cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A219365 a(n) = P(n)/(L(n)*P(n/2)*P(n/3)*P(n/7)*P(n/43)*...) with P(n) = floor(n)!, L(n) the LCM of the first n integers and where the sequence 2, 3, 7, 43, ... is A000058.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 3, 3, 3, 1, 1, 2, 2, 4, 4, 1, 5, 10, 10, 5, 5, 1, 1, 1, 3, 6, 42, 7, 7, 14, 42, 84, 84, 2, 2, 4, 12, 24, 24, 3, 3, 6, 18, 36, 36, 4, 220, 55, 165, 330, 330, 33, 33, 66, 22, 22, 1430, 130, 130, 260, 780, 156, 156, 13
Offset: 1

Views

Author

Michel Marcus, Nov 19 2012

Keywords

Comments

G. Myerson actually proved that P(n)/(P(n/2)*P(n/3)*P(n/7)*P(n/43)*...) is divisible by L(n) in a more general case. That is when n in the above expression is replaced by the terms of a sequence u(n) that satisfies GCD(u(n),u(m))=u(GCD(m,n)). And also when the sequence of quotients q(n)=2,3,7,43,... is replaced by a sequence q(n) such that sum(1/q(n))<=1.
The behavior of a(n) is quite erratic for small values of n, for instance a(26)=10, a(32)=1, a(65)=1430, a(84)=2, a(95)=542640, a(114)=3 (cf. Myerson 1994).

Crossrefs

Cf. A000058.

Programs

  • PARI
    a(n)=my(t=n!/lcm(vector(n,i,i))/(n\2)!,a1=2,a2=3);while(a2Charles R Greathouse IV, Nov 19 2012

A273056 Discriminator of Sylvester's sequence A000058.

Original entry on oeis.org

1, 2, 3, 7, 13, 13, 17, 29, 31, 61, 61, 61, 61, 61, 61, 61, 61, 109, 109, 109, 109, 109, 109, 109, 109, 128, 128, 128, 128, 128, 128, 128, 128, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 256, 512, 512, 512, 512, 512
Offset: 1

Views

Author

Jeffrey Shallit, May 14 2016

Keywords

Comments

The discriminator of a sequence is the least positive integer k such that the first n terms of the sequence are pairwise incongruent, modulo k.

Crossrefs

Cf. A000058.

A367130 a(n) is the number of divisors of A000058(n) (Sylvester's sequence).

Original entry on oeis.org

2, 2, 2, 2, 4, 2, 16, 16, 8, 32, 16
Offset: 0

Views

Author

Sean A. Irvine, Nov 05 2023

Keywords

Crossrefs

Formula

a(n) = sigma0(A000058(n)) = A000005(A000058(n)).

A367132 a(n) = phi(A000058(n)) where phi is the Euler totient function and A000058 is Sylvester's sequence.

Original entry on oeis.org

1, 2, 6, 42, 1656, 3263442, 10602458193600, 113418211994216207709696000, 12864938683276241891688635987477942287491075520480800, 164509411410546646527417252552411687615280114012485682055193015232979290788896945885448940165659663840000
Offset: 0

Views

Author

Sean A. Irvine, Nov 05 2023

Keywords

Crossrefs

Programs

  • Mathematica
    EulerPhi[a[0] = 2; a[k_] := a[k - 1]^2 - a[k - 1] + 1; Table[a[k], {k, 0, 9}]] (* Paul F. Marrero Romero, Nov 19 2023 *)

Formula

a(n) = A000010(A000058(n)).

A002061 Central polygonal numbers: a(n) = n^2 - n + 1.

Original entry on oeis.org

1, 1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, 133, 157, 183, 211, 241, 273, 307, 343, 381, 421, 463, 507, 553, 601, 651, 703, 757, 813, 871, 931, 993, 1057, 1123, 1191, 1261, 1333, 1407, 1483, 1561, 1641, 1723, 1807, 1893, 1981, 2071, 2163, 2257, 2353, 2451, 2551, 2653
Offset: 0

