A026741 -id:A026741 - cp's OEIS frontend

A165355 a(n) = 3n + 1 if n is even, or a(n) = (3n + 1)/2 if n is odd.

1, 2, 7, 5, 13, 8, 19, 11, 25, 14, 31, 17, 37, 20, 43, 23, 49, 26, 55, 29, 61, 32, 67, 35, 73, 38, 79, 41, 85, 44, 91, 47, 97, 50, 103, 53, 109, 56, 115, 59, 121, 62, 127, 65, 133, 68, 139, 71, 145, 74, 151, 77, 157, 80, 163, 83, 169, 86, 175, 89, 181, 92, 187, 95, 193, 98

Offset: 0

Paul Curtz, Sep 16 2009

Second trisection of A026741.
A111329(n+1) = A000041(a(n)). - Reinhard Zumkeller, Nov 19 2009
We observe that this sequence is a particular case of sequence for which there exists q: a(n+3) = (a(n+2)*a(n+1)+q)/a(n) for every n >= n0. Here q=-9 and n0=0. - Richard Choulet, Mar 01 2010
The entries are also encountered via the bilinear transform approximation to the natural log (unit circle). Specifically, evaluating 2(z-1)/(z+1) at z = 2, 3, 4, ..., A165355 entries stem from the pair (sums) seen 2 ahead of each new successive prime. For clarity, the evaluation output is 2, 3, 1, 1, 6, 5, 4, 3, 10, 7, 3, 2, 14, 9, 8, 5, 18, 11, ..., where (1+1), (4+3), (3+2), (8+5), ... generate the A165355 entries (after the first). As an aside, the same mechanism links A165355 to A140777. - Bill McEachen, Jan 08 2015
As a follow-up to the previous comment, it appears that the numerators and denominators of 2(z-1)/(z+1) are respectively given by A145979 and A060819, but with different offsets. - Michel Marcus, Jan 14 2015
Odd parts of the terms give A067745. E.g.: 1, 2/2, 7, 5, 13, 8/8 .... - Joe Slater, Nov 30 2016

G. C. Greubel, Table of n, a(n) for n = 0..1000
D. H. Lehmer, Continued fractions containing arithmetic progressions, Scripta Mathematica, 29 (1973): 17-24. [Annotated copy of offprint]
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A165351, A165367, A067745, A007814.

f[n_] := If[ OddQ@ n, (3n +1)/2, (3n +1)]; Array[f, 66, 0] (* Robert G. Wilson v, Jan 26 2015 *)
f[n_] := (3 (-1)^(2n) + (-1)^(1 + n)) (-2 + 3n)/4; Array[f, 66] (* or *)
CoefficientList[ Series[(x^3 + 5x^2 + 2x + 1)/(x^2 - 1)^2, {x, 0, 65}], x] (* or *)
LinearRecurrence[{0, 2, 0, -1}, {1, 2, 7, 5}, 66] (* Robert G. Wilson v, Apr 13 2017 *)

a(n)=n+=2*n+1; if(n%2,n,n/2) \\ Charles R Greathouse IV, Jan 13 2015

a(n) = A026741(3*n+1).
a(n)*A026741(n) = A005449(n).
a(n)*A022998(n+1) = A000567(n+1).
a(n) = A026741(n+1) + A022998(n).
a(2n) = A016921(n). a(2n+1) = A016789(n).
a(2n+1)*A026741(2n) = A045944(n).
G.f.: (1+2*x+5*x^2+x^3)/((x-1)^2 * (1+x)^2). - R. J. Mathar, Sep 26 2009
a(n) = (3+9*n)/4 + (-1)^n*(1+3*n)/4. - R. J. Mathar, Sep 26 2009
a(n) = 2*(3n+1)/(4-((2n+2) mod 4)). - Bill McEachen, Jan 09 2015
If a(2n-1) = x then a(2n) = 2x+3. - Robert G. Wilson v, Jan 26 2015
Let the reduced Collatz procedure be defined as Cr(n) = (3*n+1)/2. For odd n, a(n) = Cr(n). For even n, a(n) = Cr(4*n+1)/2. - Joe Slater, Nov 29 2016
a(n) = A067745(n+1) * 2^A007814((3n+1)/2). - Joe Slater, Nov 30 2016
a(n) = 2*a(n-2) - a(n-4). - G. C. Greubel, Apr 13 2017

All comments changed to formulas by R. J. Mathar, Sep 26 2009
New name from Charles R Greathouse IV, Jan 13 2015
Name corrected by Joe Slater, Nov 29 2016

A065423 Number of ordered length 2 compositions of n with at least one even summand.

