A193356 -id:A193356 - cp's OEIS frontend

A196020 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists the odd numbers interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

1, 3, 5, 1, 7, 0, 9, 3, 11, 0, 1, 13, 5, 0, 15, 0, 0, 17, 7, 3, 19, 0, 0, 1, 21, 9, 0, 0, 23, 0, 5, 0, 25, 11, 0, 0, 27, 0, 0, 3, 29, 13, 7, 0, 1, 31, 0, 0, 0, 0, 33, 15, 0, 0, 0, 35, 0, 9, 5, 0, 37, 17, 0, 0, 0, 39, 0, 0, 0, 3, 41, 19, 11, 0, 0, 1, 43, 0, 0, 7, 0, 0, 45, 21, 0, 0, 0, 0, 47, 0, 13, 0, 0, 0

Offset: 1

Omar E. Pol, Feb 02 2013

Gives an identity for sigma(n): alternating sum of row n equals the sum of divisors of n. For proof see Max Alekseyev link.
Row n has length A003056(n) hence column k starts in row A000217(k).
The number of positive terms in row n is A001227(n), the number of odd divisors of n.
If n = 2^j then the only positive integer in row n is T(n,1) = 2^(j+1) - 1.
If n is an odd prime then the only two positive integers in row n are T(n,1) = 2n - 1 and T(n,2) = n - 2.
If T(n,k) = 3 then T(n+1,k+1) = 1, the first element of the column k+1.
The partial sums of column k give the column k of A236104.
The connection with the symmetric representation of sigma is as follows: A236104 --> A235791 --> A237591 --> A237593 --> A239660 --> A237270.
Alternating sum of row n equals the number of units cubes that protrude from the n-th level of the stepped pyramid described in A245092. - Omar E. Pol, Oct 28 2015
Conjecture: T(n,k) is the difference between the square of the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the square of the total number of partitions of all positive integers < n into exactly k consecutive parts. - Omar E. Pol, Feb 14 2018
From Omar E. Pol, Nov 24 2020: (Start)
T(n,k) is also the number of steps in the first n levels of the k-th double-staircase that has at least one step in the n-th level of the "Double- staircases" diagram, otherwise T(n,k) = 0, (see the Example section).
For the connection with A280851 see also the algorithm of A280850 and the conjecture of A296508. (End)
The number of zeros in the n-th row equals A238005(n). - Omar E. Pol, Sep 11 2021
Apart from the alternating row sums and the sum of divisors function A000203 another connection with Euler's pentagonal theorem is that in the irregular triangle of A238442 the k-th column starts in the row that is the k-th generalized pentagonal number A001318(k) while here the k-th column starts in the row that is the k-th generalized hexagonal number A000217(k). Both A001318 and A000217 are successive members of the same family: the generalized polygonal numbers. - Omar E. Pol, Sep 23 2021
Other triangle with the same row lengths and alternating row sums equals sigma(n) is A252117. - Omar E. Pol, May 03 2022

			Triangle begins:
   1;
   3;
   5,  1;
   7,  0;
   9,  3;
  11,  0,  1;
  13,  5,  0;
  15,  0,  0;
  17,  7,  3;
  19,  0,  0,  1;
  21,  9,  0,  0;
  23,  0,  5,  0;
  25, 11,  0,  0;
  27,  0,  0,  3;
  29, 13,  7,  0,  1;
  31,  0,  0,  0,  0;
  33, 15,  0,  0,  0;
  35,  0,  9,  5,  0;
  37, 17,  0,  0,  0;
  39,  0,  0,  0,  3;
  41, 19, 11,  0,  0,  1;
  43,  0,  0,  7,  0,  0;
  45, 21,  0,  0,  0,  0;
  47,  0, 13,  0,  0,  0;
  49, 23,  0,  0,  5,  0;
  51,  0,  0,  9,  0,  0;
  53, 25, 15,  0,  0,  3;
  55,  0,  0,  0,  0,  0,  1;
  ...
For n = 15 the divisors of 15 are 1, 3, 5, 15, so the sum of divisors of 15 is 1 + 3 + 5 + 15 = 24. On the other hand, the 15th row of the triangle is 29, 13, 7, 0, 1, so the alternating row sum is 29 - 13 + 7 - 0 + 1 = 24, equaling the sum of divisors of 15.
If n is even then the alternating sum of the n-th row is simpler to evaluate than the sum of divisors of n. For example the sum of divisors of 24 is 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60, and the alternating sum of the 24th row of triangle is 47 - 0 + 13 - 0 + 0 - 0 = 60.
From _Omar E. Pol_, Nov 24 2020: (Start)
For an illustration of the rows of triangle consider the infinite "double-staircases" diagram defined in A335616 (see also the theorem there).
For n = 15 the diagram with first 15 levels looks like this:
.
Level                         "Double-staircases" diagram
.                                          _
1                                        _|1|_
2                                      _|1 _ 1|_
3                                    _|1  |1|  1|_
4                                  _|1   _| |_   1|_
5                                _|1    |1 _ 1|    1|_
6                              _|1     _| |1| |_     1|_
7                            _|1      |1  | |  1|      1|_
8                          _|1       _|  _| |_  |_       1|_
9                        _|1        |1  |1 _ 1|  1|        1|_
10                     _|1         _|   | |1| |   |_         1|_
11                   _|1          |1   _| | | |_   1|          1|_
12                 _|1           _|   |1  | |  1|   |_           1|_
13               _|1            |1    |  _| |_  |    1|            1|_
14             _|1             _|    _| |1 _ 1| |_    |_             1|_
15            |1              |1    |1  | |1| |  1|    1|              1|
.
The first largest double-staircase has 29 horizontal steps, the second double-staircase has 13 steps, the third double-staircase has 7 steps, and the fifth double-staircases has only one step. Note that the fourth double-staircase does not count because it does not have horizontal steps in the 15th level, so the 15th row of triangle is [29, 13, 7, 0, 1].
For a connection with the "Ziggurat" diagram and the parts and subparts of the symmetric representation of sigma(15) see also A237270. (End)

G. C. Greubel, Table of n, a(n) for the first 200 rows, flattened
Max Alekseyev, Proof of the alternating sum property of A196020, SeqFan Mailing List, Nov 17 2013.
Paul D. Hanna, About an identity for sigma, SeqFan Mailing List, Nov 18 2013.
Index entries for sequences related to sigma(n)

Columns 1-2: A005408, A193356.
Compare with A027750, A238442, A237270, A237273, A252117, A280851, A296508.
Cf. A000203, A000217, A001227, A001318, A003056, A211343, A212119, A228813, A231345, A231347, A235791, A235794, A236104, A236106, A236112, A237048, A237271, A237591, A237593, A238005, A239660, A244050, A245092, A261699, A262626, A286000, A286001, A280850, A335616, A338721.

