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 1-8 of 8 results.

A143858 Number of pairwise disjoint unions of m integer-to-integer subintervals of [0,n]; a rectangular array by antidiagonals, n>=2m-1, m>=1.

Original entry on oeis.org

1, 3, 1, 6, 5, 1, 10, 15, 7, 1, 15, 35, 28, 9, 1, 21, 70, 84, 45, 11, 1, 28, 126, 210, 165, 66, 13, 1, 36, 210, 462, 495, 286, 91, 15, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 55, 495, 1716, 3003, 3003, 1820, 680, 153, 19, 1, 66, 715, 3003, 6435, 8008, 6188, 3060
Offset: 1

Views

Author

Clark Kimberling, Sep 03 2008

Keywords

Comments

Main diagonal: A025174.

Examples

			R(2,4) counts these unions of 2 subintervals of [0,4]: [0,1]U[2,3], [0,1]U[2,4], [0,1]U[3,4], [0,2]U[3,4], [1,2]U[3,4].
   1    3    6   10   15   21   28   36   45   55   66   78
   0    0    1    5   15   35   70  126  210  330  495  715
   0    0    0    0    1    7   28   84  210  462  924 1716
   0    0    0    0    0    0    1    9   45  165  495 1287
   0    0    0    0    0    0    0    0    1   11   66  286
   0    0    0    0    0    0    0    0    0    0    1   13
		

Crossrefs

Programs

  • Haskell
    Seen as a triangle read by rows
    a143858 n k = a143858_tabl !! (n-1) !! k
    a143858_row n = a143858_tabl !! (n-1)
    a143858_tabl = map ((++ [1]) . tail) a258993_tabl
    -- Reinhard Zumkeller, Jun 22 2015
  • Maple
    A143858 := proc(m,n)
        binomial(n-1+2*m,2*m) ;
    end proc:
    seq(seq( A143858(n,d-n),n=1..d-1),d=2..8) ; # R. J. Mathar, Nov 16 2023

Formula

R(m,n) = C(n+1,2m), where n>=2m-1, m>=1. R is also given by the absolute values of terms in A109954.

A050155 Triangle T(n,k), k>=0 and n>=1, read by rows defined by: T(n,k) = (2k+3)*binomial(2n,n-k-1)/(n+k+2).

Original entry on oeis.org

1, 3, 1, 9, 5, 1, 28, 20, 7, 1, 90, 75, 35, 9, 1, 297, 275, 154, 54, 11, 1, 1001, 1001, 637, 273, 77, 13, 1, 3432, 3640, 2548, 1260, 440, 104, 15, 1, 11934, 13260, 9996, 5508, 2244, 663, 135, 17, 1, 41990, 48450, 38760, 23256, 10659, 3705, 950, 170, 19, 1
Offset: 1

Views

Author

Keywords

Comments

T(n-2k-1,k) = number of n-th generation vertices in the tree of sequences with unit increase labeled by 2k+2 (cf. Zoran Sunic reference) . - Benoit Cloitre, Oct 07 2003
Number of lattice paths from (0,0) to (n,n) with steps E=(1,0) and N=(0,1) which touch but do not cross the line x-y=k+1 . - Herbert Kociemba, May 24 2004
Number of standard tableaux of shape (n+k+1, n-k-1). - Emeric Deutsch, May 30 2004
Riordan array (c(x)^3,xc(x)^2) where c(x) is the g.f. of A000108. Inverse array is A109954. - Paul Barry, Jul 06 2005

Examples

			    1;
    3,   1;
    9,   5,   1;
   28,  20,   7,  1;
   90,  75,  35,  9,  1;
  297, 275, 154, 54, 11, 1;
  ...
		

Crossrefs

Cf. A000108, A001791 (row sums), A050144.

