A000217 -id:A000217 - cp's OEIS frontend

A141419 Triangle read by rows: T(n, k) = A000217(n) - A000217(n - k) with 1 <= k <= n.

1, 2, 3, 3, 5, 6, 4, 7, 9, 10, 5, 9, 12, 14, 15, 6, 11, 15, 18, 20, 21, 7, 13, 18, 22, 25, 27, 28, 8, 15, 21, 26, 30, 33, 35, 36, 9, 17, 24, 30, 35, 39, 42, 44, 45, 10, 19, 27, 34, 40, 45, 49, 52, 54, 55

Offset: 1

Roger L. Bagula, Aug 05 2008

As a rectangle, the accumulation array of A051340.
From Clark Kimberling, Feb 05 2011: (Start)
Here all the weights are divided by two where they aren't in Cahn.
As a rectangle, A141419 is in the accumulation chain
... < A051340 < A141419 < A185874 < A185875 < A185876 < ...
(See A144112 for the definition of accumulation array.)
row 1: A000027
col 1: A000217
diag (1,5,...): A000326 (pentagonal numbers)
diag (2,7,...): A005449 (second pentagonal numbers)
diag (3,9,...): A045943 (triangular matchstick numbers)
diag (4,11,...): A115067
diag (5,13,...): A140090
diag (6,15,...): A140091
diag (7,17,...): A059845
diag (8,19,...): A140672
(End)
Let N=2*n+1 and k=1,2,...,n. Let A_{N,n-1} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1], an n X n unit-primitive matrix (see [Jeffery]). Let M_n=[A_{N,n-1}]^4. Then t(n,k)=[M_n](1,k), that is, the n-th row of the triangle is given by the first row of M_n. - _L. Edson Jeffery, Nov 20 2011
Conjecture. Let N=2*n+1 and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n-1} be the n X n unit-primitive matrices (again see [Jeffery]) associated with N, and define the Chebyshev polynomials of the second kind by the recurrence U_0(x) = 1, U_1(x) = 2*x and U_r(x) = 2*x*U_(r-1)(x) - U_(r-2)(x)  (r>1). Define the column vectors V_(k-1) = (U_(k-1)(cos(Pi/N)), U_(k-1)(cos(3*Pi/N)), ..., U_(k-1)(cos((2*n-1)*Pi/N)))^T, where T denotes matrix transpose. Let S_N = [V_0, V_1, ..., V_(n-1)] be the n X n matrix formed by taking V_(k-1) as column k-1. Let X_N = [S_N]^T*S_N, and let [X_N](i,j) denote the entry in row i and column j of X_N, i,j in {0,...,n-1}. Then t(n,k) = [X_N](k-1,k-1), and row n of the triangle is given by the main diagonal entries of X_N. Remarks: Hence t(n,k) is the sum of squares t(n,k) = sum[m=1,...,n (U_(k-1)(cos((2*m-1)*Pi/N)))^2]. Finally, this sequence is related to A057059, since X_N = [sum_{m=1,...,n} A057059(n,m)*A_{N,m-1}] is also an integral linear combination of unit-primitive matrices from the N-th set. - L. Edson Jeffery, Jan 20 2012
Row sums: n*(n+1)*(2*n+1)/6. - L. Edson Jeffery, Jan 25 2013
n-th row = partial sums of n-th row of A004736. - Reinhard Zumkeller, Aug 04 2014
T(n,k) is the number of distinct sums made by at most k elements in {1, 2, ... n}, for 1 <= k <= n, e.g., T(6,2) = the number of distinct sums made by at most 2 elements in {1,2,3,4,5,6}. The sums range from 1, to 5+6=11. So there are 11 distinct sums. - Derek Orr, Nov 26 2014
A number n occurs in this sequence A001227(n) times, the number of odd divisors of n, see A209260. - Hartmut F. W. Hoft, Apr 14 2016
Conjecture: 2*n + 1 is composite if and only if gcd(t(n,m),m) != 1, for some m. - L. Edson Jeffery, Jan 30 2018
From Peter Munn, Aug 21 2019 in respect of the sequence read as a triangle: (Start)
A number m can be found in column k if and only if A286013(m, k) is nonzero, in which case m occurs in column k on row A286013(m, k).
The first occurrence of m is in row A212652(m) column A109814(m), which is the rightmost column in which m occurs. This occurrence determines where m appears in A209260. The last occurrence of m is in row m column 1.
Viewed as a sequence of rows, consider the subsequences (of rows) that contain every positive integer. The lexicographically latest of these subsequences consists of the rows with row numbers in A270877; this is the only one that contains its own row numbers only once.
(End)

