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.

Previous Showing 11-20 of 30 results. Next

A228576 A triangle formed like generalized Pascal's triangle. The rule is T(n,k) = 2*T(n-1,k-1) + T(n-1,k), the left border is n and the right border is n^2 instead of 1.

Original entry on oeis.org

0, 1, 1, 2, 3, 4, 3, 7, 10, 9, 4, 13, 24, 29, 16, 5, 21, 50, 77, 74, 25, 6, 31, 92, 177, 228, 173, 36, 7, 43, 154, 361, 582, 629, 382, 49, 8, 57, 240, 669, 1304, 1793, 1640, 813, 64, 9, 73, 354, 1149, 2642, 4401, 5226, 4093, 1690, 81, 10, 91, 500, 1857, 4940, 9685, 14028, 14545, 9876, 3461, 100
Offset: 1

Views

Author

Boris Putievskiy, Aug 26 2013

Keywords

Examples

			The start of the sequence as triangle array read by rows:
  0;
  1,  1;
  2,  3,  4;
  3,  7, 10,  9;
  4, 13, 24, 29, 16;
  5, 21, 50, 77, 74, 25;
...
		

Crossrefs

Cf. We denote generalized Pascal's like triangle with coefficients a, b and with L(n) on the left border and R(n) on the right border by (a,b,L(n),R(n)). The list of sequences for (1,1,L(n),R(n)) see A228196;
A038207 (1,2,2^n,1), A105728 (1, 2, 1, n+1), A112468 (1,-1,1,1), A112626 (1,2,3^n,1), A119258 (2,1,1,1), A119673 (3,1,1,1), A119725 (3,2,1,1), A119726 (4,2,1,1), A119727 (5,2,1,1), A209705 (2,1,n+1,0);
A002061 (column 2), A000244 (sums of rows r of triangle array - (r-2)(r+1)/2).

Programs

  • GAP
    T:= function(n,k)
        if k=0 then return n;
        elif k=n then return n^2;
        else return 2*T(n-1,k-1) + T(n-1,k);
        fi;
      end;
    Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 13 2019
  • Magma
    function T(n,k)
      if k eq 0 then return n;
      elif k eq n then return n^2;
      else return 2*T(n-1,k-1) + T(n-1,k);
      end if;
      return T;
    end function;
    [T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 13 2019
    
  • Maple
    T := proc(n, k) option remember;
    if k = 0 then RETURN(n) fi;
    if k = n then RETURN(n^2) fi;
    2*T(n-1, k-1) + T(n-1, k) end:
    seq(seq(T(n,k),k=0..n),n=0..9);  # Peter Luschny, Aug 26 2013
  • Mathematica
    T[n_, 0]:= n; T[n_, n_]:= n^2; T[n_, k_]:= T[n, k] = 2*T[n-1, k-1]+T[n-1, k]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 25 2014 *)
  • PARI
    T(n,k) = if(k==0, n, if(k==n, n^2, 2*T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 13 2019
    
  • Sage
    @CachedFunction
    def T(n, k):
        if (k==0): return n
        elif (k==n): return n^2
        else: return 2*T(n-1,k-1) + T(n-1, k)
    [[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 13 2019
    

Formula

T(n, k) = 2*T(n-1, k-1) + T(n-1, k) for n,k >=0, with T(n,0) = n, T(n,n) = n^2.
Closed-form formula for generalized Pascal's triangle. Let a,b be any numbers. The rule is T(n, k) = a*T(n-1, k-1) + b*T(n-1, k) for n,k >0. Let L(m) and R(m) be the left border and the right border generalized Pascal's triangle, respectively.
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} a^(n-m1) * b^k*R(m1)*C(n+k-m1-1,n-m1) + Sum_{m2=1..k} a^n*b^(k-m2)*L(m2)*C(n+k-m2-1,k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} a^(i-m1)*b^j*R(m1)*C(i+j-m1-1,i-m1) + Sum_{m2=1..j} a^i*b^(j-m2)*L(m2)*C(i+j-m2-1,j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.
Some special cases. If a=b=1, then the closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196.
If a=0, then as table read by antidiagonals T(n,k)=b*R(n), as linear sequence a(n)=b*R(i), where i=n-t*(t+1)/2-1, t=floor((-1+sqrt(8*n-7))/2); n>0. The sequence a(n) is the reluctant sequence of sequence b*R(n) - a(n) is triangle array read by rows: row number k coincides with first k elements of the sequence b*R(n). Similarly for b=0, we get T(n,k)=a*L(k).
For this sequence L(m)=m and R(m)=m^2, a=2, b=1. As table read by antidiagonals T(n,k) = Sum_{m1=1..n} 2^(n-m1)*m1^2*C(n+k-m1-1,n-m1) + Sum_{m2=1..k} 2^n*m2*C(n+k-m2-1,k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} 2^(i-m1)*m1^2*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} 2^i*m2*C(i+j-m2-1,j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.

