A000340 -id:A000340 - cp's OEIS frontend

A108286 Triangle read by rows; columns are simple recursive sequences.

1, 3, 1, 6, 4, 1, 10, 11, 5, 1, 15, 26, 18, 6, 1, 21, 57, 58, 27, 7, 1, 28, 120, 179, 112, 38, 8, 1, 36, 247, 543, 453, 194, 51, 9, 1, 45, 502, 1636, 1818, 975, 310, 66, 10, 1

Offset: 1

Gary W. Adamson, May 31 2005

Left column = triangular numbers; Col. 2, (1, 4, 11...) = A000295; Col. 3, (1, 5, 18...) = A000340; Col. 4, (1, 6, 27...) = A014825; Col.5, (1, 7, 38...) = A014827.

			First few rows of the triangle are:
1;
3, 1;
6, 4, 1;
10, 11; 5, 1;
15, 26, 18, 6, 1;
21, 57, 58, 27, 7, 1;
...
3rd offset column: (1, 5, 18, 58...) = "1", then a(r) = 3*a(r-1) + r; e.g. 58 = 3*18 + 4 since 58 is the fourth term in the third column.

Cf. A108285, A000295, A000340, A014825, A014827.

r-th term in n-th column: initial "1", then a(r) = n*a(r-1) + r. Diagonals of A108285 become the columns of A108286.

A089000 Square table, read by antidiagonals, of coefficients T(k,n) (row k; column n) defined by: T(k,n) = k*T(k,n-1)+ n; T(k,0) = 0.

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 3, 3, 1, 0, 4, 6, 4, 1, 0, 5, 10, 11, 5, 1, 0, 6, 15, 26, 18, 6, 1, 0, 7, 21, 57, 58, 27, 7, 1, 0, 8, 28, 120, 179, 112, 38, 8, 1, 0, 9, 36, 247, 543, 453, 194, 51, 9, 1, 0, 10, 45, 502, 1636, 1818, 975, 310, 66, 10, 1, 0

Offset: 0

Philippe Deléham, Nov 02 2003

Rows begin:
{0, 1, 2, 3, 4, 5, 6, 7, 8, ...}:see A001477
{0, 1, 3, 6, 10, 15, 21, 28, ...} : see A000217
{0, 1, 4, 11, 26, 57, 120, 247, 502, ...} : see A000295
{0, 1, 5, 18, 58, 179, 543, 1636, ...} : see A000340
{0, 1, 6, 27, 112, 453, 1818, 7279, ...} : see A014825
{0, 1, 7, 38, 194, 975, 4881, 24412, ...} : see A014827
{0, 1, 8, 51, 310, 1865, 11196, 67183, ...}: see diagonals of triangle A088990
Diagonal begin:
{0, 1, 4, 18, 112, 975, 11196, ... } :see A062805
{0, 1, 5, 27, 194, 1865, ...} : see A023811
Column {3, 6, 11, 18, 27, 38, 51, ...} : see A010000

Unprotect[Power]; 0^0=1; T[n_,k_]:=Sum[j*k^(n-j),{j,0,n}]; Table[T[n-k,k],{n,0,10},{k,0,n}]//Flatten (* Stefano Spezia, Apr 19 2025 *)

T(k, n)= (k^(n+1)- (k-1)*n - k)/(k-1)^2. T(k, n) = Sum(j, 0<=j<=n; j*k^(n-j)).

A108285 Triangle read by rows, generated from (1, 2, 3, ...).

Original entry on oeis.org

1, 1, 3, 1, 4, 6, 1, 5, 11, 10, 1, 6, 18, 26, 15, 1, 7, 27, 58, 57, 21, 1, 8, 38, 112, 179, 120, 28, 1, 9, 51, 194, 453, 543, 247, 36

Offset: 0

Gary W. Adamson, May 30 2005

By diagonals (d=1,2,3,...) going to the left with (1,3,6,...) = d(1), these are sequences of the form (k-th term a(k) = d*a(k-1) + k). Example: 1, 7, 38, 194, ... (the 5th diagonal) = A014827, is generated by a(k) = 5*a(k-1) + k. Diagonal 2 = (1, 4, 11, 26, ...) = A000295; Diagonal 3 = (1, 5, 18, ...) = A000340; Diagonal 4 = (1, 6, 27, ...) = A014825.