			As a triangle:
   1,
   2,  3,
   3,  5,  6,
   4,  7,  9, 10,
   5,  9, 12, 14, 15,
   6, 11, 15, 18, 20, 21,
   7, 13, 18, 22, 25, 27, 28,
   8, 15, 21, 26, 30, 33, 35, 36,
   9, 17, 24, 30, 35, 39, 42, 44, 45,
  10, 19, 27, 34, 40, 45, 49, 52, 54, 55;
As a rectangle:
   1   2   3   4   5   6   7   8   9  10
   3   5   7   9  11  13  15  17  19  21
   6   9  12  15  18  21  24  27  30  33
  10  14  18  22  26  30  34  38  42  46
  15  20  25  30  35  40  45  50  55  60
  21  27  33  39  45  51  57  63  69  75
  28  35  42  49  56  63  70  77  84  91
  36  44  52  60  68  76  84  92 100 108
  45  54  63  72  81  90  99 108 117 126
  55  65  75  85  95 105 115 125 135 145
Since the odd divisors of 15 are 1, 3, 5 and 15, number 15 appears four times in the triangle at t(3+(5-1)/2, 5) in column 5 since 5+1 <= 2*3, t(5+(3-1)/2, 3), t(1+(15-1)/2, 2*1) in column 2 since 15+1 > 2*1, and t(15+(1-1)/2, 1). - _Hartmut F. W. Hoft_, Apr 14 2016

R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 139.

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened
Johann Cigler, Some elementary observations on Narayana polynomials and related topics, arXiv:1611.05252 [math.CO], 2016. See p. 24.
Carlton Gamer, David W. Roeder, and John J. Watkins, Trapezoidal Numbers, Mathematics Magazine 58:2 (1985), pp. 108-110.
L. E. Jeffery, Unit-primitive matrices
M. A. Nyblom, On the representation of the integers as a difference of nonconsecutive triangular numbers, Fibonacci Quarterly 39:3 (2001), pp. 256-263.

Cf. A000330 (row sums), A004736, A057059, A070543.
A144112, A051340, A141419, A185874, A185875, A185876 are accumulation chain related.
A141418 is a variant.
Cf. A001227, A209260. - Hartmut F. W. Hoft, Apr 14 2016
A109814, A212652, A270877, A286013 relate to where each natural number appears in this sequence.
A000027, A000217, A000326, A005449, A045943, A059845, A115067, A140090, A140091, A140672 are rows, columns or diagonals - refer to comments.

a141419 n k =  k * (2 * n - k + 1) `div` 2
a141419_row n = a141419_tabl !! (n-1)
a141419_tabl = map (scanl1 (+)) a004736_tabl
-- Reinhard Zumkeller, Aug 04 2014

a:=(n,k)->k*n-binomial(k,2): seq(seq(a(n,k),k=1..n),n=1..12); # Muniru A Asiru, Oct 14 2018

T[n_, m_] = m*(2*n - m + 1)/2; a = Table[Table[T[n, m], {m, 1, n}], {n, 1, 10}]; Flatten[a]

t(n,m) = m*(2*n - m + 1)/2.
t(n,m) = A000217(n) - A000217(n-m). - L. Edson Jeffery, Jan 16 2013
Let v = d*h with h odd be an integer factorization, then v = t(d+(h-1)/2, h) if h+1 <= 2*d, and v = t(d+(h-1)/2, 2*d) if h+1 > 2*d; see A209260. - Hartmut F. W. Hoft, Apr 14 2016
G.f.: y*(-x + y)/((-1 + x)^2*(-1 + y)^3). - Stefano Spezia, Oct 14 2018
T(n, 2) = A060747(n) for n > 1. T(n, 3) = A008585(n - 1) for n > 2. T(n, 4) = A016825(n - 2) for n > 3. T(n, 5) = A008587(n - 2) for n > 4. T(n, 6) = A016945(n - 3) for n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7.r n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7. T(n, 9) = A008591(n - 4) for n > 8. T(n, 10) = A017329(n - 5) for n > 9. T(n, 11) = A008593(n - 5) for n > 10.  T(n, 12) = A017593(n - 6) for n > 11. T(n, 13) = A008595(n - 6) for n > 12. T(n, 14) = A147587(n - 7) for n > 13. T(n, 15) = A008597(n - 7) for n > 14. T(n, 16) = A051062(n - 8) for n > 15. T(n, 17) = A008599(n - 8) for n > 16. - Stefano Spezia, Oct 14 2018
T(2*n-k, k) = A070543(n, k). - Peter Munn, Aug 21 2019

