A055998 -id:A055998 - cp's OEIS frontend

A126890 Triangle read by rows: T(n,k) = n(n+2k+1)/2, 0 <= k <= n.

0, 1, 2, 3, 5, 7, 6, 9, 12, 15, 10, 14, 18, 22, 26, 15, 20, 25, 30, 35, 40, 21, 27, 33, 39, 45, 51, 57, 28, 35, 42, 49, 56, 63, 70, 77, 36, 44, 52, 60, 68, 76, 84, 92, 100, 45, 54, 63, 72, 81, 90, 99, 108, 117, 126, 55, 65, 75, 85, 95, 105, 115, 125, 135, 145, 155, 66, 77, 88

Offset: 0

Reinhard Zumkeller, Dec 30 2006

T(n,k) + T(n,n-k) = A014105(n);
row sums give A059270; Sum_{k=0..n-1} T(n,k) = A000578(n);
central terms give A007742; T(2*n+1,n) = A016754(n);
T(n,0)   = A000217(n);
T(n,1)   = A000096(n)   for n > 0;
T(n,2)   = A055998(n)   for n > 1;
T(n,3)   = A055999(n)   for n > 2;
T(n,4)   = A056000(n)   for n > 3;
T(n,5)   = A056115(n)   for n > 4;
T(n,6)   = A056119(n)   for n > 5;
T(n,7)   = A056121(n)   for n > 6;
T(n,8)   = A056126(n)   for n > 7;
T(n,10)  = A101859(n-1) for n > 9;
T(n,n-3) = A095794(n-1) for n > 2;
T(n,n-2) = A045943(n-1) for n > 1;
T(n,n-1) = A000326(n)   for n > 0;
T(n,n)   = A005449(n).

			From _Philippe Deléham_, Oct 03 2011: (Start)
Triangle begins:
   0;
   1,  2;
   3,  5,  7;
   6,  9, 12, 15;
  10, 14, 18, 22, 26;
  15, 20, 25, 30, 35, 40;
  21, 27, 33, 39, 45, 51, 57;
  28, 35, 42, 49, 56, 63, 70, 77; (End)

Léonard Euler, Introduction à l'analyse infinitésimale, tome premier, ACL-Editions, Paris, 1987, p. 353-354.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Émile Fourrey, Les nombres abstraits, Récreations arithmétiques, 1899 and later, Vuibert, Paris, page 86-87. Triangle without right diagonal.
Adrien-Marie Legendre, Théorie des nombres, tome 2, quatrième partie, p.131, troisième édition, Paris, 1830.

Cf. A110449.

a126890 n k = a126890_tabl !! n !! k
a126890_row n = a126890_tabl !! n
a126890_tabl = map fst $ iterate
   (\(xs@(x:_), i) -> (zipWith (+) ((x-i):xs) [2*i+1 ..], i+1)) ([0], 0)
-- Reinhard Zumkeller, Nov 10 2013

Flatten[Table[(n(n+2k+1))/2,{n,0,20},{k,0,n}]] (* Harvey P. Dale, Jun 21 2013 *)

T(n,k) = T(n,k-1) + n, for k <= n. - Philippe Deléham, Oct 03 2011

A095660 Pascal (1,3) triangle.

Original entry on oeis.org

3, 1, 3, 1, 4, 3, 1, 5, 7, 3, 1, 6, 12, 10, 3, 1, 7, 18, 22, 13, 3, 1, 8, 25, 40, 35, 16, 3, 1, 9, 33, 65, 75, 51, 19, 3, 1, 10, 42, 98, 140, 126, 70, 22, 3, 1, 11, 52, 140, 238, 266, 196, 92, 25, 3, 1, 12, 63, 192, 378, 504, 462, 288, 117, 28, 3, 1, 13, 75, 255, 570, 882, 966, 750, 405, 145, 31, 3

