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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A005994 Alkane (or paraffin) numbers l(7,n).

Original entry on oeis.org

1, 3, 9, 19, 38, 66, 110, 170, 255, 365, 511, 693, 924, 1204, 1548, 1956, 2445, 3015, 3685, 4455, 5346, 6358, 7514, 8814, 10283, 11921, 13755, 15785, 18040, 20520, 23256, 26248, 29529, 33099, 36993, 41211, 45790, 50730, 56070, 61810, 67991
Offset: 0

Views

Author

N. J. A. Sloane, Winston C. Yang (yang(AT)math.wisc.edu)

Keywords

Comments

Equals A000217 (1, 3, 6, 10, 15, ...) convolved with A193356 (1, 0, 3, 0, 5, ...). - Gary W. Adamson, Feb 16 2009
F(1,4,n) is the number of bracelets with 1 blue, 4 red and n black beads. If F(1,4,1)=3 and F(1,4,2)=9 taken as a base;
F(1,4,n) = n(n+1)(n+2)/6+F(1,2,n) + F(1,4,n-2). [F(1,2,n) is the number of bracelets with 1 blue, 2 red and n black beads. If F(1,2,1)=2 and F(1,2,2)=4 taken as a base F(1,2,n)=n+1+F(1,2,n-2)]. - Ata Aydin Uslu and Hamdi G. Ozmenekse, Jan 11 2012
a(A254338(n)) = 6 for n > 0. - Reinhard Zumkeller, Feb 27 2015

References

  • S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A006009, A005997, A005993 (first differences).

Programs

  • Haskell
    --  Following Gary W. Adamson.
    import Data.List (inits, intersperse)
    a005994 n = a005994_list !! n
    a005994_list = map (sum . zipWith (*) (intersperse 0 [1, 3 ..]) . reverse) $
                       tail $ inits $ tail a000217_list
    -- Reinhard Zumkeller, Feb 27 2015
  • Maple
    a:= n -> (Matrix([[1, 0$4, 1, 3]]). Matrix(7, (i,j)-> if (i=j-1) then 1 elif j=1 then [3, -1, -5, 5, 1, -3, 1][i] else 0 fi)^n)[1,1]: seq (a(n), n=0..40); # Alois P. Heinz, Jul 31 2008
  • Mathematica
    LinearRecurrence[{3,-1,-5,5,1,-3,1},{1,3,9,19,38,66,110},50] (* or *) CoefficientList[Series[(1+x^2)/((1-x)^3(1-x^2)^2),{x,0,50}],x] (* Harvey P. Dale, May 02 2011 *)
    nn=45;With[{a=Accumulate[Range[nn]],b=Riffle[Range[1,nn,2],0]}, Flatten[ Table[ListConvolve[Take[a,n],Take[b,n]],{n,nn}]]] (* Harvey P. Dale, Nov 11 2011 *)
  • PARI
    {a(n)=if(n<-4, n=-5-n); polcoeff( (1+x^2)/((1-x)^3*(1-x^2)^2)+x*O(x^n), n)} /* Michael Somos, Mar 08 2007 */
    

Formula

G.f.: (1+x^2)/((1-x)^3*(1-x^2)^2) = (1+x^2)/((1-x)^5*(1+x)^2).
l(c, r) = 1/2 C(c+r-3, r) + 1/2 d(c, r), where d(c, r) is C((c + r - 3)/2, r/2) if c is odd and r is even, 0 if c is even and r is odd, C((c + r - 4)/2, r/2) if c is even and r is even, C((c + r - 4)/2, (r - 1)/2) if c is odd and r is odd.
a(-5-n)=a(n). - Michael Somos, Mar 08 2007
Euler transform of length 4 sequence [3, 3, 0, -1]. - Michael Somos, Mar 08 2007
a(n) = 3a(n-1) - a(n-2) - 5a(n-3) + 5a(n-4) + a(n-5) - 3a(n-6) + a(n-7), with a(0)=1, a(1)=3, a(2)=9, a(4)=19, a(5)=38, a(6)=66, a(7)=110. - Harvey P. Dale, May 02 2011
a(n) = A006009(n)/2 - A000332(n+4) = ((1/2)*Sum_{i=1..n+1} (i+1)*floor((i+1)^2/2)) - binomial(n+4,4). - Enrique Pérez Herrero, May 11 2012
a(n) = (1/48)*(n+1)*(n+3)*((n+2)*(n+4)+3)+1/32*(2*n+5)*(1+(-1)^n). - Yosu Yurramendi, Jun 20 2013
Conjecture: a(n)+a(n+1) = A203286(n+1). - R. J. Mathar, Mar 08 2025

A237420 If n is odd, then a(n) = 0; otherwise, a(n) = n.