Simpler name by Stefano Spezia, Oct 14 2018

A183079 Tree generated by the triangular numbers: a(1) = 1; a(2n) = nontriangular(a(n)), a(2n+1) = triangular(a(n+1)), where triangular = A000217, nontriangular = A014132.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 10, 7, 21, 9, 15, 8, 55, 14, 28, 11, 231, 27, 45, 13, 120, 20, 36, 12, 1540, 65, 105, 19, 406, 35, 66, 16, 26796, 252, 378, 34, 1035, 54, 91, 18, 7260, 135, 210, 26, 666, 44, 78, 17, 1186570, 1595, 2145, 76, 5565, 119, 190, 25, 82621, 434

Offset: 1

Clark Kimberling, Dec 23 2010

A permutation of the positive integers.
In general, suppose that L and U are complementary sequences of positive integers such that
(1) L(1)=1; and
(2) if n>1, then n=L(k) or n=U(k) for some kThe tree generated by the sequence L is defined as follows:
  T(0,0)=1; T(1,0)=2; T(n,2j)=L(T(n-1,j));
  T(n,2j+1)=U(T(n-1,j)); for j=0,1,...,2^(n-1)-1, n>=2.
The numbers, taken in the order generated, form a permutation of the positive integers.

			First levels of the tree:
                                    1
                                    |
                 ...................2...................
                3                                       4
      6......../ \........5                   10......./ \........7
     / \                 / \                 / \                 / \
    /   \               /   \               /   \               /   \
   /     \             /     \             /     \             /     \
  21      9          15       8          55       14         28      11
231 27  45 13     120  20   36 12    1540  65  105  19    406  35  66  16
Beginning with 3 and 4, the numbers are generated in pairs, such as (3,4), (6,5), (10,7), (21,9),...
In all such pairs, the first number belongs to A000217; the second, to A014132.

Reinhard Zumkeller, Rows n = 1..14 of triangle, flattened
Index entries for sequences that are permutations of the natural numbers

Cf. A000217, A014132, A074049.
Cf. A220347 (inverse), A220348.
Cf. A183089, A183209 (similar permutations), also A257798.

a183079 n k = a183079_tabf !! (n-1) !! (k-1)
a183079_row n = a183079_tabf !! n
a183079_tabf = [1] : iterate (\row -> concatMap f row) [2]
   where f x = [a000217 x, a014132 x]
a183079_list = concat a183079_tabf
-- Reinhard Zumkeller, Dec 12 2012

tr[n_]:=n*(n+1)/2; nt[n_]:= n+Round@ Sqrt[2*n];a[1]=1; a[n_Integer] := a[n] = If[ EvenQ@n, nt@a[n/2], tr@ a@ Ceiling[n/2]]; a/@Range[58] (* Giovanni Resta, May 20 2015 *)

;; With memoizing definec-macro.
(definec (A183079 n) (cond ((<= n 1) n) ((even? n) (A014132 (A183079 (/ n 2)))) (else (A000217 (A183079 (/ (+ n 1) 2))))))
;; Antti Karttunen, May 18 2015

Let L(n) be the n-th triangular number (A000217).
Let U(n) be the n-th non-triangular number (A014132).
The tree-array T(n,k) is then given by rows:
T(0,0)=1; T(1,0)=2;
T(n,2j)=L(T(n-1,j));
T(n,2j+1)=U(T(n-1,j));
for j=0,1,...,2^(n-1)-1, n>=2.
a(1) = 1; after which: a(2n) = A014132(a(n)), a(2n+1) = A000217(a(n+1)). - Antti Karttunen, May 20 2015

Formula added to the name and a new tree illustration to the Example section by Antti Karttunen, May 20 2015

A072964 Number of partitions of n-th triangular number n(n+1)/2 (A000217(n)) into triangular parts.