Original entry on oeis.org

0, 0, 2, 1, 4, 2, 6, 3, 8, 4, 10, 5, 12, 6, 14, 7, 16, 8, 18, 9, 20, 10, 22, 11, 24, 12, 26, 13, 28, 14, 30, 15, 32, 16, 34, 17, 36, 18, 38, 19, 40, 20, 42, 21, 44, 22, 46, 23, 48, 24, 50, 25, 52, 26, 54, 27, 56, 28, 58, 29, 60, 30, 62, 31, 64, 32, 66, 33, 68, 34, 70, 35, 72, 36, 74

Offset: 1

Len Smiley, Nov 23 2001

			a(7) = 6 because we can write 7 = 1+6 = 2+5 = 3+4 = 4+3 = 5+2 = 6+1; a(8) = 3 because we can write 8 = 2+6 = 4+4 = 6+2.

Stefano Spezia, Table of n, a(n) for n = 1..10000
Mircea Merca and Emil Simion, n-Color Partitions into Distinct Parts as Sums over Partitions, Symmetry (2023) Vol. 15, Iss. 11.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A026741, A097140 (first differences), A030451 (absolute first differences), A210530.

int a(int n){n--;return n>>(n&1);} // Mia Boudreau, Aug 27 2025

A065423 := proc(n)
    (3*n-4-(-1)^n*n)/4 ;
end proc:
seq(A065423(n),n=1..40) ; # R. J. Mathar, Jan 24 2022

LinearRecurrence[{0,2,0,-1},{0,0,2,1},100] (* Harvey P. Dale, May 14 2014 *)

PARI
```
a(n)=n-=2;if(n%2,n+1,n/2)
```

G.f.: x^3*(x+2)/(1-x^2)^2.
a(n) = floor((n-1)/2) + (n is odd)*floor((n-1)/2).
a(n+2) = Sum_{k=0..n} (gcd(n, k) mod 2). - Paul Barry, May 02 2005
a(n) = Sum_{i=1..n-1} (-1)^i (floor(i/2) + ((i+1) mod 2)). - Olivier Gérard, Jun 21 2007
a(n) = A210530(n,4)/2 for n>2. - Boris Putievskiy, Jan 29 2013
a(n) = (3*n-4-n*(-1)^n)/4. - Boris Putievskiy, Jan 29 2013, corrected Jan 24 2022
a(n) = A026741(n)-1. - Wesley Ivan Hurt, Jun 23 2013
a(n) = floor((n-1) / 2^mod(n-1,2)). - Mia Boudreau, Aug 27 2025
E.g.f.: 1 + (x - 1)*cosh(x) + (x - 2)*sinh(x)/2. - Stefano Spezia, Dec 17 2023

A106617 Numerator of n/(n+16).

Original entry on oeis.org

0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 15, 1, 17, 9, 19, 5, 21, 11, 23, 3, 25, 13, 27, 7, 29, 15, 31, 2, 33, 17, 35, 9, 37, 19, 39, 5, 41, 21, 43, 11, 45, 23, 47, 3, 49, 25, 51, 13, 53, 27, 55, 7, 57, 29, 59, 15, 61, 31, 63, 4, 65, 33, 67, 17, 69, 35, 71, 9, 73, 37, 75, 19, 77, 39, 79

Offset: 0

N. J. A. Sloane, May 15 2005

A multiplicative sequence. Also, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n >= 1, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 21 2019

			From _Peter Bala_, Feb 21 2019: (Start)
Sum_{n >= 1} n*a(n)*x^n = G(x) - 2*G(x^2) - 4*G(x^4) - 8*G(x^8) - 16*G(x^16), where G(x) = x*(1 + x)/(1 - x)^3.
Sum_{n >= 1} n^2*a(n)*x^n = H(x) - 2^2*H(x^2) - 4^2*H(x^4) - 8^2*H(x^8) - 16^2*H(x^16), where H(x) = x*(1 + 4*x + x^2)/(1 - x)^4. In general, the o.g.f. for Sum_{n >= 1} (n^k*a(n))*x^n for positive k involves the Eulerian polynomials.
In the other direction,
Sum_{n >= 1} (a(n)/n)*x^n = J(x) - (1/2)*J(x^2) - (1/4)*J(x^4) - (1/8)*J(x^8) - (1/16)*J(x^16), where J(x) = x/(1 - x).
Sum_{n >= 1} (a(n)/n^2)*x^n = L(x) - (1/2^2)*L(x^2) - (1/4^2)*L(x^4) - (1/8^2)*L(x^8) - (1/16^2)*L(x^16), where L(x) = log(1/(1 - x)). In general, the o.g.f. for Sum_{n >= 0} (a(n)/n^k)*x^n, for k >= 3, involves the polylogarithm Li_(k-1)(x).
Sum_{n >= 1} (1/a(n))*x^n = L(x) + (1/2)*L(x^2) + (1/2)*L(x^4) + (1/2)*L(x^8) + (1/2)*L(x^16). (End)

Nathaniel Johnston, Table of n, a(n) for n = 0..10000
Peter Bala, A note on the sequence of numerators of a rational function, 2019.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1).