Views

Author

Keywords

Comments

These are Hogben's central polygonal numbers denoted by the symbol
...2....
....P...
...2.n..
(P with three attachments).
Also the maximal number of 1's that an n X n invertible {0,1} matrix can have. (See Halmos for proof.) - Felix Goldberg (felixg(AT)tx.technion.ac.il), Jul 07 2001
Maximal number of interior regions formed by n intersecting circles, for n >= 1. - Amarnath Murthy, Jul 07 2001
The terms are the smallest of n consecutive odd numbers whose sum is n^3: 1, 3 + 5 = 8 = 2^3, 7 + 9 + 11 = 27 = 3^3, etc. - Amarnath Murthy, May 19 2001
(n*a(n+1)+1)/(n^2+1) is the smallest integer of the form (n*k+1)/(n^2+1). - Benoit Cloitre, May 02 2002
For n >= 3, a(n) is also the number of cycles in the wheel graph W(n) of order n. - Sharon Sela (sharonsela(AT)hotmail.com), May 17 2002
Let b(k) be defined as follows: b(1) = 1 and b(k+1) > b(k) is the smallest integer such that Sum_{i=b(k)..b(k+1)} 1/sqrt(i) > 2; then b(n) = a(n) for n > 0. - Benoit Cloitre, Aug 23 2002
Drop the first three terms. Then n*a(n) + 1 = (n+1)^3. E.g., 7*1 + 1 = 8 = 2^3, 13*2 + 1 = 27 = 3^3, 21*3 + 1 = 64 = 4^3, etc. - Amarnath Murthy, Oct 20 2002
Arithmetic mean of next 2n - 1 numbers. - Amarnath Murthy, Feb 16 2004
The n-th term of an arithmetic progression with first term 1 and common difference n: a(1) = 1 -> 1, 2, 3, 4, 5, ...; a(2) = 3 -> 1, 3, ...; a(3) = 7 -> 1, 4, 7, ...; a(4) = 13 -> 1, 5, 9, 13, ... - Amarnath Murthy, Mar 25 2004
Number of walks of length 3 between any two distinct vertices of the complete graph K_{n+1} (n >= 1). Example: a(2) = 3 because in the complete graph ABC we have the following walks of length 3 between A and B: ABAB, ACAB and ABCB. - Emeric Deutsch, Apr 01 2004
Narayana transform of [1, 2, 0, 0, 0, ...] = [1, 3, 7, 13, 21, ...]. Let M = the infinite lower triangular matrix of A001263 and let V = the Vector [1, 2, 0, 0, 0, ...]. Then A002061 starting (1, 3, 7, ...) = M * V. - Gary W. Adamson, Apr 25 2006
The sequence 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, ... is the trajectory of 3 under repeated application of the map n -> n + 2 * square excess of n, cf. A094765.
Also n^3 mod (n^2+1). - Zak Seidov, Aug 31 2006
Also, omitting the first 1, the main diagonal of A081344. - Zak Seidov, Oct 05 2006
Ignoring the first ones, these are rectangular parallelepipeds with integer dimensions that have integer interior diagonals. Using Pythagoras: sqrt(a^2 + b^2 + c^2) = d, an integer; then this sequence: sqrt(n^2 + (n+1)^2 + (n(n+1))^2) = 2T_n + 1 is the first and most simple example. Problem: Are there any integer diagonals which do not satisfy the following general formula? sqrt((k*n)^2 + (k*(n+(2*m+1)))^2 + (k*(n*(n+(2*m+1)) + 4*T_m))^2) = k*d where m >= 0, k >= 1, and T is a triangular number. - Marco Matosic, Nov 10 2006
Numbers n such that a(n) is prime are listed in A055494. Prime a(n) are listed in A002383. All terms are odd. Prime factors of a(n) are listed in A007645. 3 divides a(3*k-1), 7 divides a(7*k-4) and a(7*k-2), 7^2 divides a(7^2*k-18) and a(7^2*k+19), 7^3 divides a(7^3*k-18) and a(7^3*k+19), 7^4 divides a(7^4*k+1048) and a(7^4*k-1047), 7^5 divides a(7^5*k+1354) and a(7^5*k-1353), 13 divides a(13*k-9) and a(13*k-3), 13^2 divides a(13^2*k+23) and a(13^2*k-22), 13^3 divides a(13^3*k+1037) and a(13^3*k-1036). - Alexander Adamchuk, Jan 25 2007
Complement of A135668. - Kieren MacMillan, Dec 16 2007
From William A. Tedeschi, Feb 29 2008: (Start)
Numbers (sorted) on the main diagonal of a 2n X 2n spiral. For example, when n=2:
.
7---8---9--10
| |
6 1---2 11
| | |
5---4---3 12
|
16--15--14--13
.
Cf. A137928. (End)
a(n) = AlexanderPolynomial[n] defined as Det[Transpose[S]-n S] where S is Seifert matrix {{-1, 1}, {0, -1}}. - Artur Jasinski, Mar 31 2008
Starting (1, 3, 7, 13, 21, ...) = binomial transform of [1, 2, 2, 0, 0, 0]; example: a(4) = 13 = (1, 3, 3, 1) dot (1, 2, 2, 0) = (1 + 6 + 6 + 0). - Gary W. Adamson, May 10 2008
Starting (1, 3, 7, 13, ...) = triangle A158821 * [1, 2, 3, ...]. - Gary W. Adamson, Mar 28 2009
Starting with offset 1 = triangle A128229 * [1,2,3,...]. - Gary W. Adamson, Mar 26 2009
a(n) = k such that floor((1/2)*(1 + sqrt(4*k-3))) + k = (n^2+1), that is A000037(a(n)) = A002522(n) = n^2 + 1, for n >= 1. - Jaroslav Krizek, Jun 21 2009
For n > 0: a(n) = A170950(A002522(n-1)), A170950(a(n)) = A174114(n), A170949(a(n)) = A002522(n-1). - Reinhard Zumkeller, Mar 08 2010
From Emeric Deutsch, Sep 23 2010: (Start)
a(n) is also the Wiener index of the fan graph F(n). The fan graph F(n) is defined as the graph obtained by joining each node of an n-node path graph with an additional node. The Wiener index of a connected graph is the sum of the distances between all unordered pairs of vertices in the graph. The Wiener polynomial of the graph F(n) is (1/2)t[(n-1)(n-2)t + 2(2n-1)]. Example: a(2)=3 because the corresponding fan graph is a cycle on 3 nodes (a triangle), having distances 1, 1, and 1.
(End)
For all elements k = n^2 - n + 1 of the sequence, sqrt(4*(k-1)+1) is an integer because 4*(k-1) + 1 = (2*n-1)^2 is a perfect square. Building the intersection of this sequence with A000225, k may in addition be of the form k = 2^x - 1, which happens only for k = 1, 3, 7, 31, and 8191. [Proof: Still 4*(k-1)+1 = 2^(x+2) - 7 must be a perfect square, which has the finite number of solutions provided by A060728: x = 1, 2, 3, 5, or 13.] In other words, the sequence A038198 defines all elements of the form 2^x - 1 in this sequence. For example k = 31 = 6*6 - 6 + 1; sqrt((31-1)*4+1) = sqrt(121) = 11 = A038198(4). - Alzhekeyev Ascar M, Jun 01 2011
a(n) such that A002522(n-1) * A002522(n) = A002522(a(n)) where A002522(n) = n^2 + 1. - Michel Lagneau, Feb 10 2012
Left edge of the triangle in A214661: a(n) = A214661(n, 1), for n > 0. - Reinhard Zumkeller, Jul 25 2012
a(n) = A215630(n, 1), for n > 0; a(n) = A215631(n-1, 1), for n > 1. - Reinhard Zumkeller, Nov 11 2012
Sum_{n > 0} arccot(a(n)) = Pi/2. - Franz Vrabec, Dec 02 2012
If you draw a triangle with one side of unit length and one side of length n, with an angle of Pi/3 radians between them, then the length of the third side of the triangle will be the square root of a(n). - Elliott Line, Jan 24 2013