Offset: 0

Wolfdieter Lang, May 21 2004

This is the third member, q=3, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with T(0,0)=2, not 1).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) = Sum_{m=0..n} T(n,m)*x^m is G(z,x) = g(z)/(1-x*z*f(z)). Here: g(x) = (3-2*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (3-2*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} T(n-1-k,k) = A000285(n-2), n>=2, with n=1 value 3. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Central terms: T(2*n,n) = A028329(n) = A100320(n) for n > 0, A028329 are the central terms of triangle A028326. - Reinhard Zumkeller, Apr 08 2012
Let P be Pascal's triangle, A007318 and R the Riordan array, A097805. Then Pascal triangle (1,q) = ((q-1) * R) + P. Example: Pascal triangle (1,3) = (2 * R) + P. - Gary W. Adamson, Sep 12 2015

			Triangle starts:
  3;
  1,  3;
  1,  4,  3;
  1,  5,  7,   3;
  1,  6, 12,  10,   3;
  1,  7, 18,  22,  13,   3;
  1,  8, 25,  40,  35,  16,   3;
  1,  9, 33,  65,  75,  51,  19,   3;
  1, 10, 42,  98, 140, 126,  70,  22,   3;
  1, 11, 52, 140, 238, 266, 196,  92,  25,   3;
  1, 12, 63, 192, 378, 504, 462, 288, 117,  28,  3;
  1, 13, 75, 255, 570, 882, 966, 750, 405, 145, 31, 3;

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Wolfdieter Lang, First 10 rows.

Row sums: A000079(n+1), n>=1, 3 if n=0. Alternating row sums are [3, -2, followed by 0's].
Column sequences (without leading zeros) give for m=1..9 with n>=0: A000027(n+3), A055998(n+1), A006503(n+1), A095661, A000574, A095662, A095663, A095664, A095665.
Cf. A097805.

a095660 n k = a095660_tabl !! n !! k
a095660_row n = a095660_tabl !! n
a095660_tabl = [3] : iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,3]
-- Reinhard Zumkeller, Apr 08 2012

A095660:= func< n,k | n eq 0 select 3 else (1+2*k/n)*Binomial(n,k) >;
[A095660(n,k): k in [0..n], n in [1..12]]; // G. C. Greubel, May 02 2021

T(n,k):=piecewise(n=0,3,0Mircea Merca, Apr 08 2012

{3}~Join~Table[(1 + 2 k/n) Binomial[n, k], {n, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Sep 14 2015 *)

def A095660(n,k): return 3 if n==0 else (1+2*k/n)*binomial(n,k)
flatten([[A095660(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 02 2021

Recursion: T(n, m)=0 if m>n, T(0, 0)= 3; T(n, 0)=1 if n>=1; T(n, m) = T(n-1, m) + T(n-1, m-1).
G.f. column m (without leading zeros): (3-2*x)/(1-x)^(m+1), m>=0.
T(n,k) = (1+2*k/n) * binomial(n,k), for n>0. - Mircea Merca, Apr 08 2012
Closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013

A185787 Sum of first k numbers in column k of the natural number array A000027; by antidiagonals.

Original entry on oeis.org

1, 7, 25, 62, 125, 221, 357, 540, 777, 1075, 1441, 1882, 2405, 3017, 3725, 4536, 5457, 6495, 7657, 8950, 10381, 11957, 13685, 15572, 17625, 19851, 22257, 24850, 27637, 30625, 33821, 37232, 40865, 44727, 48825, 53166, 57757, 62605, 67717, 73100, 78761, 84707, 90945, 97482, 104325, 111481, 118957, 126760, 134897, 143375