Cf. A106614, A106615, A106616, A106618, A106619, A106620, A106621.

Cf. Other sequences given by the formula numerator(n/(n + k)): A026741 (k = 2), A051176 (k = 3), A060819 (k = 4), A060791 (k = 5), A060789 (k = 6), A106608 thru A106612 (k = 7 thru 11), A051724 (k = 12), A106614 thru A106621 (k = 13 thru 20).

List([0..100],n->NumeratorRat(n/(n+16))); # Muniru A Asiru, Feb 19 2019

[Numerator(n/(n+16)): n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

seq(numer(n/(n+16)), n=0..100); # Nathaniel Johnston, Apr 18 2011

Numerator/@Table[n/(n+16),{n,0,10}] (* Harvey P. Dale, Apr 26 2011 *)

a(n)=n>>min(valuation(n,2),4) \\ Charles R Greathouse IV, Jun 17 2012

[lcm(n,16)/16 for n in range(0, 100)] # Zerinvary Lajos, Jun 12 2009

a(n) = 2*a(n-16) - a(n-32) for n > 31. - Paul Curtz, Apr 12 2011
Octosections: a(8*n) = A026741(n). a(2+8*n) = 1+4*n. a(4+8*n) = 1+2*n. a(6+8*n) = 3+4*n. Bisection: a(1+2*n) = 1+2*n. - Paul Curtz, Apr 12 2011
Dirichlet g.f.: zeta(s-1)*(1-1/2^s-1/4^s-1/8^s-1/16^s). - R. J. Mathar, Apr 18 2011
a(n) = numerator of n/(2^(2*n+1)). - Ralf Steiner, Feb 09 2017
The previous comment is incorrect, a(n) first differs from the numerator of n/(2^(2*n+1)) at n = 32. - Peter Bala, Feb 27 2019
From Peter Bala, Feb 21 2019: (Start)
a(n) = n/gcd(n,16), where gcd(n,16) = [1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, ...] is a periodic sequence of period 16: a(n) is thus quasi_polynomial in n.
O.g.f.: Sum_{n >= 0} a(n)*x^n = F(x) - F(x^2) - F(x^4) - F(x^8) - F(x^16), where F(x) = x/(1 - x)^2.
More generally, Sum_{n >= 0} (a(n)^m)*x^n = F(m,x) + (1 - 2^m)*( F(m,x^2) + F(m,x^4) + F(m,x^8) + F(m,x^16) ), where F(m,x) = A(m,x)/(1 - x)^(m+1) with A(m,x) the m-th Eulerian polynomial: A(1,x) = x, A(2,x) = x*(1 + x), A(3,x) = x*(1 + 4*x + x^2) - see A008292.
Repeatedly applying the Euler operator x*d/dx or its inverse operator to the o.g.f. for the sequence produces generating functions for the sequences ( (n^m)*a(n) )n>=1 for m in Z. Some examples are given below. (End)
From Amiram Eldar, Nov 25 2022: (Start)
Multiplicative with a(2^e) = 2^max(0,e-4), and a(p^e) = p^e if p>2.
Sum_{k=1..n} a(k) ~ (171/512) * n^2. (End)

A145051 Numerator of the first convergent to sqrt(n) using the recursion x = (n/x + x)/2.