a(n+1) is the number j such that j^2 = j + m + sqrt(j*m), with corresponding number m given by A100019(n). Also: sqrt(j*m) = A027444(n) = n * a(n+1). - Richard R. Forberg, Sep 03 2013
Let p(x) the interpolating polynomial of degree n-1 passing through the n points (n,n) and (1,1), (2,1), ..., (n-1,1). Then p(n+1) = a(n). - Giovanni Resta, Feb 09 2014
The number of square roots >= sqrt(n) and < n+1 (n >= 0) gives essentially the same sequence, 1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, 133, 157, 183, 211, ... . - Michael G. Kaarhus, May 21 2014
For n > 1: a(n) is the maximum total number of queens that can coexist without attacking each other on an [n+1] X [n+1] chessboard. Specifically, this will be a lone queen of one color placed in any position on the perimeter of the board, facing an opponent's "army" of size a(n)-1 == A002378(n-1). - Bob Selcoe, Feb 07 2015
a(n+1) is, for n >= 1, the number of points as well as the number of lines of a finite projective plane of order n (cf. Hughes and Piper, 1973, Theorem 3.5., pp. 79-80). For n = 3, a(4) = 13, see the 'Finite example' in the Wikipedia link, section 2.3, for the point-line matrix. - Wolfdieter Lang, Nov 20 2015
Denominators of the solution to the generalization of the Feynman triangle problem. If each vertex of a triangle is joined to the point (1/p) along the opposite side (measured say clockwise), then the area of the inner triangle formed by these lines is equal to (p - 2)^2/(p^2 - p + 1) times the area of the original triangle, p > 2. For example, when p = 3, the ratio of the areas is 1/7. The numerators of the ratio of the areas is given by A000290 with an offset of 2. [Cook & Wood, 2004.] - Joe Marasco, Feb 20 2017
n^2 equal triangular tiles with side lengths 1 X 1 X 1 may be put together to form an n X n X n triangle. For n>=2 a(n-1) is the number of different 2 X 2 X 2 triangles being contained. - Heinrich Ludwig, Mar 13 2017
For n >= 0, the continued fraction [n, n+1, n+2] = (n^3 + 3n^2 + 4n + 2)/(n^2 + 3n + 3) = A034262(n+1)/a(n+2) = n + (n+2)/a(n+2); e.g., [2, 3, 4] = A034262(3)/a(4) = 30/13 = 2 + 4/13. - Rick L. Shepherd, Apr 06 2017
Starting with b(1) = 1 and not allowing the digit 0, let b(n) = smallest nonnegative integer not yet in the sequence such that the last digit of b(n-1) plus the first digit of b(n) is equal to k for k = 1, ..., 9. This defines 9 finite sequences, each of length equal to a(k), k = 1, ..., 9. (See A289283-A289287 for the cases k = 5..9.) For k = 10, the sequence is infinite (A289288). For example, for k = 4, b(n) = 1,3,11,31,32,2,21,33,12,22,23,13,14. These terms can be ordered in the following array of size k*(k-1)+1:
1 2 3
21 22 23
31 32 33
11 12 13 14
.
The sequence ends with the term 1k, which lies outside the rectangular array and gives the term +1 (see link).- Enrique Navarrete, Jul 02 2017
The central polygonal numbers are the delimiters (in parenthesis below) when you write the natural numbers in groups of odd size 2*n+1 starting with the group {2} of size 1: (1) 2 (3) 4,5,6 (7) 8,9,10,11,12 (13) 14,15,16,17,18,19,20 (21) 22,23,24,25,26,27,28,29,30 (31) 32,33,34,35,36,37,38,39,40,41,42 (43) ... - Enrique Navarrete, Jul 11 2017
Also the number of (non-null) connected induced subgraphs in the n-cycle graph. - Eric W. Weisstein, Aug 09 2017
Since (n+1)^2 - (n+1) + 1 = n^2 + n + 1 then from 7 onwards these are also exactly the numbers that are represented as 111 in all number bases: 111(2)=7, 111(3)=13, ... - Ron Knott, Nov 14 2017
Number of binary 2 X (n-1) matrices such that each row and column has at most one 1. - Dmitry Kamenetsky, Jan 20 2018
Observed to be the squares visited by bishop moves on a spirally numbered board and moving to the lowest available unvisited square at each step, beginning at the second term (cf. A316667). It should be noted that the bishop will only travel to squares along the first diagonal of the spiral. - Benjamin Knight, Jan 30 2019
From Ed Pegg Jr, May 16 2019: (Start)
Bound for n-subset coverings. Values in A138077 covered by difference sets.
C(7,3,2), {1,2,4}
C(13,4,2), {0,1,3,9}
C(21,5,2), {3,6,7,12,14}
C(31,6,2), {1,5,11,24,25,27}
C(43,7,2), existence unresolved
C(57,8,2), {0,1,6,15,22,26,45,55}
Next unresolved cases are C(111,11,2) and C(157,13,2). (End)
"In the range we explored carefully, the optimal packings were substantially irregular only for n of the form n = k(k+1)+1, k = 3, 4, 5, 6, 7, i.e., for n = 13, 21, 31, 43, and 57." (cited from Lubachevsky, Graham link, Introduction). - Rainer Rosenthal, May 27 2020
From Bernard Schott, Dec 31 2020: (Start)
For n >= 1, a(n) is the number of solutions x in the interval 1 <= x <= n of the equation x^2 - [x^2] = (x - [x])^2, where [x] = floor(x). For n = 3, the a(3) = 7 solutions in the interval [1, 3] are 1, 3/2, 2, 9/4, 5/2, 11/4 and 3.
This sequence is the answer to the 4th problem proposed during the 20th British Mathematical Olympiad in 1984 (see link B.M.O 1984. and Gardiner reference). (End)
Called "Hogben numbers" after the British zoologist, statistician and writer Lancelot Thomas Hogben (1895-1975). - Amiram Eldar, Jun 24 2021
Minimum Wiener index of 2-degenerate graphs with n+1 vertices (n>0). A maximal 2-degenerate graph can be constructed from a 2-clique by iteratively adding a new 2-leaf (vertex of degree 2) adjacent to two existing vertices. The extremal graphs are maximal 2-degenerate graphs with diameter at most 2. - Allan Bickle, Oct 14 2022
a(n) is the number of parking functions of size n avoiding the patterns 123, 213, and 312. - Lara Pudwell, Apr 10 2023
Repeated iteration of a(k) starting with k=2 produces Sylvester's sequence, i.e., A000058(n) = a^n(2), where a^n is the n-th iterate of a(k). - Curtis Bechtel, Apr 04 2024
a(n) is the maximum number of triangles that can be traversed by starting from a triangle and moving to adjacent triangles via an edge, without revisiting any triangle, in an n X n X n equilateral triangular grid made up of n^2 unit equilateral triangles. - Kiran Ananthpur Bacche, Jan 16 2025