Programs

  • Maple
    T:= (n, k)->  (2*k+3)*binomial(2*n, n-k-1)/(n+k+2):
    seq(seq(T(n, k), k=0..n-1), n=1..10);  # Alois P. Heinz, Jan 19 2013
  • Mathematica
    T[n_, k_] :=  (2*k + 3)*Binomial[2*n, n - k - 1]/(n + k + 2);
    Table[T[n, k], {n, 1, 10}, {k, 0, n - 1}] // Flatten (* Jean-François Alcover, May 21 2016 *)

Formula

Sum_{ k = 0, .., n-1} T(n, k) = binomial(2n, n-1) = A001791(n).
G.f. of column k: x^(k+1)*C^(2*k+3) where C = (1-(1-4*x)^(1/2))/(2*x) is the g.f. of Catalan numbers A000108. - Philippe Deléham, Feb 03 2004
T(n, k) = A039599(n, k+1) = A009766(n+k+1, n-k-1) = A033184(n+k+2, 2k+3) . - Philippe Deléham, May 28 2005
Sum_{k>= 0} T(m, k)*T(n, k) = A000108(m+n) - A000108(m)*A000108(n). - Philippe Deléham, May 28 2005
T(n, k)=(2k+3)binomial(2n+2, n+k+2)/(n+k+3)=C(2n+2, n+k+2)-C(2n+2, n+k+3) [offset (0, 0)]. - Paul Barry, Jul 06 2005

Extensions

Edited by Philippe Deléham, May 22 2005

A123970 Triangle read by rows: T(0,0)=1; T(n,k) is the coefficient of x^(n-k) in the monic characteristic polynomial of the n X n matrix (min(i,j)) (i,j=1,2,...,n) (0 <= k <= n, n >= 1).

Original entry on oeis.org

1, 1, -1, 1, -3, 1, 1, -6, 5, -1, 1, -10, 15, -7, 1, 1, -15, 35, -28, 9, -1, 1, -21, 70, -84, 45, -11, 1, 1, -28, 126, -210, 165, -66, 13, -1, 1, -36, 210, -462, 495, -286, 91, -15, 1, 1, -45, 330, -924, 1287, -1001, 455, -120, 17, -1, 1, -55, 495, -1716, 3003, -3003, 1820, -680, 153, -19, 1, 1, -66, 715, -3003, 6435, -8008
Offset: 0

Views

Author

Gary W. Adamson and Roger L. Bagula, Oct 29 2006

Keywords

Comments

This sequence is the same as A129818 up to sign. - T. D. Noe, Sep 30 2011
Riordan array (1/(1-x), -x/(1-x)^2). - Philippe Deléham, Feb 18 2012

Examples

			Triangular sequence (gives the odd Tutte-Beraha constants as roots!) begins:
  1;
  1,  -1;
  1,  -3,   1;
  1,  -6,   5,   -1;
  1, -10,  15,   -7,    1;
  1, -15,  35,  -28,    9,    -1;
  1, -21,  70,  -84,   45,   -11,   1;
  1, -28, 126, -210,  165,   -66,  13,   -1;
  1, -36, 210, -462,  495,  -286,  91,  -15,  1;
  1, -45, 330, -924, 1287, -1001, 455, -120, 17, -1;
  ...
		

References

  • S. Beraha, Infinite non-trivial families of maps and chromials, Ph.D. thesis. Baltimore, MD: Johns Hopkins University, 1975.
  • Steven R. Finch, Mathematical Constants (Encyclopedia of Mathematics and its Applications), chapter 5.25.
  • W. T. Tutte, "More about Chromatic Polynomials and the Golden Ratio." In Combinatorial Structures and their Applications: Proc. Calgary Internat. Conf., Calgary, Alberta, 1969. New York: Gordon and Breach, p. 439, 1969.

Crossrefs

Cf. A109954, A129818, A143858, A165253. - R. J. Mathar, Jan 10 2011
Modulo signs, inverse matrix to A039599.