Original entry on oeis.org

0, 0, 2, 0, 4, 0, 6, 0, 8, 0, 10, 0, 12, 0, 14, 0, 16, 0, 18, 0, 20, 0, 22, 0, 24, 0, 26, 0, 28, 0, 30, 0, 32, 0, 34, 0, 36, 0, 38, 0, 40, 0, 42, 0, 44, 0, 46, 0, 48, 0, 50, 0, 52, 0, 54, 0, 56, 0, 58, 0, 60, 0, 62, 0, 64, 0, 66, 0, 68, 0, 70, 0, 72, 0, 74
Offset: 0

Views

Author

Vincenzo Librandi, Feb 24 2014

Keywords

Comments

Normally the OEIS excludes sequences in which every other term is zero. But there are exceptions for especially important sequences like this one. - N. J. A. Sloane, Feb 27 2014
Essentially the factorial expansion of exp(-1): exp(-1) = Sum_{n>=1} a(n)/(n+1)!. - Joerg Arndt, Mar 13 2014
a(n) is the number of m < n for which a(m) has the same parity as n. For instance, a(4) = 4 because 4 has the same parity as a(0), a(1), a(2), and a(3). - Alec Jones, May 16 2016
This sequence is an example of a sequence that has no limit while the Cesàro means limit is infinite. See A354280 for further information. - Bernard Schott, May 22 2022

References

  • J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, Exercice 10, pp. 14-16.

Crossrefs

About the Cesàro mean theorem: A033999, A114112.

Programs

  • Magma
    [IsOdd(n) select 0 else n: n in [1..80]];
    
  • Magma
    [(1+(-1)^n)*n/2: n in [1..80]];
    
  • Magma
    &cat [[n, 0]: n in [0..80 by 2]]; // Bruno Berselli, Nov 11 2016
    
  • Maple
    seq(op([0,2*i]),i=1..30); # Robert Israel, Aug 27 2015
  • Mathematica
    Table[If[OddQ[n], 0, n], {n, 80}]
    CoefficientList[Series[2 x /(1 - x^2)^2, {x, 0, 80}], x]
    LinearRecurrence[{0, 2, 0, -1}, {0, 0, 2, 0}, 75] (* Robert G. Wilson v, Nov 11 2016 *)
    Riffle[Range[0,80,2],0] (* Harvey P. Dale, Mar 16 2021 *)
  • PARI
    a(n)=if(n%2==0,n,0) \\ Anders Hellström, Aug 27 2015
    
  • Python
    def a(n): return 0 if n%2 else n # Michael S. Branicky, Jun 05 2022

Formula

O.g.f.: 2*x^2/(1-x^2)^2.
E.g.f.: x*sinh(x). - Robert Israel, Aug 27 2015
a(n) = 2*a(n-2) - a(n-4) for n>4.
a(n) = 2*A142150(n) = (1+(-1)^n)*n/2 = n*((n-1) mod 2).
a(n) = floor(n^(-1)^n) for n>1. - Ilya Gutkovskiy, Aug 27 2015
Sum_{i=1..n} a(i) = A110660(n). - Bruno Berselli, Feb 27 2014
a(n) = -1 + ceiling((n + 1)^(sin(Pi*n/2) + cos(Pi*n))). - Lechoslaw Ratajczak, Nov 06 2016

Extensions

Edited by Bruno Berselli, Feb 27 2014

A235800 Length of n-th vertical line segment from left to right in a diagram of a two-dimensional version of the 3x+1 (or Collatz) problem.

Original entry on oeis.org

3, 1, 7, 2, 11, 3, 15, 4, 19, 5, 23, 6, 27, 7, 31, 8, 35, 9, 39, 10, 43, 11, 47, 12, 51, 13, 55, 14, 59, 15, 63, 16, 67, 17, 71, 18, 75, 19, 79, 20, 83, 21, 87, 22, 91, 23, 95, 24, 99, 25, 103, 26, 107, 27, 111, 28, 115, 29, 119, 30, 123, 31, 127, 32
Offset: 1

Views

Author

Omar E. Pol, Jan 15 2014

Keywords

Comments

In the diagram every cycle is represented by a directed graph.
After (3x + 1) the next step is (3y + 1).
After (x/2) the next step is (y/2).
A235801(n) gives the length of n-th horizontal line segment in the same diagram.
Also A004767 and A000027 interleaved.

