cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 10 results.

A108290 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).

Original entry on oeis.org

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
Offset: 0

Views

Author

Paul D. Hanna, May 31 2005

Keywords

Comments

Row sums form A108292. Main diagonal is A000169(n) = (n+1)^n. Triangle with rows reversed is A108291.
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

Examples

			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; ...
		

Crossrefs

Programs

  • 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)}

A108288 Main diagonal of table A060543; a(n) = C((n+1)^2-1, n*(n+1)).

Original entry on oeis.org

1, 3, 28, 455, 10626, 324632, 12271512, 553270671, 28987537150, 1731030945644, 116068178638776, 8634941152058949, 705873715441872264, 62895036884524942320, 6067037854078498539696, 629921975126394617164575, 70043473196734767582082230
Offset: 0

Views

Author

Paul D. Hanna, May 31 2005

Keywords

Crossrefs

Programs

  • PARI
    a(n)=binomial((n+1)^2-1,n*(n+1))

Formula

a(n) = A060545(n+1). - R. J. Mathar, Aug 24 2008

A108289 Antidiagonal sums of table A060543.

Original entry on oeis.org

1, 2, 5, 17, 72, 357, 2022, 12900, 91448, 711180, 6004981, 54619489, 531854438, 5515551251, 60642234815, 704106298738, 8603658260904, 110306422692488, 1479905106340895, 20727595895871297, 302423908621734606
Offset: 0

Views

Author

Paul D. Hanna, May 31 2005

Keywords

Crossrefs

Programs

  • PARI
    a(n)=sum(k=0,n,binomial(n+(n-k)*k,(n-k)*k+k))

Formula

a(n)=Sum_{k=0..n} C(n+(n-k)*k, (n-k)*k+k).

A060544 Centered 9-gonal (also known as nonagonal or enneagonal) numbers. Every third triangular number, starting with a(1)=1.

Original entry on oeis.org

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
Offset: 1

Views

Author

Henry Bottomley, Apr 02 2001

Keywords

Comments

Triangular numbers not == 0 (mod 3). - Amarnath Murthy, Nov 13 2005
Shallow diagonal of triangular spiral in A051682. - Paul Barry, Mar 15 2003
Equals the triangular numbers convolved with [1, 7, 1, 0, 0, 0, ...]. - Gary W. Adamson & Alexander R. Povolotsky, May 29 2009
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

Crossrefs

Programs

  • GAP
    List([1..50],n->(2*n-1)^2+(n-1)*n/2); # Muniru A Asiru, Mar 01 2019
    
  • Magma
    [(2*n-1)^2+(n-1)*n/2: n in [1..50]]; // Vincenzo Librandi, Nov 18 2015
    
  • 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 *)
    FoldList[#1 + #2 &, 1, 9 Range @ 50] (* Robert G. Wilson v, Feb 02 2011 *)
    Table[(3 n - 1) (3 n - 2)/2, {n, 20}] (* Eric W. Weisstein, Jul 15 2021 *)
    Table[Binomial[3 n - 1, 2], {n, 20}] (* Eric W. Weisstein, Jul 15 2021 *)
    Table[PolygonalNumber[3 n - 2], {n, 20}] (* Eric W. Weisstein, Jul 15 2021 *)
  • PARI
    a(n)=(3*n-1)*(3*n-2)/2
    
  • Sage
    [(3*n-1)*(3*n-2)/2 for n in (1..50)] # G. C. Greubel, Mar 02 2019

Formula

a(n) = C(3*n, 3)/n = (3*n-1)*(3*n-2)/2 = A001504(n-1)/2.
a(n) = a(n-1) + 9*(n-1) = A060543(n, 3) = A006566(n)/n.
a(n) = A025035(n)/A025035(n-1) = A027468(n-1) + 1 = A000217(3*n-2).
a(1-n) = a(n).
From Paul Barry, Mar 15 2003: (Start)
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, ...).
a(n) = 9*A000217(n-1) + 1.
G.f.: x*(1 + 7*x + x^2)/(1-x)^3. (End)
Narayana transform (A001263) of [1, 9, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
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
From Ant King, Jun 18 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = 2*a(n-1) - a(n-2) + 9.
a(n) = A000217(n) + 7*A000217(n-1) + A000217(n-2).
Sum_{n>=1} 1/a(n) = 2*Pi/(3*sqrt(3)) = A248897.
(End)
a(n) = (2*n-1)^2 + (n-1)*n/2. - Ivan N. Ianakiev, Nov 18 2015
a(n) = A101321(9,n-1). - R. J. Mathar, Jul 28 2016
E.g.f.: (2 + 9*x^2)*exp(x)/2 - 1. - G. C. Greubel, Mar 02 2019
From Amiram Eldar, Jun 20 2020: (Start)
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) = A000567(n) + A005449(n-1) (see illustration in links). - John Elias, Nov 10 2020
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

Extensions

Additional description from Terrel Trotter, Jr., Apr 06 2002
Formulas by Paul Berry corrected for offset 1 by Wolfdieter Lang, Jan 30 2015

A060540 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.