A090843 Number of nodes on a tree with degree 11 interior nodes and degree 1 boundary nodes.

Original entry on oeis.org

1, 12, 122, 1222, 12222, 122222, 1222222, 12222222, 122222222, 1222222222, 12222222222, 122222222222, 1222222222222, 12222222222222, 122222222222222, 1222222222222222, 12222222222222222, 122222222222222222, 1222222222222222222, 12222222222222222222, 122222222222222222222
Offset: 0

Views

Author

Paul Barry, Dec 09 2003

Keywords

Comments

Sum of n-th row of triangle of powers of 10: 1; 1 10 1; 1 10 100 10 1; 1 10 100 1000 100 10 1; ... - Philippe Deléham, Feb 23 2014

Examples

			a(0) = 1;
a(1) = 1 + 10 + 1 = 12;
a(2) = 1 + 10 + 100 + 10 + 1 = 122;
a(3) = 1 + 10 + 100 + 1000 + 100 + 10 + 1 = 1222; etc. - _Philippe Deléham_, Feb 23 2014
		

Crossrefs

Programs

  • Maple
    g:=(1+z)/((1-z)* (1-10*z)): gser:=series(g, z=0, 43): seq((coeff(gser, z, n)), n=0..24); # Zerinvary Lajos, Feb 25 2009
  • Mathematica
    Table[(11 10^n - 2)/9, {n, 0, 20}] (* Vincenzo Librandi, Feb 24 2014 *)

Formula

a(n) = (11*10^n - 2)/9.
G.f.: (1+x)/((1-x)*(1-10*x)). - Zerinvary Lajos, Feb 25 2009
From Philippe Deléham, Feb 23 2014: (Start)
a(n) = 10*a(n-1) + 2, a(0) = 1.
a(n) = 11*a(n-1) - 10*a(n-2), a(0) = 1, a(1) = 12.
a(n) = Sum_{k=0..n} A112468(n,k)*11^k. (End)
E.g.f.: exp(x)*(11*exp(9*x) - 2)/9. - Elmo R. Oliveira, May 07 2025

A138894 Expansion of (1+x)/(1-10*x+9*x^2).

Original entry on oeis.org

1, 11, 101, 911, 8201, 73811, 664301, 5978711, 53808401, 484275611, 4358480501, 39226324511, 353036920601, 3177332285411, 28595990568701, 257363915118311, 2316275236064801, 20846477124583211, 187618294121248901
Offset: 0

Views

Author

Paul Barry, Apr 02 2008

Keywords

Comments

Orbit starting at 1 of A138893: a(n)=A138893^(n)(1). Partial sums of A003952.
Sum of n-th row of triangle of powers of 9: 1; 1 9 1; 1 9 81 9 1; 1 9 81 729 81 9 1; ... - Philippe Deléham, Feb 22 2014

Examples

			a(0) = 1;
a(1) = 1 + 9 + 1 = 11;
a(2) = 1 + 9 + 81 + 9 + 1 = 101;
a(3) = 1 + 9 + 81 + 729 + 81 + 9 + 1 = 911; etc. - _Philippe Deléham_, Feb 22 2014
		

Crossrefs

Cf. A096053 ((3*9^n-1)/2), a(n+1)=9a(n)-4 in A135423.

Programs

Formula

G.f.: (1+x)/((1-x)*(1-9x)).
a(n) = (5/4)*9^n - 1/4.
a(n) = A002452(n) + A002452(n+1).
Bisection of A135522/3. a(n+1)=9*a(n)+2. - Paul Curtz, Apr 22 2008
a(n) = Sum_{k=0..n} A112468(n,k)*10^k. - Philippe Deléham, Feb 22 2014

A061801 a(n) = (7*6^n - 2)/5.

