A051836 -id:A051836 - cp's OEIS frontend

A051923 Partial sums of A051836.

1, 9, 42, 140, 378, 882, 1848, 3564, 6435, 11011, 18018, 28392, 43316, 64260, 93024, 131784, 183141, 250173, 336490, 446292, 584430, 756470, 968760, 1228500, 1543815, 1923831, 2378754, 2919952, 3560040, 4312968, 5194112, 6220368, 7410249, 8783985, 10363626

Offset: 0

Barry E. Williams, Dec 19 1999

If Y is a 3-subset of an n-set X then, for n >= 8, a(n-8) is the number of 8-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007

a(n) is the n-th antidiagonal sum of the convolution array A213551. - Clark Kimberling, Jun 17 2012

			From the third formula: a(4) = 15+60+108+120+75 = 378. - _Bruno Berselli_, Sep 04 2013

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
Herbert John Ryser, Combinatorial Mathematics, "The Carus Mathematical Monographs", No. 14, John Wiley and Sons, 1963, pp. 1-8.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000217, A027801, A051836.

Cf. A093560 ((3, 1) Pascal, column m=6).

[Binomial(n+5, 5)*(n+2)/2: n in [0..40]]; // Vincenzo Librandi, Dec 27 2018

CoefficientList[Series[(1 + 2 x)/(1 - x)^7, {x, 0, 25}], x]  (* Harvey P. Dale, Mar 13 2011 *)
Nest[Accumulate,Range[1,120,3],5] (* Vladimir Joseph Stephan Orlovsky, Jan 28 2012 *)
Table[Binomial[n + 5, 5] (n + 2) / 2, {n, 0, 35}] (* Vincenzo Librandi, Dec 27 2018 *)

a(n) = binomial(n+5, 5)*(n+2)/2.
G.f.: (1+2*x)/(1-x)^7.
a(n) = Sum_{k=1..n+1} k*A000217(k)*A000217(n-k+2). - Bruno Berselli, Sep 04 2013
From Amiram Eldar, Jan 28 2022: (Start)
Sum_{n>=0} 1/a(n) = 1205/18 - 20*Pi^2/3.
Sum_{n>=0} (-1)^n/a(n) = 10*Pi^2/3 - 320*log(2)/3 + 755/18. (End)

A264428 Triangle read by rows, Bell transform of Bell numbers.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 5, 11, 6, 1, 0, 15, 45, 35, 10, 1, 0, 52, 205, 210, 85, 15, 1, 0, 203, 1029, 1330, 700, 175, 21, 1, 0, 877, 5635, 8946, 5845, 1890, 322, 28, 1, 0, 4140, 33387, 63917, 50358, 20055, 4410, 546, 36, 1, 0, 21147, 212535, 484140, 450905, 214515, 57855, 9240, 870, 45, 1

Offset: 0

Peter Luschny, Nov 13 2015

Consider the sequence S0 -> T0 -> S1 -> T1 -> S2 -> T2 -> ... Here Sn -> Tn indicates the Bell transform mapping a sequence Sn to a triangle Tn as defined in the link and Tn -> S{n+1} the operator associating a triangle with the sequence of its row sums. If
S0 = A000012 = <1,1,1,...> then
T0 = A048993 # Stirling subset numbers,
S1 = A000110 # Bell numbers,
T1 = A264428 # Bell transform of Bell numbers,
S2 = A187761 # second-order Bell numbers,
T2 = A264430 # Bell transform of second-order Bell numbers,
S3 = A264432 # third-order Bell numbers.
This construction is closely related to permutations trees and A179455. Sn is A179455_col(n+1) prepended by A179455_diag(k) = k! for k <= n. In other words, Sn 'converges' to n! for n -> oo.
Given a sequence (s(n))n>=0 with s(0) = 0 and with e.g.f. B(x) = Sum_{n >= 1} s(n)*x^n/n!, then the Bell matrix associated with s(n) equals the exponential Riordan array [1, B(x)] belonging to the Lagrange subgroup of the exponential Riordan group. Omitting the first row and column from the Bell matrix produces the exponential Riordan array [d/dx(B(x)), B(x)] belonging to the Derivative subgroup of the exponential Riordan group. - Peter Bala, Jun 07 2016

			Triangle starts:
[1]
[0,   1]
[0,   1,    1]
[0,   2,    3,    1]
[0,   5,   11,    6,    1]
[0,  15,   45,   35,   10,    1]
[0,  52,  205,  210,   85,   15,   1]
[0, 203, 1029, 1330,  700,  175,  21,  1]
[0, 877, 5635, 8946, 5845, 1890, 322, 28, 1]

G. C. Greubel, Table of n, a(n) for n = 0..1325
Peter Luschny, The Bell transform
Peter Luschny, Permutation Trees

Cf. A000012, A000110, A000217, A000914, A027801, A048993, A051836, A179455, A187761 (row sums), A264430, A264432, A265312.