Offset: 1

Clark Kimberling, Feb 03 2011

This is one of many interesting sequences and arrays that stem from the natural number array A000027, of which a northwest corner is as follows:
  1....2.....4.....7...11...16...22...29...
  3....5.....8....12...17...23...30...38...
  6....9....13....18...24...31...39...48...
  10...14...19....25...32...40...49...59...
  15...20...26....33...41...50...60...71...
  21...27...34....42...51...61...72...84...
  28...35...43....52...62...73...85...98...
Blocking out all terms below the main diagonal leaves columns whose sums comprise A185787.  Deleting the main diagonal and then summing give A185787.  Analogous treatments to the left of the main diagonal give A100182 and A101165.  Further sequences obtained directly from this array are easily obtained using the following formula for the array: T(n,k)=n+(n+k-2)(n+k-1)/2.
Examples:
row 1:  A000124
row 2:  A022856
row 3:  A016028
row 4:  A145018
row 5:  A077169
col 1:  A000217
col 2:  A000096
col 3:  A034856
col 4:  A055998
col 5:  A046691
col 6:  A052905
col 7:  A055999
diag. (1,5,...) ...... A001844
diag. (2,8,...) ...... A001105
diag. (4,12,...)...... A046092
diag. (7,17,...)...... A056220
diag. (11,23,...) .... A132209
diag. (16,30,...) .... A054000
diag. (22,38,...) .... A090288
diag. (3,9,...) ...... A058331
diag. (6,14,...) ..... A051890
diag. (10,20,...) .... A005893
diag. (15,27,...) .... A097080
diag. (21,35,...) .... A093328
antidiagonal sums:  (1,5,15,34,...)=A006003=partial sums of A002817.
Let S(n,k) denote the n-th partial sum of column k.  Then
S(n,k)=n*(n^2+3k*n+3*k^2-6*k+5)/6.
S(n,1)=n(n+1)(n+2)/6
S(n,2)=n(n+1)(n+5)/6
S(n,3)=n(n+2)(n+7)/6
S(n,4)=n(n^2+12n+29)/6
S(n,5)=n(n+5)(n+10)/6
S(n,6)=n(n+7)(n+11)/6
S(n,7)=n(n+10)(n+11)/6
Weight array of T: A144112
Accumulation array of T: A185506
Second rectangular sum array of T: A185507
Third rectangular sum array of T: A185508
Fourth rectangular sum array of T: A185509

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000027, A185788, A100182, A101165, A079824.

[n*(7*n^2-6*n+5)/6: n in [1..50]]; // Vincenzo Librandi, Jul 04 2012

f[n_,k_]:=n+(n+k-2)(n+k-1)/2;
s[k_]:=Sum[f[n,k],{n,1,k}];
Factor[s[k]]
Table[s[k],{k,1,70}]  (* A185787 *)
CoefficientList[Series[(3*x^2+3*x+1)/(1-x)^4,{x,0,50}],x] (* Vincenzo Librandi, Jul 04 2012 *)

a(n)=n*(7*n^2-6*n+5)/6.

G.f.: x*(3*x^2+3*x+1)/(1-x)^4. - Vincenzo Librandi, Jul 04 2012

Edited by Clark Kimberling, Feb 25 2023

A059438 Triangle T(n,k) (1 <= k <= n) read by rows: T(n,k) is the number of permutations of [1..n] with k components.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 13, 7, 3, 1, 71, 32, 12, 4, 1, 461, 177, 58, 18, 5, 1, 3447, 1142, 327, 92, 25, 6, 1, 29093, 8411, 2109, 531, 135, 33, 7, 1, 273343, 69692, 15366, 3440, 800, 188, 42, 8, 1, 2829325, 642581, 125316, 24892, 5226, 1146, 252, 52, 9, 1

Offset: 1

N. J. A. Sloane, Feb 01 2001

			Triangle begins:
[1] [     1]
[2] [     1,     1]
[3] [     3,     2,     1]
[4] [    13,     7,     3,    1]
[5] [    71,    32,    12,    4,   1]
[6] [   461,   177,    58,   18,   5,   1]
[7] [  3447,  1142,   327,   92,  25,   6,  1]
[8] [ 29093,  8411,  2109,  531, 135,  33,  7, 1]
[9] [273343, 69692, 15366, 3440, 800, 188, 42, 8, 1]

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 262 (#14).
Antonio Di Crescenzo, Barbara Martinucci, and Abdelaziz Rhandi, A linear birth-death process on a star graph and its diffusion approximation, arXiv:1405.4312 [math.PR], 2014.
FindStat - Combinatorial Statistic Finder, The decomposition number of a permutation.
Peter Hegarty and Anders Martinsson, On the existence of accessible paths in various models of fitness landscapes, arXiv:1210.4798 [math.PR], 2012-2014. - From _N. J. A. Sloane_, Jan 01 2013
Sergey Kitaev and Philip B. Zhang, Distributions of mesh patterns of short lengths, arXiv:1811.07679 [math.CO], 2018.

A version with reflected rows is A263484.
Diagonals give A003319, A059439, A059440, A055998.
Cf. A000007, A085771, A084938.
T(2n,n) gives A308650.