Original entry on oeis.org

1, 3, 2, 5, 3, 7, 4, 9, 5, 11, 6, 13, 7, 15, 8, 17, 9, 19, 10, 21, 11, 23, 12, 25, 13, 27, 14, 29, 15, 31, 16, 33, 17, 35, 18, 37, 19, 39, 20, 41, 21, 43, 22, 45, 23, 47, 24, 49, 25, 51, 26, 53, 27, 55, 28, 57, 29, 59, 30, 61, 31, 63, 32, 65, 33, 67, 34, 69, 35, 71, 36, 73, 37, 75

Offset: 1

Cino Hilliard, Sep 30 2008

This is the same as A026741 without the first 2 terms in A026741. The link describes the experimental derivation of the generating function.
From Jaroslav Krizek, May 28 2010: (Start)
Numerators of arithmetic means of the first n positive integers for n >= 1.
See A040001 - denominators of arithmetic means of the first n positive integers.
a(n) = A026741(n+1) = A000217(n) * A040001(n) / n. (End)
Minimum number of line segments to draw into a circle to partition the circle into n+1 congruent circular sectors, i.e., minimum number of straight cuts required to cut a circular cake into n+1 equal slices. - Felix Fröhlich, Sep 01 2015
Continued fraction expansion of A386934. - Kelvin Voskuijl, Aug 15 2025

			n=1, x=1; x = (1/1+1)/2 = 1/1;
n=2, x=1; x = (2/1+1)/2 = 3/2;
n=3, x=1; x = (3/1+1)/2 = 2/1.
G.f.: x + 3*x^2 + 2*x^3 + 5*x^4 + 3*x^5 + 7*x^6 + 4*x^7 + 9*x^8 + 5*x^9 + ...

Kelvin Voskuijl, Table of n, a(n) for n = 1..10000
Cino Hilliard, Roots by Recursion [Broken link]
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A000217, A026741, A040001.

[(n+1)*(3 - (-1)^(n-1))/4: n in [1..100]]; // Vincenzo Librandi, Sep 02 2015

lst={};Do[a=n^2+n;b=n^2-n;c=a/b;AppendTo[lst,Denominator[c]],{n,2,5!}];lst (* Vladimir Joseph Stephan Orlovsky, Oct 20 2009 *)

g(n, p) = x=1;for(j=1,p,x=(n/x+x)/2; if(j==1, print1(numerator(x), ",")))
for(k=1,100,g(k,1))

From Paul Barry, Nov 22 2009: (Start)
G.f.: x*(1 + 3*x - x^3)/(1 - x^2)^2.
a(n+1) = (n + 2)*(3 - (-1)^n)/4;
a(n+1) = Sum_{k=0..n, if(k=floor(n/2) or k=floor((n+1)/2),1,0)*(k+1)}. (End)
E.g.f.: ((x + 2)*cosh(x) + (2*x + 1)*sinh(x) - 2)/2. - Stefano Spezia, Apr 04 2024

A106619 a(n) = numerator of n/(n+18).

Original entry on oeis.org

0, 1, 1, 1, 2, 5, 1, 7, 4, 1, 5, 11, 2, 13, 7, 5, 8, 17, 1, 19, 10, 7, 11, 23, 4, 25, 13, 3, 14, 29, 5, 31, 16, 11, 17, 35, 2, 37, 19, 13, 20, 41, 7, 43, 22, 5, 23, 47, 8, 49, 25, 17, 26, 53, 3, 55, 28, 19, 29, 59, 10, 61, 31, 7, 32, 65, 11, 67, 34, 23, 35, 71, 4, 73, 37, 25, 38, 77, 13, 79

Offset: 0

N. J. A. Sloane, May 15 2005

a(n+3), n >= 0, is the denominator of the harmonic mean H(n,3) = 6*n/(n+3). a(n+3) = (n+3)/gcd(n+3,18). - Wolfdieter Lang, Jul 04 2013

G. C. Greubel, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1).

Cf. Sequences given by the formula numerator(n/(n + k)): A026741 (k = 2), A051176 (k = 3), A060819 (k = 4), A060791 (k = 5), A060789 (k = 6), A106608 thru A106612 (k = 7 thru 11), A051724 (k = 12), A106614 thru A106621 (k = 13 thru 20).

Cf. A227042.

List([0..80],n->NumeratorRat(n/(n+18))); # Muniru A Asiru, Feb 19 2019

[Numerator(n/(n+18)): n in [0..100]]; // Vincenzo Librandi, Apr 18 2011

seq(numer(n/(n+18)),n=0..80); # Muniru A Asiru, Feb 19 2019

f[n_]:=Numerator[n/(n+18)];Array[f,100,0] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2011 *)

vector(100, n, n--; numerator(n/(n+18))) \\ G. C. Greubel, Feb 19 2019

[lcm(n,18)/18 for n in range(0, 100)] # Zerinvary Lajos, Jun 12 2009

a(n) = 2*a(n-18) - a(n-36). - Paul Curtz, Feb 27 2011
Nonasection: a(9*n) = A026741(n). - Paul Curtz, Mar 21 2011
Dirichlet g.f.: zeta(s-1)*(1 - 2/3^s - 2/9^s - 1/2^s + 2/6^s + 2/18^s). - R. J. Mathar, Apr 18 2011
a(n) = n/gcd(n,18), n >= 0. See the harmonic mean comment above, and the Zerinvary Lajos program below. - Wolfdieter Lang, Jul 04 2013
a(n+3) = A227042(n+3,3), n >= 0. - Wolfdieter Lang, Jul 04 2013
From Amiram Eldar, Nov 25 2022: (Start)
Multiplicative with a(2^e) = 2^max(0, e-1), a(3^e) = 3^max(0,e-2), and a(p^e) = p^e otherwise.
Sum_{k=1..n} a(k) ~ (61/216) * n^2. (End)