# Computes sequence in matrix form.
BellMatrix := proc(f, len) local T, A; A := [seq(f(n), n=0..len-2)];
T := proc(n, k) option remember; if k=0 then k^n else
add(binomial(n-1,j-1)*T(n-j,k-1)*A[j], j=1..n-k+1) fi end;
Matrix(len, (n,k)->T(n-1,k-1), shape=triangular[lower]) end:
BellMatrix(n -> combinat:-bell(n), 9); # Peter Luschny, Jan 21 2016
# Alternative, using the recurrence of Peter Bala:
R := proc(n) option remember; if n = 0 then 1 else
t*add(binomial(n-1,k)*combinat:-bell(k)*R(n-k-1,t),k=0..n-1) fi end:
T_row := n-> seq(coeff(R(n), t, k), k=0..n):
seq(print(T_row(n)),n=0..8); # Peter Luschny, Jun 09 2016

BellMatrix[f_Function|f_Symbol, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len-1}, {k, 0, len-1}]];
rows = 11;
M = BellMatrix[BellB, rows];
Table[M[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 21 2016, updated Jul 14 2018 *)
With[{r = 8}, Flatten[Table[BellY[n, k, BellB[Range[0, r]]], {n, 0, r}, {k, 0, n}]]] (* Jan Mangaldan, May 22 2016 *)

bell_matrix(f, len) = { my( m = matrix(len, len) );  m[1, 1] = 1;
  for( n = 1, len-1, m[n+1, 2] = f(n-1) );
  for( n = 0, len-1, for( k = 1, n,
     m[n+1, k+1] = sum(j = 1, n-k+1, binomial(n-1,j-1)*m[n-j+1,k]*m[j+1,2]) ));
  return( m )
}
f(n) = polcoeff( sum( k=0, n, prod( i=1, k, x / (1 - i*x)), x^n * O(x)), n);
bell_matrix(f, 9) \\ Peter Luschny, Jan 24 2016

from functools import cache
from math import comb as binomial
def BellMatrix(f, size):
    A = [f(n) for n in range(size - 1)]
    @cache
    def T(n, k):
        if k == 0: return k ** n
        return sum(
            binomial(n - 1, j) * T(n - j - 1, k - 1) * A[j]
            for j in range(n - k + 1) )
    return [[T(n, k) for k in range(n + 1)] for n in range(size)]
@cache
def b(n, k=0): return n < 1 or k*b(n-1, k) + b(n-1, k+1)
print(BellMatrix(b, 9))  # Peter Luschny, Jun 14 2022

# The functions below are referenced in various other sequences.
def bell_transform(n, a): # partition_based
    row = []
    fn = factorial(n)
    for k in (0..n):
        result = 0
        for p in Partitions(n, length=k):
            factorial_product = 1
            power_factorial_product = 1
            for part, count in p.to_exp_dict().items():
                factorial_product *= factorial(count)
                power_factorial_product *= factorial(part)**count
            coefficient = fn//(factorial_product*power_factorial_product)
            result += coefficient*prod([a[i-1] for i in p])
        row.append(result)
    return row
def bell_matrix(generator, dim):
    G = [generator(k) for k in srange(dim)]
    row = lambda n: bell_transform(n, G)
    return matrix(ZZ, [row(n)+[0]*(dim-n-1) for n in srange(dim)])
def inverse_bell_matrix(generator, dim):
    G = [generator(k) for k in srange(dim)]
    row = lambda n: bell_transform(n, G)
    M = matrix(ZZ, [row(n)+[0]*(dim-n-1) for n in srange(dim)]).inverse()
    return matrix(ZZ, dim, lambda n,k: (-1)^(n-k)*M[n,k])
bell_numbers = [sum(bell_transform(n, [1]*10)) for n in range(11)]
for n in range(11): print(bell_transform(n, bell_numbers))

From Peter Bala, Jun 07 2016: (Start)
E.g.f.: exp(t*B(x)), where B(x) = Integral_{u = 0..x} exp(exp(u) - 1) du =  x + x^2/2! + 2*x^3/3! + 5*x^4/4! + 15*x^5/5! + 52*x^6/6! + ....
Row polynomial recurrence: R(n+1,t) = t*Sum_{k = 0 ..n} binomial(n,k)*Bell(k)* R(n-k,t) with R(0,t) = 1. (End)

A213500 Rectangular array T(n,k): (row n) = bc, where b(h) = h, c(h) = h + n - 1, n >= 1, h >= 1, and = convolution.

Original entry on oeis.org

1, 4, 2, 10, 7, 3, 20, 16, 10, 4, 35, 30, 22, 13, 5, 56, 50, 40, 28, 16, 6, 84, 77, 65, 50, 34, 19, 7, 120, 112, 98, 80, 60, 40, 22, 8, 165, 156, 140, 119, 95, 70, 46, 25, 9, 220, 210, 192, 168, 140, 110, 80, 52, 28, 10, 286, 275, 255, 228, 196, 161, 125, 90