# Uses function PMatrix from A357368. Adds column 1, 0, 0, ... to the left.
PMatrix(10, A003319); # Peter Luschny, Oct 09 2022

(* p = indecomposable permutations = A003319 *) p[n_] := p[n] = n! - Sum[ k!*p[n-k], {k, 1, n-1}]; t[n_, k_] /; n < k = 0; t[n_, 1] := p[n]; t[n_, k_] /; n >= k := t[n, k] = Sum[ t[n-j, k-1]*p[j], {j, 1, n}]; Flatten[ Table[ t[n, k], {n, 1, 10}, {k, 1, n}] ] (* Jean-François Alcover, Mar 06 2012, after Philippe Deléham *)

def A059438_triangle(dim) :
    R = PolynomialRing(ZZ, 'x')
    C = [R(0)] + [R(1) for i in range(dim+1)]
    A = [(i + 2) // 2 for i in range(dim+1)]
    A[0] = R.gen(); T = []
    for k in range(1, dim+1) :
        for n in range(k, 0, -1) :
            C[n] = C[n-1] + C[n+1] * A[n-1]
        T.append(list(C[1])[1::])
    return T
A059438_triangle(8) # Peter Luschny, Sep 10 2022

# Alternatively, using the function PartTrans from A357078:
# Adds a (0,0)-based column (1, 0, 0, ...) to the left of the triangle.
dim = 10
A = ZZ[['t']]; g = A([0]+[factorial(n) for n in range(1, 30)]).O(dim+2)
PartTrans(dim, lambda n: list(g / (1 +  g))[n]) # Peter Luschny, Sep 11 2022

Let f(x) = Sum_{n >= 0} n!*x^n, g(x) = 1 - 1/f(x). Then g(x) is g.f. for first diagonal A003319 and Sum_{n >= k} T(n, k)*x^n = g(x)^k.
Triangle T(n, k), n > 0 and k > 0, read by rows; given by [0, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, ...] DELTA A000007 where DELTA is Deléham's operator defined in A084938.
T(n+k, k) = Sum_{a_1 + a_2 + ... + a_k = n} A003319(a_1)*A003319(a_2)*...*A003319(a_k). T(n, k) = 0 if n < k, T(n, 1) = A003319(n) and for n >= k T(n, k)= Sum_{j>=1} T(n-j, k-1)* A003319(j). A059438 is formed from the self convolution of its first column (A003319). - Philippe Deléham, Feb 04 2004
Sum_{k>0} T(n, k) = n!; see A000142. - Philippe Deléham, Feb 05 2004
If g(x) = x + x^2 + 3*x^3 + 13*x^4 + ... is the generating function for the number of permutations with no global descents, then 1/(1-g(x)) is the generating function for n!. Setting t=1 in f(x, t) implies Sum_{k=1..n} T(n,k) = n!. Let g(x) be the o.g.f. for A003319. Then the o.g.f. for this table is given by f(x, t) = 1/(1 - t*g(x)) - 1 (i.e., the coefficient of x^n*t^k in f(x,t) is T(n,k)). - Mike Zabrocki, Jul 29 2004

More terms from Vladeta Jovovic, Mar 04 2001

A111774 Numbers that can be written as a sum of at least three consecutive positive integers.