Examples

			The first part of the diagram in the first quadrant looks like this:
. . . . . . . . . . . . . . . . . . . . . . . .
.              _ _|_ _|_ _|_ _|_ _|_ _|_ _|_ _.
.             |   |   |   |   |   |   |   |_|_.
.             |   |   |   |   |   |   |  _ _|_.
.             |   |   |   |   |   |   |_|_ _|_.
.             |   |   |   |   |   |  _ _|_ _|_.
.             |   |   |   |   |   |_|_ _|_ _|_.
.          _ _|_ _|_ _|_ _|_ _|_ _ _|_ _|_ _|_.
.         |   |   |   |   |   |_|_ _|_ _|_ _|_.
.         |   |   |   |   |  _ _|_ _|_ _|_ _|_.
.         |   |   |   |   |_|_ _|_ _|_ _|_ _|_.
.         |   |   |   |  _ _|_ _|_ _|_ _|_ _|_.
.         |   |   |   |_|_ _|_ _|_ _|_ _|_ _| .
.      _ _|_ _|_ _|_ _ _|_ _|_ _|_ _|_ _|     .
.     |   |   |   |_|_ _|_ _|_ _|_ _|         .
.     |   |   |  _ _|_ _|_ _|_ _|             .
.     |   |   |_|_ _|_ _|_ _|                 .
.     |   |  _ _|_ _|_ _|                     .
.     |   |_|_ _|_ _|                         .
.  _ _|_ _ _|_ _|                             .
. |   |_|_ _|                                 .
. |  _ _|                                     .
. |_|                                         .
. . . . . . . . . . . . . . . . . . . . . . . .
. 3,1,7,2,11...
From _Omar E. Pol_, Aug 25 2021: (Start)
The above diagram is the skeleton of a piping model of the 3x+1 or Collatz problem as shown below:
The model consists of pipes, 90-degree elbows and three types of pumps that propel the fluid through the pipes.
The corner of the infinite diagram looks like this:
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
                            | |         | |         | |         | |  _ _ _  .
                            | |         | |         | |         | | |     |_.
                            | |         | |         | |         | | |  12  _.
                            | |         | |         | |        _| |_|_ v _| .
                            | |         | |         | |       |  ^  |_|_|_ _.
                            | |         | |         | |       |  11  _ _ _ _.
                            | |         | |         | |  _ _ _|_ _ _| | |   .
                 _ _ _ _ _ _|_|_ _ _ _ _|_|_ _ _ _ _|_|_|     |_ _ _ _|_|_ _.
                |  _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _   10  _ _ _ _ _ _ _.
                | |         | |         | |        _| |_|_ v _|       | |   .
                | |         | |         | |       |  ^  |_| |_ _ _ _ _| |_ _.
                | |         | |         | |       |  9   _| |_ _ _ _ _| |_ _.
                | |         | |         | |  _ _ _|_ _ _| | |         | |   .
                | |         | |         | | |     |_ _ _ _| |_ _ _ _ _|_|_ _.
                | |         | |         | | |  8   _ _ _ _| |_ _ _ _ _ _ _ _.
                | |         | |        _| |_|_ v _|       | |         | |   .
                | |         | |       |  ^  |_| |_ _ _ _ _| |_ _ _ _ _|_|_ _.
                | |         | |       |  7   _| |_ _ _ _ _| |_ _ _ _ _ _ _ _.
                | |         | |  _ _ _|_ _ _| | |         | |         | |   .
                | |         | | |     |_ _ _ _| |_ _ _ _ _| |_ _ _ _ _| |   .
                | |         | | |  6   _ _ _ _| |_ _ _ _ _| |_ _ _ _ _ _|   .
                | |        _| |_|_ v _|       | |         | |               .
                | |       |  ^  |_|_|_ _ _ _ _|_|_ _ _ _ _| |               .
                | |       |  5   _ _ _ _ _ _ _ _ _ _ _ _ _ _|               .
                | |  _ _ _|_ _ _| | |         | |                           .
     _ _ _ _ _ _|_|_|     |_ _ _ _|_|_ _ _ _ _| |                           .
    |  _ _ _ _ _ _ _   4   _ _ _ _ _ _ _ _ _ _ _|                           .
    | |        _| |_|_ v _|       | |                                       .
    | |       |  ^  |_| |_ _ _ _ _| |                                       .
    | |       |  3   _| |_ _ _ _ _ _|                                       .
    | |  _ _ _|_ _ _| | |                                                   .
    | | |     |_ _ _ _| |                                                   .
    | | |  2   _ _ _ _ _|                                                   .
   _| |_|_ v _|                                                             .
  |  ^  |_| |                                                               .
  |  1   _ _|                                                               .
  |_ _ _|                                                                   .
.                                                                           .
On the main diagonal of the diagram appear the pumps labeled with the positive integers (A000027).
The pumps labeled with the numbers 2, 6, 8, 12, 14, 18, 20, 24, ... (the nonzero terms of A047238) receive the fluid from the EAST and propel it in a SOUTH direction. The fluid then passes through a 90-degree elbow and then heads WEST.
The pumps labeled with the numbers 4, 10, 16, 22, 28, 34, 40, ... (A016957) are of the type "TEE" as they have two side inlets and one outlet. These receive the fluid from the EAST and from the WEST and propel it in a SOUTH direction. The fluid then passes through a 90-degree elbow and then heads WEST.
The pumps labeled with the numbers 1, 3, 5, 7, 9, 11, 13, ... (A005408) receive the fluid from the EAST and propel it in the NORTH direction. The fluid then passes through a 90-degree elbow and then heads EAST.
Starting from the n-th pump we have that the fluid makes a path equivalent to the trayectory of the 3x+1 sequence starting at n. (End)
		

