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.

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A040000 a(0)=1; a(n)=2 for n >= 1.

Original entry on oeis.org

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

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Author

N. J. A. Sloane, Dec 11 1999

Keywords

Comments

Continued fraction expansion of sqrt(2) is 1 + 1/(2 + 1/(2 + 1/(2 + ...))).
Inverse binomial transform of Mersenne numbers A000225(n+1) = 2^(n+1) - 1. - Paul Barry, Feb 28 2003
A Chebyshev transform of 2^n: if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))A(x/(1+x^2)). - Paul Barry, Oct 31 2004
An inverse Catalan transform of A068875 under the mapping g(x)->g(x(1-x)). A068875 can be retrieved using the mapping g(x)->g(xc(x)), where c(x) is the g.f. of A000108. A040000 and A068875 may be described as a Catalan pair. - Paul Barry, Nov 14 2004
Sequence of electron arrangement in the 1s 2s and 3s atomic subshells. Cf. A001105, A016825. - Jeremy Gardiner, Dec 19 2004
Binomial transform of A165326. - Philippe Deléham, Sep 16 2009
Let m=2. We observe that a(n) = Sum_{k=0..floor(n/2)} binomial(m,n-2*k). Then there is a link with A113311 and A115291: it is the same formula with respectively m=3 and m=4. We can generalize this result with the sequence whose g.f. is given by (1+z)^(m-1)/(1-z). - Richard Choulet, Dec 08 2009
With offset 1: number of permutations where |p(i) - p(i+1)| <= 1 for n=1,2,...,n-1. This is the identical permutation and (for n>1) its reversal.
Equals INVERT transform of bar(1, 1, -1, -1, ...).
Eventual period is (2). - Zak Seidov, Mar 05 2011
Also decimal expansion of 11/90. - Vincenzo Librandi, Sep 24 2011
a(n) = 3 - A054977(n); right edge of the triangle in A182579. - Reinhard Zumkeller, May 07 2012
With offset 1: minimum cardinality of the range of a periodic sequence with (least) period n. Of course the range's maximum cardinality for a purely periodic sequence with (least) period n is n. - Rick L. Shepherd, Dec 08 2014
With offset 1: n*a(1) + (n-1)*a(2) + ... + 2*a(n-1) + a(n) = n^2. - Warren Breslow, Dec 12 2014
With offset 1: decimal expansion of gamma(4) = 11/9 where gamma(n) = Cp(n)/Cv(n) is the n-th Poisson's constant. For the definition of Cp and Cv see A272002. - Natan Arie Consigli, Sep 11 2016
a(n) equals the number of binary sequences of length n where no two consecutive terms differ. Also equals the number of binary sequences of length n where no two consecutive terms are the same. - David Nacin, May 31 2017
a(n) is the period of the continued fractions for sqrt((n+2)/(n+1)) and sqrt((n+1)/(n+2)). - A.H.M. Smeets, Dec 05 2017
Also, number of self-avoiding walks and coordination sequence for the one-dimensional lattice Z. - Sean A. Irvine, Jul 27 2020

Examples

			sqrt(2) = 1.41421356237309504... = 1 + 1/(2 + 1/(2 + 1/(2 + 1/(2 + ...)))). - _Harry J. Smith_, Apr 21 2009
G.f. = 1 + 2*x + 2*x^2 + 2*x^3 + 2*x^4 + 2*x^5 + 2*x^6 + 2*x^7 + 2*x^8 + ...
11/90 = 0.1222222222222222222... - _Natan Arie Consigli_, Sep 11 2016
		

References

  • A. Beiser, Concepts of Modern Physics, 2nd Ed., McGraw-Hill, 1973.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 186.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §4.4 Powers and Roots, p. 144.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 276-278.

Crossrefs

Convolution square is A008574.
See A003945 etc. for (1+x)/(1-k*x).
From Jaume Oliver Lafont, Mar 26 2009: (Start)
Sum_{0<=k<=n} a(k) = A005408(n).
Prod_{0<=k<=n} a(k) = A000079(n). (End)
Cf. A000674 (boustrophedon transform).
Cf. A001333/A000129 (continued fraction convergents).
Cf. A000122, A002193 (sqrt(2) decimal expansion), A006487 (Egyptian fraction).
Cf. Other continued fractions for sqrt(a^2+1) = (a, 2a, 2a, 2a....): A040002 (contfrac(sqrt(5)) = (2,4,4,...)), A040006, A040012, A040020, A040030, A040042, A040056, A040072, A040090, A040110 (contfrac(sqrt(122)) = (11,22,22,...)), A040132, A040156, A040182, A040210, A040240, A040272, A040306, A040342, A040380, A040420 (contfrac(sqrt(442)) = (21,42,42,...)), A040462, A040506, A040552, A040600, A040650, A040702, A040756, A040812, A040870, A040930 (contfrac(sqrt(962)) = (31,62,62,...)).

Programs

  • Haskell
    a040000 0 = 1; a040000 n = 2
    a040000_list = 1 : repeat 2  -- Reinhard Zumkeller, May 07 2012
  • Maple
    Digits := 100: convert(evalf(sqrt(2)),confrac,90,'cvgts'):
  • Mathematica
    ContinuedFraction[Sqrt[2],300] (* Vladimir Joseph Stephan Orlovsky, Mar 04 2011 *)
    a[ n_] := 2 - Boole[n == 0]; (* Michael Somos, Dec 28 2014 *)
    PadRight[{1},120,2] (* or *) RealDigits[11/90, 10, 120][[1]] (* Harvey P. Dale, Jul 12 2025 *)
  • PARI
    {a(n) = 2-!n}; /* Michael Somos, Apr 16 2007 */
    
  • PARI
    a(n)=1+sign(n)  \\ Jaume Oliver Lafont, Mar 26 2009
    
  • PARI
    allocatemem(932245000); default(realprecision, 21000); x=contfrac(sqrt(2)); for (n=0, 20000, write("b040000.txt", n, " ", x[n+1]));  \\ Harry J. Smith, Apr 21 2009
    