Offset: 1

Clark Kimberling, Jun 14 2012

Principal diagonal: A002412.
Antidiagonal sums: A002415.
Row 1:  (1,2,3,...)**(1,2,3,...) = A000292.
Row 2:  (1,2,3,...)**(2,3,4,...) = A005581.
Row 3:  (1,2,3,...)**(3,4,5,...) = A006503.
Row 4:  (1,2,3,...)**(4,5,6,...) = A060488.
Row 5:  (1,2,3,...)**(5,6,7,...) = A096941.
Row 6:  (1,2,3,...)**(6,7,8,...) = A096957.
...
In general, the convolution of two infinite sequences is defined from the convolution of two n-tuples:  let X(n) = (x(1),...,x(n)) and Y(n)=(y(1),...,y(n)); then X(n)**Y(n) = x(1)*y(n)+x(2)*y(n-1)+...+x(n)*y(1); this sum is the n-th term in the convolution of infinite sequences:(x(1),...,x(n),...)**(y(1),...,y(n),...), for all n>=1.
...
In the following guide to related arrays and sequences, row n of each array T(n,k) is the convolution b**c of the sequences b(h) and c(h+n-1). The principal diagonal is given by T(n,n) and the n-th antidiagonal sum by S(n). In some cases, T(n,n) or S(n) differs in offset from the listed sequence.
b(h)........ c(h)........ T(n,k) .. T(n,n) .. S(n)
h .......... h .......... A213500 . A002412 . A002415
h .......... h^2 ........ A212891 . A213436 . A024166
h^2 ........ h .......... A213503 . A117066 . A033455
h^2 ........ h^2 ........ A213505 . A213546 . A213547
h .......... h*(h+1)/2 .. A213548 . A213549 . A051836
h*(h+1)/2 .. h .......... A213550 . A002418 . A005585
h*(h+1)/2 .. h*(h+1)/2 .. A213551 . A213552 . A051923
h .......... h^3 ........ A213553 . A213554 . A101089
h^3 ........ h .......... A213555 . A213556 . A213547
h^3 ........ h^3 ........ A213558 . A213559 . A213560
h^2 ........ h*(h+1)/2 .. A213561 . A213562 . A213563
h*(h+1)/2 .. h^2 ........ A213564 . A213565 . A101094
2^(h-1) .... h .......... A213568 . A213569 . A047520
2^(h-1) .... h^2 ........ A213573 . A213574 . A213575
h .......... Fibo(h) .... A213576 . A213577 . A213578
Fibo(h) .... h .......... A213579 . A213580 . A053808
Fibo(h) .... Fibo(h) .... A067418 . A027991 . A067988
Fibo(h+1) .. h .......... A213584 . A213585 . A213586
Fibo(n+1) .. Fibo(h+1) .. A213587 . A213588 . A213589
h^2 ........ Fibo(h) .... A213590 . A213504 . A213557
Fibo(h) .... h^2 ........ A213566 . A213567 . A213570
h .......... -1+2^h ..... A213571 . A213572 . A213581
-1+2^h ..... h .......... A213582 . A213583 . A156928
-1+2^h ..... -1+2^h ..... A213747 . A213748 . A213749
h .......... 2*h-1 ...... A213750 . A007585 . A002417
2*h-1 ...... h .......... A213751 . A051662 . A006325
2*h-1 ...... 2*h-1 ...... A213752 . A100157 . A071238
2*h-1 ...... -1+2^h ..... A213753 . A213754 . A213755
-1+2^h ..... 2*h-1 ...... A213756 . A213757 . A213758
2^(n-1) .... 2*h-1 ...... A213762 . A213763 . A213764
2*h-1 ...... Fibo(h) .... A213765 . A213766 . A213767
Fibo(h) .... 2*h-1 ...... A213768 . A213769 . A213770
Fibo(h+1) .. 2*h-1 ...... A213774 . A213775 . A213776
Fibo(h) .... Fibo(h+1) .. A213777 . A001870 . A152881
h .......... 1+[h/2] .... A213778 . A213779 . A213780
1+[h/2] .... h .......... A213781 . A213782 . A005712
1+[h/2] .... [(h+1)/2] .. A213783 . A213759 . A213760
h .......... 3*h-2 ...... A213761 . A172073 . A002419
3*h-2 ...... h .......... A213771 . A213772 . A132117
3*h-2 ...... 3*h-2 ...... A213773 . A214092 . A213818
h .......... 3*h-1 ...... A213819 . A213820 . A153978
3*h-1 ...... h .......... A213821 . A033431 . A176060
3*h-1 ...... 3*h-1 ...... A213822 . A213823 . A213824
3*h-1 ...... 3*h-2 ...... A213825 . A213826 . A213827
3*h-2 ...... 3*h-1 ...... A213828 . A213829 . A213830
2*h-1 ...... 3*h-2 ...... A213831 . A213832 . A212560
3*h-2 ...... 2*h-1 ...... A213833 . A130748 . A213834
h .......... 4*h-3 ...... A213835 . A172078 . A051797
4*h-3 ...... h .......... A213836 . A213837 . A071238
4*h-3 ...... 2*h-1 ...... A213838 . A213839 . A213840
2*h-1 ...... 4*h-3 ...... A213841 . A213842 . A213843
2*h-1 ...... 4*h-1 ...... A213844 . A213845 . A213846
4*h-1 ...... 2*h-1 ...... A213847 . A213848 . A180324
[(h+1)/2] .. [(h+1)/2] .. A213849 . A049778 . A213850
h .......... C(2*h-2,h-1) A213853
...
Suppose that u = (u(n)) and v = (v(n)) are sequences having generating functions U(x) and V(x), respectively. Then the convolution u**v has generating function U(x)*V(x). Accordingly, if u and v are homogeneous linear recurrence sequences, then every row of the convolution array T satisfies the same homogeneous linear recurrence equation, which can be easily obtained from the denominator of U(x)*V(x). Also, every column of T has the same homogeneous linear recurrence as v.