Triangle A108243 is generated by analogous operations from (..., 3, 2, 1) instead of (1, 2, 3, ...).

			4th column (offset) = 10, 26, 58, 112, ...= f(x), x = 1, 2, 3; x^3 + 2x^2 + 3x + 4.
First few rows of the triangle are:
  1;
  1, 3;
  1, 4, 6;
  1, 5, 11, 10;
  1, 6, 18, 26, 15;
  1, 7, 27, 58, 57, 21;
  1, 8, 38, 112, 179, 120, 28;
  ...

Cf. A108286, A000295, A000349, A014825, A000217, A108283, A108284, A108286.

n-th column = f(x), x = 1, 2, 3, ...; x^(n) + 2*x^(n-1) + 3*x^(n-2) + ... + (n+1).

A244762 a(n) = (53^n-2n-1)/4.

Original entry on oeis.org

1, 3, 10, 32, 99, 301, 908, 2730, 8197, 24599, 73806, 221428, 664295, 1992897, 5978704, 17936126, 53808393, 161425195, 484275602, 1452826824, 4358480491, 13075441493, 39226324500, 117678973522, 353036920589, 1059110761791, 3177332285398, 9531996856220, 28595990568687, 85787971706089, 257363915118296

Offset: 0

Franklin T. Adams-Watters, Jul 06 2014

Index entries for linear recurrences with constant coefficients, signature (5,-7,3).

Cf. A060816 (first differences).

Cf. A108765, A164039, A047926, A000340.

CoefficientList[Series[(1-2*x+2*x^2)/((1-3*x)*(1-x)^2), {x, 0, 30}], x] (* Vaclav Kotesovec, Jul 06 2014 *)

a(n+1) = 3*a(n) + n.
G.f.: (1-2*x+2*x^2) / ((1-3*x)*(1-x)^2).
E.g.f.: exp(x)*(5*exp(2*x) - 2*x - 1)/4. - Stefano Spezia, Aug 28 2023

A014850 Numbers k that divide s(k), where s(1)=1, s(j)=3*s(j-1)+j.

Original entry on oeis.org

1, 3, 39, 507, 6591, 12207, 85683, 158691, 255567, 285987, 1028703, 1113879, 2062983, 3322371, 3717831, 3820791, 13373139, 14480427, 26818779, 31089279, 43190823, 48331803, 49670283, 79992471, 89513931, 173850807, 188245551, 321984039, 348644127, 404160627, 561480699, 628313439, 645713679

Offset: 1

N. J. A. Sloane, Olivier Gérard

nonn

			39 is in the sequence as A000340(39-1) = 3039416364764232180 is divisible by 39. - _David A. Corneth_, Aug 08 2021

s(n) = A000340(n-1).

lista(nn) = my(s); for(k=1, nn, s=3*s+k; if(s%k==0, print1(k, ", "))); \\ Jinyuan Wang, Aug 08 2021

is(n) = n%2==1 && lift(Mod(3, n)^(n + 1) - Mod(3, n)) == 0 \\ David A. Corneth, Aug 08 2021

a(7)-a(16) from Arkadiusz Wesolowski, Jul 03 2011

More terms from David A. Corneth, Aug 08 2021

A198895 Triangle of coefficients arising in expansion of n-th derivative of tan(x) + sec(x).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 5, 2, 1, 8, 18, 16, 5, 1, 16, 58, 88, 61, 16, 1, 32, 179, 416, 479, 272, 61, 1, 64, 543, 1824, 3111, 2880, 1385, 272, 1, 128, 1636, 7680, 18270, 24576, 19028, 7936, 1385, 1, 256, 4916, 31616, 101166, 185856, 206276