Original entry on oeis.org

1, 8, 50, 302, 1814, 10886, 65318, 391910, 2351462, 14108774, 84652646, 507915878, 3047495270, 18284971622, 109709829734, 658258978406, 3949553870438, 23697323222630, 142183939335782, 853103636014694
Offset: 0

Views

Author

Amarnath Murthy, May 28 2001

Keywords

Comments

Sum of n-th row of triangle of powers of 6: 1; 1 6 1; 1 6 36 6 1; 1 6 36 216 36 6 1; ....

Examples

			a(2) = 50 = 1 + 6 + 36 + 6 + 1.
G.f. = 1 + 8*x + 50*x^2 + 302*x^3 + 1814*x^4 + 10886*x^5 + 65318*x^6 + ...
		

Crossrefs

Programs

  • Maple
    restart:g:=(1+x)/(1-6*x)/(1-x): gser:=series(g, x=0, 43): seq(coeff(gser, x, n), n=0..30); # Zerinvary Lajos, Jan 11 2009
  • PARI
    { for (n=0, 200, write("b061801.txt", n, " ", (7*6^n - 2)/5) ) } \\ Harry J. Smith, Jul 28 2009

Formula

G.f.: (1+x)/(1-6*x)/(1-x) [Zerinvary Lajos, Jan 11 2009]
a(n) = 6*a(n-1) + 2, a(0) = 1. - Philippe Deléham, Feb 23 2014
a(n) = Sum_{k=0..n} A112468(n,k)*7^k. - Philippe Deléham, Feb 23 2014

Extensions

More terms from Larry Reeves (larryr(AT)acm.org) and Jason Earls, May 28 2001.
Better description from Dean Hickerson, Jun 06 2001
Divided g.f. by x to match the offset. - Philippe Deléham, Feb 23 2014

A112739 Array counting nodes in rooted trees of height n in which the root and internal nodes have valency k (and the leaf nodes have valency one).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 4, 5, 2, 1, 1, 5, 10, 7, 2, 1, 1, 6, 17, 22, 9, 2, 1, 1, 7, 26, 53, 46, 11, 2, 1, 1, 8, 37, 106, 161, 94, 13, 2, 1, 1, 9, 50, 187, 426, 485, 190, 15, 2, 1, 1, 10, 65, 302, 937, 1706, 1457, 382, 17, 2, 1, 1, 11, 82, 457, 1814, 4687, 6826, 4373, 766, 19
Offset: 0

Views

Author

Paul Barry, Sep 16 2005

Keywords

Comments

Rows of the square array have g.f. (1+x)/((1-x)(1-kx)). They are the partial sums of the coordination sequences for the infinite tree of valency k. Row sums are A112740.
Rows of the square array are successively: A000012, A040000, A005408, A033484, A048473, A020989, A057651, A061801, A238275, A238276, A138894, A090843, A199023. - Philippe Deléham, Feb 22 2014

Examples

			As a square array, rows begin
1,1,1,1,1,1,... (A000012)
1,2,2,2,2,2,... (A040000)
1,3,5,7,9,11,... (A005408)
1,4,10,22,46,94,... (A033484)
1,5,17,53,161,485,... (A048473)
1,6,26,106,426,1706,... (A020989)
1,7,37,187,937,4687,... (A057651)
1,8,50,302,1814,10886,... (A061801)
As a number triangle, rows start
1;
1,1;
1,2,1;
1,3,2,1;
1,4,5,2,1;
1,5,10,7,2,1;
		

References

  • L. He, X. Liu and G. Strang, (2003) Trees with Cantor Eigenvalue Distribution. Studies in Applied Mathematics 110 (2), 123-138.
  • L. He, X. Liu and G. Strang, Laplacian eigenvalues of growing trees, Proc. Conf. on Math. Theory of Networks and Systems, Perpignan (2000).