Programs

  • Magma
    /* As triangle */ [[(-1)^k*Binomial(n + k, 2*k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jan 04 2019
  • Maple
    with(linalg): m:=(i,j)->min(i,j): M:=n->matrix(n,n,m): T:=(n,k)->coeff(charpoly(M(n),x),x,n-k): 1; for n from 1 to 11 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
  • Mathematica
    An[d_] := MatrixPower[Table[Min[n, m], {n, 1, d}, {m, 1, d}], -1]; Join[{{1}}, Table[CoefficientList[CharacteristicPolynomial[An[d], x], x], {d, 1, 20}]]; Flatten[%]

Formula

f(n,x) = (2x-1)f(n-1,x)-x^2*f(n-2,x), where f(n,x) is the characteristic polynomial of the n X n matrix from the definition and f(0,x)=1. See formula in Fendley and Krushkal. - Jonathan Vos Post, Nov 04 2007
T(n,k) = (-1)^k * A085478(n,k) = (-1)^n * A129818(n,k). - Philippe Deléham, Feb 06 2012
T(n,k) = 2*T(n-1,k) - T(n-1,k-1) - T(n-2,k), T(0,0)=T(1,0)=1, T(1,1)=-1, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 29 2013

Extensions

Edited by N. J. A. Sloane, Nov 29 2006

A220670 Coefficient triangle for powers of x^2 of polynomials appearing in a generalized Melham conjecture on alternating sums of third powers of Chebyshev's S polynomials with odd indices. Coefficients in powers of x^2 of 2 + (-1)^n*S(2*n,x).

Original entry on oeis.org

3, 3, -1, 3, -3, 1, 3, -6, 5, -1, 3, -10, 15, -7, 1, 3, -15, 35, -28, 9, -1, 3, -21, 70, -84, 45, -11, 1, 3, -28, 126, -210, 165, -66, 13, -1, 3, -36, 210, -462, 495, -286, 91, -15, 1, 3, -45, 330, -924, 1287, -1001, 455, -120, 17, -1, 3, -55, 495, -1716, 3003, -3003, 1820, -680, 153, -19, 1
Offset: 0

Views

Author

Wolfdieter Lang, Jan 07 2013

Keywords

Comments

For the original Melham conjecture on sums of odd powers of even-indexed Fibonacci numbers see the references given in A217475. See especially the Wang and Zhang reference given there.
An analog conjecture stated for Chebyshev's S polynomials (see A049310) is product(tau(j,x),j=0..m)*sum(((-1)^k)*(S(2*k-1,x)/x)^(2*m+1),k=0..n)/(P(n,x^2)^2) = H(m,n,x^2), with P(n,x^2) := (1 - (-1)^n*S(2*n,x))/x^2 and certain integer polynomials H with degree (2*m-1)*n + binomial(m-1,2) in x^2. The coefficients of powers of x^2 of the monic integer polynomials tau(n,x):= 2*T(2*n+1,x/2)/x, with Chebyshev's T polynomials, are given by the signed A111125 triangle (see a comment there from Oct 23 2012). The coefficients of the powers of x^(2*j) of the polynomials P(n,x^2) are found in (-1)^(n-1)*A109954(n-1,j).
Here the conjecture is considered for m=1 (third powers): H(1,n,x^2) = sum(a(n,p)*x^(2*p),p=0..n), n >= 1. It is conjectured that in fact H(1,n,x^2) = 2 + (-1)^n*S(2*n,x). This has been checked by Maple for n=1..100. Therefore we have added a(0,0) = 3 (in the conjecture above this would be the undetermined 0/0).
The original Melham conjecture for m=1 (third powers), appears by putting x = i (the imaginary unit): 1*4*sum(F(2*k)^3)/(1-F(2*n+1))^2 = sum(a(n,p)*(-1)^p) = 2 + F(2*n+1) (the unsigned row sums of the present triangle). This m=1 identity is, of course, proved.
The row sums of this triangle are given by 2 + (-1)^n*S(2*n,1) = 2 + (-1)^n*((2/sqrt(3))*sin((2*n+1)*Pi/3)) producing the period 6 sequence periodic (3, 2, 1, 1, 2, 3).