T_row := proc(n) local T;
T := (n, k) -> if modp(n-k/2, k) = 0 and n >= k*(k+1)/2 then 2*n/k-k else 0 fi;
seq(T(n,k), k=1..floor((sqrt(8*n+1)-1)/2)) end:
seq(print(T_row(n)),n=1..24); # Peter Luschny, Oct 27 2015

T[n_, k_] := If[Mod[n - k*(k+1)/2, k] == 0 ,2*n/k - k, 0]
row[n_] := Floor[(Sqrt[8n+1]-1)/2]
line[n_] := Map[T[n, #]&, Range[row[n]]]
a196020[m_, n_] := Map[line, Range[m, n]]
Flatten[a196020[1,22]] (* data *)
(* Hartmut F. W. Hoft, Oct 26 2015 *)
A196020row = Function[n,Table[If[Divisible[Numerator[n-k/2],k] && CoprimeQ[ Denominator[n- k/2], k],2*n/k-k,0],{k,1,Floor[(Sqrt[8 n+1]-1)/2]}]]
Flatten[Table[A196020row[n], {n,1,24}]] (* Peter Luschny, Oct 28 2015 *)

def T(n,k):
    q = (2*n-k)/2
    b = k.divides(q.numerator()) and gcd(k,q.denominator()) == 1
    return 2*n/k - k if b else 0
for n in (1..24): [T(n, k) for k in (1..floor((sqrt(8*n+1)-1)/2))] # Peter Luschny, Oct 28 2015

A000203(n) = Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k).
T(n,k) = 2*A211343(n,k) - 1, if A211343(n,k) >= 1 otherwise T(n,k) = 0.
If n==k/2 (mod k) and n>=k(k+1)/2, then T(n,k) = 2*n/k - k; otherwise T(n,k) = 0. - Max Alekseyev, Nov 18 2013
T(n,k) = A236104(n,k) - A236104(n-1,k), assuming that A236104(k*(k+1)/2-1,k) = 0. - Omar E. Pol, Oct 14 2018
T(n,k) = A237048(n,k)*A338721(n,k). - Omar E. Pol, Feb 22 2022

A014682 The Collatz or 3x+1 function: a(n) = n/2 if n is even, otherwise (3n+1)/2.

Original entry on oeis.org

0, 2, 1, 5, 2, 8, 3, 11, 4, 14, 5, 17, 6, 20, 7, 23, 8, 26, 9, 29, 10, 32, 11, 35, 12, 38, 13, 41, 14, 44, 15, 47, 16, 50, 17, 53, 18, 56, 19, 59, 20, 62, 21, 65, 22, 68, 23, 71, 24, 74, 25, 77, 26, 80, 27, 83, 28, 86, 29, 89, 30, 92, 31, 95, 32, 98, 33, 101, 34, 104

Offset: 0

Mohammad K. Azarian

This is the function usually denoted by T(n) in the literature on the 3x+1 problem. See A006370 for further references and links.
Intertwining of sequence A016789 '2,5,8,11,... ("add 3")' and the nonnegative integers.
a(n) = log_2(A076936(n)). - Amarnath Murthy, Oct 19 2002
The average value of a(0), ..., a(n-1) is A004526(n). - Amarnath Murthy, Oct 19 2002
Partial sums are A093353. - Paul Barry, Mar 31 2008
Absolute first differences are essentially in A014681 and A103889. - R. J. Mathar, Apr 05 2008
Only terms of A016789 occur twice, at positions given by sequences A005408 (odd numbers) and A016957 (6n+4): (1,4), (3,10), (5,16), (7,22), ... - Antti Karttunen, Jul 28 2017
a(n) represents the unique congruence class modulo 2n+1 that is represented an odd number of times in any 2n+1 consecutive oblong numbers (A002378). This property relates to Jim Singh's 2018 formula, as n^2 + n is a relevant oblong number. - Peter Munn, Jan 29 2022

			a(3) = -3*(-1) - 2*1 - 1*(-1) - 0*1 + 1*(-1) + 2*1 + 3*(-1) + 4*1 + 5*(-1) + 6*1 = 5. - _Bruno Berselli_, Dec 14 2015

J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010.

N. J. A. Sloane (terms 0..1000) & Antti Karttunen, Table of n, a(n) for n = 0..100000
C. J. Everett, Iteration of the number-theoretic function f(2n) = n, f(2n + 1) = 3n + 2, Advances in Mathematics, Volume 25, Issue 1, July 1977, Pages 42-45.
Jeffrey C. Lagarias, The 3x+1 Problem: An Overview, arXiv:2111.02635 [math.NT], 2021.
Keenan Monks, Kenneth G. Monks, Kenneth M. Monks, and Maria Monks, Strongly sufficient sets and the distribution of arithmetic sequences in the 3x+1 graph, arXiv:1204.3904v1 [math.DS], Apr 17 2012.
R. Terras, A stopping time problem on the positive integers, Acta Arith. 30 (1976) 241-252.
Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A002378, A004526, A076936, A139391, A016116, A126241, A060412, A060413, A006370, A070168 (iterations), A005408, A016957, A064455, A153285, A008619, A193356.
Bisections: A001477, A016789.
Cf. A093353.

a014682 n = if r > 0 then div (3 * n + 1) 2 else n'
            where (n', r) = divMod n 2
-- Reinhard Zumkeller, Oct 03 2014

[IsOdd(n) select (3*n+1)/2 else n/2: n in [0..52]]; // Vincenzo Librandi, Sep 28 2018

T:=proc(n) if n mod 2 = 0 then n/2 else (3*n+1)/2; fi; end; # N. J. A. Sloane, Jan 31 2011
A076936 := proc(n) option remember ; local apr,ifr,me,i,a ; if n <=2 then n^2 ; else apr := mul(A076936(i),i=1..n-1) ; ifr := ifactors(apr)[2] ; me := -1 ; for i from 1 to nops(ifr) do me := max(me, op(2,op(i,ifr))) ; od ; me := me+ n-(me mod n) ; a := 1 ; for i from 1 to nops(ifr) do a := a*op(1,op(i,ifr))^(me-op(2,op(i,ifr))) ; od ; if a = A076936(n-1) then me := me+n ; a := 1 ; for i from 1 to nops(ifr) do a := a*op(1,op(i,ifr))^(me-op(2,op(i,ifr))) ; od ; fi ; RETURN(a) ; fi ; end: A014682 := proc(n) log[2](A076936(n)) ; end: for n from 1 to 85 do printf("%d, ",A014682(n)) ; od ; # R. J. Mathar, Mar 20 2007

