cp's OEIS Frontend

Erdős conjectured that n^2 - 1 = k! has a solution if and only if n is 5, 11 or 71 (when k is 4, 5 or 7).

Second-order linear recurrences y(m) = 2y(m-1) + a(n)*y(m-2), y(0) = y(1) = 1, have closed form solutions involving only powers of integers. - Len Smiley, Dec 08 2001

Number of edges in the join of two cycle graphs, both of order n, C_n * C_n. - Roberto E. Martinez II, Jan 07 2002

Let k be a positive integer, M_n be the n X n matrix m_(i,j) = k^abs(i-j) then det(M_n) = (-1)^(n-1)*a(k-1)^(n-1). - Benoit Cloitre, May 28 2002

Also numbers k such that 4*k + 4 is a square. - Cino Hilliard, Dec 18 2003

For each term k, the function sqrt(x^2 + 1), starting with 1, produces an integer after k iterations. - Gerald McGarvey, Aug 19 2004

a(n) mod 3 = 0 if and only if n mod 3 > 0: a(A008585(n)) = 2; a(A001651(n)) = 0; a(n) mod 3 = 2*(1-A079978(n)). - Reinhard Zumkeller, Oct 16 2006

a(n) is the number of divisors of a(n+1) that are not greater than n. - Reinhard Zumkeller, Apr 09 2007

Nonnegative X values of solutions to the equation X^3 + X^2 = Y^2. To find Y values: b(n) = n(n+1)(n+2). - Mohamed Bouhamida, Nov 06 2007

Sequence allows us to find X values of the equation: X + (X + 1)^2 + (X + 2)^3 = Y^2. To prove that X = n^2 + 2n: Y^2 = X + (X + 1)^2 + (X + 2)^3 = X^3 + 7*X^2 + 15X + 9 = (X + 1)(X^2 + 6X + 9) = (X + 1)*(X + 3)^2 it means: (X + 1) must be a perfect square, so X = k^2 - 1 with k>=1. we can put: k = n + 1, which gives: X = n^2 + 2n and Y = (n + 1)(n^2 + 2n + 3). - Mohamed Bouhamida, Nov 12 2007

From R. K. Guy, Feb 01 2008: (Start)

Toads and Frogs puzzle:

This is also the number of moves that it takes n frogs to swap places with n toads on a strip of 2n + 1 squares (or positions, or lily pads) where a move is a single slide or jump, illustrated for n = 2, a(n) = 8 by

T T - F F

T - T F F

T F T - F

T F T F -

T F - F T

- F T F T

F - T F T

F F T - T

F F - T T

I was alerted to this by the Holton article, but on consulting Singmaster's sources, I find that the puzzle goes back at least to 1867.

Probably the first to publish the number of moves for n of each animal was Edouard Lucas in 1883. (End)

a(n+1) = terms of rank 0, 1, 3, 6, 10 = A000217 of A120072 (3, 8, 5, 15). - Paul Curtz, Oct 28 2008

Row 3 of array A163280, n >= 1. - Omar E. Pol, Aug 08 2009

Final digit belongs to a periodic sequence: 0, 3, 8, 5, 4, 5, 8, 3, 0, 9. - Mohamed Bouhamida, Sep 04 2009 [Comment edited by N. J. A. Sloane, Sep 24 2009]

Let f(x) be a polynomial in x. Then f(x + n*f(x)) is congruent to 0 (mod f(x)); here n belongs to N. There is nothing interesting in the quotients f(x + n*f(x))/f(x) when x belongs to Z. However, when x is irrational these quotients consist of two parts, a) rational integers and b) integer multiples of x. The present sequence represents the non-integer part when the polynomial is x^2 + x + 1 and x = sqrt(2), f(x+n*f(x))/f(x) = A056108(n) + a(n)*sqrt(2). - A.K. Devaraj, Sep 18 2009

For n >= 1, a(n) is the number for which 1/a(n) = 0.0101... (A000035) in base (n+1). - Rick L. Shepherd, Sep 27 2009

For n > 0, continued fraction [n, 1, n] = (n+1)/a(n); e.g., [6, 1, 6] = 7/48. - Gary W. Adamson, Jul 15 2010

Starting (3, 8, 15, ...) = binomial transform of [3, 5, 2, 0, 0, 0, ...]; e.g., a(3) = 15 = (1*3 + 2*5 +1*2) = (3 + 10 + 2). - Gary W. Adamson, Jul 30 2010

a(n) is essentially the case 0 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n} ((k-2)*i-(k-3)). Thus P_0(n) = 2*n-n^2 and a(n) = -P_0(n+2). See also A067998 and for the case k=1 A080956. - Peter Luschny, Jul 08 2011

a(n) is the maximal determinant of a 2 X 2 matrix with integer elements from {1, ..., n+1}, so the maximum determinant of a 2x2 matrix with integer elements from {1, ..., 5} = 5^2 - 1 = a(4) = 24. - Aldo González Lorenzo, Oct 12 2011

Using four consecutive triangular numbers t1, t2, t3 and t4, plot the points (0, 0), (t1, t2), and (t3, t4) to create a triangle. Twice the area of this triangle are the numbers in this sequence beginning with n = 1 to give 8. - J. M. Bergot, May 03 2012