Original entry on oeis.org

6, 9, 10, 12, 14, 15, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 102

Offset: 1

Jaap Spies, Aug 15 2005

In this sequence there are no (odd) primes and there are no powers of 2.
So we have only three kinds of natural numbers: the odd primes, the powers of 2 and the numbers that can be represented as a sum of at least three consecutive integers.
Odd primes can only be written as a sum of two consecutive integers. Powers of 2 do not have a representation as a sum of k consecutive integers (other than the trivial n=n, for k=1).
Numbers of the form (x*(x+1)-y*(y+1))/2 for nonnegative integers x,y with x-y >= 3. - Bob Selcoe, Feb 21 2014
Numbers of the form (x + 1)*(x + 2*y)/2 for integers x,y with x >= 2 and y >= 1. For y = 1 only triangular numbers (A000217) >= 6 occur. - Ralf Steiner, Jun 27 2019
From Ralf Steiner, Jul 09 2019: (Start)
If k >= 1 sequences are c_k(n) = c_k(n - 1) + n + k - 1, c_k(0) = 0, means c_k(n) = n*(n + 2*k - 1)/2: A000217, A000096, A055998, A055999, A056000, ... then this sequence is the union of c_k(n), n >= 3. (End)
From Wolfdieter Lang, Oct 28 2020: (Start)
This sequence gives all positive integers that have at least one odd prime as proper divisor. The proof follows from the first two comments.
The set {a(n)}_{n>=1} equals the set {k positive integer : floor(k/2) - delta(k) >= 1}, where delta(k) = A055034(k). Proof: floor(k/2) gives the number of positive odd numbers < k, and delta(k), gives the number of positive odd numbers coprime to k. delta(1) = 1 but 1 is not < 1, therefore k = 1 is not a member of this set. Hence a member >= 2 of this set has at least one odd number > 1 and < k missing in the set of odd numbers relative prime to k. Therefore there exists at least one odd prime < k dividing k. (End)
For the multiplicity of a(n) see A338428, obtained from triangle A337940 (the array is given Bob Selcoe as example below, and in the Ralf Steiner comment above). - Wolfdieter Lang, Dec 09 2020

			a(1)=6 because 6 is the first number that can be written as a sum of three consecutive positive integers: 6 = 1+2+3.
From _Bob Selcoe_, Feb 23 2014: (Start)
Let the top row of an array be A000217(n). Let the diagonals (reading down and left) be A000217(n)-A000217(1),  A000217(n)-A000217(2), A000217(n)-A000217(3)..., A000217(n)-A000217(n-3).  This is A049777 read as a square array, starting with the third column. The array begins as follows:
   6 10 15 21 28 36 45 55 66
   9 14 20 27 35 44 54 65
  12 18 25 33 42 52 63
  15 22 30 39 49 60
  18 26 35 45 56
  21 30 40 51
  24 34 45
  27 38
  30
This is (x*(x+1)-y*(y+1))/2 for nonnegative integers x,y with x-y >= 3, because it is equivalent to 1+2+3/+4/+5/...+x/-0/-1/-2/-3/-4/-5/...-(x+3)/ for all possible strings of consecutive integers, which represents every possible way to sum three or more consecutive positive integers. So for example, 4+5+6+7 = 1+2+3+4+5+6+7-1-2-3 = 22, which is (x*(x+1)-y*(y+1))/2 when x=7, y=3. Notice that values can appear more than once in the array because some numbers can be represented as sums of more than one string of three or more consecutive positive integers. For example, 30 = (x*(x+1)-y*(y+1))/2 when (a) x=11, y=8: 9+10+11; (b) x=9, y=5: 6+7+8+9; and (c) x=8, y=3: 4+5+6+7+8. By definition, x-y is the number of integers in the string. (End)

Paul Halmos, "Problems for Mathematicians, Young and Old", Dolciani Mathematical Expositions, 1991, Solution to problem 3G p. 179.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC, Problem C, Jun 2005, p. 181-182.
Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC, Problem C: Solution of this Problem
J. Spies, Sage program for computing A111774

Cf. A000040, A000079, A066542, A174090 (complement), A000217, A049777, A055034.

Cf. A337940, A338428.

ispoweroftwo := proc(n) local a, t; t := 1; while (n > t) do t := 2*t end do; if (n = t) then a := true else a := false end if; return a; end proc; f:= proc(n) if (not isprime(n)) and (not ispoweroftwo(n)) then return n end if; end proc; seq(f(i),i = 1..150);

max=6!;lst={};Do[z=n+(n+1);Do[z+=(n+x);If[z>max,Break[]];AppendTo[lst,z],{x,2,max}],{n,max}];Union[lst] (* Vladimir Joseph Stephan Orlovsky, Mar 06 2010 *)

isok(n) = !(n == 1) && !isprime(n) && !(isprimepower(n, &p) && (p == 2)); \\ Michel Marcus, Jul 02 2019

from sympy import primepi
def A111774(n):
    def f(x): return int(n+(0 if x<=1 else primepi(x)-1)+x.bit_length())
    m, k = n, f(n)
    while m != k: m, k = k, f(k)
    return m # Chai Wah Wu, Sep 19 2024

A179986 Second 9-gonal (or nonagonal) numbers: a(n) = n(7n+5)/2.