Crossrefs

Cf. A347270 (all 3x+1 sequences).
Cf. Companion of A235801.

Programs

  • Mathematica
    LinearRecurrence[{0,2,0,-1},{3,1,7,2},70] (* Harvey P. Dale, Sep 29 2016 *)
  • Python
    from _future_ import division
    A235800_list = [4*(n//2) + 3 if n % 2 else n//2 for n in range(1,10**4)] # Chai Wah Wu, Sep 26 2016

Formula

a(n) = A006370(n) - A193356(n).
From Chai Wah Wu, Sep 26 2016: (Start)
a(n) = 2*a(n-2) - a(n-4) for n > 4.
G.f.: x*(x^2 + x + 3)/((x - 1)^2*(x + 1)^2). (End)

A274760 The multinomial transform of A001818(n) = ((2*n-1)!!)^2.

Original entry on oeis.org

1, 1, 10, 478, 68248, 21809656, 13107532816, 13244650672240, 20818058883902848, 48069880140604832128, 156044927762422185270016, 687740710497308621254625536, 4000181720339888446834235653120, 29991260979682976913756629498334208
Offset: 0

Views

Author

Johannes W. Meijer, Jul 27 2016

Keywords

Comments

The multinomial transform [MNL] transforms an input sequence b(n) into the output sequence a(n). Given the fact that the structure of the a(n) formulas, see the examples, lead to the multinomial coefficients A036039 the MNL transform seems to be an appropriate name for this transform. The multinomial transform is related to the exponential transform, see A274804 and the third formula. For the inverse multinomial transform [IML] see A274844.
To preserve the identity IML[MNL[b(n)]] = b(n) for n >= 0 for a sequence b(n) with offset 0 the shifted sequence b(n-1) with offset 1 has to be used as input for the MNL, otherwise information about b(0) will be lost in transformation.
In the a(n) formulas, see the examples, the multinomial coefficients A036039 appear.
We observe that a(0) = 1 and that this term provides no information about any value of b(n), this notwithstanding we will start the a(n) sequence with a(0) = 1.
The Maple programs can be used to generate the multinomial transform of a sequence. The first program uses the first formula which was found by Paul D. Hanna, see A158876, and Vladimir Kruchinin, see A215915. The second program uses properties of the e.g.f., see the sequences A158876, A213507, A244430 and A274539 and the third formula. The third program uses information about the inverse multinomial transform, see A274844.
Some MNL transform pairs are, n >= 1: A000045(n) and A244430(n-1); A000045(n+1) and A213527(n-1); A000108(n) and A213507(n-1); A000108(n-1) and A243953(n-1); A000142(n) and A158876(n-1); A000203(n) and A053529(n-1); A000110(n) and A274539(n-1); A000041(n) and A215915(n-1); A000035(n-1) and A177145(n-1); A179184(n) and A038205(n-1); A267936(n) and A000266(n-1); A267871(n) and A000090(n-1); A193356(n) and A088009(n-1).