Formula

G.f.: (1+x)/(1-x). - Paul Barry, Feb 28 2003
a(n) = 2 - 0^n; a(n) = Sum_{k=0..n} binomial(1, k). - Paul Barry, Oct 16 2004
a(n) = n*Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k, k)*2^(n-2*k)/(n-k). - Paul Barry, Oct 31 2004
A040000(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-1)^k*A068875(n-k). - Paul Barry, Nov 14 2004
From Michael Somos, Apr 16 2007: (Start)
Euler transform of length 2 sequence [2, -1].
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = (u-v)*(u+v) - 2*v*(u-w).
E.g.f.: 2*exp(x) - 1.
a(n) = a(-n) for all n in Z (one possible extension to n<0). (End)
G.f.: (1-x^2)/(1-x)^2. - Jaume Oliver Lafont, Mar 26 2009
G.f.: exp(2*atanh(x)). - Jaume Oliver Lafont, Oct 20 2009
a(n) = Sum_{k=0..n} A108561(n,k)*(-1)^k. - Philippe Deléham, Nov 17 2013
a(n) = 1 + sign(n). - Wesley Ivan Hurt, Apr 16 2014
10 * 11/90 = 11/9 = (11/2 R)/(9/2 R) = Cp(4)/Cv(4) = A272005/A272004, with R = A081822 (or A070064). - Natan Arie Consigli, Sep 11 2016
a(n) = A001227(A000040(n+1)). - Omar E. Pol, Feb 28 2018

A010701 Constant sequence: the all 3's sequence.

Original entry on oeis.org

3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3
Offset: 0

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Author

Keywords

Comments

Decimal expansion of 1/3. - Raymond Wang, Mar 06 2010
Continued fraction expansion of (3+sqrt(13))/2. - Bruno Berselli, Mar 15 2011

Examples

			1/3 = 0.33333333333333333333333333333333333333333333... - _Bruno Berselli_, Mar 21 2014
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 5.3.1, p. 281.

Crossrefs

Programs

Formula

G.f.: 3/(1-x). - Bruno Berselli, Mar 15 2011
E.g.f.: 3*e^x. - Vincenzo Librandi, Jan 24 2012
a(n) = A040000(n) + A054977(n). - Reinhard Zumkeller, May 06 2012
a(n) = 3*A000012(n). - Michel Marcus, Dec 18 2015
a(n) = floor(1/(n - cot(1/n))). - Clark Kimberling, Mar 10 2020
Equals Sum_{k>=1} (1/4)^k (as a constant). - Michel Marcus, Jun 11 2020
Equals Sum_{k>=2} (k-1)/binomial(2*k,k) (as a constant). - Amiram Eldar, Jun 05 2021
Equals Sum_{k>=1} (-1)^(k+1)/2^k. - Michal Paulovic, Mar 02 2023

A135278 Triangle read by rows, giving the numbers T(n,m) = binomial(n+1, m+1); or, Pascal's triangle A007318 with its left-hand edge removed.

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 5, 10, 10, 5, 1, 6, 15, 20, 15, 6, 1, 7, 21, 35, 35, 21, 7, 1, 8, 28, 56, 70, 56, 28, 8, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 12, 66, 220, 495, 792, 924, 792
Offset: 0

Views

Author

Zerinvary Lajos, Dec 02 2007

Keywords

Comments

T(n,m) is the number of m-faces of a regular n-simplex.
An n-simplex is the n-dimensional analog of a triangle. Specifically, a simplex is the convex hull of a set of (n + 1) affinely independent points in some Euclidean space of dimension n or higher, i.e., a set of points such that no m-plane contains more than (m + 1) of them. Such points are said to be in general position.
Reversing the rows gives A074909, which as a linear sequence is essentially the same as this.
From Tom Copeland, Dec 07 2007: (Start)
T(n,k) * (k+1)! = A068424. The comment on permuted words in A068424 shows that T is related to combinations of letters defined by connectivity of regular polytope simplexes.
If T is the diagonally-shifted Pascal matrix, binomial(n+m, k+m), for m=1, then T is a fundamental type of matrix that is discussed in A133314 and the following hold.
The infinitesimal matrix generator is given by A132681, so T = LM(1) of A132681 with inverse LM(-1).
With a(k) = (-x)^k / k!, T * a = [ Laguerre(n,x,1) ], a vector array with index n for the Laguerre polynomials of order 1. Other formulas for the action of T are given in A132681.
T(n,k) = (1/n!) (D_x)^n (D_t)^k Gf(x,t) evaluated at x=t=0 with Gf(x,t) = exp[ t * x/(1-x) ] / (1-x)^2.
[O.g.f. for T ] = 1 / { [ 1 - t * x/(1-x) ] * (1-x)^2 }. [ O.g.f. for row sums ] = 1 / { (1-x) * (1-2x) }, giving A000225 (without a leading zero) for the row sums. Alternating sign row sums are all 1. [Sign correction noted by Vincent J. Matsko, Jul 19 2015]
O.g.f. for row polynomials = [ (1+q)**(n+1) - 1 ] / [ (1+q) -1 ] = A(1,n+1,q) on page 15 of reference on Grassmann cells in A008292. (End)
Given matrices A and B with A(n,k) = T(n,k)*a(n-k) and B(n,k) = T(n,k)*b(n-k), then A*B = C where C(n,k) = T(n,k)*[a(.)+b(.)]^(n-k), umbrally. The e.g.f. for the row polynomials of A is {(a+t) exp[(a+t)x] - a exp(a x)}/t, umbrally. - Tom Copeland, Aug 21 2008
A007318*A097806 as infinite lower triangular matrices. - Philippe Deléham, Feb 08 2009
Riordan array (1/(1-x)^2, x/(1-x)). - Philippe Deléham, Feb 22 2012
The elements of the matrix inverse are T^(-1)(n,k)=(-1)^(n+k)*T(n,k). - R. J. Mathar, Mar 12 2013
Relation to K-theory: T acting on the column vector (-0,d,-d^2,d^3,...) generates the Euler classes for a hypersurface of degree d in CP^n. Cf. Dugger p. 168 and also A104712, A111492, and A238363. - Tom Copeland, Apr 11 2014
Number of walks of length p>0 between any two distinct vertices of the complete graph K_(n+2) is W(n+2,p)=(-1)^(p-1)*Sum_{k=0..p-1} T(p-1,k)*(-n-2)^k = ((n+1)^p - (-1)^p)/(n+2) = (-1)^(p-1)*Sum_{k=0..p-1} (-n-1)^k. This is equal to (-1)^(p-1)*Phi(p,-n-1), where Phi is the cyclotomic polynomial when p is an odd prime. For K_3, see A001045; for K_4, A015518; for K_5, A015521; for K_6, A015531; for K_7, A015540. - Tom Copeland, Apr 14 2014
Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x-1)^0 + A_1*(x-1)^1 + A_2*(x-1)^2 + ... + A_n*(x-1)^n. This sequence gives A_0, ..., A_n as the entries in the n-th row of this triangle, starting at n = 0. - Derek Orr, Oct 14 2014
See A074909 for associations among this array, the Bernoulli polynomials and their umbral compositional inverses, and the face polynomials of permutahedra and their duals (cf. A019538). - Tom Copeland, Nov 14 2014
From Wolfdieter Lang, Dec 10 2015: (Start)
A(r, n) = T(n+r-2, r-1) = risefac(n,r)/r! = binomial(n+r-1, r), for n >= 1 and r >= 1, gives the array with the number of independent components of a symmetric tensors of rank r (number of indices) and dimension n (indices run from 1 to n). Here risefac(n, k) is the rising factorial.
As(r, n) = T(n+1, r+1) = fallfac(n, r)/r! = binomial(n, r), r >= 1 and n >= 1 (with the triangle entries T(n, k) = 0 for n < k) gives the array with the number of independent components of an antisymmetric tensor of rank r and dimension n. Here fallfac is the falling factorial. (End)
The h-vectors associated to these f-vectors are given by A000012 regarded as a lower triangular matrix. Read as bivariate polynomials, the h-polynomials are the complete homogeneous symmetric polynomials in two variables, found in the compositional inverse of an e.g.f. for A008292, the h-vectors of the permutahedra. - Tom Copeland, Jan 10 2017
For a correlation between the states of a quantum system and the combinatorics of the n-simplex, see Boya and Dixit. - Tom Copeland, Jul 24 2017