Given a particle with spin S = n/2 (always a half-integer value), the quantum-mechanical expectation value of the square of the magnitude of its spin vector evaluates to = S(S+1) = n(n+2)/4, i.e., one quarter of a(n) with n = 2S. This plays an important role in the theory of magnetism and magnetic resonance. - Stanislav Sykora, May 26 2012

Twice the harmonic mean [H(x, y) = (2*x*y)/(x + y)] of consecutive triangular numbers A000217(n) and A000217(n+1). - Raphie Frank, Sep 28 2012

Number m such that floor(sqrt(m)) = floor(m/floor(sqrt(m))) - 2 for m > 0. - Takumi Sato, Oct 10 2012

The solutions of equation 1/(i - sqrt(j)) = i + sqrt(j), when i = (n+1), j = a(n). For n = 1, 2 + sqrt(3) = 3.732050.. = A019973. For n = 2, 3 + sqrt(8) = 5.828427... = A156035. - Kival Ngaokrajang, Sep 07 2013

The integers in the closed form solution of a(n) = 2*a(n-1) + a(m-2)*a(n-2), n >= 2, a(0) = 0, a(1) = 1 mentioned by Len Smiley, Dec 08 2001, are m and -m + 2 where m >= 3 is a positive integer. - Felix P. Muga II, Mar 18 2014

Let m >= 3 be a positive integer. If a(n) = 2*a(n-1) + a(m-2) * a(n-2), n >= 2, a(0) = 0, a(1) = 1, then lim_{n->oo} a(n+1)/a(n) = m. - Felix P. Muga II, Mar 18 2014

For n >= 4 the Szeged index of the wheel graph W_n (with n + 1 vertices). In the Sarma et al. reference, Theorem 2.7 is incorrect. - Emeric Deutsch, Aug 07 2014

If P_{k}(n) is the n-th k-gonal number, then a(n) = t*P_{s}(n+2) - s*P_{t}(n+2) for s=t+1. - Bruno Berselli, Sep 04 2014

For n >= 1, a(n) is the dimension of the simple Lie algebra A_n. - Wolfdieter Lang, Oct 21 2015

Finding all positive integers (n, k) such that n^2 - 1 = k! is known as Brocard's problem, (see A085692). - David Covert, Jan 15 2016

For n > 0, a(n) mod (n+1) = a(n) / (n+1) = n. - Torlach Rush, Apr 04 2016

Conjecture: When using the Sieve of Eratosthenes and sieving (n+1..a(n)), with divisors (1..n) and n>0, there will be no more than a(n-1) composite numbers. - Fred Daniel Kline, Apr 08 2016

a(n) mod 8 is periodic with period 4 repeating (0,3,0,7), that is a(n) mod 8 = 5/2 - (5/2) cos(n*Pi) - sin(n*Pi/2) + sin(3*n*Pi/2). - Andres Cicuttin, Jun 02 2016

Also for n > 0, a(n) is the number of times that n-1 occurs among the first (n+1)! terms of A055881. - R. J. Cano, Dec 21 2016

The second diagonal of composites (the only prime is number 3) from the right on the Klauber triangle (see Kival Ngaokrajang link), which is formed by taking the positive integers and taking the first 1, the next 3, the following 5, and so on, each centered below the last. - Charles Kusniec, Jul 03 2017

Also the number of independent vertex sets in the n-barbell graph. - Eric W. Weisstein, Aug 16 2017

Interleaving of A000466 and A033996. - Bruce J. Nicholson, Nov 08 2019

a(n) is the number of degrees of freedom in a triangular cell for a Raviart-Thomas or Nédélec first kind finite element space of order n. - Matthew Scroggs, Apr 22 2020

From Muge Olucoglu, Jan 19 2021: (Start)

For n > 1, a(n-2) is the maximum number of elements in the second stage of the Quine-McCluskey algorithm whose minterms are not covered by the functions of n bits. At n=3, we have a(3-2) = a(1) = 1*(1+2) = 3 and f(A,B,C) = sigma(0,1,2,5,6,7).

.

0 1 2 5 6 7

+---------------

*(0,1)| X X

(0,2)| X X

(1,5)| X X

*(2,6)| X X

*(5,7)| X X

(6,7)| X X

.

*: represents the elements that are covered. (End)

1/a(n) is the ratio of the sum of the first k odd numbers and the sum of the next n*k odd numbers. - Melvin Peralta, Jul 15 2021

For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {1, 2n}]. - Magus K. Chu, Sep 09 2022

Number of diagonals parallel to an edge in a regular (2*n+4)-gon (cf. A367204). - Paolo Xausa, Nov 21 2023

For n >= 1, also the number of minimum cyclic edge cuts in the (n+2)-trapezohedron graph. - Eric W. Weisstein, Nov 21 2024

For n >= 1, a(n) is the sum of the interior angles of a polygon with n+2 sides, in radians, multiplied by (n+2)/Pi. - Stuart E Anderson, Aug 06 2025

Also the rows of both triangles A237270 and A237593 interleaved.