Examples

			Some a(n) formulas, see A036039:
  a(0) = 1
  a(1) = 1*x(1)
  a(2) = 1*x(2) + 1*x(1)^2
  a(3) = 2*x(3) + 3*x(1)*x(2) + 1*x(1)^3
  a(4) = 6*x(4) + 8*x(1)*x(3) + 3*x(2)^2 + 6*x(1)^2*x(2) + 1*x(1)^4
  a(5) = 24*x(5) + 30*x(1)*x(4) + 20*x(2)*x(3) + 20*x(1)^2*x(3) + 15*x(1)*x(2)^2 + 10*x(1)^3*x(2) + 1*x(1)^5
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, 1995, pp. 18-23.

Crossrefs

Programs

  • Maple
    nmax:= 13: b := proc(n): (doublefactorial(2*n-1))^2 end: a:= proc(n) option remember: if n=0 then 1 else add(((n-1)!/(n-k)!) * b(k) * a(n-k), k=1..n) fi: end: seq(a(n), n = 0..nmax); # End first MNL program.
    nmax:=13: b := proc(n): (doublefactorial(2*n-1))^2 end: t1 := exp(add(b(n)*x^n/n, n = 1..nmax+1)): t2 := series(t1, x, nmax+1): a := proc(n): n!*coeff(t2, x, n) end: seq(a(n), n = 0..nmax); # End second MNL program.
    nmax:=13: b := proc(n): (doublefactorial(2*n-1))^2 end: f := series(log(1+add(s(n)*x^n/n!, n=1..nmax)), x, nmax+1): d := proc(n): n*coeff(f, x, n) end: a(0) := 1: a(1) := b(1): s(1) := b(1): for n from 2 to nmax do s(n) := solve(d(n)-b(n), s(n)): a(n):=s(n): od: seq(a(n), n=0..nmax); # End third MNL program.
  • Mathematica
    b[n_] := (2*n - 1)!!^2;
    a[0] = 1; a[n_] := a[n] = Sum[((n-1)!/(n-k)!)*b[k]*a[n-k], {k, 1, n}];
    Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Nov 17 2017 *)

Formula

a(n) = Sum_{k=1..n} ((n-1)!/(n-k)!)*b(k)*a(n-k), n >= 1 and a(0) = 1, with b(n) = A001818(n) = ((2*n-1)!!)^2.
a(n) = n!*P(n), with P(n) = (1/n)*(Sum_{k=0..n-1} b(n-k)*P(k)), n >= 1 and P(0) = 1, with b(n) = A001818(n) = ((2*n-1)!!)^2.
E.g.f.: exp(Sum_{n >= 1} b(n)*x^n/n) with b(n) = A001818(n) = ((2*n-1)!!)^2.
denom(a(n)/2^n) = A001316(n); numer(a(n)/2^n) = [1, 1, 5, 239, 8531, 2726207, ...].

A328203 Expansion of Sum_{k>=1} k * x^k / (1 - x^(2*k))^2.

Original entry on oeis.org

1, 2, 5, 4, 8, 10, 11, 8, 20, 16, 17, 20, 20, 22, 42, 16, 26, 40, 29, 32, 58, 34, 35, 40, 53, 40, 74, 44, 44, 84, 47, 32, 90, 52, 94, 80, 56, 58, 106, 64, 62, 116, 65, 68, 174, 70, 71, 80, 102, 106, 138, 80, 80, 148, 146, 88, 154, 88, 89, 168, 92, 94, 241, 64, 172
Offset: 1

Views

Author

Ilya Gutkovskiy, Oct 07 2019

Keywords

Crossrefs

Programs

  • Magma
    a:=[]; for k in [1..65] do if IsOdd(k) then a[k]:=(k * #Divisors(k) + DivisorSigma(1,k)) / 2; else a[k]:=(k * (#Divisors(k) - #Divisors(k div 2)) + DivisorSigma(1,k) - DivisorSigma(1,k div 2)) / 2;  end if; end for; a; // Marius A. Burtea, Oct 07 2019
    
  • Mathematica
    nmax = 65; CoefficientList[Series[Sum[k x^k/(1 - x^(2 k))^2, {k, 1, nmax}], {x, 0, nmax}], x] // Rest
    a[n_] := DivisorSum[n, (n Mod[#, 2] + Boole[OddQ[n/#]] #)/2 &]; Table[a[n], {n, 1, 65}]
  • PARI
    A328203(n) = if(n%2,(1/2)*(sigma(n)+(n*numdiv(n))),2*A328203(n/2)); \\ Antti Karttunen, Nov 13 2021