Examples

			The triangle T(n, k) begins:
   n\k  0  1   2   3   4   5   6   7   8  9 10 11 ...
   0:   1
   1:   2  1
   2:   3  3   1
   3:   4  6   4   1
   4:   5 10  10   5   1
   5:   6 15  20  15   6   1
   6:   7 21  35  35  21   7   1
   7:   8 28  56  70  56  28   8   1
   8:   9 36  84 126 126  84  36   9   1
   9:  10 45 120 210 252 210 120  45  10  1
  10:  11 55 165 330 462 462 330 165  55 11  1
  11:  12 66 220 495 792 924 792 495 220 66 12  1
  ... reformatted by _Wolfdieter Lang_, Mar 23 2015
Production matrix begins
   2   1
  -1   1   1
   1   0   1   1
  -1   0   0   1   1
   1   0   0   0   1   1
  -1   0   0   0   0   1   1
   1   0   0   0   0   0   1   1
  -1   0   0   0   0   0   0   1   1
   1   0   0   0   0   0   0   0   1   1
- _Philippe Deléham_, Jan 29 2014
From _Wolfdieter Lang_, Nov 08 2018: (Start)
Recurrence [_Philippe Deléham_]: T(7, 3) = 2*35 + 35 - 15 - 20 = 70.
Recurrence from Riordan A- and Z-sequences: [1,1,repeat(0)] and [2, repeat(-1, +1)]: From Z: T(5, 0) = 2*5 - 10 + 10 - 5 + 1 = 6. From A: T(7, 3) = 35 + 35 = 70.
Boas-Buck column k=3 recurrence: T(7, 3) = (5/4)*(1 + 5 + 15 + 35) = 70. (End)
		

Crossrefs

Programs

  • Maple
    for i from 0 to 12 do seq(binomial(i, j)*1^(i-j), j = 1 .. i) od;
  • Mathematica
    Flatten[Table[CoefficientList[D[1/x ((x + 1) Exp[(x + 1) z] - Exp[z]), {z, k}] /. z -> 0, x], {k, 0, 11}]]
    CoefficientList[CoefficientList[Series[1/((1 - x)*(1 - x - x*y)), {x, 0, 10}, {y, 0, 10}], x], y] // Flatten (* G. C. Greubel, Nov 22 2017 *)
  • PARI
    for(n=0, 20, for(k=0, n, print1(1/k!*sum(i=0, n, (prod(j=0, k-1, i-j))), ", "))) \\ Derek Orr, Oct 14 2014
    