Crossrefs

Formula

As a square array read by antidiagonals, T(n, k)=sum{j=0..k, (2-0^j)*(n-1)^(k-j)}; T(n, k)=(n(n-1)^k-2)/(n-2), n<>2, T(2, n)=2n+1; T(n, k)=sum{j=0..k, (n(n-1)^j-0^j)/(n-1)}, j<>1. As a triangle read by rows, T(n, k)=if(k<=n, sum{j=0..k, (2-0^j)*(n-k-1)^(k-j)}, 0).

A239473 Triangle read by rows: signed version of A059260: coefficients for expansion of partial sums of sequences a(n,x) in terms of their binomial transforms (1+a(.,x))^n ; Laguerre polynomial expansion of the truncated exponential.

Original entry on oeis.org

1, 0, 1, 1, -1, 1, 0, 2, -2, 1, 1, -2, 4, -3, 1, 0, 3, -6, 7, -4, 1, 1, -3, 9, -13, 11, -5, 1, 0, 4, -12, 22, -24, 16, -6, 1, 1, -4, 16, -34, 46, -40, 22, -7, 1, 0, 5, -20, 50, -80, 86, -62, 29, -8, 1, 1, -5, 25, -70, 130, -166, 148, -91, 37, -9, 1, 0, 6, -30, 95, -200, 296, -314, 239, -128, 46, -10, 1
Offset: 0

Views

Author

Tom Copeland, Mar 19 2014

Keywords

Comments

With T the lower triangular array above and the Laguerre polynomials L(k,x) = Sum_{j=0..k} (-1)^j binomial(k, j) x^j/j!, the following identities hold:
(A) Sum_{k=0..n} (-1)^k L(k,x) = Sum_{k=0..n} T(n,k) x^k/k!;
(B) Sum_{k=0..n} x^k/k! = Sum_{k=0..n} T(n,k) L(k,-x);
(C) Sum_{k=0..n} x^k = Sum_{k=0..n} T(n,k) (1+x)^k = (1-x^(n+1))/(1-x).
More generally, for polynomial sequences,
(D) Sum_{k=0..n} P(k,x) = Sum_{k=0..n} T(n,k) (1+P(.,x))^k,
where, e.g., for an Appell sequence, such as the Bernoulli polynomials, umbrally, (1+ Ber(.,x))^k = Ber(k,x+1).
Identity B follows from A through umbral substitution of j!L(j,-x) for x^j in A. Identity C, related to the cyclotomic polynomials for prime index, follows from B through the Laplace transform.
Integrating C gives Sum_{k=0..n} T(n,k) (2^(k+1)-1)/(k+1) = H(n+1), the harmonic numbers.
Identity A >= 0 for x >= 0 (see MathOverflow link for evaluation in terms of Hermite polynomials).
From identity C, W(m,n) = (-1)^n Sum_{k=0..n} T(n,k) (2-m)^k = number of walks of length n+1 between any two distinct vertices of the complete graph K_m for m > 2.
Equals A112468 with the first column of ones removed. - Georg Fischer, Jul 26 2023

Examples

			Triangle begins:
   1
   0    1
   1   -1    1
   0    2   -2    1
   1   -2    4   -3    1
   0    3   -6    7   -4    1
   1   -3    9  -13   11   -5    1
   0    4  -12   22  -24   16   -6    1
   1   -4   16  -34   46  -40   22   -7    1
   0    5  -20   50  -80   86  -62   29   -8    1
   1   -5   25  -70  130 -166  148  -91   37   -9    1
		

Crossrefs

For column 2: A001057, A004526, A008619, A140106.
Column 3: A002620, A087811.
Column 4: A002623, A173196.
Column 5: A001752.
Column 6: A001753.
Cf. Bottomley's cross-references in A059260.
Embedded in alternating antidiagonals of T are the reversals of arrays A071921 (A225010) and A210220.