Original entry on oeis.org

0, 6, 19, 39, 66, 100, 141, 189, 244, 306, 375, 451, 534, 624, 721, 825, 936, 1054, 1179, 1311, 1450, 1596, 1749, 1909, 2076, 2250, 2431, 2619, 2814, 3016, 3225, 3441, 3664, 3894, 4131, 4375, 4626, 4884, 5149, 5421, 5700, 5986, 6279, 6579, 6886

Offset: 0

Bruno Berselli, Jan 13 2011

This sequence is a bisection of A118277 (even part).
Sequence found by reading the line from 0, in the direction 0, 19... and the line from 6, in the direction 6, 39,..., in the square spiral whose vertices are the generalized 9-gonal numbers A118277. - Omar E. Pol, Jul 24 2012
The early part of this sequence is a strikingly close approximation to the early part of A100752. - Peter Munn, Nov 14 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. second k-gonal numbers: A005449 (k=5), A014105 (k=6), A147875 (k=7), A045944 (k=8), this sequence (k=9), A033954 (k=10), A062728 (k=11), A135705 (k=12).

Cf. A001106, A022264, A022265, A024966, A055998, A100752, A174738, A186029, A218471.

[n*(7*n+5)/2: n in [0..50]]; // Bruno Berselli, Sep 23 2016

I:=[0, 6, 19]; [n le 3 select I[n] else 3*Self(n-1) -3*Self(n-2) +Self(n-3): n in [1..60]]; // Vincenzo Librandi, Oct 15 2012

f[n_] := n (7 n + 5)/2; f[Range[0, 60]] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011*)
LinearRecurrence[{3, -3, 1}, {0, 6, 19}, 60] (* or *) Array[(#(7# + 5))/2&, 60, 0] (* Harvey P. Dale, Aug 19 2011 *)
CoefficientList[Series[x (6 + x)/(1 - x)^3, {x, 0, 60}], x] (* Vincenzo Librandi, Oct 15 2012 *)

a(n)=n*(7*n+5)/2 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: x*(6 + x)/(1 - x)^3.
a(n) = Sum_{i=0..(n-1)} A017053(i) for n>0.
a(-n) = A001106(n).
Sum_{i=0..n} (a(n)+i)^2 = ( Sum_{i=(n+1)..2*n} (a(n)+i)^2 ) + 21*A000217(n)^2 for n>0.
a(n) = a(n-1)+7*n-1 for n>0, with a(0)=0. - Vincenzo Librandi, Feb 05 2011
a(0)=0, a(1)=6, a(2)=19; for n>2, a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). - Harvey P. Dale, Aug 19 2011
a(n) = A174738(7n+5). - Philippe Deléham, Mar 26 2013
a(n) = A001477(n) + 2*A000290(n) + 3*A000217(n). - J. M. Bergot, Apr 25 2014
a(n) = A055998(4*n) - A055998(3*n). - Bruno Berselli, Sep 23 2016
E.g.f.: (x/2)*(12 + 7*x)*exp(x). - G. C. Greubel, Aug 19 2017

A255961 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is Euler transform of (j->j*k).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 3, 0, 1, 3, 7, 6, 0, 1, 4, 12, 18, 13, 0, 1, 5, 18, 37, 47, 24, 0, 1, 6, 25, 64, 111, 110, 48, 0, 1, 7, 33, 100, 215, 303, 258, 86, 0, 1, 8, 42, 146, 370, 660, 804, 568, 160, 0, 1, 9, 52, 203, 588, 1251, 1938, 2022, 1237, 282, 0

Offset: 0

Alois P. Heinz, Mar 11 2015

A(n,k) is the number of partitions of n when parts i are of k*i kinds. A(2,2) = 7: [2a], [2b], [2c], [2d], [1a,1a], [1a,1b], [1b,1b].

