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Triangle, read by rows, such that row n equals the inverse binomial transform of row n of table A060543, where A060543(n,k) = C(n+n*k+k, n*k+k).
+20
4
1, 1, 2, 1, 9, 9, 1, 34, 96, 64, 1, 125, 750, 1250, 625, 1, 461, 5265, 16470, 19440, 7776, 1, 1715, 35329, 184877, 386561, 352947, 117649, 1, 6434, 232288, 1913408, 6307840, 9863168, 7340032, 2097152, 1, 24309, 1513656, 18921924, 92365758, 220016574
COMMENTS
The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial (1+n*x)*(2+n*x)*...*(n-1+n*x)/(n-1)! in the basis made of the binomial(x+i,i). - F. Chapoton, Nov 01 2022
EXAMPLE
BINOMIAL[1, 9, 9] = {1, 10, 28, 55, 91, 136, 190, 253, ...}.
BINOMIAL[1, 34, 96, 64] = {1, 35, 165, 455, 969, 1771, 2925, ...}.
BINOMIAL[1, 125, 750, 1250, 625] = {1, 126, 1001, 3876, 10626, ...}.
Triangle begins:
1;
1, 2;
1, 9, 9;
1, 34, 96, 64;
1, 125, 750, 1250, 625;
1, 461, 5265, 16470, 19440, 7776;
1, 1715, 35329, 184877, 386561, 352947, 117649;
1, 6434, 232288, 1913408, 6307840, 9863168, 7340032, 2097152; ...
PROG
(PARI) {T(n, k)=local(X=x+x*O(x^k)); polcoeff(sum(j=0, n, binomial(n+n*j+j, n*j+j)*(x/(1+X))^j)/(1+X), k)}
Main diagonal of table A060543; a(n) = C((n+1)^2-1, n*(n+1)).
+20
2
1, 3, 28, 455, 10626, 324632, 12271512, 553270671, 28987537150, 1731030945644, 116068178638776, 8634941152058949, 705873715441872264, 62895036884524942320, 6067037854078498539696, 629921975126394617164575, 70043473196734767582082230
PROG
(PARI) a(n)=binomial((n+1)^2-1, n*(n+1))
Antidiagonal sums of table A060543.
+20
2
1, 2, 5, 17, 72, 357, 2022, 12900, 91448, 711180, 6004981, 54619489, 531854438, 5515551251, 60642234815, 704106298738, 8603658260904, 110306422692488, 1479905106340895, 20727595895871297, 302423908621734606
FORMULA
a(n)=Sum_{k=0..n} C(n+(n-k)*k, (n-k)*k+k).
PROG
(PARI) a(n)=sum(k=0, n, binomial(n+(n-k)*k, (n-k)*k+k))
Centered 9-gonal (also known as nonagonal or enneagonal) numbers. Every third triangular number, starting with a(1)=1.
+10
57
1, 10, 28, 55, 91, 136, 190, 253, 325, 406, 496, 595, 703, 820, 946, 1081, 1225, 1378, 1540, 1711, 1891, 2080, 2278, 2485, 2701, 2926, 3160, 3403, 3655, 3916, 4186, 4465, 4753, 5050, 5356, 5671, 5995, 6328, 6670, 7021, 7381, 7750, 8128, 8515, 8911, 9316
COMMENTS
a(n) is congruent to 1 (mod 9) for all n. The sequence of digital roots of the a(n) is A000012(n). The sequence of units' digits of the a(n) is period 20: repeat [1, 0, 8, 5, 1, 6, 0, 3, 5, 6, 6, 5, 3, 0, 6, 1, 5, 8, 0, 1]. - Ant King, Jun 18 2012 Divide each side of any triangle ABC with area (ABC) into 2n + 1 equal segments by 2n points: A_1, A_2, ..., A_(2n) on side a, and similarly for sides b and c. If the hexagon with area (Hex(n)) delimited by AA_n, AA_(n+1), BB_n, BB_(n+1), CC_n and CC_(n+1) cevians, we have a(n+1) = (ABC)/(Hex(n)) for n >= 1, (see link with java applet). - Ignacio Larrosa Cañestro, Jan 02 2015; edited by Wolfdieter Lang, Jan 30 2015 For the case n = 1 see the link for Marion's Theorem (actually Marion Walter's Theorem, see the Cugo et al, reference). Also, the generalization considered here has been called there (Ryan) Morgan's Theorem. - Wolfdieter Lang, Jan 30 2015 Pollock states that every number is the sum of at most 11 terms of this sequence, but note that "1, 10, 28, 35, &c." has a typo (35 should be 55). - Michel Marcus, Nov 04 2017 a(n) is also the number of (nontrivial) paths as well as the Wiener sum index of the (n-1)-alkane graph. - Eric W. Weisstein, Jul 15 2021
FORMULA
a(n) = C(3*n, 3)/n = (3*n-1)*(3*n-2)/2 = A001504(n-1)/2. a(1-n) = a(n).