Programs

  • Magma
    [[(&+[(-1)^(j+k)*Binomial(j,k): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 06 2018
    
  • Maple
    A239473 := proc(n,k)
        add(binomial(j,k)*(-1)^(j+k),j=k..n) ;
    end proc; # R. J. Mathar, Jul 21 2016
  • Mathematica
    Table[Sum[(-1)^(j+k)*Binomial[j,k], {j,0,n}], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 06 2018 *)
  • PARI
    for(n=0,10, for(k=0,n, print1(sum(j=0,n, (-1)^(j+k)*binomial(j, k)), ", "))) \\ G. C. Greubel, Feb 06 2018
    
  • Sage
    Trow = lambda n: sum((x-1)^j for j in (0..n)).list()
    for n in (0..10): print(Trow(n)) # Peter Luschny, Jul 09 2019

Formula

T(n, k) = Sum_{j=0..n} (-1)^(j+k) * binomial(j, k).
E.g.f: (exp(t) - (x-1)*exp((x-1)*t))/(2-x).
O.g.f. (n-th row): (1-(x-1)^(n+1))/(2-x).
Associated operator identities:
With D=d/dx, :xD:^n=x^n*D^n, and :Dx:^n=D^n*x^n, then bin(xD,n)= binomial(xD,n)=:xD:^n/n! and L(n,-:xD:)=:Dx:^n/n!=bin(xD+n,n)=(-1)^n bin(-xD-1,n),
A-o) Sum_{k=0..n} (-1)^k L(k,-:xD:) = Sum_{k=0..n} :-Dx:^k/k!
= Sum_{k=0..n} T(n,k) :-xD:^k/k! = Sum_{k=0..n} (-1)^k T(n,k)bin(xD,k)
B-o) Sum_{k=0..n} :xD:^k/k! = Sum_{k=0..n}, T(n,k) L(k,-:xD:)
= Sum_{k=0..n} T(n,k) :Dx:^k/k! = Sum_{k=0..n}, bin(xD,k).
Associated binomial identities:
A-b) Sum_{k=0..n} (-1)^k bin(s+k,k) = Sum_{k=0..n} (-1)^k T(n,k) bin(s,k)
= Sum_{k=0..n} bin(-s-1,k) = Sum{k=0..n} T(n,k) bin(-s-1+k,k)
B-b) Sum_{k=0..n} bin(s,k) = Sum_{k=0..n} T(n,k) bin(s+k,k)
= Sum_{k=0..n} (-1)^k bin(-s-1+k,k)
= Sum_{k=0..n} (-1)^k T(n,k) bin(-s-1,k).
In particular, from B-b with s=n, Sum_{k=0..n} T(n,k) bin(n+k,k) = 2^n. From B-b with s=0, row sums are all 1.
From identity C with x=-2, the unsigned row sums are the Jacobsthal sequence, i.e., Sum_{k=0..n} T(n,k) (1+(-2))^k = (-1)^n A001045(n+1); for x=2, the Mersenne numbers A000225; for x=-3, A014983 or signed A015518; for x=3, A003462; for x=-4, A014985 or signed A015521; for x=4, A002450; for x=-5, A014986 or signed A015531; and for x=5, A003463; for x=-6, A014987 or signed A015540; and for x=6, A003464.
With -s-1 = m = 0,1,2,..., B-b gives finite differences (recursions):
Sum_{k=0..n} (-1)^k T(n,k) bin(m,k) = Sum_{k=0..n} (-1)^k bin(m+k,k) = T(n+m,m), i.e., finite differences of the columns of T generate shifted columns of T. The columns of T are signed, shifted versions of sequences listed in the cross-references. Since the finite difference is an involution, T(n,k) = Sum_{j=0..k} (-1)^j T(n+j,j) bin(k,j)}. Gauss-Newton interpolation can be applied to give a generalized T(n,s) for s noninteger.
