cp's OEIS Frontend

The square roots of these numbers have some remarkable properties - see the link to Schizophrenic numbers.

Partial sums of A002275. - Jonathan Vos Post, Apr 25 2010

This sequence is the particular case of a(0) = 0, a(n) = r*a(n-1) + n, when r = 10. If now the first N terms are computed for (r > N) then the resulting set of numbers is readable as the smallest k-digits permutations (1 <= k <= N): Those built from the concatenation of the first k digits in base-r (see links). - R. J. Cano, Jan 09 2013

There is also an interesting structure to the decimal expansion of 1/sqrt(a(2*n+1)), which has long strings of 0's (that gradually shorten in length until they disappear) interspersed with strings of what, at first sight, appear to be random digits. However, if we factorize these blocks of 'random' digits we find they are related to each other. An example illustrating this is given below. - Peter Bala, Sep 13 2015

From Peter Bala, Sep 15 2015: (Start)

Extending the previous empirical observation, it appears that the decimal expansions of the numbers 1/ (a(4*n+1))^(1/2), 1/a((4*n+3))^(1/4), 1/(10*a(4*n))^(1/2), 1/(10*a(4*n+2))^(1/4), and their powers, have the same pattern of beginning with long strings of 0's (that gradually shorten in length until they disappear) interlaced with strings of digits which, when read as integers and factorized, turn out to be related to each other.

The following result, which is a consequence of Bottomley's explicit formula for a(n), should be helpful in explaining these observations: the decimal expansion of 1/a(2*n-1) for n >= 5 begins with long strings of 0's interlaced successively with the digits of the numbers 81*(18*n + 1)^k for k going from 0 up to approximately n/log_10(18*n). For example, for n = 7 we have 1/a(13) = 0.00000000000081000000000102870000000130644900000 165919023..., with 10287 = 81*127, 1306449 = 81*127^2 and 165919023 = 81*127^3. A similar result holds for the decimal expansion of the number 1/a(2*n).

It appears that a(4*n+3)^(1/4), a(4*n+3)^(3/4), (10*a(4*n))^(1/2), (10*a(4*n+2))^(1/4) and (10*a(4*n+2))^(3/4) are further examples of Brown's schizophrenic numbers.

A theorem of Kuzmin in the measure theory of continued fractions says that for a random real number alpha, the probability that some given partial quotient of alpha is equal to a positive integer k is given by 1/log(2)*( log(1 + 1/k) - log(1 + 1/(k+1)) ). Thus large partial quotients are the exception in continued fraction expansions. Empirically, we observe the presence of unexpectedly large partial quotients early in the continued fraction expansions of the numbers (a(4*n+1))^(1/2), (a(4*n+3))^(1/4), (10*a(4*n))^(1/2), (10*a(4*n+2))^(1/4) and their powers. An example is given below. (End)

From Ya-Ping Lu, Dec 21 2024: (Start)

To get a(n), concatenate the first n digits in the cyclic string '123456790' and subtract the number of occurrences of '9' from the concatenated number. For example, a(8) = 12345679 - 1 = 12345678.

There are 2 prime terms for n <= 20000: a(2497) and a(3301). (End)

The number -2 can be used as a base of numeration (see the Weisstein link). - Alonso del Arte, Mar 30 2014

Contribution from M. F. Hasler, Oct 21 2014: (Start)

This is the inverse binomial transform of A033999 = n->(-1)^n, and the binomial transform of A033999*A000244 = n->(-3)^n, see also A141413.

Prefixed with one 0, i.e., (0,1,-2,4,...) = -A033999*A131577, it is the binomial transform of (0, 1, -4, 13, -40, 121,...) = -A033999*A003462, and inverse binomial transform of (0,1,0,1,0,1,...) = A000035.