Collatz[n_?OddQ] := (3n + 1)/2; Collatz[n_?EvenQ] := n/2; Table[Collatz[n], {n, 0, 79}] (* Alonso del Arte, Apr 21 2011 *)
LinearRecurrence[{0, 2, 0, -1}, {0, 2, 1, 5}, 70] (* Jean-François Alcover, Sep 23 2017 *)
Table[If[OddQ[n], (3 n + 1) / 2, n / 2], {n, 0, 60}] (* Vincenzo Librandi, Sep 28 2018 *)

a(n)=if(n%2,3*n+1,n)/2 \\ Charles R Greathouse IV, Sep 02 2015

a(n)=if(n<2,2*n,(n^2-n-1)%(2*n+1)) \\ Jim Singh, Sep 28 2018

def a(n): return n//2 if n%2==0 else (3*n + 1)//2
print([a(n) for n in range(101)]) # Indranil Ghosh, Jul 29 2017

From Paul Barry, Mar 31 2008: (Start)
G.f.: x*(2 + x + x^2)/(1-x^2)^2.
a(n) = (4*n+1)/4 - (2*n+1)*(-1)^n/4. (End)
a(n) = -a(n-1) + a(n-2) + a(n-3) + 4. - John W. Layman
For n > 1 this is the image of n under the modified "3x+1" map (cf. A006370): n -> n/2 if n is even, n -> (3*n+1)/2 if n is odd. - Benoit Cloitre, May 12 2002
O.g.f.: x*(2+x+x^2)/((-1+x)^2*(1+x)^2). - R. J. Mathar, Apr 05 2008
a(n) = 5/4 + (1/2)*((-1)^n)*n + (3/4)*(-1)^n + n. - Alexander R. Povolotsky, Apr 05 2008
a(n) = Sum_{i=-n..2*n} i*(-1)^i. - Bruno Berselli, Dec 14 2015
a(n) = Sum_{k=0..n-1} Sum_{i=0..k} C(i,k) + (-1)^k. - Wesley Ivan Hurt, Sep 20 2017
a(n) = (n^2-n-1) mod (2*n+1) for n > 1. - Jim Singh, Sep 26 2018
The above formula can be rewritten to show a pattern: a(n) = (n*(n+1)) mod (n+(n+1)). - Peter Munn, Jan 29 2022
Binary: a(n) = (n shift left (n AND 1)) - (n shift right 1) = A109043(n) - A004526(n). - Rudi B. Stranden, Jun 15 2021
From Rudi B. Stranden, Mar 21 2022: (Start)
a(n) = A064455(n+1) - 1, relating the number ON cells in row n of cellular automaton rule 54.
a(n) = 2*n - A071045(n).
(End)
E.g.f.: (1 + x)*sinh(x)/2 + 3*x*cosh(x)/2 = ((4*x+1)*e^x + (2*x-1)*e^(-x))/4. - Rénald Simonetto, Oct 20 2022
a(n) = n*(n mod 2) + ceiling(n/2) = A193356(n) + A008619(n+1). - Jonathan Shadrach Gilbert, Mar 12 2023
a(n) = 2*a(n-2) - a(n-4) for n > 3. - Chai Wah Wu, Apr 17 2024

Edited by N. J. A. Sloane, Apr 26 2008, at the suggestion of Artur Jasinski

Edited by N. J. A. Sloane, Jan 31 2011

A059260 Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-y-x*y-x^2) = 1/((1+x)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 1, 2, 4, 3, 1, 0, 3, 6, 7, 4, 1, 1, 3, 9, 13, 11, 5, 1, 0, 4, 12, 22, 24, 16, 6, 1, 1, 4, 16, 34, 46, 40, 22, 7, 1, 0, 5, 20, 50, 80, 86, 62, 29, 8, 1, 1, 5, 25, 70, 130, 166, 148, 91, 37, 9, 1, 0, 6, 30, 95, 200, 296, 314, 239, 128, 46, 10, 1

Offset: 0

N. J. A. Sloane, Jan 23 2001

Coefficients of the (left, normalized) shifted cyclotomic polynomial. Or, coefficients of the basic n-th q-series for q=-2. Indeed, let Y_n(x) = Sum_{k=0..n} x^k, having as roots all the n-th roots of unity except for 0; then coefficients in x of (-1)^n Y_n(-x-1) give exactly the n-th row of A059260 and a practical way to compute it. - Olivier Gérard, Jul 30 2002
The maximum in the (2n)-th row is T(n,n), which is A026641; also T(n,n) ~ (2/3)*binomial(2n,n). The maximum in the (2n-1)-th row is T(n-1,n), which is A014300 (but T does not have the same definition as in A026637); also T(n-1,n) ~ (1/3)*binomial(2n,n). Here is a generalization of the formula given in A026641: T(i,j) = Sum_{k=0..j} binomial(i+k-x,j-k)*binomial(j-k+x,k) for all x real (the proof is easy by induction on i+j using T(i,j) = T(i-1,j) + T(i,j-1)). - Claude Morin, May 21 2002
The second greatest term in the (2n)-th row is T(n-1,n+1), which is A014301; the second greatest term in the (2n+1)-th row is T(n+1,n) = 2*T(n-1,n+1), which is 2*A014301. - Claude Morin
Diagonal sums give A008346. - Paul Barry, Sep 23 2004
Riordan array (1/(1-x^2), x/(1-x)). As a product of Riordan arrays, factors into the product of (1/(1+x),x) and (1/(1-x),1/(1-x)) (binomial matrix). - Paul Barry, Oct 25 2004
Signed version is A239473 with relations to partial sums of sequences. - Tom Copeland, Mar 24 2014
From Robert Coquereaux, Oct 01 2014: (Start)
Columns of the triangle (cf. Example below) give alternate partial sums along nw-se diagonals of the Pascal triangle, i.e., sequences A000035, A004526, A002620 (or A087811), A002623 (or A173196), A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808, etc.
The dimension of the space of closed currents (distributional forms) of degree p on Gr(n), the Grassmann algebra with n generators, equivalently, the dimension of the space of Gr(n)-valued symmetric multilinear forms with vanishing graded divergence, is V(n,p) = 2^n T(p,n-1) - (-1)^p.
If p is odd V(n,p) is also the dimension of the cyclic cohomology group of order p of the Z2 graded algebra Gr(n).
If p is even the dimension of this cohomology group is V(n,p)+1.
Cf. A193844. (End)
From Peter Bala, Feb 07 2024: (Start)
The following remarks assume the row indexing starts at n = 1.
The sequence of row polynomials R(n,x), beginning R(1,x) = 1, R(2,x) = x, R(3,x) = 1 + x + x^2 , ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd( R(n,x), R(m,x)) = R(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the polynomial sequence {R(n,x): n >= 1} is a divisibility sequence; that is, if n divides m then R(n,x) divides R(m,x) in Z[x]. (End)
From Miquel A. Fiol, Oct 04 2024: (Start)
For j>=1, T(i,j) is the independence number of the (i-j)-supertoken graph FF_(i-j)(S_j) of the star graph S_j with j points.
(Given a graph G on n vertices and an integer k>=1, the k-supertoken (or reduced k-th power) FF_k(G) of G has vertices representing configurations of k indistinguishable tokens in the (not necessarily different) vertices of G, with two configurations being adjacent if one can be obtained from the other by moving one token along an edge. See an example below.)
Following the suggestion of Peter Munn, the k-supertoken graph FF_k(S_j) can also be defined as follows: Consider the Lattice graph L(k,j), whose vertices are the k^j j-vectors with elements in the set {0,..,k-1}, two being adjacent if they differ in just one coordinate by one unity. Then, FF_k(S_j) is the subgraph of L(k+1,j) induced by the vertices at distance at most k from (0,..,0). (End)