A209297 Triangle read by rows: T(n,k) = k*n + k - n, 1 <= k <= n.

Original entry on oeis.org

1, 1, 4, 1, 5, 9, 1, 6, 11, 16, 1, 7, 13, 19, 25, 1, 8, 15, 22, 29, 36, 1, 9, 17, 25, 33, 41, 49, 1, 10, 19, 28, 37, 46, 55, 64, 1, 11, 21, 31, 41, 51, 61, 71, 81, 1, 12, 23, 34, 45, 56, 67, 78, 89, 100, 1, 13, 25, 37, 49, 61, 73, 85, 97, 109, 121, 1, 14, 27

Offset: 1

Reinhard Zumkeller, Jan 19 2013

From Michel Marcus, May 18 2021: (Start)
The n-th row of the triangle is the main diagonal of an n X n square array whose elements are the numbers from 1..n^2, listed in increasing order by rows.
                                             [1   2  3  4  5]
                             [1   2  3  4]   [6   7  8  9 10]
                   [1 2 3]   [5   6  7  8]   [11 12 13 14 15]
          [1 2]    [4 5 6]   [9  10 11 12]   [16 17 18 19 20]
  [1]     [3 4]    [7 8 9]   [13 14 15 16]   [21 22 23 24 25]
  -----------------------------------------------------------
   1       1 4      1 5 9     1  6  11 16     1  7  13 19 25
(End)

			From _Muniru A Asiru_, Oct 31 2017: (Start)
Triangle begins:
  1;
  1,  4;
  1,  5,  9;
  1,  6, 11, 16;
  1,  7, 13, 19, 25;
  1,  8, 15, 22, 29, 36;
  1,  9, 17, 25, 33, 41, 49;
  1, 10, 19, 28, 37, 46, 55, 64;
  1, 11, 21, 31, 41, 51, 61, 71, 81;
  1, 12, 23, 34, 45, 56, 67, 78, 89, 100;
  ... (End)

Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened

Cf. A162610; A000012 (left edge), A000290 (right edge), A006003 (row sums), A001844 (central terms), A026741 (number of odd terms per row), A142150 (number of even terms per row), A221490 (number of primes per row).

Flat(List([1..10^3], n -> List([1..n], k -> k * n + k - n))); # Muniru A Asiru, Oct 31 2017

a209297 n k = k * n + k - n
a209297_row n = map (a209297 n) [1..n]
a209297_tabl = map a209297_row [1..]

Array[Range[1, #^2, #+1]&,10] (* Paolo Xausa, Feb 08 2024 *)

T(n,k) = (k-1)*(n+1)+1.

A107711 Triangle read by rows: T(0,0)=1, T(n,m) = binomial(n,m) * gcd(n,m)/n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 5, 10, 5, 1, 1, 1, 1, 3, 5, 5, 3, 1, 1, 1, 1, 7, 7, 35, 7, 7, 1, 1, 1, 1, 4, 28, 14, 14, 28, 4, 1, 1, 1, 1, 9, 12, 42, 126, 42, 12, 9, 1, 1, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 1, 1, 11, 55, 165, 66, 462, 66, 165, 55, 11, 1, 1

Offset: 0

Leroy Quet, Jun 10 2005

T(0,0) is an indeterminate, but 1 seems a logical value to assign it. T(n,0) = T(n,1) = T(n,n-1) = T(n,n) = 1.

T(2n,n) = A001700(n-1) (n>=1). - Emeric Deutsch, Jun 13 2005

			T(6,2)=5 because binomial(6,2)*gcd(6,2)/6 = 15*2/6 = 5.
The triangle T(n,m) begins:
n\m 0  1  2   3   4    5   6   7  8  9  10...
0:  1
1:  1  1
2:  1  1  1
3:  1  1  1   1
4:  1  1  3   1   1
5:  1  1  2   2   1    1
6:  1  1  5  10   5    1   1
7:  1  1  3   5   5    3   1   1
8:  1  1  7   7  35    7   7   1  1
9:  1  1  4  28  14   14  28   4  1  1
10: 1  1  9  12  42  126  42  12  9  1   1
n\m 0  1  2   3   4    5   6   7  8  9  10...
... reformatted - _Wolfdieter Lang_, Feb 23 2014

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened
Wolfdieter Lang, On Collatz' Words, Sequences and Trees, arXiv:1404.2710 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.11.7.