			Northwest corner (the array is read by southwest falling antidiagonals):
  1,  4, 10, 20,  35,  56,  84, ...
  2,  7, 16, 30,  50,  77, 112, ...
  3, 10, 22, 40,  65,  98, 140, ...
  4, 13, 28, 50,  80, 119, 168, ...
  5, 16, 34, 60,  95, 140, 196, ...
  6, 19, 40, 70, 110, 161, 224, ...
T(6,1) = (1)**(6) = 6;
T(6,2) = (1,2)**(6,7) = 1*7+2*6 = 19;
T(6,3) = (1,2,3)**(6,7,8) = 1*8+2*7+3*6 = 40.

Clark Kimberling, Antidiagonals n = 1..60, flattened

Cf. A000027.

b[n_] := n; c[n_] := n
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213500 *)

t(n,k) = sum(i=0, k - 1, (k - i) * (n + i));
tabl(nn) = {for(n=1, nn, for(k=1, n, print1(t(k,n - k + 1),", ");); print(););};
tabl(12) \\ Indranil Ghosh, Mar 26 2017

def t(n, k): return sum((k - i) * (n + i) for i in range(k))
for n in range(1, 13):
    print([t(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 26 2017

T(n,k) = 4*T(n,k-1) - 6*T(n,k-2) + 4*T(n,k-3) - T(n,k-4).
T(n,k) = 2*T(n-1,k) - T(n-2,k).
G.f. for row n:  x*(n - (n - 1)*x)/(1 - x)^4.

A005585 5-dimensional pyramidal numbers: a(n) = n(n+1)(n+2)(n+3)(2n+3)/5!.

Original entry on oeis.org

1, 7, 27, 77, 182, 378, 714, 1254, 2079, 3289, 5005, 7371, 10556, 14756, 20196, 27132, 35853, 46683, 59983, 76153, 95634, 118910, 146510, 179010, 217035, 261261, 312417, 371287, 438712, 515592, 602888, 701624, 812889, 937839, 1077699, 1233765, 1407406

Offset: 1

N. J. A. Sloane

Convolution of triangular numbers (A000217) and squares (A000290) (n>=1). - Graeme McRae, Jun 07 2006
p^k divides a(p^k-3), a(p^k-2), a(p^k-1) and a(p^k) for prime p > 5 and integer k > 0. p^k divides a((p^k-3)/2) for prime p > 5 and integer k > 0. - Alexander Adamchuk, May 08 2007
If a 2-set Y and an (n-3)-set Z are disjoint subsets of an n-set X then a(n-5) is the number of 6-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007
5-dimensional square numbers, fourth partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} binomial(n+4, i+4)*b(i), where b(i)=[1,2,0,0,0,...]. - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
Antidiagonal sums of the convolution array A213550. - Clark Kimberling, Jun 17 2012
Binomial transform of (1, 6, 14, 16, 9, 2, 0, 0, 0, ...). - Gary W. Adamson, Jul 28 2015
2*a(n) is number of ways to place 4 queens on an (n+3) X (n+3) chessboard so that they diagonally attack each other exactly 6 times. The maximal possible attack number, p=binomial(k,2)=6 for k=4 queens, is achievable only when all queens are on the same diagonal. In graph-theory representation they thus form a corresponding complete graph. - Antal Pinter, Dec 27 2015
While adjusting for offsets, add A000389 to find the next in series A000389, A005585, A051836, A034263, A027800, A051843, A051877, A051878, A051879, A051880, A056118, A271567. (See Bruno Berselli's comments in A271567.) - Bruce J. Nicholson, Jun 21 2018
Coefficients in the terminating series identity 1 - 7*n/(n + 6) + 27*n*(n - 1)/((n + 6)*(n + 7)) - 77*n*(n - 1)*(n - 2)/((n + 6)*(n + 7)*(n + 8)) + ... = 0 for n = 1,2,3,.... Cf. A002415 and A040977. - Peter Bala, Feb 18 2019

			G.f. = x + 7*x^2 + 27*x^3 + 77*x^4 + 182*x^5 + 378*x^6 + 714*x^7 + 1254*x^8 + ... - _Michael Somos_, Jun 24 2018

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 797.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000 (first 121 terms from Alexander Adamchuk)
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Paul Barry, On the Gap-sum and Gap-product Sequences of Integer Sequences, arXiv:2104.05593 [math.CO], 2021.
R. K. Guy, Letter to N. J. A. Sloane, Feb 1988
Milan Janjic, Two Enumerative Functions
C. H. Karlson and N. J. A. Sloane, Correspondence, 1974
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 15.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
R. P. Stanley and F. Zanello, The Catalan case of Armstrong's conjecture on core partitions, arXiv preprint arXiv:1312.4352 [math.CO], 2013.
Index entries for sequences related to Chebyshev polynomials.
Index to sequences related to pyramidal numbers
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

a(n) = ((-1)^(n+1))*A053120(2*n+3, 5)/16, (1/16 of sixth unsigned column of Chebyshev T-triangle, zeros omitted).
Partial sums of A002415.
Cf. A006542, A040977, A047819, A111125 (third column).
Cf. a(n) = ((-1)^(n+1))*A084960(n+1, 2)/16 (compare with the first line). - Wolfdieter Lang, Aug 04 2014
Cf. A000389, A051836, A034263, A027800, A051843, A051877, A051878, A051879.