			Triangle begins
  1;
  0,  1;
  1,  1,  1;
  0,  2,  2,  1;
  1,  2,  4,  3,  1;
  0,  3,  6,  7,  4,  1;
  1,  3,  9, 13, 11,  5,  1;
  0,  4, 12, 22, 24, 16,  6,  1;
  1,  4, 16, 34, 46, 40, 22,  7,  1;
  0,  5, 20, 50, 80, 86, 62, 29,  8,  1;
Sequences obtained with _Miquel A. Fiol_'s Sep 30 2024 formula of A(n,c1,c2) for other values of (c1,c2). (In the table, rows are indexed by c1=0..6 and columns by c2=0..6):
A000007  A000012  A000027  A025747  A000292* A000332* A000389*
A059841  A008619  A087811* A002623  A001752  A001753  A001769
A193356  A008794* A005993  A005994  -------  -------  -------
-------  -------  -------  A005995  A018210  -------  A052267
-------  -------  -------  -------  A018211  A018212  -------
-------  -------  -------  -------  -------  A018213  A018214
-------  -------  -------  -------  -------  -------  A062136
*requires offset adjustment.
The 2-supertoken FF_2(S_3) of the star graph S_3 with central vertex 1 and peripheral vertices 2,3,4. (The vertex `ij' of FF_2(S_3) represents the configuration of one token in `ì' and the other token in `j'). The T(5,3)=7 independent vertices are 22, 24, 44, 23, 11, 34, and 33.
     22--12---24---14---44
          | \    / |
         23   11   34
            \  |  /
              13
               |
              33

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Roland Bacher, Chebyshev polynomials, quadratic surds and a variation of Pascal's triangle, arXiv:1509.09054 [math.CO], 2015.
Joseph Briggs, Alex Parker, Coy Schwieder, and Chris Wells, Frogs, hats and common subsequences, arXiv:2404.07285 [math.CO], 2024. See p. 28.
Robert Coquereaux and Éric Ragoucy, Currents on Grassmann algebras, J. of Geometry and Physics, 1995, Vol 15, pp 333-352.
Robert Coquereaux and Éric Ragoucy, Currents on Grassmann algebras, arXiv:hep-th/9310147, 1993.
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. II. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See p. 8.
R. H. Hammack and G. D. Smith, Cycle bases of reduced powers of graphs, Ars Math. Contemp. 12 (2017) 183-203.
Christian Kassel, A Künneth formula for the cyclic cohomology of Z2-graded algebras, Math. Ann. 275 (1986) 683.
Ana Filipa Loureiro and Pascal Maroni, Polynomial sequences associated with the classical linear functionals, Numerical Algorithms, June 2012, Volume 60, Issue 2, pp 297-314. - From _N. J. A. Sloane_, Oct 12 2012
Ana Filipa Loureiro and Pascal Maroni, Polynomial sequences associated with the classical linear functionals, preprint, Centro de Matemática da Universidade do Porto.
MathOverflow, Cyclotomic Polynomials in Combinatorics
Mark Norfleet, Characterization of second-order strong divisibility sequences of polynomials, The Fibonacci Quarterly, 43(2) (2005), 166-169.

Cf. A059259. Row sums give A001045.
Seen as a square array read by antidiagonals this is the coefficient of x^k in expansion of 1/((1-x^2)*(1-x)^n) with rows A002620, A002623, A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808 etc. (allowing for signs). A058393 would then effectively provide the table for nonpositive n. - Henry Bottomley, Jun 25 2001
Cf. A026641, A014300.
Cf. A007318, A097805, A097808, A130595, A200139, A062135.

read transforms; 1/(1-y-x*y-x^2); SERIES2(%,x,y,12); SERIES2TOLIST(%,x,y,12);

t[n_, k_] := Sum[ (-1)^(n-j)*Binomial[j, k], {j, 0, n}]; Flatten[ Table[t[n, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Oct 20 2011, after Paul Barry *)

T(n, k) = sum(j=0, n, (-1)^(n - j)*binomial(j, k));
for(n=0, 12, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 11 2017

from sympy import binomial
def T(n, k): return sum((-1)**(n - j)*binomial(j, k) for j in range(n + 1))
for n in range(13): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 11 2017

def A059260_row(n):
    @cached_function
    def prec(n, k):
        if k==n: return 1
        if k==0: return 0
        return -prec(n-1,k-1)-sum(prec(n,k+i-1) for i in (2..n-k+1))
    return [(-1)^(n-k+1)*prec(n+1, n-k+1) for k in (1..n)]
for n in (1..9): print(A059260_row(n)) # Peter Luschny, Mar 16 2016