  • 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=k..n} binomial(j,k) = binomial(n+1, k+1), n >= k >= 0, else 0. (Partial sum of column k of A007318 (Pascal), or summation on the upper binomial index (Graham et al. (GKP), eq. (5.10). For the GKP reference see A007318.) - Wolfdieter Lang, Aug 22 2012
E.g.f.: 1/x*((1 + x)*exp(t*(1 + x)) - exp(t)) = 1 + (2 + x)*t + (3 + 3*x + x^2)*t^2/2! + .... The infinitesimal generator for this triangle has the sequence [2,3,4,...] on the main subdiagonal and 0's elsewhere. - Peter Bala, Jul 16 2013
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=1, T(1,0)=2, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013
T(n,k) = A193862(n,k)/2^k. - Philippe Deléham, Jan 29 2014
G.f.: 1/((1-x)*(1-x-x*y)). - Philippe Deléham, Mar 13 2014
From Tom Copeland, Mar 26 2014: (Start)
[From Copeland's 2007 and 2008 comments]
A) O.g.f.: 1 / { [ 1 - t * x/(1-x) ] * (1-x)^2 } (same as Deleham's).
B) The infinitesimal generator for T is given in A132681 with m=1 (same as Bala's), which makes connections to the ubiquitous associated Laguerre polynomials of integer orders, for this case the Laguerre polynomials of order one L(n,-t,1).
C) O.g.f. of row e.g.f.s: Sum_{n>=0} L(n,-t,1) x^n = exp[t*x/(1-x)]/(1-x)^2 = 1 + (2+t)x + (3+3*t+t^2/2!)x^2 + (4+6*t+4*t^2/2!+t^3/3!)x^3+ ... .
D) E.g.f. of row o.g.f.s: ((1+t)*exp((1+t)*x)-exp(x))/t (same as Bala's).
E) E.g.f. for T(n,k)*a(n-k): {(a+t) exp[(a+t)x] - a exp(a x)}/t, umbrally. For example, for a(k)=2^k, the e.g.f. for the row o.g.f.s is {(2+t) exp[(2+t)x] - 2 exp(2x)}/t.
(End)
From Tom Copeland, Apr 28 2014: (Start)
With different indexing
A) O.g.f. by row: [(1+t)^n-1]/t.
B) O.g.f. of row o.g.f.s: {1/[1-(1+t)*x] - 1/(1-x)}/t.
C) E.g.f. of row o.g.f.s: {exp[(1+t)*x]-exp(x)}/t.
These generating functions are related to row e.g.f.s of A111492. (End)
From Tom Copeland, Sep 17 2014: (Start)
A) U(x,s,t)= x^2/[(1-t*x)(1-(s+t)x)] = Sum_{n >= 0} F(n,s,t)x^(n+2) is a generating function for bivariate row polynomials of T, e.g., F(2,s,t)= s^2 + 3s*t + 3t^2 (Buchstaber, 2008).
B) dU/dt=x^2 dU/dx with U(x,s,0)= x^2/(1-s*x) (Buchstaber, 2008).
C) U(x,s,t) = exp(t*x^2*d/dx)U(x,s,0) = U(x/(1-t*x),s,0).
D) U(x,s,t) = Sum[n >= 0, (t*x)^n L(n,-:xD:,-1)] U(x,s,0), where (:xD:)^k=x^k*(d/dx)^k and L(n,x,-1) are the Laguerre polynomials of order -1, related to normalized Lah numbers. (End)
E.g.f. satisfies the differential equation d/dt(e.g.f.(x,t)) = (x+1)*e.g.f.(x,t) + exp(t). - Vincent J. Matsko, Jul 18 2015
The e.g.f. of the Norlund generalized Bernoulli (Appell) polynomials of order m, NB(n,x;m), is given by exponentiation of the e.g.f. of the Bernoulli numbers, i.e., multiple binomial self-convolutions of the Bernoulli numbers, through the e.g.f. exp[NB(.,x;m)t] = (t/(e^t - 1))^(m+1) * e^(xt). Norlund gave the relation to the factorials (x-1)!/(x-1-n)! = (x-1) ... (x-n) = NB(n,x;n), so T(n,m) = NB(m+1,n+2;m+1)/(m+1)!. - Tom Copeland, Oct 01 2015
From Wolfdieter Lang, Nov 08 2018: (Start)
Recurrences from the A- and Z- sequences for the Riordan triangle (see the W. Lang link under A006232 with references), which are A(n) = A019590(n+1), [1, 1, repeat (0)] and Z(n) = (-1)^(n+1)*A054977(n), [2, repeat(-1, 1)]:
T(0, 0) = 1, T(n, k) = 0 for n < k, and T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and T(n, k) = T(n-1, k-1) + T(n-1, k), for n >= m >= 1.
Boas-Buck recurrence for columns (see the Aug 10 2017 remark in A036521 also for references):
T(n, k) = ((2 + k)/(n - k))*Sum_{j=k..n-1} T(j, k), for n >= 1, k = 0, 1, ..., n-1, and input T(n, n) = 1, for n >= 0, (the BB-sequences are alpha(n) = 2 and beta(n) = 1). (End)
T(n, k) = [x^k] Sum_{j=0..n} (x+1)^j. - Peter Luschny, Jul 09 2019

Extensions

Edited by Tom Copeland and N. J. A. Sloane, Dec 11 2007

A062395 a(n) = 8^n + 1.

Original entry on oeis.org

2, 9, 65, 513, 4097, 32769, 262145, 2097153, 16777217, 134217729, 1073741825, 8589934593, 68719476737, 549755813889, 4398046511105, 35184372088833, 281474976710657, 2251799813685249, 18014398509481985, 144115188075855873
Offset: 0

Views

Author

Henry Bottomley, Jun 22 2001

Keywords

Comments

Any number of the form b^k+1 is composite for b>2 and k odd since b+1 algebraically divides b^k+1. - Robert G. Wilson v, Aug 25 2002

References

  • D. M. Burton, Elementary Number Theory, Allyn and Bacon, Boston, MA, 1976, pp. 51.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.

Crossrefs

Programs

  • Magma
    [8^n + 1: n in [0..40] ]; // Vincenzo Librandi, Apr 30 2011
  • Mathematica
    Table[8^n + 1, {n, 0, 20}]
    LinearRecurrence[{9,-8},{2,9},20] (* Harvey P. Dale, Jan 24 2019 *)
  • PARI
    for(n=0,22,print(8^n+1))
    

Formula

a(n) = 8a(n-1)-7 = A001018(n)+1 = 9a(n-1) - 8a(n-2).
G.f.: -(-2+9*x)/(-1+x)/(-1+8*x). - R. J. Mathar, Nov 16 2007
E.g.f.: e^x+e^(8*x). - Mohammad K. Azarian, Jan 02 2009

A062396 a(n) = 9^n + 1.

Original entry on oeis.org

2, 10, 82, 730, 6562, 59050, 531442, 4782970, 43046722, 387420490, 3486784402, 31381059610, 282429536482, 2541865828330, 22876792454962, 205891132094650, 1853020188851842, 16677181699666570, 150094635296999122
Offset: 0

Views

Author

Henry Bottomley, Jun 22 2001

Keywords

Crossrefs

Programs

Formula

a(n) = 9*a(n-1) - 8 = A001019(n) + 1 = 10*a(n-1) - 9*a(n-2).
From Mohammad K. Azarian, Jan 02 2009: (Start)
G.f.: 1/(1-x) + 1/(1-9*x).
E.g.f.: e^x + e^(9*x). (End)

A062397 a(n) = 10^n + 1.