Formula

a(n) = (n * d(n) + sigma(n)) / 2 if n odd, (n * (d(n) - d(n/2)) + sigma(n) - sigma(n/2)) / 2 if n even.
a(n) = (n * A001227(n) + A002131(n)) / 2.
a(2*n) = 2 * a(n).
From Antti Karttunen, Nov 13 2021: (Start)
The following two convolutions were found by Jon Maiga's Sequence Machine search algorithm. Both are easy to prove:
a(n) = Sum_{d|n} A003602(d) * A026741(n/d).
a(n) = Sum_{d|n} A109168(d) * A193356(n/d), where A109168(d) = A140472(d) = (d+A006519(d))/2.
(End)

A309337 a(n) = n^3 if n odd, 3*n^3/4 if n even.

Original entry on oeis.org

0, 1, 6, 27, 48, 125, 162, 343, 384, 729, 750, 1331, 1296, 2197, 2058, 3375, 3072, 4913, 4374, 6859, 6000, 9261, 7986, 12167, 10368, 15625, 13182, 19683, 16464, 24389, 20250, 29791, 24576, 35937, 29478, 42875, 34992, 50653, 41154, 59319, 48000, 68921, 55566, 79507, 63888, 91125
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 24 2019

Keywords

Comments

Moebius transform of A078307.

Crossrefs

Programs

  • Mathematica
    a[n_] := If[OddQ[n], n^3, 3 n^3/4]; Table[a[n], {n, 0, 45}]
    nmax = 45; CoefficientList[Series[x (1 + 6 x + 23 x^2 + 24 x^3 + 23 x^4 + 6 x^5 + x^6)/(1 - x^2)^4, {x, 0, nmax}], x]
    LinearRecurrence[{0, 4, 0, -6, 0, 4, 0, -1}, {0, 1, 6, 27, 48, 125, 162, 343}, 46]
    Table[n^3 (7 - (-1)^n)/8, {n, 0, 45}]

Formula

G.f.: x * (1 + 6*x + 23*x^2 + 24*x^3 + 23*x^4 + 6*x^5 + x^6)/(1 - x^2)^4.
G.f.: Sum_{k>=1} J_3(k) * x^k/(1 + x^k), where J_3() is the Jordan function (A059376).
Dirichlet g.f.: zeta(s-3) * (1 - 2^(1-s)).
a(n) = n^3 * (7 - (-1)^n)/8.
a(n) = Sum_{d|n} (-1)^(n/d + 1) * J_3(d).
Sum_{n>=1} 1/a(n) = 25*zeta(3)/24 = 1.252142607457910713958...
Multiplicative with a(2^e) = 3*2^(3*e-2), and a(p^e) = p^(3*e) for odd primes p. - Amiram Eldar, Oct 26 2020
a(n) = Sum_{1 <= i, j, k <= n} (-1)^(1 + gcd(i,j,k,n)) = Sum_{d | n} (-1)^(d+1) * J_3(n/d). Cf. A129194. - Peter Bala, Jan 16 2024

A332794 a(n) = Sum_{d|n} (-1)^(d + 1) * d * phi(n/d).

Original entry on oeis.org

1, -1, 5, -4, 9, -5, 13, -12, 21, -9, 21, -20, 25, -13, 45, -32, 33, -21, 37, -36, 65, -21, 45, -60, 65, -25, 81, -52, 57, -45, 61, -80, 105, -33, 117, -84, 73, -37, 125, -108, 81, -65, 85, -84, 189, -45, 93, -160, 133, -65, 165, -100, 105, -81, 189
Offset: 1

Views

Author

Ilya Gutkovskiy, Feb 24 2020

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[(-1)^(d + 1) d EulerPhi[n/d], {d, Divisors[n]}]; Table[a[n], {n, 1, 55}]
    nmax = 55; CoefficientList[Series[Sum[EulerPhi[k] x^k/(1 + x^k)^2, {k, 1, nmax}], {x, 0, nmax}], x] // Rest
    a[n_] := If[OddQ[n], Sum[GCD[n, k], {k, 1, n}], Sum[(-1)^(k + 1) GCD[n, k], {k, 1, n}]]; Table[a[n], {n, 1, 55}]
    f[p_, e_] := (e*(p-1) + p)*p^(e-1); f[2, e_] := -e*2^(e-1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Nov 04 2022 *)
  • PARI
    a(n) = sumdiv(n, d, (-1)^(d+1)*d*eulerphi(n/d)); \\ Michel Marcus, Feb 24 2020