Original entry on oeis.org

1, 1, 2, 4, 7, 15, 32, 66, 141, 295, 619, 1286, 2654, 5460, 11066, 22357, 44962, 89258, 176459, 347103, 675846, 1309903, 2525893, 4830943, 9196093, 17418788, 32772432, 61375543, 114401182, 212026732, 391231769, 718710706, 1313781686

Offset: 0

Paul D. Hanna, Aug 13 2002

nonn

What is limit_{n->inf} a(n)^(1/n)? [This limit is equal to 1. - Vaclav Kotesovec, May 21 2018]

T. D. Noe, Table of n, a(n) for n=0..100

Cf. A007294, A073420, A114738.

c = CoefficientList[ Series[1/Product[1 - x^(i(i + 1)/2), {i, 1, 50}], {x, 0, 565}], x]; c[[Range[33]*(Range[33] + 1)/2 + 1]] (* Robert G. Wilson v *)

a(n) = A007294[n(n+1)/2] = coefficient of x^[n(n+1)/2] in the expansion of product_{k=1..inf} 1/(1 - x^(k(k+1)/2)).
a(n) = A007294(A000217(n)).
a(n) ~ exp(3*Pi^(1/3) * Zeta(3/2)^(2/3) * (n*(n+1))^(1/3) / 2^(4/3)) * Zeta(3/2) / (4*Pi*sqrt(3)*n^3). - Vaclav Kotesovec, May 21 2018

Entry revised by N. J. A. Sloane, Jan 28 2007

A226254 Number of ways of writing n as the sum of 10 triangular numbers from A000217.

Original entry on oeis.org

1, 10, 45, 130, 300, 612, 1105, 1830, 2925, 4420, 6341, 9000, 12325, 16290, 21645, 27932, 34980, 44370, 54900, 66430, 81702, 98050, 115440, 138330, 162565, 187800, 220545, 254800, 289265, 334890, 382058, 427350, 488700, 550420, 609960, 691812, 770185, 845750, 949365, 1049400, 1145580, 1274580

Offset: 0

N. J. A. Sloane, Jun 01 2013

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..10000
K. Ono, S. Robins and P. T. Wahl, On the representation of integers as sums of triangular numbers, Aequationes mathematicae, August 1995, Volume 50, Issue 1-2, pp 73-94. Case k=10, Theorem 6.

Cf. A000217, A030212, A050456.

Number of ways of writing n as a sum of k triangular numbers, for k=1,...: A010054, A008441, A008443, A008438, A008439, A008440, A226252, A007331, A226253, A226254, A226255, A014787, A014809.

G.f. is 10th power of g.f. for A010054.
a(n) = (A050456(4*n+5) - A030212(4*n+5))/640. See the Ono et al. link, case k=10, Theorem 6. - Wolfdieter Lang, Jan 13 2017
a(0) = 1, a(n) = (10/n)*Sum_{k=1..n} A002129(k)*a(n-k) for n > 0. - Seiichi Manyama, May 06 2017
G.f.: exp(Sum_{k>=1} 10*(x^k/k)/(1 + x^k)). - Ilya Gutkovskiy, Jul 31 2017

A231599 T(n,k) is the coefficient of x^k in Product_{i=1..n} (1-x^i); triangle T(n,k), n >= 0, 0 <= k <= A000217(n), read by rows.

Original entry on oeis.org

1, 1, -1, 1, -1, -1, 1, 1, -1, -1, 0, 1, 1, -1, 1, -1, -1, 0, 0, 2, 0, 0, -1, -1, 1, 1, -1, -1, 0, 0, 1, 1, 1, -1, -1, -1, 0, 0, 1, 1, -1, 1, -1, -1, 0, 0, 1, 0, 2, 0, -1, -1, -1, -1, 0, 2, 0, 1, 0, 0, -1, -1, 1, 1, -1, -1, 0, 0, 1, 0, 1, 1, 0, -1, -1, -2, 0

Offset: 0

Marc Bogaerts, Nov 11 2013

From Tilman Piesk, Feb 21 2016: (Start)
The sum of each row is 0. The even rows are symmetric; in the odd rows numbers with the same absolute value and opposed signum are symmetric to each other.
The odd rows where n mod 4 = 3 have the central value 0.
The even rows where n mod 4 = 0 have positive central values. They form the sequence A269298 and are also the rows maximal values.
A086376 contains the maximal values of each row, A160089 the maximal absolute values, and A086394 the absolute parts of the minimal values.
Rows of this triangle can be used to efficiently calculate values of A026807.
(End)