From identity C, S(n,m) = Sum_{k=0..n} T(n,k) bin(k,m) = 1 for m < n+1 and 0 otherwise, i.e., S = T*P, where S = A000012, as a lower triangular matrix and P = Pascal = A007318, so T = S*P^(-1), where P^(-1) = A130595, the signed Pascal array (see A132440), the inverse of P, and T^(-1) = P*S^(-1) = P*A167374 = A156644.
U(n,cos(x)) = e^(-n*i*x)*Sum_{k=0..n} T(n,k)*(1+e^(2*i*x))^k = sin((n+1)x)/sin(x), where U is the Chebyschev polynomial of the second kind A053117 and i^2 = -1. - Tom Copeland, Oct 18 2014
From Tom Copeland, Dec 26 2015: (Start)
With a(n,x) = e^(nx), the partial sums are 1+e^x+...+e^(nx) = Sum_{k=0..n} T(n,k) (1+e^x)^k = [ x / (e^x-1) ] [ e^((n+1)x) -1 ] / x = [ (x / (e^x-1)) e^((n+1)x) - (x / (e^x-1)) ] / x = Sum_{k>=0} [ (Ber(k+1,n+1) - Ber(k+1,0)) / (k+1) ] * x^k/k!, where Ber(n,x) are the Bernoulli polynomials (cf. Adams p. 140). Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of the m-th powers of the integers, their binomial transforms, and the Bernoulli polynomials.
With a(n,x) = (-1)^n e^(nx), the partial sums are 1-e^x+...+(-1)^n e^(nx) = Sum_{k=0..n} T(n,k) (1-e^x)^k = [ (-1)^n e^((n+1)x) + 1 ] / (e^x+1) = [ (-1)^n (2 / (e^x+1)) e^((n+1)x) + (2 / (e^x+1)) ] / 2 = (1/2) Sum_{k>=0} [ (-1)^n Eul(k,n+1) + Eul(k,0) ] * x^k/k!, where Eul(n,x) are the Euler polynomials. Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of signed m-th powers of the integers; their binomial transforms, related to the Stirling numbers of the second kind and face numbers of the permutahedra; and the Euler polynomials. (End)
As in A059260, a generator in terms of bivariate polynomials with the coefficients of this entry is given by (1/(1-y))*1/(1 + (y/(1-y))*x - (1/(1-y))*x^2) = 1 + y + (x^2 - x*y + y^2) + (2*x^2*y - 2*x*y^2 + y^3) + (x^4 - 2*x^3*y + 4*x^2*y^2 - 3*x*y^3 + y^4) + ... . This is of the form -h2 * 1 / (1 + h1*x + h2*x^2), related to the bivariate generator of A049310 with h1 = y/(1-y) and h2 = -1/(1-y) = -(1+h1). - Tom Copeland, Feb 16 2016
From Tom Copeland, Sep 05 2016: (Start)
Letting P(k,x) = x in D gives Sum_{k=0..n} T(n,k)*Sum_{j=0..k} binomial(k,j) = Sum_{k=0..n} T(n,k) 2^k = n + 1.
The quantum integers [n+1]q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)) = q^(-n)*(1 - q^(2*(n+1))) / (1 - q^2) = q^(-n)*Sum{k=0..n} q^(2k) = q^(-n)*Sum_{k=0..n} T(n,k)*(1 + q^2)^k. (End)
T(n, k) = [x^k] Sum_{j=0..n} (x-1)^j. - Peter Luschny, Jul 09 2019
a(n) = -n + Sum_{k=0..n} A341091(k). - Thomas Scheuerle, Jun 17 2022