a(n) = C(n-1, 0) + 9*C(n-1, 1) + 9*C(n-1, 2); binomial transform of (1, 9, 9, 0, 0, 0, ...).
G.f.: x*(1 + 7*x + x^2)/(1-x)^3. (End)
a(n-1) = Pochhammer(4,3*n)/(Pochhammer(2,n)*Pochhammer(n+1,2*n)).
a(n-1) = 1/Hypergeometric([-3*n,3*n+3,1],[3/2,2],3/4). - Peter Luschny, Jan 09 2012 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = 2*a(n-1) - a(n-2) + 9.
Sum_{n>=1} 1/a(n) = 2*Pi/(3*sqrt(3)) = A248897. (End)
Sum_{n>=1} a(n)/n! = 11*e/2 - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 11/(2*e) - 1. (End)
a(n) = P(2*n,4)*P(3*n,3)/24 for n>=2, where P(s,k) = ((s - 2)*k^2 - (s - 4)*k)/2 is the k-th s-gonal number. - Lechoslaw Ratajczak, Jul 18 2021
MAPLE
H := n -> simplify(1/hypergeom([-3*n, 3*n+3, 1], [3/2, 2], 3/4)); A060544 := n -> H(n-1); seq( A060544(i), i=1..19); # Peter Luschny, Jan 09 2012
MATHEMATICA
Take[Accumulate[Range[150]], {1, -1, 3}] (* Harvey P. Dale, Mar 11 2013 *) LinearRecurrence[{3, -3, 1}, {1, 10, 28}, 50] (* Harvey P. Dale, Mar 11 2013 *)
PROG
(PARI) a(n)=(3*n-1)*(3*n-2)/2
(PARI) for (n=1, 100, write("b060544.txt", n, " ", (3*n - 1)*(3*n - 2)/2); ) \\ Harry J. Smith, Jul 06 2009 (GAP) List([1..50], n->(2*n-1)^2+(n-1)*n/2); # Muniru A Asiru, Mar 01 2019 (Sage) [(3*n-1)*(3*n-2)/2 for n in (1..50)] # G. C. Greubel, Mar 02 2019
EXTENSIONS
Formulas by Paul Berry corrected for offset 1 by Wolfdieter Lang, Jan 30 2015
Square array read by antidiagonals downwards: T(n,k) = (n*k)!/(k!^n*n!), (n>=1, k>=1), the number of ways of dividing nk labeled items into n unlabeled boxes with k items in each box.
+10
29
1, 1, 1, 1, 3, 1, 1, 10, 15, 1, 1, 35, 280, 105, 1, 1, 126, 5775, 15400, 945, 1, 1, 462, 126126, 2627625, 1401400, 10395, 1, 1, 1716, 2858856, 488864376, 2546168625, 190590400, 135135, 1, 1, 6435, 66512160, 96197645544, 5194672859376, 4509264634875, 36212176000, 2027025, 1
COMMENTS
The Copeland link gives the associations of this entry with the operator calculus of Appell Sheffer polynomials, the combinatorics of simple set partitions encoded in the Faa di Bruno formula for composition of analytic functions (formal Taylor series), the Pascal matrix, and the geometry of the n-dimensional simplices (hypertriangles, or hypertetrahedra). These, in turn, are related to simple instances of the application of the exponential formula / principle / schema giving the number of not-necessarily-connected objects composed from an ensemble of connected objects. - Tom Copeland, Jun 09 2021
FORMULA
T(n,k) = Product_{j=2..n} binomial(j*k-1,k-1). - M. F. Hasler, Aug 22 2014
EXAMPLE
Array begins:
1, 1, 1, 1, 1, 1, ...