Row sums A159554.
Cf. A007318, A001700, A000108, A050169, A234040, A026741, A234042.
Cf. A109004.

a107711 n k = a107711_tabl !! n !! k
a107711_row n = a107711_tabl !! n
a107711_tabl = [1] : zipWith (map . flip div) [1..]
               (tail $ zipWith (zipWith (*)) a007318_tabl a109004_tabl)
-- Reinhard Zumkeller, Feb 28 2014

a:=proc(n,k) if n=0 and k=0 then 1 elif k<=n then binomial(n,k)*gcd(n,k)/n else 0 fi end: for n from 0 to 13 do seq(a(n,k),k=0..n) od; # yields sequence in triangular form. - Emeric Deutsch, Jun 13 2005

T[0, 0] = 1; T[n_, m_] := Binomial[n, m] * GCD[n, m]/n;
Table[T[n, m], {n, 1, 13}, {m, 1, n}] // Flatten (* Jean-François Alcover, Nov 16 2017 *)

From Wolfdieter Lang, Feb 28 2014 (Start)
T(n, m) = T(n-1,m)*(n-1)*gcd(n,m)/((n-m)*gcd(n-1,m)), n > m >= 1, T(n, 0) = 1, T(n, n) = 1, otherwise 0.
T(n, m) = binomial(n-1,m-1)*gcd(n,m)/m for n >= m >= 1, T(n,0) = 1, otherwise 0 (from iteration of the preceding recurrence).
T(n, m) = T(n-1, m-1)*(n-1)*gcd(n,m)/(m*gcd(n-1,m-1)) for n >= m >= 2, T(n, 0) = 1, T(n, 1) = 0, otherwise 0 (from the preceding formula).
T(2*n, n) = A001700(n-1) (n>=1) (see the Emeric Deutsch comment above), T(2*n, n-1) = A234040(n), T(2*n+1,n) = A000108(n), n >= 0 (Catalan numbers).
Column sequences: T(n+2, 2) = A026741(n+1), T(n+3, 3) = A234041(n), T(n+4, 4) = A208950(n+2), T(n+5, 5) = A234042, n >= 0. (End)

More terms from Emeric Deutsch, Jun 13 2005

A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+1).

Original entry on oeis.org

0, 1, 1, 9, 4, 25, 9, 49, 16, 81, 25, 121, 36, 169, 49, 225, 64, 289, 81, 361, 100, 441, 121, 529, 144, 625, 169, 729, 196, 841, 225, 961, 256, 1089, 289, 1225, 324, 1369, 361, 1521, 400, 1681, 441, 1849, 484, 2025, 529, 2209, 576, 2401, 625, 2601

Offset: 0

Paul Curtz, Nov 18 2009

From Paul Curtz, Mar 26 2011: (Start)
Successive A026741(n) * A026741(n+p):
  p=0:  0, 1,  1,  9,  4, 25,  9,     a(n),
  p=1:  0, 1,  3,  6, 10, 15, 21,  A000217,
  p=2:  0, 3,  2, 15,  6, 35, 12,  A142705,
  p=3:  0, 2,  5,  9, 14, 20, 27,  A000096,
  p=4:  0, 5,  3, 21,  8, 45, 15,  A171621,
  p=5:  0, 3,  7, 12, 18, 25, 33,  A055998,
  p=6:  0, 7,  4, 27, 10, 55, 18,
  p=7:  0, 4,  9, 15, 22, 30, 39,  A055999,
  p=8:  0, 9,  5, 33, 12, 65, 21,  (see A061041),
  p=9:  0, 5, 11, 18, 26, 35, 45,  A056000. (End)
The moment generating function of p(x, m=2, n=1, mu=2) = 4*x*E(x, 2, 1), see A163931 and A274181, is given by M(a) = (-4 * log(1-a) - 4 * polylog(2, a))/a^2. The series expansion of M(a) leads to the sequence given above. - Johannes W. Meijer, Jul 03 2016
Multiplicative because both A129194 and A040001 are. - Andrew Howroyd, Jul 26 2018

G. C. Greubel, Table of n, a(n) for n = 0..1000
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A040001, A168068, A181318.