G.f.: 1/(1-y-x*y-x^2) = 1 + y + x^2 + xy + y^2 + 2x^2y + 2xy^2 + y^3 + ...
E.g.f: (exp(-t)+(x+1)*exp((x+1)*t))/(x+2). - Tom Copeland, Mar 19 2014
O.g.f. (n-th row): ((-1)^n+(x+1)^(n+1))/(x+2). - Tom Copeland, Mar 19 2014
T(i, 0) = 1 if i is even or 0 if i is odd, T(0, i) = 1 and otherwise T(i, j) = T(i-1, j) + T(i, j-1); also T(i, j) = Sum_{m=j..i+j} (-1)^(i+j+m)*binomial(m, j). - Robert FERREOL, May 17 2002
T(i, j) ~ (i+j)/(2*i+j)*binomial(i+j, j); more precisely, abs(T(i, j)/binomial(i+j, j) - (i+j)/(2*i+j) )<=1/(4*(i+j)-2); the proof is by induction on i+j using the formula 2*T(i, j) = binomial(i+j, j)+T(i, j-1). - Claude Morin, May 21 2002
T(n, k) = Sum_{j=0..n} (-1)^(n-j)binomial(j, k). - Paul Barry, Aug 25 2004
T(n, k) = Sum_{j=0..n-k} binomial(n-j, j)*binomial(j, n-k-j). - Paul Barry, Jul 25 2005
Equals A097807 * A007318. - Gary W. Adamson, Feb 21 2007
Equals A128173 * A007318 as infinite lower triangular matrices. - Gary W. Adamson, Feb 17 2007
Equals A130595*A097805*A007318 = (inverse Pascal matrix)*(padded Pascal matrix)*(Pascal matrix) = A130595*A200139. Inverse is A097808 = A130595*(padded A130595)*A007318. - Tom Copeland, Nov 14 2016
T(i, j) = binomial(i+j, j)-T(i-1, j). - Laszlo Major, Apr 11 2017
Recurrence for row polynomials (with row indexing starting at n = 1): R(n,x) = x*R(n-1,x) + (x + 1)*R(n-2,x) with R(1,x) = 1 and R(2,x) = x. - Peter Bala, Feb 07 2024
From Miquel A. Fiol, Sep 30 2024: (Start)
The triangle can be seen as a slice of a 3-dimensional table that links it to well-known sequences as follows.
The j-th column of the triangle, T(i,j) for i >= j, equals A(n,c1,c2) = Sum_{k=0..floor(n/2)} binomial(c1+2*k-1,2*k)*binomial(c2+n-2*k-1,n-2*k) when c1=1, c2=j, and n=i-j.
This gives T(i,j) = Sum_{k=0..floor((i-j)/2)} binomial(i-2*k-1, j-1). For other values of (c1,c2), see the example below. (End)

Formula corrected by Philippe Deléham, Jan 11 2014

A317614 a(n) = (1/2)(n^3 + n(n mod 2)).

Original entry on oeis.org

1, 4, 15, 32, 65, 108, 175, 256, 369, 500, 671, 864, 1105, 1372, 1695, 2048, 2465, 2916, 3439, 4000, 4641, 5324, 6095, 6912, 7825, 8788, 9855, 10976, 12209, 13500, 14911, 16384, 17985, 19652, 21455, 23328, 25345, 27436, 29679, 32000, 34481, 37044, 39775, 42592

Offset: 1

Stefano Spezia, Aug 01 2018

Terms are obtained as partial sums in an algorithm for the generation of the sequence of the fourth powers (A000583). Starting with the sequence of the positive integers (A000027), it is necessary to delete every 4th term and to consider the partial sums of the obtained sequence, then to delete every 3rd term, and lastly to consider again the partial sums (see References).
a(n) is the trace of an n X n square matrix M(n) formed by writing the numbers 1, ..., n^2 successively forward and backward along the rows in zig-zag pattern as shown in the examples below. Specifically, M(n) is defined as M[i,j,n] = j + n*(i-1) if i is odd and M[i,j,n] = n*i - j + 1 if i is even, and it has det(M(n)) = 0 for n > 2 (proved).
From Saeed Barari, Oct 31 2021: (Start)
Also the sum of the entries in an n X n matrix whose elements start from 1 and increase as they approach the center. For instance, in case of n=5, the entries of the following matrix sum to 65:
  1 2 3 2 1
  2 3 4 3 2
  3 4 5 4 3
  2 3 4 3 2
  1 2 3 2 1. (End)
The n X n square matrix of the preceding comment is defined as: A[i,j,n] = n - abs((n + 1)/2 - j) - abs((n + 1)/2 - i). - Stefano Spezia, Nov 05 2021

			For n = 1 the matrix M(1) is
  1
with trace Tr(M(1)) = a(1) = 1.
For n = 2 the matrix M(2) is
  1, 2
  4, 3
with Tr(M(2)) = a(2) = 4.
For n = 3 the matrix M(3) is
  1, 2, 3
  6, 5, 4
  7, 8, 9
with Tr(M(3)) = a(3) = 15.

Edward A. Ashcroft, Anthony A. Faustini, Rangaswami Jagannathan, and William W. Wadge, Multidimensional Programming, Oxford University Press 1995, p. 12.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 64.
G. Polya, Mathematics and Plausible Reasoning: Induction and analogy in mathematics, Princeton University Press 1990, p. 118.
Shailesh Shirali, A Primer on Number Sequences, Universities Press (India) 2004, p. 106.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 3.7.3 on pages 122-123.

Stefano Spezia, Table of n, a(n) for n = 0..10000
Alfred Moessner, Eine Bemerkung über die Potenzen der natürlichen Zahlen, München 1952. Sitzungsberichte: 1951,3.
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Cf. A000583, A000027, A186424 (first differences).
Cf. A000578, A000035, A193356, A210378.
Cf. A006003, A317297, A001489, A322844.
Cf. related to the M matrices: A074147 (antidiagonals), A130130 (rank), A241016 (row sums), A317617 (column sums), A322277 (permanent), A323723 (subdiagonal sums), A323724 (superdiagonal sums).

a_n:=List([1..nmax], n->(1/2)*(n^3 + n*RemInt(n, 2)));

List([1..50],n->(1/2)*(n^3+n*(n mod 2))); # Muniru A Asiru, Aug 24 2018

[IsEven(n) select n^3/2 else (n^3+n)/2: n in [1..50]]; // Vincenzo Librandi, Aug 07 2018

a:=n->(1/2)*(n^3+n*modp(n,2)): seq(a(n),n=1..50); # Muniru A Asiru, Aug 24 2018

CoefficientList[Series[1/4 E^-x (1 + 3 E^(2 x) + 6 E^(2 x) x + 2 E^(2 x) x^2), {x, 0, 45}], x]*Table[(k + 1)!, {k, 0, 45}]
CoefficientList[Series[-(1 + x^2)/((-1 + x)*(1 + x)^3), {x, 0, 45}], x]*Table[(k + 1)*(-1)^k, {k, 0, 45}]
CoefficientList[Series[-(1 + x^2)/((-1 + x)^3*(1 + x)), {x, 0, 45}], x]*Table[(k + 1), {k, 0, 45}]
From Robert G. Wilson v, Aug 01 2018: (Start)
a[i_, j_, n_] := If[OddQ@ i, j + n (i - 1), n*i - j + 1]; f[n_] := Tr[Table[a[i, j, n], {i, n}, {j, n}]]; Array[f, 45]
CoefficientList[Series[(x^4 + 2x^3 + 6x^2 + 2x + 1)/((x - 1)^4 (x + 1)^2), {x, 0, 45}], x]
LinearRecurrence[{2, 1, -4, 1, 2, -1}, {1, 4, 15, 32, 65, 108}, 45]
(End)

a(n):=(1/2)*(n^3 + n*mod(n,2))$ makelist(a(n), n, 1, nmax);