1, 3, 10, 35, 126, 462, ...
1, 15, 280, 5775, 126126, 2858856, ...
1, 105, 15400, 2627625, 488864376, 96197645544, ...
1, 945, 1401400, 2546168625, 5194672859376, 11423951396577720, ...
...
MATHEMATICA
T[n_, k_] := (n*k)!/(k!^n*n!);
PROG
(PARI) { i=0; for (m=1, 20, for (n=1, m, k=m - n + 1; write("b060540.txt", i++, " ", (n*k)!/(k!^n*n!))); ) } \\ Harry J. Smith, Jul 06 2009
CROSSREFS
Cf. A000217, A000292, A000332, A000389, A000579, A000580, A007318, A036040, A099174, A133314, A132440, A135278 (associations in Copeland link).
a(n) = binomial(n^2, n)/n.
+10
9
1, 3, 28, 455, 10626, 324632, 12271512, 553270671, 28987537150, 1731030945644, 116068178638776, 8634941152058949, 705873715441872264, 62895036884524942320, 6067037854078498539696, 629921975126394617164575, 70043473196734767582082230
FORMULA
a(p^r) == 1 (mod p^(3+r)) for all positive integers r and all primes p >= 5 (apply Meštrović, Remark 17, p. 12).
Conjecture: a(2*p^r) == 4*p^r - 1 (mod p^(3+r)) for all positive integers r and all primes p >= 5. (End)
Triangle read by rows, T(n, k) = [x^k] (1-x)^(n+1)*Sum_{j=0..n} binomial(n + n*j + j, n*j + j)*x^j.
+10
9
1, 1, 1, 1, 7, 1, 1, 31, 31, 1, 1, 121, 381, 121, 1, 1, 456, 3431, 3431, 456, 1, 1, 1709, 26769, 60691, 26769, 1709, 1, 1, 6427, 193705, 848443, 848443, 193705, 6427, 1, 1, 24301, 1343521, 10350421, 19610233, 10350421, 1343521, 24301, 1
COMMENTS
G.f. of row n divided by (1-x)^(n+1) equals g.f. of row n of table A060543. Matrix product of this triangle with Pascal's triangle ( A007318) equals A108291. Seeing each row as a polynomial, all roots seem to be negative reals. - F. Chapoton, Nov 01 2022 Consider the set [m] := {1, 2, 3, ..., m} ordered cyclically, and then mapped into itself via f. Let us consider a in [m] as the (a-1)th m-th root of unity e^(2*Pi*i*(a-1)/m). Then f may be extended to a continuous map f':S^1 -> S^1 as follows:
For a immediately before b in the cyclic order, map the interval between a and b to S^1 so that a point in it moving clockwise at constant speed has a value moving clockwise at constant speed, and the map travels the shortest distance possible given this condition.
T(n, k) gives the number of f for m = n-1 such that f(1) = 1 and f' has degree k. This is trivially one n-th of the number of f with degree k when f(1) is arbitrary.
Equivalent to having degree k is that there are k values a immediately before b in the cyclic order such that f(a) > f(b) (in the standard order of N).
If we change things so that a immediately before b satisfies f(a) = f(b) corresponds to a full rotation (this is equivalent to using the condition f(a) >= f(b) in the last paragraph), then T(n, k) is the number of f with degree k+1.
T(n, k) is the (k+1)*(n-1)th (n-1)-nomial coefficient of power n - 1.
(End)
LINKS
M. Bayer, B. Goeckner, S. J. Hong, T. McAllister, M. Olsen, C. Pinckney, J. Vega and M. Yip, Lattice polytopes from Schur and symmetric Grothendieck polynomials, Electronic Journal of Combinatorics, Volume 28, Issue 2 (2021). See Proposition 53 and Table 1.
FORMULA
T(n, 1) = A048775(n) = binomial(2*n + 1, n + 1) - (n + 1). Sum_{k=0..n} T(n, k) = A000169(n) = (n + 1)^n. Sum_{k=0..n} T(n, k)*2^k = A108292(n). T(n, k) = Sum_{i=0..k} (-1)^i*binomial(n + 1, i)*binomial(n+(n+1)*(k-i), n).
T(n, k) = T(n, n-k).