Original entry on oeis.org

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
Offset: 1

Views

Author

Henry Bottomley, Apr 02 2001

Keywords

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

Examples

			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, ...
  ...
		

Crossrefs

Main diagonal is A057599.
Related to A057599, see also A096126 and A246048.
Cf. A060358, A361948 (includes row/col 0).
Cf. A000217, A000292, A000332, A000389, A000579, A000580, A007318, A036040, A099174, A133314, A132440, A135278 (associations in Copeland link).

Programs

  • Mathematica
    T[n_, k_] := (n*k)!/(k!^n*n!);
    Table[T[n-k+1, k], {n, 1, 10}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jun 29 2018 *)
  • 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

Formula

T(n,k) = (n*k)!/(k!^n*n!) = T(n-1,k)*A060543(n,k) = A060538(n,k)/k!.
T(n,k) = Product_{j=2..n} binomial(j*k-1,k-1). - M. F. Hasler, Aug 22 2014

Extensions

Definition reworded by M. F. Hasler, Aug 23 2014

A060545 a(n) = binomial(n^2, n)/n.

Original entry on oeis.org

1, 3, 28, 455, 10626, 324632, 12271512, 553270671, 28987537150, 1731030945644, 116068178638776, 8634941152058949, 705873715441872264, 62895036884524942320, 6067037854078498539696, 629921975126394617164575, 70043473196734767582082230
Offset: 1

Views

Author

Henry Bottomley, Apr 02 2001

Keywords

Crossrefs

Programs

Formula

a(n) = A060543(n, n) = A014062(n)/n.
a(n+1) = C(A005563(n), n) for n >= 0. - Fred Daniel Kline, Sep 27 2016
From Peter Bala, Oct 22 2023: (Start)
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)

Extensions

More terms from Fred Daniel Kline, Sep 28 2016

A108267 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.

Original entry on oeis.org

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
Offset: 0

Views

Author

Paul D. Hanna, May 29 2005 and May 31 2005

Keywords

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
From Thomas Anton, Jan 05 2023: (Start)
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)

Examples

			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 + ...).
		

Crossrefs

Programs

  • 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];
    Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 23 2021 *)
  • PARI
    T(n,k)=polcoeff((1-x)^(n+1)*sum(j=0,n,binomial(n+n*j+j,n*j+j)*x^j),k)

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).
From Thomas Anton, Jan 05 2023: (Start)
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)

A108292 Row sums of triangle A108290.

Original entry on oeis.org

1, 3, 19, 195, 2751, 49413, 1079079, 27760323, 822299383, 27565191753, 1031671508495, 42643092165765, 1929325374428791, 94835735736471369, 5032700868665421519, 286770182910733076163, 17463186681730290301671
Offset: 0

Views

Author

Paul D. Hanna, May 31 2005

Keywords

Crossrefs

Programs

  • 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))

Formula

a(n) = Sum_{k=0..n} A108267(n, k)*2^k.

A108291 Triangle, read by rows, resulting from the matrix product of triangle A108267 with Pascal's triangle (A007318).

Original entry on oeis.org

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
Offset: 0

Views

Author

Paul D. Hanna, May 31 2005

Keywords

Comments

Row sums form A108292. Column 0 is A000169(n) = (n+1)^n. Triangle with rows reversed is A108290.

Examples

			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; ...
		

Crossrefs

Programs

  • 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)}

A173622 Triangle T(n,m) read by rows: The number of m-Schroeder paths of order n with 2 diagonal steps.

Original entry on oeis.org

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
Offset: 2

Views

Author

R. J. Mathar, Nov 08 2010

Keywords

Comments

The case with 1 diagonal step is A060543.

Examples

			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,.. # A081266
30, 180, 546, 1224, 2310, 3900, 6090, 8976,.. # bisection A055112
140, 1430, 6120, 17710, 40950, 81840, 147630, 246820, 389160,.. # 5-section A034827
630, 10920, 65835, 245700, 695640, 1645020, 3426885, 6497400, ...
2772, 81396, 690690, 3322704, 11515140, 32212719, 77481495, ...
12012, 596904, 7125300, 44170896, 187336380, 619851960, ...
		

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) .
Showing 1-10 of 10 results.