Offset: 0

N. J. A. Sloane, Oct 31 2011

From Petros Hadjicostas, Aug 10 2019: (Start)
The recurrence about T(n, k) and the equation that connects T(n, k) to P(n, k) = A059427(n,k), which are given below, appear on p. 159 of the book by David and Barton (1962). The initial conditions, however, for their triangular array S^*{N,t} are slightly different, but there is an agreement starting at t = k = 1. They do not provide tables for S^*{N,t} (that matches the current array T(n, k) for N = n >= 0 and t = k >= 1).
Despite the slightly different initial conditions between T(n, k) and S^*_{N,t} (from p. 159 in the book), the recurrence given below can be proved very easily from the recurrence for the row polynomials R_n(x) given in Shi-Mei Ma (2011, 2012). (End)

			Triangle T(n,k) (with rows n >= 0 and columns k >= 0) begins as follows:
  1
  1   1
  1   2    1
  1   4    5     2
  1   8   18    16      5
  1  16   58    88     61     16
  1  32  179   416    479    272     61
  1  64  543  1824   3111   2880   1385    272
  1 128 1636  7680  18270  24576  19028   7936  1385
  1 256 4916 31616 101166 185856 206276 137216 50521 7936
  ...

Florence Nightingale David and D. E. Barton, Combinatorial Chance, Charles Griffin, 1962; see pp. 159-162.

Shi-Mei Ma, Derivative polynomials and permutations by numbers of interior peaks and left peaks, arXiv:1106.5781 [math.CO], 2011.
Shi-Mei Ma, Derivative polynomials and enumeration of permutations by number of interior and left peaks , Discrete Mathematics 312(2) (2012), 405-412.
Wikipedia, Florence Nightingale David.

Cf. A059427, A098558 (row sums), A000111 (diagonal and 1st subdiagonal), A000340 (column 3) A000431 (column 4), A000363 (column 5)

n-th row represents the coefficients of the polynomial R_n(x) defined by the recurrence: R_0(x) = 1, R_1(x) = 1 + x, and for n >= 1, R_{n+1}(x) = (1 + n*x^2)*R_n(x) + x*(1 - x^2)*R'_n(x).
From Petros Hadjicostas, Aug 10 2019: (Start)
T(n, k) = (k + 1) * T(n-1, k) + (n - k + 1) * T(n-1, k-2) for n >= 0 and 2 <= k <= n with initial conditions T(n, k=0) = 1 for n >= 0, T(n, k=1) = 2^(n-1) for n >= 1, and T(n, k) = 0 for n < 0 or n < k.
Setting x = 1 in the equation R_{n+1}(x) = (1 + n*x^2)*R_n(x) + x*(1 - x^2)*R'n(x) (valid for n >= 1), we get R{n+1}(1) = (n + 1)*R_n(1) for n >= 1. Since R_1(1) = 2, we have that R_n(1) = 2*n! for n >= 1. Since also R_0(1) = 1, we conclude that Sum_{k = 0..n} T(n,k) = R_n(1) = 2*n! - 0^n = A098558(n) for n >= 0.
Let P(n, k) = A059427(n,k) with P(n, k) = 0 for n <= 1 or n <= k. Then T(n, k) = (1/2)*P(n, k-1) + P(n, k) + (1/2) * P(n, k+1) for n >= 2 and 0 <= k <= n (but this is not true for n = 0 and n = 1). (End)

More terms from Max Alekseyev, Feb 17 2012

A229463 Expansion of g.f. 1/((1-x)^2(1-26x)).

Original entry on oeis.org

1, 28, 731, 19010, 494265, 12850896, 334123303, 8687205886, 225867353045, 5872551179180, 152686330658691, 3969844597125978, 103215959525275441, 2683614947657161480, 69773988639086198495, 1814123704616241160886, 47167216320022270183053, 1226347624320579024759396

Offset: 0

Yahia Kahloune, Sep 24 2013

This sequence was chosen to illustrate a method of solution.

			a(3) = (26^5 - 25*3 - 51)/625 = 19010.