			For n=2 the corresponding polynomial is (1-x)*(1-x^2) = 1 -x - x^2 + x^3.
Irregular triangle starts:
  k    0   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15
n
0      1
1      1  -1
2      1  -1  -1   1
3      1  -1  -1   0   1   1  -1
4      1  -1  -1   0   0   2   0   0  -1  -1   1
5      1  -1  -1   0   0   1   1   1  -1  -1  -1   0   0   1   1  -1

Alois P. Heinz, Rows n = 0..40, flattened
Dorin Andrica and Ovidiu Bagdasar, On some results concerning the polygonal polynomials, Carpathian Journal of Mathematics (2019) Vol. 35, No. 1, 1-11.
Tilman Piesk, Rows n = 0..40 as left-aligned and centered table

Cf. A000217 (triangular numbers).
Cf. A086376, A160089, A086394 (maxima, etc.).
Cf. A269298 (central nonzero values).

T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))
        (expand(mul(1-x^i, i=1..n))):
seq(T(n), n=0..10);  # Alois P. Heinz, Dec 22 2013

Table[If[k == 0, 1, Coefficient[Product[(1 - x^i), {i, n}], x^k]], {n, 0, 6}, {k, 0, (n^2 + n)/2}] // Flatten (* Michael De Vlieger, Mar 04 2018 *)

row(n) = pol = prod(i=1, n, 1 - x^i); for (i=0, poldegree(pol), print1(polcoeff(pol, i), ", ")); \\ Michel Marcus, Dec 21 2013

from sympy import poly, symbols
def a231599_row(n):
    if n == 0:
        return [1]
    x = symbols('x')
    p = 1
    for i in range(1, n+1):
        p *= poly(1-x**i)
    p = p.all_coeffs()
    return p[::-1]
# Tilman Piesk, Feb 21 2016

T(n,k) = [x^k] Product_{i=1..n} (1-x^i).

T(n,k) = T(n-1, k) + (-1)^n*T(n-1, n*(n+1)/2-k), n > 1. - Gevorg Hmayakyan, Feb 09 2017 [corrected by Giuliano Cabrele, Mar 02 2018]

A195836 Triangle read by rows which arises from A195826 in the same way as A175003 arises from A195310. Column k starts at row A000217(k).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 3, 1, 4, 2, -1, 5, 3, -1, 7, 4, -1, 10, 5, -2, 13, 7, -3, -1, 16, 10, -4, -1, 21, 13, -5, -1, 28, 16, -7, -2, 35, 21, -10, -3, 43, 28, -13, -4, 1, 55, 35, -16, -5, 1, 70, 43, -21, -7, 1, 86, 55, -28, -10, 2, 105, 70, -35, -13, 3, 130, 86, -43, -16, 4

Offset: 1

Omar E. Pol, Sep 24 2011

The sum of terms of row n is equal to the leftmost term of row n+1. It appears that this sequence is related to the generalized hexagonal numbers (A000217), A195826 and A006950 in the same way as A175003 is related to the generalized pentagonal numbers A001318, A195310 and A000041. It appears that row sums give A006950. See comments in A195825.

			Written as a triangle:
.  1,
.  1,
.  1,   1,
.  2,   1,
.  3,   1,
.  4,   2,  -1,
.  5,   3,  -1,
.  7,   4,  -1,
. 10,   5,  -2,
. 13,   7,  -3,   -1,
. 16,  10,  -4,   -1,
. 21,  13,  -5,   -1,
. 28,  16,  -7,   -2,
. 35,  21,  -10,  -3,
. 43,  28,  -13,  -4,   1,
. 55,  35,  -16,  -5,   1,
. 70,  43,  -21,  -7,   1,
. 86,  55,  -28, -10,   2,

Cf. A000217, A006950, A175003, A195825, A195826, A195837, A195838, A195939, A195840, A195841, A195842, A195843.

A086602 a(n) = A000217(A000217(n))-n^2.