Original entry on oeis.org

2, 11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001, 1000000001, 10000000001, 100000000001, 1000000000001, 10000000000001, 100000000000001, 1000000000000001, 10000000000000001, 100000000000000001
Offset: 0

Views

Author

Henry Bottomley, Jun 22 2001

Keywords

Comments

The first three terms (indices 0, 1 and 2) are the only known primes. Moreover, the terms not of the form a(2^k) are all composite, except for a(0). Indeed, for all n >= 0, a(2n+1) is divisible by 11, a(4n+2) is divisible by 101, a(8n+4) is divisible by 73, a(16n+8) is divisible by 17, a(32n+16) is divisible by 353, a(64n+32) is divisible by 19841, etc. - M. F. Hasler, Nov 03 2018 [Edited based on the comment by Jeppe Stig Nielsen, Oct 17 2019]
This sequence also results when each term is generated by converting the previous term into a Roman numeral, then replacing each letter with its corresponding decimal value, provided that the vinculum is used and numerals are written in a specific way for integers greater than 3999, e.g., IV with a vinculum over the I and V for 4000. - Jamie Robert Creasey, Apr 14 2021
By Mihăilescu's theorem, a(n) can never be a perfect power (see "Catalan's conjecture" in Links). - Marco Ripà, Mar 10 2025

Crossrefs

Except for the initial term, essentially the same as A000533. Cf. A054977, A007395, A000051, A034472, A052539, A034474, A062394, A034491, A062395, A062396, A007689, A063376, A063481, A074600-A074624, A034524, A178248, A228081 for numbers one more than powers, i.e., this sequence translated from base n (> 2) to base 10.
Cf. A038371 (smallest prime factor), A185121.

Programs

Formula

a(n) = 10*a(n-1) - 9 = A011557(n) + 1 = A002283(n) + 2.
From Mohammad K. Azarian, Jan 02 2009: (Start)
G.f.: 1/(1-x) + 1/(1-10*x).
E.g.f.: exp(x) + exp(10*x). (End)

A036262 Array of numbers read by upward antidiagonals, arising from Gilbreath's conjecture: leading row lists the primes; the following rows give absolute values of differences of previous row.

Original entry on oeis.org

2, 1, 3, 1, 2, 5, 1, 0, 2, 7, 1, 2, 2, 4, 11, 1, 2, 0, 2, 2, 13, 1, 2, 0, 0, 2, 4, 17, 1, 2, 0, 0, 0, 2, 2, 19, 1, 2, 0, 0, 0, 0, 2, 4, 23, 1, 2, 0, 0, 0, 0, 0, 2, 6, 29, 1, 0, 2, 2, 2, 2, 2, 2, 4, 2, 31, 1, 0, 0, 2, 0, 2, 0, 2, 0, 4, 6, 37, 1, 0, 0, 0, 2, 2, 0, 0, 2, 2, 2, 4, 41, 1, 0, 0, 0, 0, 2, 0, 0, 0
Offset: 0

Views

Author

Keywords

Comments

The conjecture is that the leading term is always 1.
Odlyzko has checked it for primes up to pi(10^13) = 3*10^11.
From M. F. Hasler, Jun 02 2012: (Start)
The second column, omitting the initial 3, is given in A089582. The number of "0"s preceding the first term > 1 in the n-th row is given in A213014. The first term > 1 in any row must equal 2, else the conjecture is violated: Obviously all terms except for the first one are even. Thus, if the 2nd term in some row is > 2, it is >= 4, and the first term of the subsequent row is >= 3. If there is a positive number of zeros preceding a first term > 2 (thus >= 4), this "jump" will remain constant and "propagate" (in subsequent rows) to the beginning of the row, and the previously discussed case applies.
The previous statement can also be formulated as: Gilbreath's conjecture is equivalent to: A036277(n) > A213014(n)+2 for all n.
CAVEAT: While table A036261 starts with the first absolute differences of the primes in its first row, the present sequence has the primes themselves in its uppermost row, which is sometimes referred to as "row 0". Thus, "first row" of this table A036262 may either refer to row 1 (1,2,2,...), or to row 0 (2,3,5,7,...), while the latter might, however, as well be referred to "row 1 of A036262" in other sequences or papers.
(End)
From Clark Kimberling, Nov 27 2022: (Start)
Suppose that S = (s(k)), for k >= 1, is a sequence of real numbers. For n >= 1, let g(1,n) = |s(n+1)-s(n)| and g(k,n) = |g(k-1,n+1) - g(k-1,n)| for k >= 2.
Call (g(k,n)) the Gilbreath array of S. Call the first column of this array the Gilbreath transform of S. Denote this transform by G(S), so that G(S) is the sequence (g(n,1)). If S is the sequence of primes, then the Gilbreath conjecture holds that G(S) consists exclusively of 1's. More generally, it appears that there are many S such that G(S) is eventually periodic. See A358691 for conjectured examples. (End)

Examples

			The array begins (conjecture is leading term is always 1):
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101
1 2 2 4  2  4  2  4  6  2  6  4  2  4  6  6  2  6  4  2  6  4  6  8  4   2
1 0 2 2  2  2  2  2  4  4  2  2  2  2  0  4  4  2  2  4  2  2  2  4  2   2
1 2 0 0  0  0  0  2  0  2  0  0  0  2  4  0  2  0  2  2  0  0  2  2  0   0
1 2 0 0  0  0  2  2  2  2  0  0  2  2  4  2  2  2  0  2  0  2  0  2  0   0
1 2 0 0  0  2  0  0  0  2  0  2  0  2  2  0  0  2  2  2  2  2  2  2  0   8
1 2 0 0  2  2  0  0  2  2  2  2  2  0  2  0  2  0  0  0  0  0  0  2  8   8
1 2 0 2  0  2  0  2  0  0  0  0  2  2  2  2  2  0  0  0  0  0  2  6  0   8
1 2 2 2  2  2  2  2  0  0  0  2  0  0  0  0  2  0  0  0  0  2  4  6  8   6
1 0 0 0  0  0  0  2  0  0  2  2  0  0  0  2  2  0  0  0  2  2  2  2  2   4
...
		