Examples

			G.f. = 1 + x + 3*x^2 + 7*x^3 + 13*x^4 + 21*x^5 + 31*x^6 + 43*x^7 + ...
		

References

  • Archimedeans Problems Drive, Eureka, 22 (1959), 15.
  • Steve Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 1 of the British Mathematical Olympiad 2007, page 160.
  • Anthony Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Problem 4 pp. 64 and 173 (1984).
  • Paul R. Halmos, Linear Algebra Problem Book, MAA, 1995, pp. 75-6, 242-4.
  • Ross Honsberger, Ingenuity in Mathematics, Random House, 1970, p. 87.
  • Daniel R. Hughes and Frederick Charles Piper, Projective Planes, Springer, 1973.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Cf. A010000 (minimum Weiner index of 3-degenerate graphs).

Programs

  • GAP
    List([0..50], n->n^2-n+1); # Muniru A Asiru, May 27 2018
  • Haskell
    a002061 n = n * (n - 1) + 1  -- Reinhard Zumkeller, Dec 18 2013
    
  • Magma
    [ n^2 - n + 1 : n in [0..50] ]; // Wesley Ivan Hurt, Jun 12 2014
    
  • Maple
    A002061 := proc(n)
        numtheory[cyclotomic](6,n) ;
    end proc:
    seq(A002061(n), n=0..20); # R. J. Mathar, Feb 07 2014
  • Mathematica
    FoldList[#1 + #2 &, 1, 2 Range[0, 50]] (* Robert G. Wilson v, Feb 02 2011 *)
    LinearRecurrence[{3, -3, 1}, {1, 1, 3}, 60] (* Harvey P. Dale, May 25 2011 *)
    Table[n^2 - n + 1, {n, 0, 50}] (* Wesley Ivan Hurt, Jun 12 2014 *)
    CoefficientList[Series[(1 - 2x + 3x^2)/(1 - x)^3, {x, 0, 52}], x] (* Robert G. Wilson v, Feb 18 2018 *)
    Cyclotomic[6, Range[0, 100]] (* Paolo Xausa, Feb 09 2024 *)
  • Maxima
    makelist(n^2 - n + 1,n,0,55); /* Martin Ettl, Oct 16 2012 */
    
  • PARI
    a(n) = n^2 - n + 1
    

Formula

G.f.: (1 - 2*x + 3*x^2)/(1-x)^3. - Simon Plouffe in his 1992 dissertation
a(n) = -(n-5)*a(n-1) + (n-2)*a(n-2).
a(n) = Phi_6(n) = Phi_3(n-1), where Phi_k is the k-th cyclotomic polynomial.
a(1-n) = a(n). - Michael Somos, Sep 04 2006
a(n) = a(n-1) + 2*(n-1) = 2*a(n-1) - a(n-2) + 2 = 1+A002378(n-1) = 2*A000124(n-1) - 1. - Henry Bottomley, Oct 02 2000 [Corrected by N. J. A. Sloane, Jul 18 2010]
a(n) = A000217(n) + A000217(n-2) (sum of two triangular numbers).
From Paul Barry, Mar 13 2003: (Start)
x*(1+x^2)/(1-x)^3 is g.f. for 0, 1, 3, 7, 13, ...
a(n) = 2*C(n, 2) + C(n-1, 0).
E.g.f.: (1+x^2)*exp(x). (End)
a(n) = ceiling((n-1/2)^2). - Benoit Cloitre, Apr 16 2003. [Hence the terms are about midway between successive squares and so (except for 1) are not squares. - N. J. A. Sloane, Nov 01 2005]
a(n) = 1 + Sum_{j=0..n-1} (2*j). - Xavier Acloque, Oct 08 2003
a(n) = floor(t(n^2)/t(n)), where t(n) = A000217(n). - Jon Perry, Feb 14 2004
a(n) = leftmost term in M^(n-1) * [1 1 1], where M = the 3 X 3 matrix [1 1 1 / 0 1 2 / 0 0 1]. E.g., a(6) = 31 since M^5 * [1 1 1] = [31 11 1]. - Gary W. Adamson, Nov 11 2004
a(n+1) = n^2 + n + 1. a(n+1)*a(n) = (n^6-1)/(n^2-1) = n^4 + n^2 + 1 = a(n^2+1) (a product of two consecutive numbers from this sequence belongs to this sequence). (a(n+1) + a(n))/2 = n^2 + 1. (a(n+1) - a(n))/2 = n. a((a(n+1) + a(n))/2) = a(n+1)*a(n). - Alexander Adamchuk, Apr 13 2006
a(n+1) is the numerator of ((n + 1)! + (n - 1)!)/ n!. - Artur Jasinski, Jan 09 2007
a(n) = A132111(n-1, 1), for n > 1. - Reinhard Zumkeller, Aug 10 2007
a(n) = Det[Transpose[{{-1, 1}, {0, -1}}] - n {{-1, 1}, {0, -1}}]. - Artur Jasinski, Mar 31 2008
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), n >= 3. - Jaume Oliver Lafont, Dec 02 2008
a(n) = A176271(n,1) for n > 0. - Reinhard Zumkeller, Apr 13 2010
a(n) == 3 (mod n+1). - Bruno Berselli, Jun 03 2010
a(n) = (n-1)^2 + (n-1) + 1 = 111 read in base n-1 (for n > 2). - Jason Kimberley, Oct 18 2011
a(n) = A228643(n, 1), for n > 0. - Reinhard Zumkeller, Aug 29 2013
a(n) = sqrt(A058031(n)). - Richard R. Forberg, Sep 03 2013
G.f.: 1 / (1 - x / (1 - 2*x / (1 + x / (1 - 2*x / (1 + x))))). - Michael Somos, Apr 03 2014
a(n) = A243201(n - 1) / A003215(n - 1), n > 0. - Mathew Englander, Jun 03 2014
For n >= 2, a(n) = ceiling(4/(Sum_{k = A000217(n-1)..A000217(n) - 1}, 1/k)). - Richard R. Forberg, Aug 17 2014
A256188(a(n)) = 1. - Reinhard Zumkeller, Mar 26 2015
Sum_{n>=0} 1/a(n) = 1 + Pi*tanh(Pi*sqrt(3)/2)/sqrt(3) = 2.79814728056269018... . - Vaclav Kotesovec, Apr 10 2016
a(n) = A101321(2,n-1). - R. J. Mathar, Jul 28 2016
a(n) = A000217(n-1) + A000124(n-1), n > 0. - Torlach Rush, Aug 06 2018
Sum_{n>=1} arctan(1/a(n)) = Pi/2. - Amiram Eldar, Nov 01 2020
Sum_{n=1..M} arctan(1/a(n)) = arctan(M). - Lee A. Newberg, May 08 2024
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(7)*Pi/2)*sech(sqrt(3)*Pi/2).
Product_{n>=2} (1 - 1/a(n)) = Pi*sech(sqrt(3)*Pi/2). (End)
For n > 1, sqrt(a(n)+sqrt(a(n)-sqrt(a(n)+sqrt(a(n)- ...)))) = n. - Diego Rattaggi, Apr 17 2021
a(n) = (1 + (n-1)^4 + n^4) / (1 + (n-1)^2 + n^2) [see link B.M.O. 2007 and Steve Dinh reference]. - Bernard Schott, Dec 27 2021