I:=[0,1,1,9,4,25]; [n le 6 select I[n] else 3*Self(n-2)-3*Self(n-4)+Self(n-6): n in [1..60]]; // Vincenzo Librandi, Jul 10 2016

a := proc(n): n^2*(5-3*(-1)^n)/8 end: seq(a(n), n=0..46); # Johannes W. Meijer, Jul 03 2016

LinearRecurrence[{0,3,0,-3,0,1},{0,1,1,9,4,25},60] (* Harvey P. Dale, May 14 2011 *)
f[n_] := Numerator[(n/2)^2]; Array[f, 60, 0] (* Robert G. Wilson v, Dec 18 2012 *)
CoefficientList[Series[x(1+x+6x^2+x^3+x^4)/((1-x)^3(1+x)^3), {x,0,60}], x] (* Vincenzo Librandi, Jul 10 2016 *)

concat(0, Vec(x*(1+x+6*x^2+x^3+x^4)/((1-x)^3*(1+x)^3) + O(x^60))) \\ Altug Alkan, Jul 04 2016

a(n) = lcm(4, n^2)/4; \\ Andrew Howroyd, Jul 26 2018

(x*(1+x+6*x^2+x^3+x^4)/(1-x^2)^3).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 20 2019

From R. J. Mathar, Jan 22 2011: (Start)
G.f.: x*(1 + x + 6*x^2 + x^3 + x^4) / ((1-x)^3*(1+x)^3).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
a(n) = n^2*(5 - 3*(-1)^n)/8. (End)
a(n) = A026741(n)^2.
a(2*n) = A000290(n); a(2*n+1) = A016754(n).
a(n) - a(n-4) = 4*A064680(n+2). - Paul Curtz, Mar 27 2011
4*a(n) = A061038(n) * A010121(n+2) = A109043(n)^2, n >= 2. - Paul Curtz, Apr 07 2011
a(n) = A129194(n) / A040001(n). - Andrew Howroyd, Jul 26 2018
From Peter Bala, Feb 19 2019: (Start)
a(n) = numerator(n^2/(n^2 + 4)) = n^2/(gcd(n^2,4)) = (n/gcd(n,2))^2.
a(n) = n^2/b(n), where b(n) = [1, 4, 1, 4, ...] is a purely periodic sequence of period 2. Thus a(n) is a quasi-polynomial in n.
O.g.f.: x*(1 + x)/(1 - x)^3 - 3*x^2*(1 + x^2)/(1 - x^2)^3.
Cf. A181318. (End)
From Werner Schulte, Aug 30 2020: (Start)
Multiplicative with a(2^e) = 2^(2*e-2) for e > 0, and a(p^e) = p^(2*e) for prime p > 2.
Dirichlet g.f.: zeta(s-2) * (1 - 3/2^s).
Dirichlet convolution with A259346 equals A000290.
Sum_{n>0} 1/a(n) = Pi^2 * 7 / 24. (End)
Sum_{k=1..n} a(k) ~ (5/24) * n^3. - Amiram Eldar, Nov 28 2022

A337377 Primorial deflation (denominator) of Doudna-tree.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 4, 1, 5, 3, 2, 1, 9, 2, 8, 1, 7, 5, 10, 3, 3, 1, 4, 1, 25, 9, 6, 1, 27, 4, 16, 1, 11, 7, 14, 5, 21, 5, 20, 3, 5, 3, 2, 1, 9, 2, 8, 1, 49, 25, 50, 9, 15, 3, 4, 1, 125, 27, 18, 2, 81, 8, 32, 1, 13, 11, 22, 7, 33, 7, 28, 5, 55, 21, 14, 5, 63, 10, 40, 3, 7, 5, 10, 3, 3, 1, 4, 1, 25, 9, 6, 1, 27, 4, 16, 1, 121

Offset: 0

Antti Karttunen, Michael De Vlieger and Peter Munn, Aug 25 2020

Like A005940, also this irregular table can be represented as a binary tree:
                                      1
                                      |
                   ...................1...................
                  2                                       1
        3......../ \........1                   4......../ \........1
       / \                 / \                 / \                 / \
      /   \               /   \               /   \               /   \
     /     \             /     \             /     \             /     \
    5       3           2       1           9       2           8       1
   7 5    10 3         3 1     4 1        25 9     6 1        27 4    16 1
etc.
A194602 gives the positions of nodes that have value 1. They correspond to terms of A005940 that are products of primorials (A025487). The first 2^k nodes contain A000041(k+1) 1's.
a(n) is even if and only if A005940(1+n) occurs in A277569.