Extensions

Inverse array added by Tom Copeland, Mar 26 2014
Formula re Euler polynomials corrected by Tom Copeland, Mar 08 2024

A112465 Riordan array (1/(1+x), x/(1-x)).

Original entry on oeis.org

1, -1, 1, 1, 0, 1, -1, 1, 1, 1, 1, 0, 2, 2, 1, -1, 1, 2, 4, 3, 1, 1, 0, 3, 6, 7, 4, 1, -1, 1, 3, 9, 13, 11, 5, 1, 1, 0, 4, 12, 22, 24, 16, 6, 1, -1, 1, 4, 16, 34, 46, 40, 22, 7, 1, 1, 0, 5, 20, 50, 80, 86, 62, 29, 8, 1, -1, 1, 5, 25, 70, 130, 166, 148, 91, 37, 9, 1, 1, 0, 6, 30, 95, 200, 296, 314, 239, 128, 46, 10, 1
Offset: 0

Views

Author

Paul Barry, Sep 06 2005

Keywords

Comments

Inverse is A112466. Note that C(n,k) = Sum_{j = 0..n-k} C(j+k-1, j).

Examples

			Triangle starts
   1;
  -1, 1;
   1, 0, 1;
  -1, 1, 1,  1;
   1, 0, 2,  2,  1;
  -1, 1, 2,  4,  3,  1;
   1, 0, 3,  6,  7,  4,  1;
  -1, 1, 3,  9, 13, 11,  5, 1;
   1, 0, 4, 12, 22, 24, 16, 6, 1;
Production matrix begins
  -1, 1;
   0, 1, 1;
   0, 0, 1, 1;
   0, 0, 0, 1, 1;
   0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 0, 0, 1, 1; - _Paul Barry_, Apr 08 2011
		

Crossrefs

Columns: A033999(n) (k=0), A000035(n) (k=1), A004526(n) (k=2), A002620(n-1) (k=3), A002623(n-4) (k=4), A001752(n-5) (k=5), A001753(n-6) (k=6), A001769(n-7) (k=7), A001779(n-8) (k=8), A001780(n-9) (k=9), A001781(n-10) (k=10), A001786(n-11) (k=11), A001808(n-12) (k=12).
Diagonals: A000012(n) (k=n), A023443(n) (k=n-1), A152947(n-1) (k=n-2), A283551(n-3) (k=n-3).
Main diagonal: A072547.
Sums: A078008 (row), A078024 (diagonal), A092220 (signed diagonal), A280560 (signed row).

Programs

  • Haskell
    a112465 n k = a112465_tabl !! n !! k
    a112465_row n = a112465_tabl !! n
    a112465_tabl = iterate f [1] where
       f xs'@(x:xs) = zipWith (+) ([-x] ++ xs ++ [0]) ([0] ++ xs')
    -- Reinhard Zumkeller, Jan 03 2014
    
  • Magma
    A112465:= func< n,k | (-1)^(n+k)*(&+[(-1)^j*Binomial(j+k-1,j): j in [0..n-k]]) >;
    [A112465(n,k): k in [0..n], n in [0..13]]; // G. C. Greubel, Apr 18 2025
    
  • Mathematica
    T[n_, k_]:= Sum[Binomial[j+k-1, j]*(-1)^(n-k-j), {j, 0, n-k}];
    Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, Jul 23 2018 *)
  • SageMath
    def A112465(n,k): return (-1)^(n+k)*sum((-1)^j*binomial(j+k-1,j) for j in range(n-k+1))
    print(flatten([[A112465(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Apr 18 2025

Formula

Number triangle T(n, k) = Sum_{j=0..n-k} (-1)^(n-k-j)*C(j+k-1, j).
T(2*n, n) = A072547(n) (main diagonal). - Paul Barry, Apr 08 2011
From Reinhard Zumkeller, Jan 03 2014: (Start)
T(n, k) = T(n-1, k-1) + T(n-1, k), 0 < k < n, with T(n, 0) = (-1)^n and T(n, n) = 1.
T(n, k) = A108561(n, n-k). (End)
T(n, k) = T(n-1, k-1) + T(n-2, k) + T(n-2, k-1), T(0, 0) = 1, T(1, 0) = -1, T(1, 1) = 1, T(n, k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 11 2014
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(-1 + x + x^2/2! + x^3/3!) = -1 + 2*x^2/2! + 6*x^3/3! + 13*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014

A174296 Row sums of A174294.

Original entry on oeis.org

1, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2
Offset: 0

Views

Author

Mats Granvik, Mar 15 2010

Keywords

Crossrefs

Programs

  • Magma
    [n lt 2 select (n+1) else 2 + (n mod 2): n in [0..110]]; // G. C. Greubel, Nov 25 2021
    
  • Mathematica
    Table[If[n<2, n+1, (5-(-1)^n)/2], {n,0,110}] (* G. C. Greubel, Nov 25 2021 *)
  • Sage
    [1,2]+[(5-(-1)^n)/2 for n in (2..110)] # G. C. Greubel, Nov 25 2021

Formula

a(A004280(n)) = 3 for n > 2.
From G. C. Greubel, Nov 25 2021: (Start)
a(n) = a(n-2) for n > 3, with a(0) = 1, a(1) = 2, a(2) = 2, a(3) = 3.
a(n) = (5 - (-1)^n)/2 for n > 1, with a(0) = 1, a(1) = 2.
a(n) = (n+1)*[n<2] + A010693(n)*[n>1].
G.f.: (1_+ 2*x + x^2 + x^3)/(1 - x^2).
E.g.f.: (1/2)*( -exp(-x) - 2*(1+x) + 5*exp(x) ). (End)

A199023 a(n) = (6*11^n - 1)/5.