Prefixed with two 0's, i.e., (0,0,1,-2,4,-8,...), it is the binomial transform of (0,0,1,-5,18,-58,179,-543,...) (cf. A000340) and inverse binomial transform of (0,0,1,1,2,2,3,3,...) = A004526. (End)

Prefixed with three 0's, this is the inverse binomial difference of (0, 0, 0, 1, 2, 4, 6, 9, 12, 16,...) = concat(0, A002620), which has as successive differences (0, 0, 1, 1, 2, 2,...) = A004526, then (0, 1, 0, 1,...) = A000035, then (1, -1, 1, -1,...) = A033999, and then (-2)^k*A033999 with k=1,2,3,... - Paul Curtz, Oct 16 2014, edited by M. F. Hasler, Oct 21 2014

Stirling-Bernoulli transform of triangular numbers: 1, 3, 6, 10, 15, 21, 28, ... - Philippe Deléham, May 25 2015

Essentially the same as A185411. Row reverse of A185410. - Peter Bala, Jul 24 2012

The SF(z; n) formulas, see below, were discovered while studying certain properties of the Dirichlet eta function.

From Peter Bala, Apr 03 2011: (Start)

Let D be the differential operator 2*x*d/dx. The row polynomials of this table come from repeated application of the operator D to the function g(x) = 1/sqrt(1 - x). For example,

D(g) = x*g^3

D^2(g) = x*(2 + x)*g^5

D^3(g) = x*(4 + 10*x + x^2)*g^7

D^4(g) = x*(8 + 60*x + 36*x^2 + x^3)*g^9.

Thus this triangle is analogous to the triangle of Eulerian numbers A008292, whose row polynomials come from the repeated application of the operator x*d/dx to the function 1/(1 - x). (End)

The originator sequences are A142963 and A156919.

The Flower Triangle seems to be an appropriate name for the triangular array of this sequence. The zero patterns of the Flower Polynomials of the first, see A156921, the second, see A156925, the third, see A156927, and the fourth kind, see A156933, look like flowers.

The first Maple program generates the Flower Triangle sequence.

The second program generates the Right Hand Columns sequences and the third one generates the Left Hand Column sequences. For an explanation of these two algorithms see A142963.

Previous name: Table of coefficients of row polynomials of certain o.g.f.s.

The o.g.f.s G(k, x) for the k-family of sequences S(k, n):= Sum_{p=0..n} p^k*binomial(2*p, p)*binomial(2*(n-p), n-p), k=0,1,... (convolution of two sequences involving the central binomial coefficients) are 1/(1-4*x) for k=0 and 2*x*P(k, x)/(1-4*x)^(k+1) for k=1,2,..., with the row polynomials P(k, x) = Sum_{m=0..k-1} a(n,m)*x^m).

The author was led to compute the sums S(k, n) by a question asked by M. Greiter, Jun 27 2008.

In order to keep the index k>=1 of Sigma(k, n) also for the polynomials P(k, x), their degree is then k-1.

The FP1 polynomials appear in the numerators of the GF1 o.g.f.s. of the right hand columns of A156920. The FP1 can be calculated with the formula for the RHC sequence, see A156920, and the formula for the general structure of the generating function GF1, see below.

An appropriate name for the FP1 polynomials seems to be the flower polynomials of the first kind because the zero patterns of these polynomials look like flowers. The zero patterns of the FP2, see A156925, and the FP1 resemble each other closely.

A Maple program that generates for a right hand column with a certain RHCnr its GF1 and FP1 can be found below. RHCnr stands for right hand column number and starts from 1.

Row sums are A027649. Antidiagonal sums are A106517.

From Wolfdieter Lang, Jan 09 2015: (Start)

Alternating row sums give A025192. The A-sequence of this Riordan lower triangular matrix is [1, 1, repeat(0, )] (leading to the Pascal recurrence for T(n,k) for n >= k >= 1. The Z-sequence is [3, repeat(0, )] (leading to the recurrence T(n,0) = 3*T(n-1,0), n >= 1. For A- and Z-sequences see the W. Lang link under A006232.

The inverse of this Riordan matrix is Tinv = ((1 - 2*x)/(1 + x), x/(1 + x)) given as a signed version of A093560: Tinv(n,m) = (-1)^(n-m)*A093560(n,m). (End)