Extensions

Partially edited by Joerg Arndt, Mar 11 2010
Partially edited by Bruno Berselli, Dec 19 2013

A000215 Fermat numbers: a(n) = 2^(2^n) + 1.

Original entry on oeis.org

3, 5, 17, 257, 65537, 4294967297, 18446744073709551617, 340282366920938463463374607431768211457, 115792089237316195423570985008687907853269984665640564039457584007913129639937
Offset: 0

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Comments

It is conjectured that just the first 5 numbers in this sequence are primes.
An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
For n>0, Fermat numbers F(n) have digital roots 5 or 8 depending on whether n is even or odd (Koshy). - Lekraj Beedassy, Mar 17 2005
This is the special case k=2 of sequences with exact mutual k-residues. In general, a(1)=k+1 and a(n)=min{m | m>a(n-1), mod(m,a(i))=k, i=1,...,n-1}. k=1 gives Sylvester's sequence A000058. - Seppo Mustonen, Sep 04 2005
For n>1 final two digits of a(n) are periodically repeated with period 4: {17, 57, 37, 97}. - Alexander Adamchuk, Apr 07 2007
For 1 < k <= 2^n, a(A007814(k-1)) divides a(n) + 2^k. More generally, for any number k, let r = k mod 2^n and suppose r != 1, then a(A007814(r-1)) divides a(n) + 2^k. - T. D. Noe, Jul 12 2007
From Daniel Forgues, Jun 20 2011: (Start)
The Fermat numbers F_n are F_n(a,b) = a^(2^n) + b^(2^n) with a = 2 and b = 1.
For n >= 2, all factors of F_n = 2^(2^n) + 1 are of the form k*(2^(n+2)) + 1 (k >= 1).
The products of distinct Fermat numbers (in their binary representation, see A080176) give rows of Sierpiński's triangle (A006943). (End)
Let F(n) be a Fermat number. For n > 2, F(n) is prime if and only if 5^((F(n)-1)/4) == sqrt(F(n)-1) (mod F(n)). - Arkadiusz Wesolowski, Jul 16 2011
Conjecture: let the smallest prime factor of Fermat number F(n) be P(F(n)). If F(n) is composite, then P(F(n)) < 3*2^(2^n/2 - n - 2). - Arkadiusz Wesolowski, Aug 10 2012
The Fermat primes are not Brazilian numbers, so they belong to A220627, but the Fermat composites are Brazilian numbers so they belong to A220571. For a proof, see Proposition 3 page 36 on "Les nombres brésiliens" in Links. - Bernard Schott, Dec 29 2012
It appears that this sequence is generated by starting with a(0)=3 and following the rule "Write in binary and read in base 4". For an example of "Write in binary and read in ternary", see A014118. - John W. Layman, Jul 30 2013
Conjecture: the numbers > 5 in this sequence, i.e., 2^2^k + 1 for k>1, are exactly the numbers n such that (n-1)^4-1 divides 2^(n-1)-1. - M. F. Hasler, Jul 24 2015

Examples

			a(0) = 1*2^1 + 1 = 3 = 1*(2*1) + 1.
a(1) = 1*2^2 + 1 = 5 = 1*(2*2) + 1.
a(2) = 1*2^4 + 1 = 17 = 2*(2*4) + 1.
a(3) = 1*2^8 + 1 = 257 = 16*(2*8) + 1.
a(4) = 1*2^16 + 1 = 65537 = 2048*(2*16) + 1.
a(5) = 1*2^32 + 1 = 4294967297 = 641*6700417 = (10*(2*32) + 1)*(104694*(2*32) + 1).
a(6) = 1*2^64 + 1 = 18446744073709551617 = 274177*67280421310721 = (2142*(2*64) + 1)*(525628291490*(2*64) + 1).
		

References

  • M. Aigner and G. M. Ziegler, Proofs from The Book, Springer-Verlag, Berlin, 2nd. ed., 2001; see p. 3.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 7.
  • P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 87.
  • James Gleick, Faster, Vintage Books, NY, 2000 (see pp. 259-261).
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.2 Prime Numbers, pp. 78-79.
  • R. K. Guy, Unsolved Problems in Number Theory, A3.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 14.
  • E. Hille, Analytic Function Theory, Vol. I, Chelsea, N.Y., see p. 64.
  • T. Koshy, "The Digital Root Of A Fermat Number", Journal of Recreational Mathematics Vol. 32 No. 2 2002-3 Baywood NY.
  • M. Krizek, F. Luca & L. Somer, 17 Lectures on Fermat Numbers, Springer-Verlag NY 2001.
  • C. S. Ogilvy and J. T. Anderson, Excursions in Number Theory, Oxford University Press, NY, 1966, p. 36.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see pp. 18, 59.
  • C. A. Pickover, The Math Book, Sterling, NY, 2009; see p. 202.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 6-7, 70-75.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 136-137.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 148-149.

Crossrefs

a(n) = A001146(n) + 1 = A051179(n) + 2.
See A004249 for a similar sequence.
Cf. A080176 for binary representation of Fermat numbers.