References

  • R. K. Guy, Unsolved Problems Number Theory, A10.
  • H. L. Montgomery, Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis, Amer. Math. Soc., 1996, p. 208.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 192.
  • W. Sierpiński, L'induction incomplète dans la théorie des nombres, Scripta Math. 28 (1967), 5-13.
  • C. A. Pickover, The Math Book, Sterling, NY, 2009; see p. 410.

Crossrefs

See A255483 for an interesting generalization.

Programs

  • Haskell
    a036262 n k = delta !! (n - k) !! (k - 1) where delta = iterate
       (\pds -> zipWith (\x y -> abs (x - y)) (tail pds) pds) a000040_list
    -- Reinhard Zumkeller, Jan 23 2011
  • Maple
    A036262 := proc(n, k)
        option remember ;
        if n = 0 then
            ithprime(k) ;
        else
            abs(procname(n-1, k+1)-procname(n-1, k)) ;
        end if;
    end proc:
    seq(seq( A036262(d-k,k),k=1..d),d=1..13) ; # R. J. Mathar, May 10 2023
  • Mathematica
    max = 14; triangle = NestList[ Abs[ Differences[#]] &, Prime[ Range[max]], max]; Flatten[ Table[ triangle[[n - k + 1, k]], {n, 1, max}, {k, 1, n}]] (* Jean-François Alcover, Nov 04 2011 *)

Formula

T(0,k) = A000040(k). T(n,k) = |T(n-1,k+1) - T(n-1,k)|, n > 0. - R. J. Mathar, Sep 19 2013

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 23 2003
Definition edited by N. J. A. Sloane, May 03 2023

A059481 Triangle read by rows. T(n, k) = binomial(n+k-1, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 126, 1, 6, 21, 56, 126, 252, 462, 1, 7, 28, 84, 210, 462, 924, 1716, 1, 8, 36, 120, 330, 792, 1716, 3432, 6435, 1, 9, 45, 165, 495, 1287, 3003, 6435, 12870, 24310, 1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, 92378
Offset: 0

Views

Author

Fabian Rothelius, Feb 04 2001

Keywords

Comments

T(n,k) is the number of ways to distribute k identical objects in n distinct containers; containers may be left empty.
T(n,k) is the number of nondecreasing functions f from {1,...,k} to {1,...,n}. - Dennis P. Walsh, Apr 07 2011
Coefficients of Faber polynomials for function x^2/(x-1). - Michael Somos, Sep 09 2003
Consider k-fold Cartesian products CP(n,k) of the vector A(n)=[1,2,3,...,n].
An element of CP(n,k) is a n-tuple T_t of the form T_t=[i_1,i_2,i_3,...,i_k] with t=1,...,n^k.
We count members T of CP(n,k) which satisfy some condition delta(T_t), so delta(.) is an indicator function which attains values of 1 or 0 depending on whether T_t is to be counted or not; the summation sum_{CP(n,k)} delta(T_t) over all elements T_t of CP produces the count.
For the triangle here we have delta(T_t) = 0 if for any two i_j, i_(j+1) in T_t one has i_j > i_(j+1), T(n,k) = Sum_{CP(n,k)} delta(T_t) = Sum_{CP(n,k)} delta(i_j > i_(j+1)).
The indicator function which tests on i_j = i_(j+1) generates A158497, which contains further examples of this type of counting.
Triangle of the numbers of combinations of k elements with repetitions from n elements {1,2,...,n} (when every element i, i=1,...,n, appears in a k-combination either 0, or 1, or 2, ..., or k times). - Vladimir Shevelev, Jun 19 2012
G.f. for Faber polynomials is -log(-t*x-(1-sqrt(1-4*t))/2+1)=sum(n>0, T(n,k)*t^k/n). - Vladimir Kruchinin, Jul 04 2013
Values of complete homogeneous symmetric polynomials with all arguments equal to 1, or, equivalently, the number of monomials of degree k in n variables. - Tom Copeland, Apr 07 2014
Row k >= 0 of the infinite square array A[k,n] = C(n,k), n >= 0, would start with k zeros in front of the first nonzero element C(k,k) = 1; this here is the triangle obtained by taking the first k+1 nonzero terms C(k .. 2k, k) of rows k = 0, 1, 2, ... of that array. - M. F. Hasler, Mar 05 2017

Examples

			The triangle T(n,k), n >= 0, 0 <= k <= n, begins
  1      A000217
  1 1   /     A000292
  1 2  3    /    A000332
  1 3  6  10    /    A000389
  1 4 10  20  35    /     A000579
  1 5 15  35  70 126     /
  1 6 21  56 126 252  462
  1 7 28  84 210 462  924 1716
  1 8 36 120 330 792 1716 3432 6435
.
T(3,2)=6 considers the CP with the 3^2=9 elements (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3), and does not count the 3 of them which are (2,1),(3,1) and (3,2).
T(3,3) = 10 because the ways to distribute the 3 objects into the three containers are: (3,0,0) (0,3,0) (0,0,3) (2,1,0) (1,2,0) (2,0,1) (1,0,2) (0,1,2) (0,2,1) (1,1,1), for a total of 10 possibilities.
T(3,3)=10 since (x^2/(x-1))^3 = (x+1+1/x+O(1/x^2))^3 = x^3+3x^2+6x+10+O(x).
T(4,2)=10 since there are 10 nondecreasing functions f from {1,2} to {1,2,3,4}. Using <f(1),f(2)> to denote such a function, the ten functions are <1,1>, <1,2>, <1,3>, <1,4>, <2,2>, <2,3>, <2,4>, <3,3>, <3,4>, and <4,4>. - _Dennis P. Walsh_, Apr 07 2011
T(4,0) + T(4,1) + T(4,2) + T(4,3) = 1 + 4 + 10 + 20 = 35 = T(4,4). - _Jonathan Sondow_, Jun 28 2014
From _Paul Curtz_, Jun 18 2018: (Start)
Consider the array
2,    1,    1,    1,    1,    1,     ... = A054977(n)
1,    1/2,  1/3,  1/4,  1/5,  1/6,   ... = 1/(n+1) = 1/A000027(n)
1/3,  1/6,  1/10, 1/15, 1/21, 1/28,  ... = 2/((n+2)*(n+3)) = 1/A000217(n+2)
1/10, 1/20, 1/35, 1/56, 1/84, 1/120, ... = 6/((n+3)*(n+4)*(n+5)) =1/A000292(n+2) (see the triangle T(n,k)).
Every row is an autosequence of the second kind. (See OEIS Wiki, Autosequence.)
By decreasing antidiagonals the denominator of the array is a(n).
Successive vertical denominators: A088218(n), A000984(n), A001700(n), A001791(n+1), A002054(n), A002694(n).
Successive diagonal denominators: A165817(n), A005809(n), A045721(n), A025174(n+1), A004319(n). (End)
Without the first row (2, 1, 1, 1, ... ), the array leads to A165257(n) instead of a(n). - _Paul Curtz_, Jun 19 2018
		