Vec(x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4) / ((1 - x)^4*(1 + x)^2) + O(x^40)) \\ Colin Barker, Aug 02 2018

M(i, j, n) = if (i % 2, j + n*(i-1), n*i - j + 1);
a(n) = sum(k=1, n, M(k, k, n)); \\ Michel Marcus, Aug 07 2018

for (n in 1:nmax){
   a <- (n^3+n*n%%2)/2
   output <- c(n, a)
   cat(output, "\n")
}
(MATLAB and FreeMat)
for(n=1:nmax); a=(n^3+n*mod(n,2))/2; fprintf('%d\t%0.f\n',n,a); end

a(n) = (1/2)*(A000578(n) + n*A000035(n)).
a(n) = A006003(n) - (n/2)*(1 - (n mod 2)).
a(n) = Sum_{k=1..n} T(n,k), where T(n,k) = ((n + 1)*k - n)*(n mod 2) + ((n - 1)*k + 1)*(1 - (n mod 2)).
E.g.f.: E(x) = (1/4)*exp(-x)*x*(1 + 3*exp(2*x) + 6*exp(2*x)*x + 2*exp(2*x)*x^2).
L.g.f.: L(x) = -x*(1 + x^2)/((-1 + x)*(1 + x)^3).
H.l.g.f.: LH(x) = -x*(1 + x^2)/((-1 + x)^3*(1 + x)).
Dirichlet g.f.: (1/2)*(Zeta(-3 + s) + 2^(-s)*(-2 + 2^s)*Zeta(-1 + s)).
From Colin Barker, Aug 02 2018: (Start)
G.f.: x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4) / ((1 - x)^4*(1 + x)^2).
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) for n>6.
a(n) = n^3/2 for n even.
a(n) = (n^3+n)/2 for n odd. (End)
a(2*n) = A317297(n+1) + A001489(n). - Stefano Spezia, Dec 28 2018
Sum_{n>0} 1/a(n) = (1/2)*(-2*polygamma(0, 1/2) + polygamma(0, (1-i)/2)+ polygamma(0, (1+i)/2)) + zeta(3)/4 approximately equal to 1.3959168891658447368440622669882813003351669... - Stefano Spezia, Feb 11 2019
a(n) = (A000578(n) + A193356(n))/2. - Stefano Spezia, Jun 27 2022
a(n) = A210378(n-1)/n. - Stefano Spezia, Jul 15 2024

A057211 Alternating runs of ones and zeros, where the n-th run has length n.

Original entry on oeis.org

1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Ben Tyner (tyner(AT)phys.ufl.edu), Sep 27 2000

Seen as a triangle read by rows: T(n,k) = n mod 2, 1<=k<=n. - Reinhard Zumkeller, Mar 18 2011
a(A007607(n)) = 0; a(A007606(n)) = 1. - Reinhard Zumkeller, Dec 30 2011
Row sums give A193356. - Omar E. Pol, Mar 05 2014

K. H. Rosen, Discrete Mathematics and its Applications, 1999, Fourth Edition, page 79, exercise 10 (g).

Reinhard Zumkeller, Rows n=1..125 of triangle, flattened
Index entries for characteristic functions

Cf. A057212, A059841.

a057211 n = a057211_list !! (n-1)
a057211_list = concat $ zipWith ($) (map replicate [1..]) a059841_list
-- Reinhard Zumkeller, Mar 18 2011

A002024 := n->round(sqrt(2*n)):A057211 := n->(1-(-1)^A002024(n))/2;
# alternative Maple program:
T:= n-> [irem(n, 2)$n][]:
seq(T(n), n=1..14);  # Alois P. Heinz, Oct 06 2021

Flatten[Table[{PadRight[{},n,1],PadRight[{},n+1,0]},{n,1,21,2}]] (* Harvey P. Dale, Jun 07 2015 *)

from math import isqrt
def A057211(n): return int(bool(isqrt(n<<3)+1&2)) # Chai Wah Wu, Jun 19 2024

a(n) = (1-(-1)^A002024(n))/2, where A002024(n)=round(sqrt(2*n)). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 23 2003
Also a(n) = A000035(A002024(n)) = A002024(n) mod 2 = A002024(n)-2*floor(A002024(n)/2). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 23 2003
G.f.: x/(1-x)*sum_{n>=0} (-1)^n*x^(n*(n+1)/2). - Mircea Merca, Mar 05 2014
a(n) = 1 - A057212(n). - Alois P. Heinz, Oct 06 2021

Definition amended by Georg Fischer, Oct 06 2021

A129194 a(n) = (n/2)^2*(3 - (-1)^n).

Original entry on oeis.org

0, 1, 2, 9, 8, 25, 18, 49, 32, 81, 50, 121, 72, 169, 98, 225, 128, 289, 162, 361, 200, 441, 242, 529, 288, 625, 338, 729, 392, 841, 450, 961, 512, 1089, 578, 1225, 648, 1369, 722, 1521, 800, 1681, 882, 1849, 968, 2025, 1058, 2209, 1152, 2401, 1250, 2601, 1352

Offset: 0

Paul Barry, Apr 02 2007

The numerator of the integral is 2,1,2,1,2,1,...; the moments of the integral are 2/(n+1)^2. See 2nd formula.
The sequence alternates between twice a square and an odd square, A001105(n) and A016754(n).
Partial sums of the positive elements give the absolute values of A122576. - Omar E. Pol, Aug 22 2011
Partial sums of the positive elements give A212760. - Omar E. Pol, Dec 28 2013
Conjecture: denominator of 4/n - 2/n^2. - Wesley Ivan Hurt, Jul 11 2016
Multiplicative because both A000290 and A040001 are. - Andrew Howroyd, Jul 25 2018

G. Pólya and G. Szegő, Problems and Theorems in Analysis II (Springer 1924, reprinted 1976), Part Eight, Chap. 1, Sect. 7, Problem 73.

Vincenzo Librandi, Table of n, a(n) for n = 0..960
Olivier Bordelles, A Multidimensional Cesaro Type Identity and Applications, J.
Int. Seq. 18 (2015) # 15.3.7.
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. A000290, A001105, A007434, A010713, A016742, A016754, A040001.