I:=[1, 7, 27, 77, 182, 378]; [n le 6 select I[n] else 6*Self(n-1)-15*Self(n-2)+20*Self(n-3)-15*Self(n-4)+6*Self(n-5)-Self(n-6): n in [1..40]]; // Vincenzo Librandi, Jun 09 2013

[seq(binomial(n+2,6)-binomial(n,6), n=4..45)]; # Zerinvary Lajos, Jul 21 2006
A005585:=(1+z)/(z-1)**6; # Simon Plouffe in his 1992 dissertation

With[{c=5!},Table[n(n+1)(n+2)(n+3)(2n+3)/c,{n,40}]] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1},{1,7,27,77,182,378},40] (* Harvey P. Dale, Oct 04 2011 *)
CoefficientList[Series[(1 + x) / (1 - x)^6, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)

a(n)=binomial(n+3,4)*(2*n+3)/5 \\ Charles R Greathouse IV, Jul 28 2015

G.f.: x*(1+x)/(1-x)^6.
a(n) = 2*C(n+4, 5) - C(n+3, 4). - Paul Barry, Mar 04 2003
a(n) = C(n+3, 5) + C(n+4, 5). - Paul Barry, Mar 17 2003
a(n) = C(n+2, 6) - C(n, 6), n >= 4. - Zerinvary Lajos, Jul 21 2006
a(n) = Sum_{k=1..n} T(k)*T(k+1)/3, where T(n) = n(n+1)/2 is a triangular number. - Alexander Adamchuk, May 08 2007
a(n-1) = (1/4)*Sum_{1 <= x_1, x_2 <= n} |x_1*x_2*det V(x_1,x_2)| = (1/4)*Sum_{1 <= i,j <= n} i*j*|i-j|, where V(x_1,x_2) is the Vandermonde matrix of order 2. First differences of A040977. - Peter Bala, Sep 21 2007
a(n) = C(n+4,4) + 2*C(n+4,5). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6), a(1)=1, a(2)=7, a(3)=27, a(4)=77, a(5)=182, a(6)=378. - Harvey P. Dale, Oct 04 2011
a(n) = (1/6)*Sum_{i=1..n+1} (i*Sum_{k=1..i} (i-1)*k). - Wesley Ivan Hurt, Nov 19 2014
E.g.f.: x*(2*x^4 + 35*x^3 + 180*x^2 + 300*x + 120)*exp(x)/120. - Robert Israel, Nov 19 2014
a(n) = A000389(n+3) + A000389(n+4). - Bruce J. Nicholson, Jun 21 2018
a(n) = -a(-3-n) for all n in Z. - Michael Somos, Jun 24 2018
From Amiram Eldar, Jun 28 2020: (Start)
Sum_{n>=1} 1/a(n) = 40*(16*log(2) - 11)/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 20*(8*Pi - 25)/3. (End)
a(n) = A004302(n+1) - A207361(n+1). - J. M. Bergot, May 20 2022
a(n) = Sum_{i=0..n+1} Sum_{j=i..n+1} i*j*(j-i)/2. - Darío Clavijo, Oct 11 2023
a(n) = (A000538(n+1) - A000330(n+1))/12. - Yasser Arath Chavez Reyes, Feb 21 2024

A093560 (3,1) Pascal triangle.

Original entry on oeis.org

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

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(3;n,m) gives in the columns m >= 1 the figurate numbers based on A016777, including the pentagonal numbers A000326 (see the W. Lang link).
This is the third member, d=3, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal (d=1), A029653 (d=2).
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} a(n,m)*x^m is G(z,x)=g(z)/(1-x*z*f(z)). Here: g(x)=(1+2*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(1+2*z)/(1-(1+x)*z).
The SW-NE diagonals give the Lucas numbers A000032: L(n) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with L(0)=2. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Triangle T(n,k), read by rows, given by [3,-2,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 17 2009
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
From Wolfdieter Lang, Jan 09 2015: (Start)
The signed lower triangular matrix (-1)^(n-1)*a(n,m) is the inverse of the Riordan matrix A106516; that is Riordan ((1-2*x)/(1+x),x/(1+x)).
See the Peter Bala comment from Dec 23 2014 in A106516 for general Riordan triangles of the type (g(x), x/(1-x)): exp(x)*r(n,x) = d(n,x) with the e.g.f. r(n,x) of row n and the e.g.f. of diagonal n.
Similarly, for general Riordan triangles of the type (g(x), x/(1+x)): exp(x)*r(n,-x) = d(n,x). (End)
The n-th row polynomial is (3 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018
Binomial(n-2,k)+2*Binomial(n-3,k) is also the number of permutations avoiding both 123 and 132 with k double descents, i.e., positions with w[i]>w[i+1]>w[i+2]. - Lara Pudwell, Dec 19 2018

			Triangle begins
  1,
  3,  1,
  3,  4,  1,
  3,  7,  5,   1,
  3, 10, 12,   6,   1,
  3, 13, 22,  18,   7,   1,
  3, 16, 35,  40,  25,   8,   1,
  3, 19, 51,  75,  65,  33,   9,  1,
  3, 22, 70, 126, 140,  98,  42, 10,  1,
  3, 25, 92, 196, 266, 238, 140, 52, 11, 1,

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Peter Bala, A note on the diagonals of a proper Riordan Array
M. Bukata, R. Kulwicki, N. Lewandowski, L. Pudwell, J. Roth, and T. Wheeland, Distributions of Statistics over Pattern-Avoiding Permutations, arXiv preprint arXiv:1812.07112 [math.CO], 2018.
W. Lang, First 10 rows and array of figurate numbers .