(End)
EXAMPLE
Triangle begins:
1;
1, 1;
1, 7, 1;
1, 31, 31, 1;
1, 121, 381, 121, 1;
1, 456, 3431, 3431, 456, 1;
1, 1709, 26769, 60691, 26769, 1709, 1;
1, 6427, 193705, 848443, 848443, 193705, 6427, 1;
...
G.f. of row 3: (1 + 31*x + 31*x^2 + x^3) = (1-x)^4*(1 + 35*x + 165*x^2 + 455*x^3 + ... + C(4*j+3,4*j)*x^j + ...).
MAPLE
p := n -> (1-x)^(n+1)*add(binomial(n + n*j + j, n*j + j)*x^j, j = 0..n):
seq(print(seq(coeff(p(n), x, k), k = 0..n)), n = 0..8); # Peter Luschny, Nov 02 2022
MATHEMATICA
T[n_, k_] := Coefficient[(1 - x)^(n + 1)*
Sum[Binomial[n + n*j + j, n*j + j]*x^j, {j, 0, n}], x, k];
PROG
(PARI) T(n, k)=polcoeff((1-x)^(n+1)*sum(j=0, n, binomial(n+n*j+j, n*j+j)*x^j), k)
1, 3, 19, 195, 2751, 49413, 1079079, 27760323, 822299383, 27565191753, 1031671508495, 42643092165765, 1929325374428791, 94835735736471369, 5032700868665421519, 286770182910733076163, 17463186681730290301671
FORMULA
a(n) = Sum_{k=0..n} A108267(n, k)*2^k.
PROG
(PARI) a(n)=local(X=x+x*O(x^n)); sum(k=0, n, polcoeff(sum(j=0, n, binomial(n+n*j+j, n*j+j)*(x/(1+X))^j)/(1+X), k))
(PARI) a(n)=sum(k=0, n, 2^k*polcoeff( (1-x)^(n+1)*sum(j=0, n, binomial(n+n*j+j, n*j+j)*x^j), k))
Triangle, read by rows, resulting from the matrix product of triangle A108267 with Pascal's triangle ( A007318).
+10
2
1, 2, 1, 9, 9, 1, 64, 96, 34, 1, 625, 1250, 750, 125, 1, 7776, 19440, 16470, 5265, 461, 1, 117649, 352947, 386561, 184877, 35329, 1715, 1, 2097152, 7340032, 9863168, 6307840, 1913408, 232288, 6434, 1, 43046721, 172186884, 274223556, 220016574
EXAMPLE
Triangle begins:
1;
2,1;
9,9,1;
64,96,34,1;
625,1250,750,125,1;
7776,19440,16470,5265,461,1;
117649,352947,386561,184877,35329,1715,1;
2097152,7340032,9863168,6307840,1913408,232288,6434,1; ...
PROG
(PARI) {T(n, k)=local(X=x+x*O(x^(n-k))); polcoeff(sum(j=0, n, binomial(n+n*j+j, n*j+j)*(x/(1+X))^j)/(1+X), n-k)}
Triangle T(n,m) read by rows: The number of m-Schroeder paths of order n with 2 diagonal steps.
+10
2
1, 6, 21, 30, 180, 546, 140, 1430, 6120, 17710, 630, 10920, 65835, 245700, 695640, 2772, 81396, 690690, 3322704, 11515140, 32212719, 12012, 596904, 7125300, 44170896, 187336380, 619851960, 1721286532, 51480, 4326300, 72624816
COMMENTS
The case with 1 diagonal step is A060543.
REFERENCES
Chunwei Song, The Generalized Schroeder Theory, El. J. Combin. 12 (2005) #R53 Theorem 2.1.
FORMULA
T(n,m) = trinomial(m*n+n-2; m*n-2,n-2,2)/(m*n-1) .
EXAMPLE
This is the left-lower portion of the array which starts in row n=2, columns m>=1 as:
1, 2, 3, 4, 5, 6,..
6, 21, 45, 78, 120, 171, 231,.. # A08126630, 180, 546, 1224, 2310, 3900, 6090, 8976,.. # bisection A055112140, 1430, 6120, 17710, 40950, 81840, 147630, 246820, 389160,.. # 5-section A034827630, 10920, 65835, 245700, 695640, 1645020, 3426885, 6497400, ...
2772, 81396, 690690, 3322704, 11515140, 32212719, 77481495, ...
12012, 596904, 7125300, 44170896, 187336380, 619851960, ...
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