Original entry on oeis.org

0, 0, 2, 12, 39, 95, 195, 357, 602, 954, 1440, 2090, 2937, 4017, 5369, 7035, 9060, 11492, 14382, 17784, 21755, 26355, 31647, 37697, 44574, 52350, 61100, 70902, 81837, 93989, 107445, 122295, 138632, 156552, 176154, 197540, 220815, 246087

Offset: 0

Jon Perry, Jul 23 2003

			a(3) = t(t(3))-3^2 = t(6)-9 = 21-9 = 12.

Q. T. Bach, R. Paudyal, J. B. Remmel, A Fibonacci analogue of Stirling numbers, arXiv preprint arXiv:1510.04310 [math.CO], 2015-2016.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000330, A001296, A005581.

[n*(n-1)*(n^2+3*n-2)/8: n in [0..40]]; // Vincenzo Librandi, Jun 26 2016

seq(3*binomial(n+2,4)-binomial(n,2), n=0..35); # Zerinvary Lajos, May 02 2007

Table[n (n - 1) (n^2 + 3 n - 2)/8, {n, 0, 40}] (* Bruno Berselli, Aug 27 2014 *)
LinearRecurrence[{5,-10,10,-5,1},{0,0,2,12,39},60] (* Harvey P. Dale, Apr 04 2023 *)

t(i)=i*(i+1)/2
vector(40,i,t(t(i))-i^2)

a(n) = A000330(n-1)+A001295(n-1). - Alford Arnold, Jun 29 2005
a(n) = 3*C(n+2,4) - C(n,2). - Zerinvary Lajos, May 02 2007, corrected Jun 12 2018
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) = n*(n-1)*(n^2+3*n-2)/8. [R. J. Mathar, Oct 30 2009]
G.f.: x^2*(-2-2*x+x^2)/(x-1)^5. [R. J. Mathar, Oct 30 2009]
a(n) = (n-1)*A005581(n) - Sum_{i=0..n-1} A005581(i). [Bruno Berselli, Aug 27 2014]

A195826 Triangle read by rows with T(n,k) = n - A000217(k), n>=1, k>=1, if (n - A000217(k))>=0.

Original entry on oeis.org

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

Offset: 1

Omar E. Pol, Sep 24 2011

Also triangle read by rows in which column k lists the nonnegative integers A001477 starting at the row A000217(k).

This sequence is related to the generalized hexagonal numbers (A000217), A195836 and A006950 in the same way as A195310 is related to the generalized pentagonal numbers A001318, A175003 and A000041. See comments in A195825.

			Written as a triangle:
.  0;
.  1;
.  2,  0;
.  3,  1;
.  4,  2;
.  5,  3,  0;
.  6,  4,  1;
.  7,  5,  2;
.  8,  6,  3;
.  9,  7,  4,  0;
. 10,  8,  5,  1;
. 11,  9,  6,  2;
. 12, 10,  7,  3;
. 13, 11,  8,  4;
. 14, 12,  9,  5,  0;
. 15, 13,  10, 6,  1;
. 16, 14,  11, 7,  2;
. 17, 15,  12, 8,  3;

Cf. A000217, A006950, A195310, A195825, A195827, A195828, A195829, A195830, A195831, A195832, A195833, A195836.

A105479 a(n) = C(n,2)*Bell(n-2) (cf. A000217, A000110).

Original entry on oeis.org

0, 0, 1, 3, 12, 50, 225, 1092, 5684, 31572, 186300, 1163085, 7654350, 52928460, 383437327, 2902665885, 22907918640, 188082362120, 1603461748491, 14169892736484, 129594593170210, 1224875863061970, 11948280552370932, 120142063487658003, 1243853543811461148

Offset: 0

Augustine O. Munagi, Apr 10 2005

Number of blocks of size 2 in all set partitions of {1,2,...,n}. Example: a(3)=3 because the set partitions of {1,2,3} are 1|2|3, 1|23, 12|3, 13|2 and 123, containing exactly 3 blocks of size 2. a(n) = Sum_{k>=0} k*A124498(n-1,k). - Emeric Deutsch, Nov 06 2006
Number of partitions of {1...n} containing 2 pairs of consecutive integers, where each pair is counted within a block and a string of more than 2 consecutive integers are counted two at a time. E.g. a(4) = 3 because the partitions of {1,2,3,4} with 2 pairs of consecutive integers are 123/4,12/34,1/234. - Augustine O. Munagi, Apr 10 2005
a(n) is the total sum of singletons in all set partitions of [n-1]. a(4) = 12 = 0+1+2+3+6: 123, 1|23, 13|2, 12|3, 1|2|3. - Alois P. Heinz, Mar 06 2024

Augustine O. Munagi, Set Partitions with Successions and Separations, Int. J. Math and Math. Sc. 2005, no. 3 (2005), 451-463.