Formula

G.f.: Sum_{k>=1} phi(k) * x^k / (1 + x^k)^2.
Dirichlet g.f.: zeta(s-1)^2 * (1 - 2^(2 - s)) / zeta(s).
a(n) = Sum_{k=1..n} gcd(n, k) if n odd, Sum_{k=1..n} (-1)^(k + 1) * gcd(n, k) if n even.
From Amiram Eldar, Nov 04 2022: (Start)
Multiplicative with a(2^e) = -e*2^(e-1), and a(p^e) = (e*(p-1) + p)*p^(e-1) for p > 2.
Sum_{k=1..n} a(k) ~ c * n^2, where c = 3*log(2)/Pi^2 = 0.210691... . (End)
a(2*n) = - Sum_{k = 1..n} gcd(2*k, n) = - A344372(n); a(2*n+1) = A018804(2*n+1). - Peter Bala, Jan 11 2024
a(n) = Sum_{k = 1..n} (-1)^(1 + gcd(k, n)) * gcd(k, n) (follows from an identity of Cesàro. See, for example, Bordelles, Lemma 1). - Peter Bala, Jan 16 2024

A382995 Square array A(n,k), n >= 1, k >= 1, read by antidiagonals downwards, where A(n,k) = Sum_{d|n} phi(n/d) * (-k)^(d-1).

Original entry on oeis.org

1, 1, 0, 1, -1, 3, 1, -2, 6, 0, 1, -3, 11, -8, 5, 1, -4, 18, -28, 20, 0, 1, -5, 27, -66, 85, -30, 7, 1, -6, 38, -128, 260, -238, 70, 0, 1, -7, 51, -220, 629, -1014, 735, -136, 9, 1, -8, 66, -348, 1300, -3108, 4102, -2216, 270, 0, 1, -9, 83, -518, 2405, -7750, 15631, -16452, 6585, -500, 11
Offset: 1

Views

Author

Seiichi Manyama, Apr 12 2025

Keywords

Examples

			Square array begins:
  1,   1,    1,     1,     1,     1,      1, ...
  0,  -1,   -2,    -3,    -4,    -5,     -6, ...
  3,   6,   11,    18,    27,    38,     51, ...
  0,  -8,  -28,   -66,  -128,  -220,   -348, ...
  5,  20,   85,   260,   629,  1300,   2405, ...
  0, -30, -238, -1014, -3108, -7750, -16770, ...
  7,  70,  735,  4102, 15631, 46662, 117655, ...
		

Crossrefs

Columns k=1..3 give A193356, A382999, A383000.
Main diagonal gives A382998.

Programs

  • PARI
    a(n, k) = sumdiv(n, d, eulerphi(n/d)*(-k)^(d-1));

Formula

A(n,k) = (1/k) * A382994(n,k).
A(n,k) = Sum_{j=1..n} (-k)^(gcd(n,j) - 1).
G.f. of column k: Sum_{j>=1} phi(j) * x^j / (1 + k*x^j).

A271344 Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the odd numbers multiplied by -1, interleaved with k-1 zeros, but T(n,1) = 1 and the first element of column k is in row k(k+1)/2.

Original entry on oeis.org

1, 1, 1, -1, 1, 0, 1, -3, 1, 0, -1, 1, -5, 0, 1, 0, 0, 1, -7, -3, 1, 0, 0, -1, 1, -9, 0, 0, 1, 0, -5, 0, 1, -11, 0, 0, 1, 0, 0, -3, 1, -13, -7, 0, -1, 1, 0, 0, 0, 0, 1, -15, 0, 0, 0, 1, 0, -9, -5, 0, 1, -17, 0, 0, 0, 1, 0, 0, 0, -3, 1, -19, -11, 0, 0, -1, 1, 0, 0, -7, 0, 0, 1, -21, 0, 0, 0, 0, 1, 0, -13, 0, 0, 0
Offset: 1

Views

Author

Omar E. Pol, Apr 19 2016

Keywords

Comments

Gives an identity for the deficiency of n. Alternating sum of row n equals the deficiency of n, i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A033879(n).
Row n has length A003056(n) hence the first element of column k is in row A000217(k).
The number of nonzero elements of row n is A001227(n).
If T(n,k) is the second nonzero term in column k then T(n+1,k+1) = -1 is the first element of column k+1.

