)/2, leading to T(n, k) = (k+2)*(2*n-k+3)/2.

Therefore this triangle is related to the problem of giving the numbers which are sums of at least three consecutive positive integers given as sequence A111774. It allows us to find the multiplicities for the numbers of A111774. They are given in A338428(n).

To obtain the multiplicity for number N (>= 6) from A111774 one has to consider only the triangle rows n = 1, 2, ..., floor((N-3)/3).

The row reversed triangle, considered by Bob Selcoe in A111774, is T(n, n-k+1) = T(n+2) - T(k-1), for n >= 1, and k=1, 2, ..., n.

This triangle contains no odd prime numbers and no exact powers 2^m, for m >= 0. This can be seen by considering the diagonal sequences D(d, k), for d >= 1, k >= 1 or the row sequences of the array A(n, m), for n >= 1 and m >= 1. The result is A(r+1, s-2) = s*(s + 2*r + 1)/2, for r >= 0 and s >= 3 (from the g.f. of the diagonals of T given below). This is also given in the Jul 09 2019 comment by Ralf Steiner in A111774. Therefore A(r+1, s-2) is a product of two numbers >= 2, hence not a prime. And in both cases (i) s/2 integer or (ii) (s + 2*r + 1)/2 integer not both numbers can be powers of 2 by simple parity arguments.

The previous comment means that each T(n, k) has at least one odd prime as a proper divisor.

A number N appears in this triangle, or in A111774, if and only if floor(N/2) - delta(N) >= 1, where delta(N) = A055034(N). For the sequence b(n) := floor(n/2) - delta(n), for n >= 2, see A219839(n), b(1) = -1. See a W. Lang comment in A111774 for the proof.

See A066272 for definition of anti-divisor.

The following conjectures have been proved by Bob Selcoe. - Michael Somos, Feb 28 2014

Additional conjectures suggested by computational experiments:

1) Numbers all of whose anti-divisors (AD's) are odd => {2^k} (A000079).

2) Numbers with AD 2, all other AD's odd => primes (A000040).

3) Numbers none of whose AD's are multiples of 3 => 3*2^k (A007283).

4) Numbers all of whose AD's are even => 3*A002822 = A040040 (except for a(0)=1), both related to twin prime pairs.

Calculations suggest the following conjecture. This sequence consists of all odd primes and nonnegative powers of 2 and no other terms. This has been verified for to n=100000. Robert G. Wilson v extended the conjecture out to 2^20.

From Bob Selcoe, Feb 24 2014: (Start)

The sequence consists of all odd primes and powers of two (>=2^2) and no other terms.

Proof: Denote the even anti-divisors of n as ADe(n). ADe(n) is defined as the set of numbers x satisfying the equation n(mod x)=x/2. Substitute x = 2n/y, since it can be shown that ADe(n) => 2n divided by the odd divisors of n when n>1 (This is because 2j anti-divides only numbers of the form 3j+2j*k; j>=1, k>=0. For example: j=7; 14 anti-divides only 21,35,49,63.... So in other words, even numbers anti-divide only odd multiples (>=3) of themselves, divided by 2). Therefore, ADe(n) is n(mod [2n/y])=n/y, and y must be an odd divisor of n and 2n, y>1. Since y is the only odd divisor of n when y>1 iff n is prime, then ADe(n) => 2 when n is prime. Since 2n has no odd divisors when n=2^k, then ADe(n) is null when n=2^k. Therefore, the only numbers whose anti-divisors are either 2 or odd must be primes and powers of 2.

Similarly, for odd anti-divisors (ADo(n)): Given 2j+1 (odd numbers) anti-divide only numbers of the forms [(3j+1)+(2j+1)*k] and [(3j+2)+(2j+1)*k]; j>=1, k>=0. (For example: j=6; 13 anti-divides only 19,20, 32,33, 45,46...). Since odd n divided by its odd divisors ARE its odd divisors, then ADo(n) => the divisors of 2n-1 and 2n+1 (except 1, 2n-1 and 2n+1).

By extension:

1) Numbers all of whose anti-divisors (AD's) are odd => {2^k} (A000079).

2) Numbers with ADe(n)=2, all other AD's odd => primes (A000040).

3) Numbers none of whose AD's are multiples of j => j*2^k.

4) When 2n-1 and 2n+1 are twin primes, (A040040, except for a(0)=1) then n has only even AD's.

(End)

If 1 and 2 are included, this sequence contains all positive integers not contained in A111774. - Bob Selcoe, Sep 09 2014 [corrected by Wolfdieter Lang, Nov 06 2020]

a(n)=0 if n is an odd prime or a power of 2. For numbers of the third kind we proceed as follows: suppose n is to be written as sum of k consecutive integers starting with m, then 2n = k(2m + k - 1). Let p be the smallest odd prime divisor of n then a(n) = min(p,2n/p).

The length of row n is given in A338431(n), for n >= 1.

The length of the sublists t(n, k) of the power basis coefficients of DSR(n, k), for k = 1, 2, ..., floor(n/2), is 1 if n = 1, for n >= 2 it is k except for n = n(j) = A111774(j) for which the final A219839(n) sublists have fewer than k members.

Trailing vanishing coefficients of the delta(n) = A055034(n) power base elements <1 = rho(n)^0, rho(n)^1, ..., rho(n)^{delta(n)-1}> are not recorded. The coefficients of the minimal polynomial C(n, x) of rho(n) = 2*cos(Pi/n) of degree delta(n) are given in A187360. C(n, rho(n)) = 0 is used to eliminate all powers of rho with exponent >= delta(n).

The length ratios DSR(n, k) := diagonal(n, k)/side(n) of regular n-gons, for n >= 2, and k = 1, 2, ..., floor(n/2) (distinct diagonals, starting with the side for k = 1, in increasing order) are given by DSR(n, k) = S(k-1, rho(n)), with the Chebyshev S polynomials (A049310). See the W. Lang link.

For n = 2, the degenerate case, diagonal/side = side/side = 1 for k = 1. For n = 1 (a point) diagonal/side is undetermined, and T(1, 1) is set to 1.

For the power basis sublists t(n, k), for k = 1, 2, ..., delta(n), only the k coefficients of S(k-1, x) are present (trailing vanishing coefficients are not recorded). For k = delta(n)+1, ..., floor(n/2) less than k coefficients appear due to elimination via C(n, rho(n)) = 0. E.g., for n = 6 with delta(6) = 2 the only coefficient for k = 3 is 2 (coefficient of rho^0). This appears for n = n(j) = A111774(j), because then floor(n/2) - delta(n) = A219839(n) > 0.

Because A219839(n) = 0 means that n is from A174090, i.e., a prime or a power of 2 (complement of A111774), these rows n have all the sublists t(n, k) with the k coefficients of S(k-1, x), hence they are identical (but the basis differs). See especially the table for the pairs of consecutive numbers n with identical coefficients, like (2, 3), (4, 5), (16, 17), (256, 267), (65536, 65537), ?... (cf. Fermat primes A019434).