References

  • R. Grimaldi, Discrete and Combinatorial Mathematics, Addison-Wesley, 4th edition, chapter 1.4.

Crossrefs

Columns: T(n,1) = A000027(n), n >= 1. T(n,2) = A000217(n) = A161680(n+1), n >= 2. T(n,3) = A000292(n), n >= 3. T(n,4) = A000332(n+3), n >= 4. T(n,5) = A000389(n+4), n >= 5. T(n,6) = A000579(n+5), n >=6. T(n,k) = A001405(n+k-1) for k <= n <= k+2. [Corrected and extended by M. F. Hasler, Mar 05 2017]
Rows: T(5,k) = A000332(k+4). T(6,k) = A000389(k+5). T(7,k) = A000579(k+6).
Diagonals: T(n,n) = A001700(n-1). T(n,n-1) = A000984(n-1).
T(n,k) = A046899(n-1,k). - R. J. Mathar, Mar 26 2009
Take Pascal's triangle A007318, delete entries to the right of a vertical line just right of center, then scan the diagonals.
For a signed version of this triangle see A027555.
Row sums give A000984.
Cf. A007318, A158497, A100100 (mirrored), A009766.

Programs

  • GAP
    Flat(List([0..10], n->List([0..n], k->Binomial(n+k-1, k)))); # Stefano Spezia, Oct 30 2018
    
  • Haskell
    a059481 n k = a059481_tabl !! n !! n
    a059481_row n = a059481_tabl !! n
    a059481_tabl = map reverse a100100_tabl
    -- Reinhard Zumkeller, Jan 15 2014
    
  • Magma
    &cat [[&*[ Binomial(n+k-1,k)]: k in [0..n]]: n in [0..30] ]; // Vincenzo Librandi, Apr 08 2011
    
  • Maple
    for n from 0 to 10 do for k from 0 to n do print(binomial(n+k-1,k)) ; od: od: # R. J. Mathar, Mar 31 2009
  • Mathematica
    t[n_, k_] := Binomial[n+k-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 09 2013 *)
    (* The combinatorial objects defined in the first comment can, for n >= 1, be generated by: *) r[n_, k_] := FrobeniusSolve[ConstantArray[1,n],k]; (* Peter Luschny, Jan 24 2019 *)
  • Maxima
    sjoin(v, j) := apply(sconcat, rest(join(makelist(j, length(v)), v)))$ display_triangle(n) := for i from 0 thru n do disp(sjoin(makelist(binomial(i+j-1, j), j, 0, i), " ")); display_triangle(10); /* triangle output */ /* Stefano Spezia, Oct 30 2018 */
    
  • PARI
    {T(n, k) = binomial( n+k-1, k)}; \\ Michael Somos, Sep 09 2003, edited by M. F. Hasler, Mar 05 2017
    
  • PARI
    {T(n, k) = if( n<0, 0, polcoeff( Pol(((1 / (x - x^2) + x * O(x^n))^n + O(x)) * x^n), k))}; /* Michael Somos, Sep 09 2003 */
    
  • Sage
    [[binomial(n+k-1,k) for k in range(n+1)] for n in range(11)] # G. C. Greubel, Nov 21 2018

Formula

T(n,0) + T(n,1) + . . . + T(n,n-1) = T(n,n). - Jonathan Sondow, Jun 28 2014
From Peter Bala, Jul 21 2015: (Start)
T(n, k) = Sum_{j = k..n} (-1)^(k+j)*binomial(2*n,n+j)*binomial(n+j-1,j)* binomial(j,k) (gives the correct value T(n,k) = 0 for k > n).
O.g.f.: 1/2*( x*(2*x - 1)/(sqrt(1 - 4*t*x)*(1 - x - t)) + (1 + 2*x)/sqrt(1 - 4*t*x) + (1 - t)/(1 - x - t) ) = 1 + (1 + t)*x + (1 + 2*t + 3*t^2)*x^2 + (1 + 3*t + 6*t^2 + 10*t^3)*x^3 + ....
n-th row polynomial R(n,t) = [x^n] ( (1 + x)^2/(1 + x(1 - t)) )^n.
exp( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (1 + t)*x + (1 + 2*t + 2*t^2)*x^2 + (1 + 3*t + 5*t^2 + 5*t^3)*x^3 + ... is the o.g.f for A009766. (End)
a(n) = abs(A027555(n)). - M. F. Hasler, Mar 05 2017
For n >= k > 0, T(n, k) = Sum_{j=1..n} binomial(k + j - 2, k - 1) = Sum_{j=1..n} A007318(k + j - 2, k - 1). - Stefano Spezia, Oct 30 2018
T(n, k) = RisingFactorial(n, k) / k!. - Peter Luschny, Nov 24 2023

Extensions

Offset changed from 1 to 0 by R. J. Mathar, Jan 15 2013
Edited by M. F. Hasler, Mar 05 2017

A271466 Number T(n,k) of set partitions of [n] such that k is the largest element of the last block; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 2, 0, 1, 4, 0, 1, 4, 10, 0, 1, 6, 15, 30, 0, 1, 10, 29, 59, 104, 0, 1, 18, 63, 139, 250, 406, 0, 1, 34, 149, 365, 692, 1145, 1754, 0, 1, 66, 375, 1039, 2110, 3627, 5649, 8280, 0, 1, 130, 989, 3149, 6932, 12521, 20085, 29874, 42294, 0, 1, 258, 2703, 10039, 24190, 46299, 77133, 117488, 168509, 231950
Offset: 1

Views

Author

Alois P. Heinz, Apr 08 2016

Keywords

Comments

Each set partition is written as a sequence of blocks, ordered by the smallest elements in the blocks.