Original entry on oeis.org

1, 13, 145, 1597, 17569, 193261, 2125873, 23384605, 257230657, 2829537229, 31124909521, 342374004733, 3766114052065, 41427254572717, 455699800299889, 5012697803298781, 55139675836286593, 606536434199152525, 6671900776190677777, 73390908538097455549, 807299993919072011041
Offset: 0

Views

Author

Vincenzo Librandi, Nov 02 2011

Keywords

Comments

Sum of n-th row of triangle of powers of 11: 1; 1 11 1; 1 11 121 11 1; 1 11 121 1331 121 11 1; ... - Philippe Deléham, Feb 23 2014

Examples

			a(0) = 1;
a(1) = 1 + 11 + 1 = 13;
a(2) = 1 + 11 + 121 + 11 + 1 = 145;
a(3) = 1 + 11 + 121 + 1331 + 121 + 11 + 1 = 1597; etc. - _Philippe Deléham_, Feb 23 2014
		

Crossrefs

Programs

  • Magma
    [(6*11^n-1)/5 : n in [0..20]];
    
  • Mathematica
    CoefficientList[Series[(1 + x)/(1 - 12*x + 11*x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jan 04 2013 *)
  • PARI
    a(n)=(6*11^n-1)/5 \\ Charles R Greathouse IV, Oct 07 2015

Formula

a(n) = 11*a(n-1) + 2.
a(n) = 12*a(n-1) - 11*a(n-2), n > 1.
G.f.: (1 + x)/(1 - 12*x + 11*x^2). - Vincenzo Librandi, Jan 04 2013
a(n) = Sum_{k=0..n} A112468(n,k)*12^k. - Philippe Deléham, Feb 23 2014
From Elmo R. Oliveira, Mar 02 2025: (Start)
E.g.f.: exp(x)*(6*exp(10*x) - 1)/5.
a(n) = A199024(n)/5. (End)

A174294 Triangle T(n,k), read by rows, T(n,k) = (T(n-1,k-1) + T(n-2,k-1)) - (T(n-1,k) + T(n-2,k)), with T(n, 0) = T(n, k) = 1 and T(n, 1) = (n mod 2).

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, -1, 0, 1, 1, 0, 0, 2, -2, 0, 1, 1, 1, 0, 0, 3, -3, 0, 1, 1, 0, 1, -2, 1, 4, -4, 0, 1, 1, 1, 0, 3, -6, 3, 5, -5, 0, 1, 1, 0, 0, 0, 6, -12, 6, 6, -6, 0, 1, 1, 1, 1, -3, 3, 9, -20, 10, 7, -7, 0, 1, 1, 0, 0, 4, -12, 12, 11, -30, 15, 8, -8, 0, 1
Offset: 0

Views

Author

Mats Granvik, Mar 15 2010

Keywords

Examples

			Table begins:
  n\k|...0...1...2...3...4...5...6...7...8...9..10
  ---|--------------------------------------------
  0..|...1
  1..|...1...1
  2..|...1...0...1
  3..|...1...1...0...1
  4..|...1...0...0...0...1
  5..|...1...1...1..-1...0...1
  6..|...1...0...0...2..-2...0...1
  7..|...1...1...0...0...3..-3...0...1
  8..|...1...0...1..-2...1...4..-4...0...1
  9..|...1...1...0...3..-6...3...5..-5...0...1
  10.|...1...0...0...0...6.-12...6...6..-6...0...1
		

Crossrefs

Programs

  • Mathematica
    T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[k==0 || k==n, 1, If[k==1, Mod[n, 2], T[n-1, k-1] +T[n-2, k-1] -T[n-1, k] -T[n-2, k] ]]];
    Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 25 2021 *)
  • Sage
    @CachedFunction
    def T(n,k): # A174294
        if (k<0 or k>n): return 0
        elif (k==0 or k==n): return 1
        elif (k==1): return n%2
        else: return T(n-1, k-1) + T(n-2, k-1) - T(n-1, k) - T(n-2, k)
    flatten([[T(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Nov 25 2021

Formula

T(n,k) = (T(n-1,k-1) + T(n-2,k-1)) - (T(n-1,k) + T(n-2,k)), with T(n, 0) = T(n, k) = 1 and T(n, 1) = (n mod 2).
Previous Showing 11-20 of 30 results. Next