Cf. Column sequences for m=1..9: A016777, A000326 (pentagonal), A002411, A001296, A051836, A051923, A050494, A053367, A053310;

A007318 (Pascal's triangle), A029653 ((2,1) Pascal triangle), A093561 ((4,1) Pascal triangle), A228196, A228576.

Concatenation([1],Flat(List([1..11],n->List([0..n],k->Binomial(n,k)+2*Binomial(n-1,k))))); # Muniru A Asiru, Dec 20 2018

a093560 n k = a093560_tabl !! n !! k
a093560_row n = a093560_tabl !! n
a093560_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [3, 1]
-- Reinhard Zumkeller, Aug 31 2014

from math import comb, isqrt
def A093560(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+(r-a<<1))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m)=F(3;n-m, m) for 0<= m <= n, otherwise 0, with F(3;0, 0)=1, F(3;n, 0)=3 if n>=1 and F(3;n, m):=(3*n+m)*binomial(n+m-1, m-1)/m if m>=1.
G.f. column m (without leading zeros): (1+2*x)/(1-x)^(m+1), m>=0.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=3 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
T(n, k) = C(n, k) + 2*C(n-1, k). - Philippe Deléham, Aug 28 2005
Equals M * A007318, where M = an infinite triangular matrix with all 1's in the main diagonal and all 2's in the subdiagonal. - Gary W. Adamson, Dec 01 2007
Sum_{k=0..n} T(n,k) = A151821(n+1). - Philippe Deléham, Sep 17 2009
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(3 + 7*x + 5*x^2/2! + x^3/3!) = 3 + 10*x + 22*x^2/2! + 40*x^3/3! + 65*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014
G.f.: (-1-2*x)/(-1+x+x*y). - R. J. Mathar, Aug 11 2015

Incorrect connection with A046055 deleted by N. J. A. Sloane, Jul 08 2009

A038763 Triangular matrix arising in enumeration of catafusenes, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 4, 3, 1, 7, 15, 9, 1, 10, 36, 54, 27, 1, 13, 66, 162, 189, 81, 1, 16, 105, 360, 675, 648, 243, 1, 19, 153, 675, 1755, 2673, 2187, 729, 1, 22, 210, 1134, 3780, 7938, 10206, 7290, 2187, 1, 25, 276, 1764, 7182, 19278, 34020, 37908, 24057, 6561, 1, 28, 351, 2592, 12474, 40824, 91854, 139968, 137781, 78732, 19683

Offset: 0

N. J. A. Sloane, May 03 2000

Triangle T(n,k), 0<=k<=n, read by rows, given by [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 10 2005
Triangle read by rows, n-th row = X^(n-1) * [1, 1, 0, 0, 0, ...] where X = an infinite bidiagonal matrix with (1,1,1,...) in the main diagonal and (3,3,3,...) in the subdiagonal; given row 0 = 1. - Gary W. Adamson, Jul 19 2008
Fusion of polynomial sequences P and Q given by p(n,x)=(x+2)^n and q(n,x)=(2x+1)^n; see A193722 for the definition of fusion of two sequences of polynomials or triangular arrays. - Clark Kimberling, Aug 04 2011

			Triangle begins:
  1;
  1,  1;
  1,  4,   3;
  1,  7,  15,   9;
  1, 10,  36,  54,   27;
  1, 13,  66, 162,  189,   81;
  1, 16, 105, 360,  675,  648,  243;
  1, 19, 153, 675, 1755, 2673, 2187, 729;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
S. J. Cyvin, B. N. Cyvin, and J. Brunvoll, Unbranched catacondensed polygonal systems containing hexagons and tetragons, Croatica Chem. Acta, 69 (1996), 757-774.

Cf. A000007, A000012, A001296, A006138, A006234, A016777, A024462.
Cf. A051836, A062741, A080420, A080421, A080422, A080423, A081294.
Cf. A084938, A098399, A110523, A133494, A136158, A193722.

A038763:= func< n,k | n eq 0 select 1 else 3^(k-1)*(3*n-2*k)*Binomial(n,k)/n >;
[A038763(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 27 2023

A038763[n_,k_]:= If[n==0, 1, 3^(k-1)*(3*n-2*k)*Binomial[n,k]/n];
Table[A038763[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 27 2023 *)

T(n,k) = if ((n<0) || (k<0), return(0)); if ((n==0) && (k==0), return(1)); if (n==1, if (k<=1, return(1))); T(n-1,k) + 3*T(n-1,k-1);
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", "))); \\ Michel Marcus, Jul 25 2023

def A038763(n,k): return 1 if (n==0) else 3^(k-1)*(3*n-2*k)*binomial(n,k)/n
flatten([[A038763(n, k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Dec 27 2023