Cf. A105480, A105489, A105484, A124498.

Column k=2 of A193297.

[seq(binomial(n,2)*combinat[bell](n-2),n=0..50)];

Join[{0,0},Table[Binomial[n,2]BellB[n-2],{n,2,30}]] (* Harvey P. Dale, May 06 2014 *)

from itertools import count, accumulate, islice
def A105479_gen(): # generator of terms
    yield from (0,0,1)
    blist, b, c = (1,), 1, 1
    for n in count(2):
        c += n
        blist = list(accumulate(blist, initial=(b:=blist[-1])))
        yield b*c
A105479_list = list(islice(A105479_gen(),20)) # Chai Wah Wu, Jun 22 2022

a(n) = binomial(n-1, 2)*Bell(n-3), the case r = 2 of the general case of r pairs: c(n, r) = binomial(n-1, r)*Bell(n-r-1).
E.g.f.: z^2/2 * e^(e^z-1) - Frank Ruskey, Dec 26 2006
G.f.: exp(-1)*Sum_{n>=0} (x^2/(n!*(1-n*x)^3)). - Vladeta Jovovic, Feb 05 2008
Let A be the upper Hessenberg matrix of order n defined by: A[i,i-1]=-1, A[i,j]=binomial(j-1,i-1), (i<=j), and A[i,j]=0 otherwise. Then, for n>=2, a(n)=(-1)^(n-2)coeff(charpoly(A,x),x^2). - Milan Janjic, Jul 08 2010
G.f.: x^2/exp(1)*G(0), where G(k) = 1 + (2*k*x-1)^3/((2*k+1)*(2*k*x+x-1)^3 - (2*k+1)*(2*k*x+x-1)^6/((2*k*x+x-1)^3 + 2*(k+1)*(2*k*x+2*x-1)^3/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 14 2013

Edited by N. J. A. Sloane, Jan 01 2007

A129467 Orthogonal polynomials with all zeros integers from 2*A000217.

Original entry on oeis.org

1, 0, 1, 0, -2, 1, 0, 12, -8, 1, 0, -144, 108, -20, 1, 0, 2880, -2304, 508, -40, 1, 0, -86400, 72000, -17544, 1708, -70, 1, 0, 3628800, -3110400, 808848, -89280, 4648, -112, 1, 0, -203212800, 177811200, -48405888, 5808528, -349568, 10920, -168, 1, 0, 14631321600, -13005619200, 3663035136, -466619904, 30977424, -1135808, 23016, -240, 1

Offset: 0

Wolfdieter Lang, May 04 2007

The row polynomials p(n,x) = Sum_{k=0..n} T(n,k)*x^k have the n integer zeros 2*A000217(j), j=0..n-1.
The row polynomials satisfy a three-term recurrence relation which qualify them as orthogonal polynomials w.r.t. some (as yet unknown) positive measure.
Column sequences (without leading zeros) give A000007, A010790(n-1)*(-1)^(n-1), A084915(n-1)*(-1)^(n-2), A130033 for m=0..3.
Apparently this is the triangle read by rows of Legendre-Stirling numbers of the first kind. See the Andrews-Gawronski-Littlejohn paper, table 2. The mirror version is the triangle A191936. - Omar E. Pol, Jan 10 2012

			Triangle starts:
  1;
  0,    1;
  0,   -2,     1;
  0,   12,    -8,   1;
  0, -144,   108, -20,   1;
  0, 2880, -2304, 508, -40,  1;
  ...
n=3: [0,12,-8,1]. p(3,x) = x*(12-8*x+x^2) = x*(x-2)*(x-6).
n=5: [0,2880,-2304,508,-40,1]. p(5,x) = x*(2880-2304*x+508*x^2-40*x^3 +x^4) = x*(x-2)*(x-6)*(x-12)*(x-20).