Examples

			Triangle begins:
  1;
  1;
  1,  -1;
  1,   0;
  1,  -3;
  1,   0,  -1;
  1,  -5,   0;
  1,   0,   0;
  1,  -7,  -3;
  1,   0,   0,  -1;
  1,  -9,   0,   0;
  1,   0,  -5,   0;
  1, -11,   0,   0;
  1,   0,   0,  -3;
  1, -13,  -7,   0,  -1;
  1,   0,   0,   0,   0;
  1, -15,   0,   0,   0;
  1,   0,  -9,  -5,   0;
  1, -17,   0,   0,   0;
  1,   0,   0,   0,  -3;
  1, -19, -11,   0,   0,  -1;
  1,   0,   0,  -7,   0,   0;
  1, -21,   0,   0,   0,   0;
  1,   0, -13,   0,   0,   0;
  1, -23,   0,   0,  -5,   0;
  1,   0    0,  -9,   0,   0;
  1, -25, -15,   0,   0,  -3;
  1,   0,   0,   0,   0,   0,  -1;
  ...
For n = 24 the divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24 so the deficiency of 24 is 24 - 12 - 8 - 6 - 4 - 3 - 2 - 1 = -12. On the other hand the 24th row of triangle is 1, 0, -13, 0, 0, 0, and the alternating row sum is 1 - 0 +(-13) - 0 + 0 - 0 = -12, equaling the deficiency of 24; A033879(24) = -12, so 24 is an abundant number (A005101).
For n = 27 the divisors of 27 are 1, 3, 9, 27 so the deficiency of 27 is 27 - 9 - 3 - 1 = 14. On the other hand the 27th row of triangle is 1, -25, -15, 0, 0, -3, and the alternating row sum is 1 -(-25) +(-15) - 0 + 0 -(-3) = 14, equalling the deficiency of 27; A033879(27) = 14, so 27 is a deficient number (A005100).
For n = 28 the divisors of 28 are 1, 2, 4, 7, 14, 28 so the deficiency of 28 is 28 - 14 - 7 - 4 - 2 - 1 = 0. On the other hand the 28th row of triangle is 1, 0, 0, 0, 0, 0, -1, and the alternating row sum is 1 - 0 + 0 - 0 + 0 - 0 +(-1) = 0, equaling the deficiency of 28; A033879(28) = 0, so 28 is a perfect number (A000396).
		

Crossrefs

Formula

T(n,k) = -1*A231345(n,k).
T(n,k) = -1*A196020(n,k), if k >= 2.

A296063 a(n) is the smallest number (in absolute value) not yet in the sequence such that the arithmetic mean of the first n terms a(1), a(2), ..., a(n) is an integer. Preference is given to positive values of a(n); a(1)=1; 0 not allowed.

Original entry on oeis.org

1, -1, 3, -3, 5, -5, 7, -7, 9, -9, 11, -11, 13, -13, 15, -15, 17, -17, 19, -19, 21, -21, 23, -23, 25, -25, 27, -27, 29, -29, 31, -31, 33, -33, 35, -35, 37, -37, 39, -39, 41, -41, 43, -43, 45, -45, 47, -47, 49, -49, 51, -51
Offset: 1

Views

Author

Enrique Navarrete, Dec 04 2017

Keywords

Crossrefs

Cf. A193356 (partial sums), A059841 (a(n)/n), A109613.

Programs

  • Mathematica
    Array[(2 Floor[(# + 1)/2] - 1) (2 Boole@ OddQ@ # - 1) &, 52] (* or *)
    Nest[Append[#, Block[{k = 1, s = 1}, While[Nand[FreeQ[#, s k], IntegerQ@ Mean[Append[#, s k]]], If[s == 1, s = -1, k++; s = 1]]; s k]] &, {1}, 51] (* Michael De Vlieger, Dec 12 2017 *)
  • PARI
    Vec(x*(1 + x^2) / ((1 - x)*(1 + x)^2) + O(x^50)) \\ Colin Barker, Mar 14 2020

Formula

a(n) = (-1)^(n+1)*A109613(n+1). - Michel Marcus, Dec 05 2017
From Colin Barker, Mar 14 2020: (Start)
G.f.: x*(1 + x^2) / ((1 - x)*(1 + x)^2).
a(n) = -a(n-1) + a(n-2) + a(n-3) for n>3.
(End)
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