T(n, 0)=1; T(1, 1)=1; T(n, k)=0 for k>n; T(n, k) = T(n-1, k-1)*3 + T(n-1, k) for n >= 2.
Sum_{k=0..n} T(n,k) = A081294(n). - Philippe Deléham, Sep 22 2006
T(n, k) = A136158(n, n-k). - Philippe Deléham, Dec 17 2007
G.f.: (1-2*x*y)/(1-(3*y+1)*x). - R. J. Mathar, Aug 11 2015
From G. C. Greubel, Dec 27 2023: (Start)
T(n, 0) = A000012(n).
T(n, 1) = A016777(n-1).
T(n, 2) = A062741(n-1).
T(n, 3) = 9*A002411(n-2).
T(n, 4) = 27*A001296(n-3).
T(n, 5) = 81*A051836(n-4).
T(n, n) = A133494(n).
T(n, n-1) = A006234(n+2).
T(n, n-2) = A080420(n-2).
T(n, n-3) = A080421(n-3).
T(n, n-4) = A080422(n-4).
T(n, n-5) = A080423(n-5).
T(2*n, n) = 4*A098399(n-1) + (2/3)*[n=0].
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A006138(n-1) + (2/3)*[n=0].
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = A110523(n-1) + (4/3)*[n=0]. (End)

More terms from Michel Marcus, Jul 25 2023

A125232 Triangle T(n,k) read by rows: the (n-k)-th term of the k-fold iterated partial sum of the pentagonal numbers.

Original entry on oeis.org

1, 5, 1, 12, 6, 1, 22, 18, 7, 1, 35, 40, 25, 8, 1, 51, 75, 65, 33, 9, 1, 70, 126, 140, 98, 42, 10, 1, 92, 196, 266, 238, 140, 52, 11, 1, 117, 288, 462, 504, 378, 192, 63, 12, 1, 145, 405, 750, 966, 882, 570, 255, 75, 13, 1, 176, 550, 1155, 1716, 1848, 1452, 825, 330, 88, 14, 1

Offset: 1

Gary W. Adamson, Nov 24 2006

			First few rows of the triangle are:
   1;
   5,   1;
  12,   6,   1;
  22,  18,   7,   1;
  35,  40,  25,   8,   1;
  51,  75,  65,  33,   9,   1;
  70, 126, 140,  98,  42,  10,   1;
  ...
Example: (5,3) = 65 = 25 + 40 = (4,3) + (4,2).

Albert H. Beiler, "Recreations in the Theory of Numbers", Dover, 1966, p 189.

Robert Israel, Table of n, a(n) for n = 1..10011(rows 0 to 140, flattened)

Columns: A000326 (pentagonal numbers), A002411, A001296, A051836, A051923.

Cf. A095264 (row sums).

A125232 := proc(n,k) option remember ; if k = 0 then A000326(n) ; elif k = n-1 then 1 ; else procname(n-1,k)+procname(n-1,k-1) ; fi : end: # R. J. Mathar, Jun 09 2008

nmax = 11; col[1] = Table[n(3n-1)/2, {n, 1, nmax}]; col[k_] := col[k] = Prepend[Accumulate[col[k-1]], 0]; Table[col[k][[n]], {n, 1, nmax}, {k, 1, n}] // Flatten (* Jean-François Alcover, Mar 25 2019 *)

T(n,0)=A000326(n). T(n,k)=T(n-1,k) + T(n-1,k-1), k>0. - R. J. Mathar, Jun 09 2008

G.f. as triangle: (1+2*x)/((1-x)^2*(1-x-x*y)). - Robert Israel, Nov 07 2016

Edited and extended by R. J. Mathar, Jun 09 2008

A213548 Rectangular array: (row n) = bc, where b(h) = h, c(h) = m(m+1)/2, m = n-1+h, n>=1, h>=1, and = convolution.

Original entry on oeis.org

1, 5, 3, 15, 12, 6, 35, 31, 22, 10, 70, 65, 53, 35, 15, 126, 120, 105, 81, 51, 21, 210, 203, 185, 155, 115, 70, 28, 330, 322, 301, 265, 215, 155, 92, 36, 495, 486, 462, 420, 360, 285, 201, 117, 45, 715, 705, 678, 630, 560, 470, 365, 253, 145, 55, 1001

Offset: 1

Clark Kimberling, Jun 16 2012

Principal diagonal: A213549.
Antidiagonal sums: A051836.
Row 1, (1,2,3,...)**(1,3,6,...): A000332.
Row 2, (1,2,3,...)**(3,6,10,...): A005718.
Row 3, (1,2,3,...)**(6,10,15,...): k*(k+1)*(k^2 + 13*k + 58)/24.
For a guide to related arrays, see A213500.

			Northwest corner (the array is read by falling antidiagonals):
.  1,  5,  15,  35,  70, ...
.  3, 12,  31,  65, 120, ...
.  6, 22,  53, 105, 185, ...
. 10, 35,  81, 155, 265, ...
. 15, 51, 115, 215, 360, ...
. 21, 70, 155, 285, 470, ...
...
T(5,1) = (1)**(15) = 15;
T(5,2) = (1,2)**(15,21) = 1*21 + 2*15 = 51;
T(5,3) = (1,2,3)**(15,21,28) = 1*28 + 2*21 + 3*15 = 115;
T(4,4) = (1,2,3,4)**(10,15,21,28) = 1*28 + 2*21 + 3*15 + 4*10 = 155.