			Square array A(n,k) begins:
  1,  1,   1,    1,    1,     1,     1,     1, ...
  0,  1,   2,    3,    4,     5,     6,     7, ...
  0,  3,   7,   12,   18,    25,    33,    42, ...
  0,  6,  18,   37,   64,   100,   146,   203, ...
  0, 13,  47,  111,  215,   370,   588,   882, ...
  0, 24, 110,  303,  660,  1251,  2160,  3486, ...
  0, 48, 258,  804, 1938,  4005,  7459, 12880, ...
  0, 86, 568, 2022, 5400, 12150, 24354, 44885, ...

Alois P. Heinz, Rows n = 0..140, flattened

Columns k=0-10 give: A000007, A000219, A161870, A255610, A255611, A255612, A255613, A255614, A193427, A316461, A316462.
Rows n=0-3 give: A000012, A001477, A055998, A101853.
Main diagonal gives A255672.
Antidiagonal sums give A299166.
Cf. A144064, A257673.

A:= proc(n, k) option remember; `if`(n=0, 1, k*add(
      A(n-j, k)*numtheory[sigma][2](j), j=1..n)/n)
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);

A[n_, k_] := A[n, k] = If[n==0, 1, k*Sum[A[n-j, k]*DivisorSigma[2, j], {j, 1, n}]/n]; Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Feb 02 2016, after Alois P. Heinz *)

G.f. of column k: Product_{j>=1} 1/(1-x^j)^(j*k).

T(n,k) = Sum_{i=0..k} C(k,i) * A257673(n,k-i).

A238858 Triangle T(n,k) read by rows: T(n,k) is the number of length-n ascent sequences with exactly k descents.

Original entry on oeis.org

1, 1, 0, 2, 0, 0, 4, 1, 0, 0, 8, 7, 0, 0, 0, 16, 33, 4, 0, 0, 0, 32, 131, 53, 1, 0, 0, 0, 64, 473, 429, 48, 0, 0, 0, 0, 128, 1611, 2748, 822, 26, 0, 0, 0, 0, 256, 5281, 15342, 9305, 1048, 8, 0, 0, 0, 0, 512, 16867, 78339, 83590, 21362, 937, 1, 0, 0, 0, 0, 1024, 52905, 376159, 647891, 307660, 35841, 594, 0, 0, 0, 0, 0

Offset: 0

Joerg Arndt and Alois P. Heinz, Mar 06 2014

Columns k=0-10 give: A011782, A066810(n-1), A241872, A241873, A241874, A241875, A241876, A241877, A241878, A241879, A241880.
The sequence of column k satisfies a linear recurrence with constant coefficients of order k*(k+5)/2 = A055998(k) for k>0.
T(2n,n) gives A241871(n).
Last nonzero elements of rows give A241881(n).
Row sums give A022493.