Cf. A061038, A061040, A061050, A105398, A129204, A152020, A222171, A309337.

[n^2*(3-(-1)^n)/4: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011

A129194:=n->n^2*(3-(-1)^n)/4: seq(A129194(n), n=0..80); # Wesley Ivan Hurt, Jul 11 2016

Table[n^2*(3-(-1)^n)/4, {n,0,60}] (* Wesley Ivan Hurt, Jul 11 2016 *)
LinearRecurrence[{0,3,0,-3,0,1},{0,1,2,9,8,25},60] (* Harvey P. Dale, Dec 27 2023 *)

a(n) = lcm(2, n^2)/2; \\ Andrew Howroyd, Jul 25 2018

[n^2*(1+(n%2))/2 for n in range(61)] # G. C. Greubel, Apr 04 2023

G.f.: x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4)/(1-x^2)^3.
a(n+1) = denominator((1/(2*Pi))*Integral_{t=0..2*Pi} exp(i*n*t)(-((Pi-t)/i)^2)), i=sqrt(-1).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n > 5. - Paul Curtz, Mar 07 2011
a(n) is the numerator of the coefficient of x^4 in the Maclaurin expansion of exp(-n*x^2). - Francesco Daddi, Aug 04 2011
O.g.f. as a Lambert series: x*Sum_{n >= 1} J_2(n)*x^n/(1 + x^n), where J_2(n) denotes the Jordan totient function A007434(n). See Pólya and Szegő. - Peter Bala, Dec 28 2013
From Ilya Gutkovskiy, Jul 11 2016: (Start)
E.g.f.: x*((2*x + 1)*sinh(x) + (x + 2)*cosh(x))/2.
Sum_{n>=1} 1/a(n) = 5*Pi^2/24. [corrected by Amiram Eldar, Sep 11 2022] (End)
a(n) = A000290(n) / A040001(n). - Andrew Howroyd, Jul 25 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/24 (A222171). - Amiram Eldar, Sep 11 2022
From Peter Bala, Jan 16 2024: (Start)
a(n) = Sum_{1 <= i, j <= n} (-1)^(1 + gcd(i,j,n)) = Sum_{d | n} (-1)^(d+1) * J_2(n/d), that is, the Dirichlet convolution of the pair of multiplicative functions f(n) = (-1)^(n+1) and the Jordan totient function J_2(n) = A007434(n). Hence this sequence is multiplicative. Cf. A193356 and A309337.
Dirichlet g.f.: (1 - 2/2^s)*zeta(s-2). (End)
a(n) = Sum_{1 <= i, j <= n} (-1)^(n + gcd(i, n)*gcd(j, n)) = Sum_{d|n, e|n} (-1)^(n+e*d) * phi(n/d)*phi(n/e). - Peter Bala, Jan 22 2024

More terms from Michel Marcus, Dec 28 2013

A231345 Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the odd numbers 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

Offset: 1

Omar E. Pol, Dec 26 2013

Gives an identity for the abundance of n. Alternating sum of row n equals the abundance of n, i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A033880(n).

Row n has length A003056(n) hence the first element of column k is in row A000217(k).

			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;
  ...
For n = 15 the divisors of 15 are 1, 3, 5, 15 hence the abundance of 15 is 1 + 3 + 5 + 15 - 2*15 = 1 + 3 + 5 - 15 = -6. On the other hand the 15th row of triangle is -1, 13, 7, 0, 1, hence the alternating row sum is -1 - 13 + 7 - 0 + 1 = -6, equalling the abundance of 15.
If n is even then the alternating sum of the n-th row of triangle is simpler than the sum of divisors of n minus 2*n. Example: the sum of divisors of 24 minus 2*24 is 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 - 2*24 = 60 - 48 = 12, and the alternating sum of the 24th row of triangle is -1 - 0 + 13 - 0 + 0 - 0 = 12.

Eric Weisstein's World of Mathematics, Abundance
Eric Weisstein's World of Mathematics, Quasiperfect Number

Column 2 is A193356.

Cf. A000203, A000217, A000396, A003056, A005100, A005843, A033879, A033880, A069283, A196020, A212119, A228813, A231347, A235791, A235794, A236104, A236106, A236112, A237593.

T(n,1) = -1; T(n,k) = A196020(n,k), for k >= 2.

A231347 Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the odd numbers interleaved with k-1 zeros but T(n,1) = n - 1 and the first element of column k is in row k(k+1)/2.

Original entry on oeis.org

0, 1, 2, 1, 3, 0, 4, 3, 5, 0, 1, 6, 5, 0, 7, 0, 0, 8, 7, 3, 9, 0, 0, 1, 10, 9, 0, 0, 11, 0, 5, 0, 12, 11, 0, 0, 13, 0, 0, 3, 14, 13, 7, 0, 1, 15, 0, 0, 0, 0, 16, 15, 0, 0, 0, 17, 0, 9, 5, 0, 18, 17, 0, 0, 0, 19, 0, 0, 0, 3, 20, 19, 11, 0, 0, 1, 21, 0, 0, 7, 0, 0

Offset: 1

Omar E. Pol, Dec 28 2013

Alternating sum of row n equals the sum of aliquot divisors of n, i.e., sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A001065(n).
Row n has length A003056(n).
Column k starts in row A000217(k).
The number of positive terms in row n is A001227(n), for n >= 2.
If n = 2^j then the only positive integer in row n is T(n,1) = n - 1, for j >= 1.
If n is an odd prime then the only two positive integers in row n are T(n,1) = n - 1 and T(n,2) = n - 2.

			Triangle begins:
  0;
  1;
  2,   1;
  3,   0;
  4,   3;
  5,   0,  1;
  6,   5,  0;
  7,   0,  0;
  8,   7,  3;
  9,   0,  0,  1;
  10,  9,  0,  0;
  11,  0,  5,  0;
  12, 11,  0,  0;
  13,  0,  0,  3;
  14, 13,  7,  0,  1;
  15,  0,  0,  0,  0;
  16, 15,  0,  0,  0;
  17,  0,  9,  5,  0;
  18, 17,  0,  0,  0;
  19,  0,  0,  0,  3;
  20, 19, 11,  0,  0,  1;
  21,  0,  0,  7,  0,  0;
  22, 21,  0,  0,  0,  0;
  23,  0, 13,  0,  0,  0;
  ...
For n = 15 the aliquot divisors of 15 are 1, 3, 5, therefore the sum of aliquot divisors of 15 is 1 + 3 + 5 = 9. On the other hand the 15th row of triangle is 14, 13, 7, 0, 1, hence the alternating row sum is 14 - 13 + 7 - 0 + 1 = 9, equalling the sum of aliquot divisors of 15.
If n is even then the alternating sum of the n-th row of triangle is simpler than the sum of aliquot divisors of n. Example: the sum of aliquot divisors of 24 is 1 + 2 + 3 + 4 + 6 + 8 + 12 = 36, and the alternating sum of the 24th row of triangle is 23 - 0 + 13 - 0 + 0 - 0 = 36.