Programs

  • Haskell
    a000215 = (+ 1) . (2 ^) . (2 ^)  -- Reinhard Zumkeller, Feb 13 2015
    
  • Maple
    A000215 := n->2^(2^n)+1;
  • Mathematica
    Table[2^(2^n) + 1, {n, 0, 8}] (* Alonso del Arte, Jun 07 2011 *)
  • Maxima
    A000215(n):=2^(2^n)+1$ makelist(A000215(n),n,0,10); /* Martin Ettl, Dec 10 2012 */
    
  • PARI
    a(n)=if(n<1,3*(n==0),(a(n-1)-1)^2+1)
    
  • Python
    def a(n): return 2**(2**n) + 1
    print([a(n) for n in range(9)]) # Michael S. Branicky, Apr 19 2021

Formula

a(0) = 3; a(n) = (a(n-1)-1)^2 + 1, n >= 1.
a(n) = a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get the empty product, i.e., 1, plus 2, giving 3 = a(0). - Benoit Cloitre, Sep 15 2002 [edited by Daniel Forgues, Jun 20 2011]
The above formula implies that the Fermat numbers (being all odd) are coprime.
Conjecture: F is a Fermat prime if and only if phi(F-2) = (F-1)/2. - Benoit Cloitre, Sep 15 2002
A000120(a(n)) = 2. - Reinhard Zumkeller, Aug 07 2010
If a(n) is composite, then a(n) = A242619(n)^2 + A242620(n)^2 = A257916(n)^2 - A257917(n)^2. - Arkadiusz Wesolowski, May 13 2015
Sum_{n>=0} 1/a(n) = A051158. - Amiram Eldar, Oct 27 2020
From Amiram Eldar, Jan 28 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = A249119.
Product_{n>=0} (1 - 1/a(n)) = 1/2. (End)
a(n) = 2*A077585(n) + 3. - César Aguilera, Jul 26 2023
a(n) = 2*2^A000225(n) + 1. - César Aguilera, Jul 11 2024

A001146 a(n) = 2^(2^n).

Original entry on oeis.org

2, 4, 16, 256, 65536, 4294967296, 18446744073709551616, 340282366920938463463374607431768211456, 115792089237316195423570985008687907853269984665640564039457584007913129639936
Offset: 0

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Or, write previous term in base 2, read in base 4.
a(1) = 2, a(n) = smallest power of 2 which does not divide the product of all previous terms.
Number of truth tables generated by Boolean expressions of n variables. - C. Bradford Barber (bradb(AT)shore.net), Dec 27 2005
From Ross Drewe, Feb 13 2008: (Start)
Or, number of distinct n-ary operators in a binary logic. The total number of n-ary operators in a k-valued logic is T = k^(k^n), i.e., if S is a set of k elements, there are T ways of mapping an ordered subset of n elements from S to an element of S. Some operators are "degenerate": the operator has arity p, if only p of the n input values influence the output. Therefore the set of operators can be partitioned into n+1 disjoint subsets representing arities from 0 to n.
For n = 2, k = 2 gives the familiar Boolean operators or functions, C = F(A,B). There are 2^2^2 = 16 operators, composed of: arity 0: 2 operators (C = 0 or 1), arity 1: 4 operators (C = A, B, not(A), not(B)), arity 2: 10 operators (including well-known pairs AND/NAND, OR/NOR, XOR/EQ). (End)
From José María Grau Ribas, Jan 19 2012: (Start)
Or, numbers that can be formed using the number 2, the power operator (^), and parenthesis. (End) [The paper by Guy and Selfridge (see also A003018) shows that this is the same as the current sequence. - N. J. A. Sloane, Jan 21 2012]
a(n) is the highest value k such that A173419(k) = n+1. - Charles R Greathouse IV, Oct 03 2012
Let b(0) = 8 and b(n+1) = the smallest number not in the sequence such that b(n+1) - Product_{i=0..n} b(i) divides b(n+1)*Product_{i=0..n} b(i). Then b(n) = a(n) for n > 0. - Derek Orr, Jan 15 2015
Twice the number of distinct minimal toss sequences of a coin to obtain all sequences of length n, which is 2^(2^n-1). This derives from the 2^n ways to cut each of the de Bruijn sequences B(2,n). - Maurizio De Leo, Feb 28 2015
I conjecture that { a(n) ; n>1 } are the numbers such that n^4-1 divides 2^n-1, intersection of A247219 and A247165. - M. F. Hasler, Jul 25 2015
Erdős has shown that it is an irrationality sequence (see Guy reference). - Stefano Spezia, Oct 13 2024

References

  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E24.
  • D. E. Knuth, The Art of Computer Programming, Vol. 4A, Section 7.1.1, p. 79.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

Formula

a(n+1) = (a(n))^2.
1 = Sum_{n>=0} a(n)/A051179(n+1) = 2/3 + 4/15 + 16/255 + 256/65535, ..., with partial sums: 2/3, 14/15, 254/255, 65534/65535, ... - Gary W. Adamson, Jun 15 2003
a(n) = A000079(A000079(n)). - Robert Israel, Jan 15 2015
Sum_{n>=0} 1/a(n) = A007404. - Amiram Eldar, Oct 14 2020
From Amiram Eldar, Jan 28 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = 2.
Product_{n>=0} (1 - 1/a(n)) = A215016. (End)

A003095 a(n) = a(n-1)^2 + 1 for n >= 1, with a(0) = 0.