Examples

			The triangle a(n,p) begins:
n\p 0    1    2     3    4     5    6    7   8   9 10 ...
0:  3
1:  3   -1
2:  3   -3    1
3:  3   -6    5    -1
4:  3  -10   15    -7    1
5:  3  -15   35   -28    9    -1
6:  3  -21   70   -84   45   -11    1
7:  3  -28  126  -210  165   -66   13  -1
8:  3  -36  210  -462  495  -286   91  -15   1
9:  3  -45  330  -924 1287 -1001  455 -120  17  -1
10: 3  -55  495 -1716 3003 -3003 1820 -680 153 -19  1
...
Row n=2: H(1,2,x^2) := (-3+x^2)*(0 - (S(1,x)/x)^3 + (S(3,x)/x)^3)/((1 - S(4,x))/x^2)^2 = 3 - 3*x^2 + x^4 =
  2 + S(4,x).
Row n=3:  H(1,3,x^2) := (-3+x^2)*(0 - (S(1,x)/x)^3 + (S(3,x)/x)^3 - (S(5,x)/x)^3 )/((1 + S(6,x))/x^2)^2 =  3-6*x^2+5*x^4-x^6 = 2 - S(6,x).
		

Crossrefs

Cf, A049310, A111125 (signed), A109954 (signed), A217475, A220671 (fifth powers).

Formula

a(n,p) = [x^(2*p)] H(1,n,x^2), with H(1,n,x^2) := (-3+x^2)*sum(((-1)^k)*(S(2*k-1,x)/x)^3,k=0..n)/((1 - (-1)^n*S(2*n,x))/x^2)^2, n >= 1, p = 0..n, and a(0,0):=3.
a(n,p) = [x^(2*p)] (2 + (-1)^n*S(2*n,x)), n >= 0, p = 0..n.

A092879 Triangle of coefficients of the product of two consecutive Fibonacci polynomials.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 5, 7, 2, 1, 7, 16, 13, 3, 1, 9, 29, 40, 22, 3, 1, 11, 46, 91, 86, 34, 4, 1, 13, 67, 174, 239, 166, 50, 4, 1, 15, 92, 297, 541, 553, 296, 70, 5, 1, 17, 121, 468, 1068, 1461, 1163, 496, 95, 5, 1, 19, 154, 695, 1912, 3300, 3544, 2269, 791, 125, 6, 1, 21, 191
Offset: 0

Views

Author

Michael Somos, Mar 10 2004

Keywords

Comments

The Fibonacci polynomials are defined by F(0,x) = 1, F(1,x) = 1 and F(n, x) = F(n-1, x) + x*F(n-2, x).
This is also the reflected triangle of coefficients of the polynomials defined by the recursion: c0=-1; p(x, n) = (2 + c0 - x)*p(x, n - 1) + (-1 - c0*(2 - x))*p(x, n - 2) + c0*p(x, n - 3). - Roger L. Bagula, Apr 09 2008

Examples

			Triangle begins;
  1;
  1,1;
  1,3,2;
  1,5,7,2;
  1,7,16,13,3;
  1,9,29,40,22,3;
  ...
F(3,x) = 1 + 2*x and F(4,x) = 1 + 3*x + x^2 so F(3,x)*F(4,x)=(1 + 3*x + x^2)*(1 + 2*x) = 1 + 5*x + 7*x^2 + 2*x^3 leads to T(3,k) = [1,5,7,2].
		

Crossrefs

Row sums are A001654(n+1).

Programs

  • Maple
    T:=proc(n,k): add((-1)^(i+k)*binomial(i+2*n-2*k+1,i), i=0..k) end: seq(seq(T(n,k), k=0..n), n=0..10); # Johannes W. Meijer, Jul 20 2011
    T:=proc(n,k): coeff(F(n, x)*F(n+1, x), x, k) end: F:=proc(n, x) option remember: if n=0 then 1 elif n=1 then 1 else procname(n-1, x) + x*procname(n-2, x) fi: end: seq(seq(T(n,k), k=0..n), n=0..10); # Johannes W. Meijer, Jul 20 2011
  • Mathematica
    c0 = -1; p[x, -1] = 0; p[x, 0] = 1; p[x, 1] = 2 - x + c0; p[x_, n_] :=p[x, n] = (2 + c0 -x)*p[x, n - 1] + (-1 - c0 (2 - x))*p[x, n - 2] + c0*p[x, n - 3]; Table[ExpandAll[p[x, n]], {n, 0, 10}]; a = Table[Reverse[CoefficientList[p[x, n], x]], {n, 0, 10}]; Flatten[a] (* Roger L. Bagula, Apr 09 2008 *)
  • PARI
    T(n,k)=local(m);if(k<0 || k>n,0,n++; m=contfracpnqn(matrix(2,n,i,j,x)); polcoeff(m[1,1]*m[2,1]/x^n,n-k))