Columns 1-2: A001477, A193356.

Cf. A000217, A001065, A001227, A000203, A003056, A196020, A211343, A212119, A228813, A231345, A235791, A235794, A236104, A236106, A236112, A237591, A237593, A286001.

T(n,1) = n - 1.

T(n,k) = A196020(n,k), for k >= 2.

A274804 The exponential transform of sigma(n).

Original entry on oeis.org

1, 1, 4, 14, 69, 367, 2284, 15430, 115146, 924555, 7991892, 73547322, 718621516, 7410375897, 80405501540, 914492881330, 10873902417225, 134808633318271, 1738734267608613, 23282225008741565, 323082222240744379, 4638440974576329923, 68794595993688306903

Offset: 0

Johannes W. Meijer, Jul 27 2016

nonn

The exponential transform [EXP] transforms an input sequence b(n) into the output sequence a(n). The EXP transform is the inverse of the logarithmic transform [LOG], see the Weisstein link and the Sloane and Plouffe reference. This relation goes by the name of Riddell's formula. For information about the logarithmic transform see A274805. The EXP transform is related to the multinomial transform, see A274760 and the second formula.
The definition of the EXP transform, see the second formula, shows that n >= 1. To preserve the identity LOG[EXP[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 exponential transform, otherwise information about b(0) will be lost in transformation.
In the a(n) formulas, see the examples, the multinomial coefficients A178867 appear.
We observe that a(0) = 1 and provides no information about any value of b(n), this notwithstanding it is customary to start the a(n) sequence with a(0) = 1.
The Maple programs can be used to generate the exponential transform of a sequence. The first program uses a formula found by Alois P. Heinz, see A007446 and the first formula. The second program uses the definition of the exponential transform, see the Weisstein link and the second formula. The third program uses information about the inverse of the exponential transform, see A274805.
Some EXP transform pairs are, n >= 1: A000435(n) and A065440(n-1); 1/A000027(n) and A177208(n-1)/A177209(n-1); A000670(n) and A075729(n-1); A000670(n-1) and A014304(n-1); A000045(n) and A256180(n-1); A000290(n) and A033462(n-1); A006125(n) and A197505(n-1); A053549(n) and A198046(n-1); A000311(n) and A006351(n); A030019(n) and A134954(n-1); A038048(n) and A053529(n-1); A193356(n) and A003727(n-1).

			Some a(n) formulas, see A178867:
a(0) = 1
a(1) = x(1)
a(2) = x(1)^2 + x(2)
a(3) = x(1)^3 + 3*x(1)*x(2) + x(3)
a(4) = x(1)^4 + 6*x(1)^2*x(2) + 4*x(1)*x(3) + 3*x(2)^2 + x(4)
a(5) = x(1)^5 + 10*x(1)^3*x(2) + 10*x(1)^2*x(3) + 15*x(1)*x(2)^2 + 5*x(1)*x(4) + 10*x(2)*x(3) + x(5)

Frank Harary and Edgar M. Palmer, Graphical Enumeration, 1973.
Robert James Riddell, Contributions to the theory of condensation, Dissertation, University of Michigan, Ann Arbor, 1951.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, 1995, pp. 18-23.

Alois P. Heinz, Table of n, a(n) for n = 0..531
M. Bernstein and N. J. A. Sloane, Some Canonical Sequences of Integers, Linear Algebra and its Applications, Vol. 226-228 (1995), pp. 57-72. Erratum 320 (2000), 210. [Link to arXiv version]
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to Lin. Alg. Applic. version together with omitted figures]
N. J. A. Sloane, Transforms.
Eric W. Weisstein MathWorld, Exponential Transform.

Cf. A274805, A274760, A178867, A000203, A007446.
Cf. A177208, A177209, A006351, A197505, A144180, A256180, A033462, A198046, A134954, A145460, A188489, A005432, A029725, A124213, A002801.
Cf. A036040, another version.

nmax:=21: with(numtheory): b := proc(n): sigma(n) end: a:= proc(n) option remember; if n=0 then 1 else add(binomial(n-1, j-1) * b(j) *a(n-j), j=1..n) fi: end: seq(a(n), n=0..nmax); # End first EXP program.
nmax:= 21: with(numtheory): b := proc(n): sigma(n) 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 EXP program.
nmax:=21: with(numtheory): b := proc(n): sigma(n) end: f := series(log(1+add(q(n)*x^n/n!, n=1..nmax+1)), x, nmax+1): d := proc(n): n!*coeff(f, x, n) end: a(0):=1: q(0):=1: a(1):=b(1): q(1):=b(1): for n from 2 to nmax+1 do q(n) := solve(d(n)-b(n), q(n)): a(n):=q(n): od: seq(a(n), n=0..nmax); # End third EXP program.

a[0] = 1; a[n_] := a[n] = Sum[Binomial[n-1, j-1]*DivisorSigma[1, j]*a[n-j], {j, 1, n}]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Feb 22 2017 *)
nmax = 20; CoefficientList[Series[Exp[Sum[DivisorSigma[1, k]*x^k/k!, {k, 1, nmax}]], {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Jun 08 2021 *)

a(n) = Sum_{j=1..n} (binomial(n-1,j-1) * b(j) * a(n-j)), n >= 1 and a(0) = 1, with b(n) = A000203(n) = sigma(n).

E.g.f.: exp(Sum_{n >= 1} b(n)*x^n/n!) with b(n) = sigma(n) = A000203(n).

cp's OEIS Frontend

A349343 Dirichlet inverse of A193356, which is defined as n if n is odd, 0 if n is even.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A196020 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists the odd numbers interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Sage

Formula

A014682 The Collatz or 3x+1 function: a(n) = n/2 if n is even, otherwise (3n+1)/2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

PARI

Python

Formula

Extensions

A059260 Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-y-x*y-x^2) = 1/((1+x)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Sage

Formula

Extensions

A317614 a(n) = (1/2)*(n^3 + n*(n mod 2)).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

GAP

GAP

Magma

Maple

Mathematica

Maxima

PARI

PARI

R

Formula

A057211 Alternating runs of ones and zeros, where the n-th run has length n.

A317614 a(n) = (1/2)(n^3 + n(n mod 2)).