Also, irregular triangle read by rows in which T(n,k) is the area of the k-th region (from left to right in ascending diagonal) of the n-th symmetric set of regions (from the top to the bottom in descending diagonal) in the two-dimensional diagram of the perspective view of the infinite stepped pyramid described in A245092 (see the diagram in the Links section).

The diagram of the symmetric representation of sigma is also the top view of the pyramid, see Links section. For more information about the diagram see also A237593 and A237270.

The number of cubes at the n-th level is also A024916(n), the sum of all divisors of all positive integers <= n.

Note that this pyramid is also a quarter of the pyramid described in A244050. Both pyramids have infinitely many levels.

Odd-indexed rows are also the rows of the irregular triangle A237270.

Even-indexed rows are also the rows of the triangle A237593.

Lengths of the odd-indexed rows are in A237271.

Lengths of the even-indexed rows give 2*A003056.

Row sums of the odd-indexed rows gives A000203, the sum of divisors function.

Row sums of the even-indexed rows give the positive even numbers (see A005843).

Row sums give A245092.

From the front view of the stepped pyramid emerges a geometric pattern which is related to A001227, the number of odd divisors of the positive integers.

The connection with the odd divisors of the positive integers is as follows: A261697 --> A261699 --> A237048 --> A235791 --> A237591 --> A237593 --> A237270 --> this sequence.

Row n+1 = partial sums of row n.

T(n,1) = A002522(n+1); T(n,2) = A144396(n+1); T(n,3) = A002522(n+2).

This square array A(row,col) is read by downwards antidiagonals as: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), etc.

Ludic sieve starts with natural numbers larger than one: 2, 3, 4, 5, 6, 7, ... and in each subsequent stage one sets k = (which will be one of Ludic numbers) and removes both k and every k-th term after it, from column positions 1, 1+k, 1+2k, 1+3k, etc. of the preceding row to produce the next row of this array.

For a(2*n) see A341730.

Theorem: a(2*n+1) = 2*n+1. For the proof see A342222.

It would be nice to have a bigger b-file.

Normally the '3x+1 problem' or 'Collatz problem' asks for the number of steps to go from n to 1 (A006577). Here we ask for the number of iterations of the mapping, msa, to go from n to less than n; the mapping of x is either -> (3x+1)/2 if x is odd or -> x/2 if x is even.

Since the number of iterations of msa for an even number is always 1, we will only investigate the odd numbers greater than one.

a(n) = 1 for no values of n;

a(n) = 2 for n = 2 + 2k (k=0,1,2,3,...);

a(n) = 3 for no values of n;

a(n) = 4 for n = 1 + 8k (k=0,1,2,3,...);

a(n) = 5 for n = 5 + 16k and 11 + 16k (k=0,1,2,3,...);

a(n) = 6 for no values of n;

a(n) = 7 for n = 3 + 64k, 7 + 64k, 29 + 64k, etc. (k=0,1,2,3,...).

Possible values for a(n) are: 2, 4, 5, 7, 8, 10, 12, 13, 15, 16, 18, 20, 21, 23, 24, 26, 27, 29, ... (A260593, sorted). Density is ~ 5/8.

Record values: 4, 7, 59, 81, 105, 135, 164, 165, 173, 176, 183, 224, 246, 287, 292, 298, 308, 376, 395, 398, 433, 447, 547, ....

And the records occur for n: 1, 3, 13, 351, 5043, 17827, 135135, 181171, 190863, 313165, 513715, 563007, 4044031, 6710835, 10319167, 13358335, 28462477, 31864063, 108870007, 600495895, 913698783, 1394004493, ....

Remember these n-values are the indices of odd numbers (A005408).

This can be computed using a recursion formula discovered by an algorithm called "The Ramanujan Machine":

1*3

4/(Pi-2) = 3 + --------------------

2*4

5 + ----------------

3*5

7 + ------------

4*6

9 + --------

11 + ... .

For a proof by humans see the arXiv:1907.00205 preprint linked below.

Consider the Lyman spectrum of Hydrogen A005563(n)/A000290(n+1) = n*(n+2)/(n+1)^2 = 0/1, 3/4, 8/9, 15/16, ...

The first differences of these fractions are 3/4, 5/36, 7/144, 9/400, 11/900, 13/1764, 15/3136, ... = (2n+1)/(n*(n+1))^2.

Adding numerator and denominator of these first differences yields 1 + 2n + n^2 + 2n^3 + n^4 = A165563(n) = 3+4, 5+36, 7+144, ... = 1 + 2n + n^2*(n+1)^2 = A144396(n) + A035287(n+1) = A005408(n) + A035287(n+1).

Subtracting numerator from denominator, on the other hand, yields this sequence here: a(n) = A035287(n+1) - A005408(n).

Complement of A257521 in A144396 (odd numbers > 1).

The terms are only odd primes or squares of odd primes.

Most odd primes are present except those in A085104.

All terms which are not primes are squares of odd primes except 121 = 11^2.

Nonnegative k such that k/9 + 1 is a square. - Bruno Berselli, Apr 10 2018