Formula

From Johannes W. Meijer, Jul 20 2011: (Start)
T(n, k) = Sum_{i=0..k} (-1)^(i+k)*binomial(i+2*n-2*k+1, i).
T(n, k) = A035317(2*n-k, k) = A158909(n, n-k.) (End)
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k-1) + T(n-2,k-2) - T(n-3,k-3), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 12 2013

Extensions

Edited and information added by Johannes W. Meijer, Jul 20 2011

A186024 Inverse of eigentriangle of triangle A085478.

Original entry on oeis.org

1, -1, 1, -1, -1, 1, -1, -3, -1, 1, -1, -6, -5, -1, 1, -1, -10, -15, -7, -1, 1, -1, -15, -35, -28, -9, -1, 1, -1, -21, -70, -84, -45, -11, -1, 1, -1, -28, -126, -210, -165, -66, -13, -1, 1, -1, -36, -210, -462, -495, -286, -91, -15, -1, 1, -1, -45, -330, -924, -1287, -1001, -455, -120, -17, -1, 1
Offset: 0

Views

Author

Paul Barry, Feb 10 2011

Keywords

Comments

Row sums are A186025.

Examples

			Triangle begins
1,
-1, 1,
-1, -1, 1,
-1, -3, -1, 1,
-1, -6, -5, -1, 1,
-1, -10, -15, -7, -1, 1,
-1, -15, -35, -28, -9, -1, 1,
-1, -21, -70, -84, -45, -11, -1, 1,
-1, -28, -126, -210, -165, -66, -13, -1, 1,
-1, -36, -210, -462, -495, -286, -91, -15, -1, 1,
-1, -45, -330, -924, -1287, -1001, -455, -120, -17, -1, 1
		

Crossrefs

Formula

T(n,k)=if(k

A220671 Coefficient array for powers of x^2 of polynomials appearing in a generalized Melham conjecture on alternating sums of fifth powers of Chebyshev S polynomials with odd indices.

Original entry on oeis.org

-14, 15, -20, 8, -1, 55, -170, 221, -153, 59, -12, 1, 115, -670, 1773, -2696, 2549, -1538, 589, -138, 18, -1, 195, -1850, 8215, -21530, 36330, -41110, 31865, -17080, 6314, -1579, 255, -24, 1, 295, -4150, 27735, -110795, 289540, -518290, 654595, -595805, 396316, -193906, 69641, -18129, 3327, -408, 30, -1
Offset: 1

Author

Wolfdieter Lang, Jan 11 2013

Keywords

Comments

The row lengths sequence is 3*n + 1 = A016777(n).
For the generalized Melham conjecture and links to the references concerned with the Melham conjecture on sums of fifth powers of even-indexed Fibonacci numbers see a comment under A220670.
Here the conjecture is considered for m=2 (fifth powers): H(2,n,x^2):= product(tau(j,x), j=0..2) * sum(((-1)^k)*(S(2*k-1,x)/x)^5, k=0..n) / (P(n,x^2)^2), with P(n,x^2):= (1 - (-1)^n*S(2*n,x))/x^2. For tau(j,x):= 2*T(2*j+1,x/2)/x, with Chebyshev's T polynomials see a Oct 23 2012 comment on A111125. For the polynomials P see signed A109954. The conjecture is that H(2,n,x^2) is an integer polynomial of degree 3*n: H(2,n,x^2) = sum(a(n,p)*x^(2*p), p=0..3*n), n >= 1.
If one puts x = i (the imaginary unit) one obtains the original Melham conjecture for Fibonacci numbers F = A000045.
H(2,n,-1) = +44*sum(F(2*k)^5,k=0..n)/(1+F(2*n+1))^2, n>=1, which is conjectured to be -14 - 3*y(n) + 8*y(n)^2 + 4*y(n)^3, with y(n):=F(2*n+1) (see row m=2 of A217475).
It is conjectured that H(2,n,x^2) = h(2,n,x^2) - 3*z(n) + 8*z(n)^2 + 4*z(n)^3, with z(n):= ((-1)^n)*S(2*n,x), with h an integer polynomial of degree 3*n. See A220672 for the coefficients of h(2,n,x^2) for n = 1..5. Because h(2,n,-1) = -14 by the usual Melham conjecture, we put h(2,0,x^2) = -14.