Original entry on oeis.org

0, 1, 2, 5, 26, 677, 458330, 210066388901, 44127887745906175987802, 1947270476915296449559703445493848930452791205, 3791862310265926082868235028027893277370233152247388584761734150717768254410341175325352026
Offset: 0

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Number of binary trees of height less than or equal to n. [Corrected by Orson R. L. Peters, Jan 03 2020]
The rightmost digits cycle (0,1,2,5,6,7,0,1,2,5,6,7,...). - Jonathan Vos Post, Jul 21 2005
Apart from the initial term, a subsequence of A008318. - Reinhard Zumkeller, Jan 17 2008
Partial sums of A001699. - Jonathan Vos Post, Feb 17 2010
Corresponds to the second and second last diagonals of A119687. - John M. Campbell, Jul 25 2011
This is a divisibility sequence. - Michael Somos, Jan 01 2013
Sum_{n>=1} 1/a(n) = 1.739940825174794649210636285335916041018367182486941... . - Vaclav Kotesovec, Jan 30 2015
From Vladimir Vesic, Oct 03 2015: (Start)
Forming Herbrand's domains of formula: (∃x)(∀y)(∀z)(∃k)(P(x)∨Q(y)∧R(k))
where: x->a
k->f(y,z)
we get:
H0 = {a}
H1 = {a, f(a,a)}
H2 = {a, f(a,a), f(a,f(a,a)), f(f(a,a),a), f(f(a,a),f(a,a))}
...
The number of elements in each domain follows this sequence.
(End)
It is an open question whether or not this sequence satisfies Benford's law [Berger-Hill, 2017] - N. J. A. Sloane, Feb 07 2017
This is a strong divisibility sequence; see A329429. - Clark Kimberling, Nov 13 2019
From Peter Bala, Oct 31 2022: (Start)
Let k be a positive integer. Clearly, the sequence obtained by reducing a(n) modulo k is eventually periodic. Conjectures:
1) The sequence obtained by reducing a(n) modulo 2^k is eventually periodic with period 2.
2) The sequence obtained by reducing a(n) modulo 10^k is eventually periodic with period 6 (the case k = 1 is noted above).
3) The sequence obtained by reducing a(n) modulo 20^k is eventually periodic with period 6.
4) For n >= floor(k/2) and for 1 <= i <= 6, the value of a(6*n+i) mod 10^k is a constant independent of n. The digits of these 6 constant integers, when read from right to left, are the first k digits of the 10-adic numbers A318135 (i = 1), A318136 (i = 2), A318137 (i = 3), A318138 (i = 4), A318139 (i = 5) and A318140 (i = 6), respectively. An example is given below.
n a(6*n+1) mod 10^11
1 10066388901
2 72084948901
3 67988948901
4 61588948901
5 01588948901
6 01588948901
7 01588948901
... ...
A318135 begins 1, 0, 9, 8, 4, 9, 8, 8, 5, 1, 0, 2, .... (End)

References

  • Mordechai Ben-Ari, Mathematical Logic for Computer Science, Third edition, 173-203.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 443-448.
  • R. K. Guy, How to factor a number, Proc. 5th Manitoba Conf. Numerical Math., Congress. Num. 16 (1975), 49-89.
  • R. Penrose, The Emperor's New Mind, Oxford, 1989, p. 122.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A137560, which enumerates binary trees of height less than n and exactly j leaf nodes. - Robert Munafo, Nov 03 2009

Programs

Formula

a(n) = B_{n-1}(1) where B_n(x) = 1 + x*B_{n-1}(x)^2 is the generating function of trees of height <= n.
a(n) is asymptotic to c^(2^n) where c=1.2259024435287485386279474959130085213... (see A076949). - Benoit Cloitre, Nov 27 2002
c = b^(1/4) where b is the constant in Bottomley's formula in A004019. a(n) appears very asymptotic to c^(2^n) - Sum_{k>=1} A088674(k)/(2*c^(2^n))^(2*k-1). - Gerald McGarvey, Nov 17 2007
a(n) = Sum_{i=1..n} A001699(i). - Jonathan Vos Post, Feb 17 2010
G.f. = x + 2*x^2 + 5*x^3 + 26*x^4 + 677*x^5 + 458330*x^6 + 210066388901*x^7 + ... . - Michael Somos, Jan 01 2013
a(2n) mod 2 = 0 ; a(2n+1) mod 2 = 1. - Altug Alkan, Oct 04 2015
a(n) + a(n-1) = A213437(n). - Peter Bala, Feb 03 2017
0 = a(n)^2*(+a(n+1) + a(n+2)) + a(n+1)^2*(-a(n+1) - a(n+2) - a(n+3)) + a(n+2)^3 for all n>=0. - Michael Somos, Feb 10 2017
a(n) = A091980(2^(n-1)) for n > 0. - Alois P. Heinz, Jul 11 2019

Extensions

Additional comments from Cyril Banderier, Jun 05 2000
Minor edits by Vaclav Kotesovec, Oct 04 2014
Initial term clarified by Clark Kimberling, Nov 13 2019

A002966 Egyptian fractions: number of solutions of 1 = 1/x_1 + ... + 1/x_n where 0 < x_1 <= ... <= x_n.

Original entry on oeis.org

1, 1, 3, 14, 147, 3462, 294314, 159330691
Offset: 1

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All denominators in the expansion 1 = 1/x_1 + ... + 1/x_n are bounded by A000058(n-1), i.e., 0 < x_1 <= ... <= x_n < A000058(n-1). Furthermore, for a fixed n, x_i <= (n+1-i)*(A000058(i-1)-1). - Max Alekseyev, Oct 11 2012
From R. J. Mathar, May 06 2010: (Start)
This is the leading edge of the triangle A156869. This is also the row n=1 of an array T(n,m) which gives the number of ways to write 1/n as a sum over m (not necessarily distinct) unit fractions:
1, 1, 3, 14, 147, 3462, 294314, ...
1, 2, 10, 108, 2892, 270332, ...
1, 2, 21, 339, 17253, ...
1, 3, 28, 694, 51323, ...
...
T(.,2) = A018892. T(.,3) = A004194. T(.,4) = A020327, T(.,5) = A020328. T(2,6) is computed by D. S. McNeil, who conjectures that the 2nd row is A003167. (End)
If on the other hand, all x_k must be unique, see A006585. - Robert G. Wilson v, Jul 17 2013

Examples

			For n=3 the 3 solutions are {2,3,6}, {2,4,4}, {3,3,3}.
For n=4 the solutions are: {2,3,7,42}, {2,3,8,24}, {2,3,9,18}, {2,3,10,15}, {2,3,12,12}, {2,4,5,20}, {2,4,6,12}, {2,4,8,8}, {2,5,5,10}, {2,6,6,6}, {3,3,4,12}, {3,3,6,6}, {3,4,4,6}, {4,4,4,4}. [Neven Juric, May 14 2008]
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, D11.
  • D. Singmaster, The number of representations of one as a sum of unit fractions, unpublished manuscript, 1972.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • PARI
    a(n,rem=1,mn=1)=if(n==1,return(numerator(rem)==1)); sum(k=max(1\rem+1,mn), n\rem, a(n-1,rem-1/k,k)) \\ Charles R Greathouse IV, Jan 04 2015

Formula

a(n) <= binomial(A007018(n), n-1). - Charles R Greathouse IV, Jul 29 2024

Extensions

a(7) from Jud McCranie, Nov 15 1999. Confirmed by Marc Paulhus.
a(8) from John Dethridge (jcd(AT)ms.unimelb.edu.au) and Jacques Le Normand (jacqueslen(AT)sympatico.ca), Jan 06 2004

A007018 a(n) = a(n-1)^2 + a(n-1), a(0)=1.