Antti Karttunen, Table of n, a(n) for n = 0..8191
Antti Karttunen, Data supplement: n, a(n) computed for n = 0..65537
Index entries for fraction trees

Cf. A337376 (numerators).
A003961, A005940, A006519, A026741, A064989, A319627 are used in a formula defining this sequence.
Cf. A000041, A025487, A277569.
Positions of 1's: A194602.
Cf. also A329886, A346097.

Array[#2/GCD[#1, #2] & @@ {#, Apply[Times, Map[If[#1 <= 2, 1, NextPrime[#1, -1]]^#2 & @@ # &, FactorInteger[#]]]} &@ Function[p, Times @@ Flatten@ Table[Prime[Count[Flatten[#], 0] + 1]^#[[1, 1]] &@ Take[p, -i], {i, Length[p]}]]@ Partition[Split[Join[IntegerDigits[# - 1, 2], {2}]], 2] &[# + 1] &, 96] (* Michael De Vlieger, Aug 27 2020 *)

A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); (t); };
A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
A319627(n) = (A064989(n) / gcd(n, A064989(n)));
A337377(n) = A319627(A005940(1+n));

a(n) = A319627(A005940(1+n)).
For n >= 1, a(2*n) = A003961(a(n)) * A006519(n+1).
a(2*n+1) = A026741(a(n)).

A092092 Back and Forth Summant S(n, 3): a(n) = Sum{i=0..floor(2n/3)} (n-3i).

Original entry on oeis.org

1, 1, 0, 3, 2, 0, 5, 3, 0, 7, 4, 0, 9, 5, 0, 11, 6, 0, 13, 7, 0, 15, 8, 0, 17, 9, 0, 19, 10, 0, 21, 11, 0, 23, 12, 0, 25, 13, 0, 27, 14, 0, 29, 15, 0, 31, 16, 0, 33, 17, 0, 35, 18, 0, 37, 19, 0, 39, 20, 0, 41, 21, 0, 43, 22, 0, 45, 23, 0, 47, 24, 0, 49, 25, 0, 51, 26, 0, 53, 27, 0, 55, 28

Offset: 1

Jahan Tuten (jahant(AT)indiainfo.com), Mar 29 2004

The terms for n>1 can also be defined by: a(n)=0 if n==0 (mod 3), and otherwise a(n) equals the inverse of 3 in Z/nZ*. - José María Grau Ribas, Jun 18 2013

The subsequence of nonzero terms is essentially the same as A026741. - Giovanni Resta, Jun 18 2013

F. Smarandache, Back and Forth Summants, Arizona State Univ., Special Collections, 1972.

Robert Israel, Table of n, a(n) for n = 1..10000
J. Dezert, editor, Smarandacheials, Mathematics Magazine, Aurora, Canada
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A092094, A092397.
Other values of k: A000004 (k = 1, 2), A027656 (k = 4), A092093 (k = 5).
Cf. A226782 - A226787 for inverses of 4,5,6,.. in Z/nZ*.

f:= proc(n) local t;
t:= n mod 3;
if t = 0 then 0 elif t = 1 then 2/3*(n+1/2) else (n+1)/3 fi
end proc:
map(f, [$1..100]); # Robert Israel, May 19 2016

LinearRecurrence[{0, 0, 2, 0, 0, -1}, {1, 1, 0, 3, 2, 0}, 100] (* Jean-François Alcover, Jun 04 2020 *)

S(n, k=3) = local(s, x); s = n; x = n - k; while (x >= -n, s = s + x; x = x - k); s;

a(3n) = 0; a(3n+1) = 2n+1; a(3n+2) = n+1.
G.f.: x*(1+x+x^3) / ( (x-1)^2*(1+x+x^2)^2 ). - R. J. Mathar, Jun 26 2013
a(n) = Sum_{k=1..n} k*( floor((3k-1)/n)-floor((3k-2)/n) ). - Anthony Browne, May 17 2016

Edited and extended by David Wasserman, Dec 19 2005

cp's OEIS Frontend

A165355 a(n) = 3n + 1 if n is even, or a(n) = (3n + 1)/2 if n is odd.

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A065423 Number of ordered length 2 compositions of n with at least one even summand.

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A106617 Numerator of n/(n+16).

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A145051 Numerator of the first convergent to sqrt(n) using the recursion x = (n/x + x)/2.

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A106619 a(n) = numerator of n/(n+18).

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A209297 Triangle read by rows: T(n,k) = k*n + k - n, 1 <= k <= n.

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