Examples

			The array a(n,p) begins:
  n\p   0     1     2      3      4      5     6      7   8     9   10 11 12
  0:  -14
  1:   15   -20     8     -1
  2:   55  -170   221   -153     59    -12     1
  3:  115  -670  1773  -2696   2549  -1538   589   -138   18    -1
  4:  195 -1850  8215 -21530  36330 -41110 31865 -17080 6314 -1579 255 -24 1
...
Row n=5: [295, -4150, 27735, -110795, 289540, -518290, 654595, -595805, 396316, -193906, 69641, -18129, 3327, -408, 30, -1],
Row n=6: [415, -8120, 76118, -429531, 1599441, -4125672, 7621983, -10350335, 10539787, -8164410, 4853792, -2222153, 781514, -209172, 41823, -6047, 597, -36, 1].
		

Crossrefs

Formula

a(n,p) = [x^(2*p)] H(2,n,x^2), n>=1, with H(2,n,x^2) defined in a comment above. a(0,0) has been put to -14 ad hoc.

A135552 Riordan array (1/((1-2x)(1-x)^2), -x/(1-x)^2).

Original entry on oeis.org

1, 4, -1, 11, -6, 1, 26, -22, 8, -1, 57, -64, 37, -10, 1, 120, -163, 130, -56, 12, -1, 247, -382, 386, -232, 79, -14, 1, 502, -848, 1024, -794, 378, -106, 16, -1, 1013, -1816, 2510, -2380, 1471, -576, 137, -18, 1, 2036, -3797, 5812, -6476, 4944, -2517, 834, -172, 20, -1, 4083, -7814, 12911, -16384, 14893, -9402, 4048, -1160, 211, -22, 1
Offset: 1

Author

Roger L. Bagula, Apr 08 2008

Keywords

Comments

Row sums are A024495. First column is essentially A000295, second column is essentially -A002663. - Ralf Stephan, Jan 03 2014

Examples

			{1},
{4, -1},
{11, -6, 1},
{26, -22, 8, -1},
{57, -64, 37, -10, 1},
{120, -163, 130, -56, 12, -1},
{247, -382, 386, -232, 79, -14, 1},
{502, -848, 1024, -794, 378, -106, 16, -1},
{1013, -1816, 2510, -2380, 1471, -576, 137, -18, 1},
{2036, -3797, 5812, -6476, 4944, -2517, 834, -172, 20, -1},
{4083, -7814, 12911, -16384, 14893, -9402, 4048, -1160, 211, -22, 1}
		

Crossrefs

Programs

  • Mathematica
    Clear[p, x, n, a]; p[x, -1] = 0; p[x, 0] = 1; p[x, 1] = 4 - x; p[x_, n_] := p[x, n] = (4 - x)*p[x, n - 1] + (-5 + 2*x)*p[x, n - 2] + 2*p[x, n - 3]; Table[ExpandAll[p[x, n]], {n, 0, 10}]; a = Table[CoefficientList[p[x, n], x], {n, 0, 10}]; Flatten[a]

Formula

Coefficients of polynomials defined by recurrence: c0=2; p(x, n) = (2 + c0 - x)*p(x, n - 1) + (-1 - c0 (2 - x))*p(x, n - 2) + c0*p(x, n - 3).

Extensions

New name from Ralf Stephan, Jan 03 2014
Showing 1-8 of 8 results.