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
G. E. Andrews, W. Gawronski, and L. L. Littlejohn, The Legendre-Stirling Numbers, Discrete Mathematics, Volume 311, Issue 14, 28 July 2011, Pages 1255-1272.
M. Bruschi, F. Calogero and R. Droghei, Proof of certain Diophantine conjectures and identification of remarkable classes of orthogonal polynomials, J. Physics A, 40(2007), pp. 3815-3829.
José L. Cereceda, A refinement of Lang's formula for the sum of powers of integers, arXiv:2301.02141 [math.NT], 2023.
José L. Cereceda, Sums of powers of integers and the sequence A304330, arXiv:2405.05268 [math.GM], 2024. See p. 14.
Mark W. Coffey and Matthew C. Lettington, On Fibonacci Polynomial Expressions for Sums of mth Powers, their implications for Faulhaber's Formula and some Theorems of Fermat, arXiv:1510.05402 [math.NT], 2015.
Wolfdieter Lang, First ten rows and more.

Cf. A129462 (v=2 member), A129065 (v=1 member), A191936 (row reversed?).
Cf. A000217, A130031 (row sums), A130032 (unsigned row sums), A191936.
Column sequences (without leading zeros): A000007 (k=0), (-1)^(n-1)*A010790(n-1) (k=1), (-1)^n*A084915(n-1) (k=2), A130033 (k=3).
Cf. A008275.

f:= func< n,k | (&+[Binomial(2*k+j,j)*StirlingFirst(2*n,2*k+j)*n^j: j in [0..2*(n-k)]]) >;
function T(n,k) // T = A129467
  if k eq n then return 1;
  else return f(n,k) -  (&+[Binomial(j,2*(j-k))*T(n,j): j in [k+1..n]]);
end if;
end function;
[[T(n,k): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Feb 09 2024

T[n_, k_, m_]:= T[n,k,m]= If[k<0 || k>n, 0, If[n==0, 1, (2*(n-1)*(n-m) -(m-1))*T[n-1,k,m] -((n-1)*(n-m-1))^2*T[n-2,k,m] +T[n-1,k-1,m]]]; (* T=A129467 *)
Table[T[n,k,n], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 09 2024 *)

@CachedFunction
def f(n,k): return sum(binomial(2*k+j,j)*(-1)^j*stirling_number1(2*n,2*k+j)*n^j for j in range(2*n-2*k+1))
def T(n,k): # T = A129467
    if n==0: return 1
    else: return - sum(binomial(j,2*j-2*k)*T(n,j) for j in range(k+1,n+1)) + f(n,k)
flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Feb 09 2024

Row polynomials p(n,x) = Product_{m=1..n} (x - m*(m-1)), n>=1, with p(0,x) = 1.
Row polynomials p(n,x) = p(n, v=n, x) with the recurrence: p(n,v,x) = (x + 2*(n-1)^2 - 2*(v-1)*(n-1) - v + 1)*p(n-1,v,x) - (n-1)^2*(n-1-v)^2*p(n-2,v,x) with p(-1,v,x) = 0 and p(0,v,x) = 1.
T(n, k) = [x^k] p(n, n, x), n >= k >= 0, otherwise 0.
T(n, k) = Sum_{j=0..2*(n-k)} ( binomial(2*k+j, j)*s(n,k)*n^j ) - Sum_{j=k+1..n} binomial(j, 2*(j-k))*T(n, j) (See Coffey and Lettington formula (4.7)). - G. C. Greubel, Feb 09 2024

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A141419 Triangle read by rows: T(n, k) = A000217(n) - A000217(n - k) with 1 <= k <= n.

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A183079 Tree generated by the triangular numbers: a(1) = 1; a(2n) = nontriangular(a(n)), a(2n+1) = triangular(a(n+1)), where triangular = A000217, nontriangular = A014132.

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A072964 Number of partitions of n-th triangular number n(n+1)/2 (A000217(n)) into triangular parts.

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A226254 Number of ways of writing n as the sum of 10 triangular numbers from A000217.

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A231599 T(n,k) is the coefficient of x^k in Product_{i=1..n} (1-x^i); triangle T(n,k), n >= 0, 0 <= k <= A000217(n), read by rows.

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A195836 Triangle read by rows which arises from A195826 in the same way as A175003 arises from A195310. Column k starts at row A000217(k).

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A086602 a(n) = A000217(A000217(n))-n^2.

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