Index entries for linear recurrences with constant coefficients, signature (28,-53,26).

Cf. A000295, A000340, A014825, A014827, A014936.

my(x='x+O('x^18)); Vec(1/((1-26*x)*(1-x)^2)) \\ Elmo R. Oliveira, May 24 2025

a(n) = (26^(n+2) - 25*n - 51)/625.
In general, for the expansion of 1/((1-s*x)^2*(1-r*x)) with r>s>=1 we have the formula: a(n) = (r^(n+2)- s^(n+1)*((r-s)*n +(2*r-s)))/(r-s)^2.
From Elmo R. Oliveira, May 24 2025: (Start)
E.g.f.: exp(x)*(-51 - 25*x + 676*exp(25*x))/625.
a(n) = 28*a(n-1) - 53*a(n-2) + 26*a(n-3). (End)

A378962 First differences of A378726.

Original entry on oeis.org

1, 1, 1, 5, 1, 1, 5, 1, 1, 5, 1, 1, 18, 1, 1, 5, 1, 1, 5, 1, 1, 18, 1, 1, 5, 1, 1, 5, 1, 1, 18, 1, 1, 5, 1, 1, 5, 1, 1, 58, 1, 1, 5, 1, 1, 5, 1, 1, 18, 1, 1, 5, 1, 1, 5, 1, 1, 18, 1, 1, 5, 1, 1, 5, 1, 1, 58, 1, 1, 5, 1, 1, 5, 1

Offset: 1

Tanya Khovanova and the MIT PRIMES STEP senior group, Dec 12 2024

nonn

Sequence A378726(n) is defined to be the total number of fires on a rooted undirected infinite ternary tree with a self-loop at the root, when a chip-firing process starts with 3n chips at the root. The total number of fires for 3n, 3n-1, and 3n-2 chips are the same, so the sequence is defined for one of these three values to remove duplicates.
The corresponding sequence for binary trees is A376132; its distinct values are Eulerian numbers A000295.
The distinct values of this sequence form sequence A000340.

			The total number of fires when starting with 12 chips at the root is 3, and the total number of fires when starting with 15 chips at the root is 8. This means that a(4) = 5.

Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See p. 17.
Wikipedia, Chip-firing game.

Cf. A000295, A000340, A376132, A378724, A378726.

import math
def F(N, k):
    n = int(math.log(N * (k - 1) + 1, k))
    a = [0] * (n + 1)
    num = N - ((k ** n) - 1)/(k - 1)
    for i in range(n, -1, -1):
        if k ** i <= num:
            a[i] = int(num/(k ** i))
            num %= (k ** i)
    res = 0
    for j in range(1, n):
        x = int((j * (k ** (j + 1)) - (j + 1) * (k ** j) + 1)/((k - 1) ** 2))
        res += x * (a[j] + 1)
    return res
s = ""
for N in range(1, 200):
    s += str(int(F(3 * N + 3, 3) - F(3 * N, 3)))
    s += ", "
print(s)

a(n) = A000340(A378724(n+1)-A378724(n)-1).

cp's OEIS Frontend

A108286 Triangle read by rows; columns are simple recursive sequences.

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A089000 Square table, read by antidiagonals, of coefficients T(k,n) (row k; column n) defined by: T(k,n) = k*T(k,n-1)+ n; T(k,0) = 0.

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A108285 Triangle read by rows, generated from (1, 2, 3, ...).

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A244762 a(n) = (5*3^n-2*n-1)/4.

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A014850 Numbers k that divide s(k), where s(1)=1, s(j)=3*s(j-1)+j.

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A198895 Triangle of coefficients arising in expansion of n-th derivative of tan(x) + sec(x).

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Extensions

A229463 Expansion of g.f. 1/((1-x)^2*(1-26*x)).

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PARI

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A378962 First differences of A378726.

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A244762 a(n) = (53^n-2n-1)/4.

A229463 Expansion of g.f. 1/((1-x)^2(1-26x)).