Clark Kimberling, Antidiagonals n = 1..60

Cf. A213500.

b[n_] := n; c[n_] := n (n + 1)/2
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213548 *)
d = Table[t[n, n], {n, 1, 40}] (* A213549 *)
s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A051836 *)

T(n,k) = 5*T(n,k-1) - 10*T(n,k-2) + 10*T(n,k-3) - 5*T(n,k-4) + T(n,k-5).

G.f. for row n: f(x)/g(x), where f(x) = n*(n+1) - 2*((n-1)^2)*x + n*(n-1)*x^2 and g(x) = 2*(1 - x)^5.

A135857 Partial sums triangle based on A016777. Riordan convolution triangle ((1 + 2*x)/(1-x)^2, x/(1-x)).

Original entry on oeis.org

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

Offset: 0

Gary W. Adamson, Dec 01 2007

A007318 * a bidiagonal matrix with all 1's in the main diagonal and all 3's in the subdiagonal.
Row sums give A036563(n+2), n >= 0.
From Wolfdieter Lang, Mar 23 2015: (Start)
This is the triangle of iterated partial sums of A016777. Such iterated partial sums of arithmetic progression sequences have been considered by Narayana Pandit (see the Mar 20 2015 comment on A000580 where the MacTutor History of Mathematics archive link and the Gottwald et al. reference, p. 338, are given).
This is therefore the Riordan triangle ((1+2*x)/(1-x)^2, x/(1-x)) with o.g.f. of the columns ((1+2*x)/(1-x)^2)*(x/(1-x))^k, k >= 0.
The column sequences are A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310, for k = 0..8.
The alternating row sums are A122553(n) = {1, repeat(3)}.
The Riordan A-sequence is A(y) = 1 + y (implying the Pascal triangle recurrence for k >= 1).
The Riordan Z-sequence is A256096, leading to a recurrence for T(n,0) given in the formula section. See the link "Sheffer a- and z-sequences" under A006232 also for Riordan A- and Z-sequences with references. (End)
When the first column (k = 0) is removed from this triangle, the result is A125232. - Georg Fischer, Jul 26 2023

			The triangle T(n, k) begins:
n\k  0   1   2    3    4    5    6   7   8  9 10 11
0:   1
1:   4   1
2:   7   5   1
3:  10  12   6    1
4:  13  22  18    7    1
5:  16  35  40   25    8    1
6:  19  51  75   65   33    9    1
7:  22  70 126  140   98   42   10   1
8:  25  92 196  266  238  140   52  11   1
9:  28 117 288  462  504  378  192  63  12  1
10: 31 145 405  750  966  882  570 255  75 13  1
11: 34 176 550 1155 1716 1848 1452 825 330 88 14  1
... reformatted and extended by _Wolfdieter Lang_, Mar 23 2015
From _Wolfdieter Lang_, Mar 23 2015: (Start)
T(3, 1) = T(2, 0) + T(2, 1) = 7 + 5 = 12 (Pascal, from the A-sequence given above).
T(4, 0) = 4*T(3, 0) - 9*T(3, 1) + 27*T(3, 2) - 81* T(3, 3) = 4*10 - 9*12 + 27*6 - 81*1 = 13, from the Z-sequence given above and in A256096.
T(4, 0) = 2*T(3, 0) - T(2, 0) = 2*10 - 7 = 13.
(End)

Columns k=0-8 give: A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310.
Cf. A036563, A125232.
Cf. A110813. - Philippe Deléham, Feb 08 2009

Binomial transform of an infinite lower triangular matrix with all 1's in the main diagonal and all 3's in the subdiagonal; i.e., by columns - every column = (1, 3, 0, 0, 0, ...).
T(n,k) = (3n-2k+1)*binomial(n+1,k+1)/(n+1). - Philippe Deléham, Feb 08 2009
From Wolfdieter Lang, Mar 23 2015: (Start)
O.g.f. for row polynomials: (1 + 2*z)/((1- z*(1 + x))*(1 - z)) (see the Riordan property from the comment).
O.g.f. for column k (without leading zeros): (1 + 2*x)/(1-x)^(2+k), k >= 0, (Riordan property).
T(n, k) = T(n-1, k-1) + T(n-1, k) for k >= 1. From the Riordan A-sequence given above in a comment.
T(n, 0) = Sum_{j=0..n} Z(j)*T(n-1, j), for n >= 1, from the Riordan Z-sequence A256096 mentioned above in a comment. Of course, T(n, 0) = 2*T(n-1, 0) - T(n-2, 0) for n >= 2 (see A016777).
(End)

Edited. Offset is 0 from the old name and the Philippe Deléham formula. New name, old name as first comment. - Wolfdieter Lang, Mar 23 2015

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A038763 Triangular matrix arising in enumeration of catafusenes, read by rows.

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A213500 Rectangular array T(n,k): (row n) = bc, where b(h) = h, c(h) = h + n - 1, n >= 1, h >= 1, and = convolution.

A005585 5-dimensional pyramidal numbers: a(n) = n(n+1)(n+2)(n+3)(2n+3)/5!.

A213548 Rectangular array: (row n) = bc, where b(h) = h, c(h) = m(m+1)/2, m = n-1+h, n>=1, h>=1, and = convolution.