			Triangle starts:
00:     1;
01:     1,      0;
02:     2,      0,       0;
03:     4,      1,       0,       0;
04:     8,      7,       0,       0,       0;
05:    16,     33,       4,       0,       0,      0;
06:    32,    131,      53,       1,       0,      0,     0;
07:    64,    473,     429,      48,       0,      0,     0,   0;
08:   128,   1611,    2748,     822,      26,      0,     0,   0, 0;
09:   256,   5281,   15342,    9305,    1048,      8,     0,   0, 0, 0;
10:   512,  16867,   78339,   83590,   21362,    937,     1,   0, 0, 0, 0;
11:  1024,  52905,  376159,  647891,  307660,  35841,   594,   0, 0, 0, 0, 0;
12:  2048, 163835, 1728458, 4537169, 3574869, 834115, 45747, 262, 0, 0, 0, 0, 0;
...
The 53 ascent sequences of length 5 together with their numbers of descents are (dots for zeros):
01:  [ . . . . . ]   0      28:  [ . 1 1 . 1 ]   1
02:  [ . . . . 1 ]   0      29:  [ . 1 1 . 2 ]   1
03:  [ . . . 1 . ]   1      30:  [ . 1 1 1 . ]   1
04:  [ . . . 1 1 ]   0      31:  [ . 1 1 1 1 ]   0
05:  [ . . . 1 2 ]   0      32:  [ . 1 1 1 2 ]   0
06:  [ . . 1 . . ]   1      33:  [ . 1 1 2 . ]   1
07:  [ . . 1 . 1 ]   1      34:  [ . 1 1 2 1 ]   1
08:  [ . . 1 . 2 ]   1      35:  [ . 1 1 2 2 ]   0
09:  [ . . 1 1 . ]   1      36:  [ . 1 1 2 3 ]   0
10:  [ . . 1 1 1 ]   0      37:  [ . 1 2 . . ]   1
11:  [ . . 1 1 2 ]   0      38:  [ . 1 2 . 1 ]   1
12:  [ . . 1 2 . ]   1      39:  [ . 1 2 . 2 ]   1
13:  [ . . 1 2 1 ]   1      40:  [ . 1 2 . 3 ]   1
14:  [ . . 1 2 2 ]   0      41:  [ . 1 2 1 . ]   2
15:  [ . . 1 2 3 ]   0      42:  [ . 1 2 1 1 ]   1
16:  [ . 1 . . . ]   1      43:  [ . 1 2 1 2 ]   1
17:  [ . 1 . . 1 ]   1      44:  [ . 1 2 1 3 ]   1
18:  [ . 1 . . 2 ]   1      45:  [ . 1 2 2 . ]   1
19:  [ . 1 . 1 . ]   2      46:  [ . 1 2 2 1 ]   1
20:  [ . 1 . 1 1 ]   1      47:  [ . 1 2 2 2 ]   0
21:  [ . 1 . 1 2 ]   1      48:  [ . 1 2 2 3 ]   0
22:  [ . 1 . 1 3 ]   1      49:  [ . 1 2 3 . ]   1
23:  [ . 1 . 2 . ]   2      50:  [ . 1 2 3 1 ]   1
24:  [ . 1 . 2 1 ]   2      51:  [ . 1 2 3 2 ]   1
25:  [ . 1 . 2 2 ]   1      52:  [ . 1 2 3 3 ]   0
26:  [ . 1 . 2 3 ]   1      53:  [ . 1 2 3 4 ]   0
27:  [ . 1 1 . . ]   1
There are 16 ascent sequences with no descent, 33 with one, and 4 with 2, giving row 4 [16, 33, 4, 0, 0, 0].

Joerg Arndt and Alois P. Heinz, Rows n = 0..140, flattened

Cf. A137251 (ascent sequences with k ascents), A242153 (ascent sequences with k flat steps).

# b(n, i, t): polynomial in x where the coefficient of x^k is   #
#             the number of postfixes of these sequences of     #
#             length n having k descents such that the prefix   #
#             has rightmost element i and exactly t ascents     #
b:= proc(n, i, t) option remember; `if`(n=0, 1, expand(add(
      `if`(ji, 1, 0)), j=0..t+1)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n, -1$2)):
seq(T(n), n=0..12);

b[n_, i_, t_] := b[n, i, t] = If[n == 0, 1, Sum[If[ji, 1, 0]], {j, 0, t+1}]]; T[n_] := Function[{p}, Table[Coefficient[p, x, i], {i, 0, n}]][b[n, -1, -1]]; Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Jan 06 2015, translated from Maple *)

# Transcription of the Maple program
R. = QQ[]
@CachedFunction
def b(n,i,t):
    if n==0: return 1
    return sum( ( x if ji) ) for j in range(t+2) )
def T(n): return b(n, -1, -1)
for n in range(0,10): print(T(n).list())

cp's OEIS Frontend

A126890 Triangle read by rows: T(n,k) = n*(n+2*k+1)/2, 0 <= k <= n.

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A095660 Pascal (1,3) triangle.

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A185787 Sum of first k numbers in column k of the natural number array A000027; by antidiagonals.

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A059438 Triangle T(n,k) (1 <= k <= n) read by rows: T(n,k) is the number of permutations of [1..n] with k components.

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A111774 Numbers that can be written as a sum of at least three consecutive positive integers.

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A179986 Second 9-gonal (or nonagonal) numbers: a(n) = n*(7*n+5)/2.

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A126890 Triangle read by rows: T(n,k) = n(n+2k+1)/2, 0 <= k <= n.

A179986 Second 9-gonal (or nonagonal) numbers: a(n) = n(7n+5)/2.

A226492 a(n) = n(11n-5)/2.

A140673 a(n) = 3n(n + 5)/2.