Original entry on oeis.org

1, 2, 6, 42, 1806, 3263442, 10650056950806, 113423713055421844361000442, 12864938683278671740537145998360961546653259485195806
Offset: 0

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Number of ordered trees having nodes of outdegree 0,1,2 and such that all leaves are at level n. Example: a(2)=6 because, denoting by I a path of length 2 and by Y a Y-shaped tree with 3 edges, we have I, Y, I*I, I*Y, Y*I, Y*Y, where * denotes identification of the roots. - Emeric Deutsch, Oct 31 2002
Equivalently, the number of acyclic digraphs (dags) that unravel to a perfect binary tree of height n. - Nachum Dershowitz, Jul 03 2022
a(n) has at least n different prime factors. [Saidak]
Subsequence of squarefree numbers (A005117). - Reinhard Zumkeller, Nov 15 2004 [This has been questioned, see MathOverflow link. - Charles R Greathouse IV, Mar 30 2015]
For prime factors see A007996.
Curtiss shows that if the reciprocal sum of the multiset S = {x_1, x_2, ..., x_n} is 1, then max(S) <= a(n). - Charles R Greathouse IV, Feb 28 2007
The number of reduced ZBDDs for Boolean functions of n variables in which there is no zero sink. (ZBDDs are "zero-suppressed binary decision diagrams.") For example, a(2)=6 because of the 2-variable functions whose truth tables are 1000, 1010, 1011, 1100, 1110, 1111. - Don Knuth, Jun 04 2007
Using the methods of Aho and Sloane, Fibonacci Quarterly 11 (1973), 429-437, it is easy to show that a(n) is the integer just a tiny bit below the real number theta^{2^n}-1/2, where theta =~ 1.597910218 is the exponential of the rapidly convergent series Sum_{n>=0} log(1+1/a_n)/2^{n+1}. For example, theta^32 - 1/2 =~ 3263442.0000000383. - Don Knuth, Jun 04 2007 [Corrected by Darryl K. Nester, Jun 19 2017]
The next term has 209 digits. - Harvey P. Dale, Sep 07 2011
Urquhart shows that a(n) is the minimum size of a tableau refutation of the clauses of the complete binary tree of depth n, see pp. 432-434. - Charles R Greathouse IV, Jan 04 2013
For any positive a(0), the sequence a(n) = a(n-1) * (a(n-1) + 1) gives a constructive proof that there exists integers with at least n distinct prime factors, e.g. a(n). As a corollary, this gives a constructive proof of Euclid's theorem stating that there are an infinity of primes. - Daniel Forgues, Mar 03 2017
Lower bound for A100016 (with equality for the first 5 terms), where a(n)+1 is replaced by nextprime(a(n)). - M. F. Hasler, May 20 2019

References

  • R. Honsberger, Mathematical Gems III, M.A.A., 1985, p. 94.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Lower bound for A100016.
Row sums of A122888.

Programs

  • Haskell
    a007018 n = a007018_list !! n
    a007018_list = iterate a002378 1  -- Reinhard Zumkeller, Dec 18 2013
    
  • Magma
    [n eq 1 select 1 else Self(n-1)^2 + Self(n-1): n in [1..10]]; // Vincenzo Librandi, May 19 2015
    
  • Maple
    A007018 := proc(n)
        option remember;
        local aprev;
        if n = 0 then
            1;
        else
            aprev := procname(n-1) ;
            aprev*(aprev+1) ;
        end if;
    end proc: # R. J. Mathar, May 06 2016
  • Mathematica
    FoldList[#^2 + #1 &, 1, Range@ 8] (* Robert G. Wilson v, Jun 16 2011 *)
    NestList[#^2 + #&, 1, 10] (* Harvey P. Dale, Sep 07 2011 *)
  • Maxima
    a[1]:1$
    a[n]:=(a[n-1] + (a[n-1]^2))$
    A007018(n):=a[n]$
    makelist(A007018(n),n,1,10); /* Martin Ettl, Nov 08 2012 */
    
  • PARI
    a(n)=if(n>0,my(x=a(n-1));x^2+x,1) \\ Edited by M. F. Hasler, May 20 2019 and Jason Yuen, Mar 01 2025
    
  • Python
    from itertools import islice
    def A007018_gen(): # generator of terms
        a = 1
        while True:
            yield a
            a *= a+1
    A007018_list = list(islice(A007018_gen(),9)) # Chai Wah Wu, Mar 19 2024

Formula

a(n) = A000058(n)-1 = A000058(n-1)^2 - A000058(n-1) = 1/(1-Sum_{jA000058(j)) where A000058 is Sylvester's sequence. - Henry Bottomley, Jul 23 2001
a(n) = floor(c^(2^n)) where c = A077125 = 1.597910218031873178338070118157... - Benoit Cloitre, Nov 06 2002
a(1)=1, a(n) = Product_{k=1..n-1} (a(k)+1). - Benoit Cloitre, Sep 13 2003
a(n) = A139145(2^(n+1) - 1). - Reinhard Zumkeller, Apr 10 2008
If an (additional) initial 1 is inserted, a(n) = Sum_{kFranklin T. Adams-Watters, Jun 11 2009
a(n+1) = a(n)-th oblong (or promic, pronic, or heteromecic) numbers (A002378). a(n+1) = A002378(a(n)) = A002378(a(n-1)) * (A002378(a(n-1)) + 1). - Jaroslav Krizek, Sep 13 2009
a(n) = A053631(n)/2. - Martin Ettl, Nov 08 2012
Sum_{n>=0} (-1)^n/a(n) = A118227. - Amiram Eldar, Oct 29 2020
Sum_{n>=0} 1/a(n) = A371321. - Amiram Eldar, Mar 19 2024
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