Examples

			T(1,1) = 1: 1.
T(2,2) = 2: 12, 1|2.
T(3,2) = 1: 13|2.
T(3,3) = 4: 123, 12|3, 1|23, 1|2|3.
T(4,2) = 1: 134|2.
T(4,3) = 4: 124|3, 14|23, 14|2|3, 1|24|3.
T(4,4) = 10: 1234, 123|4, 12|34, 12|3|4, 13|24, 13|2|4, 1|234, 1|23|4, 1|2|34, 1|2|3|4.
T(5,2) = 1: 1345|2.
T(5,3) = 6: 1245|3, 145|23, 145|2|3, 14|25|3, 15|24|3, 1|245|3.
T(5,4) = 15: 1235|4, 125|34, 125|3|4, 12|35|4, 135|24, 135|2|4, 13|25|4, 15|234, 15|23|4, 1|235|4, 15|2|34, 1|25|34, 15|2|3|4, 1|25|3|4, 1|2|35|4.
Triangle T(n,k) begins:
  1;
  0, 2;
  0, 1,   4;
  0, 1,   4,  10;
  0, 1,   6,  15,   30;
  0, 1,  10,  29,   59,  104;
  0, 1,  18,  63,  139,  250,   406;
  0, 1,  34, 149,  365,  692,  1145,  1754;
  0, 1,  66, 375, 1039, 2110,  3627,  5649,  8280;
  0, 1, 130, 989, 3149, 6932, 12521, 20085, 29874, 42294;
  ...
		

Crossrefs

Columns k=1-10 give: A000007(n-1), A054977(n-2), A052548(n-3) for n>3, A271743, A271744, A271745, A271746, A271747, A271748, A271749.
Main diagonal gives A186021(n-1).
Lower diagonals d=1-10 give: A271752, A271753, A271754, A271755, A271756, A271757, A271758, A271759, A271760, A271761.
Row sums give A000110.
T(2n,n) gives A271467.
T(2n+1,n+1) gives A271607.
Cf. A095149 (k is maximum of the first block), A113547 (k is minimum of the last block).

Programs

  • Maple
    b:= proc(n, m, c) option remember; `if`(n=0, x^c, add(
          b(n-1, max(m, j), `if`(j>=m, n, c)), j=1..m+1))
        end:
    T:= n-> (p-> seq(coeff(p, x, n-i), i=0..n-1))(b(n, 0$2)):
    seq(T(n), n=1..12);
  • Mathematica
    b[n_, m_, c_] := b[n, m, c] = If[n == 0, x^c, Sum[b[n-1, Max[m, j], If[j >= m, n, c]], {j, 1, m+1}]];
    T[n_] := Function[p, Table[Coefficient[p, x, n-i], {i, 0, n-1}]][b[n, 0, 0]];
    Table[T[n], {n, 1, 12}] // Flatten (* Jean-François Alcover, Apr 24 2016, translated from Maple *)

Formula

T(n,n) = 2 * A000110(n-1) = 2 * Sum_{j=0..n-1} T(n-1,j) for n>1.

A161739 The RSEG2 triangle.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 4, 1, 0, 0, 13, 10, 1, 0, -4, 30, 73, 20, 1, 0, 0, -14, 425, 273, 35, 1, 0, 120, -504, 1561, 3008, 798, 56, 1, 0, 0, 736, -2856, 25809, 14572, 1974, 84, 1, 0, -12096, 44640, -73520, 125580, 218769, 55060, 4326, 120, 1
Offset: 0

Views

Author

Johannes W. Meijer & Nico Baken (n.h.g.baken(AT)tudelft.nl), Jun 18 2009

Keywords

Comments

The EG2[2*m,n] matrix coefficients were introduced in A008955. We discovered that EG2[2m,n] = Sum_{k = 1..n} (-1)^(k+n)*t1(n-1,k-1)*2*eta(2*m-2*n+2*k)/((n-1)!)^2 with t1(n,m) the central factorial numbers A008955 and eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function.
A different way to define these matrix coefficients is EG2[2*m,n] = (1/m)*Sum_{k = 0..m-1} ZETA(2*m-2*k, n-1)*EG2[2*k, n] with ZETA(2*m, n-1) = zeta(2*m) - Sum_{k = 1..n-1} (k)^(-2*m) and EG2[0, n] = 1, for m = 0, 1, 2, ..., and n = 1, 2, 3, ... .
We define the row sums of the EG2 matrix rs(2*m,p) = Sum_{n >= 1} (n^p)*EG2(2*m,n) for p = -2, -1, 0, 1, ... and m >= p+2. We discovered that rs(2*m,p=-2) = 2*eta(2*m+2) = (1 - 2^(1-(2*m+2)))*zeta(2*m+2). This formula is quite unlike the other rs(2*m,p) formulas, see the examples.
The series expansions of the row generators RGEG2(z,2*m) about z = 0 lead to the EG2[2*m,n] coefficients while the series expansions about z = 1 lead to the ZG1[2*m-1,n] coefficients, see the formulas.
The first Maple program gives the triangle coefficients. Adding the second program to the first one gives information about the row sums rs(2*m,p).
The a(n) formulas of the right hand columns are related to sequence A036283, see also A161740 and A161741.

Examples

			The first few expressions for the ZG1[2*m-1,p+1] coefficients are:
  ZG1[2*m-1, 1] = (zeta(2*m-1))/(1/2)
  ZG1[2*m-1, 2] = (zeta(2*m-3) - zeta(2*m-1))/1
  ZG1[2*m-1, 3] = (zeta(2*m-5) - 5*zeta(2*m-3) + 4*zeta(2*m-1))/6
  ZG1[2*m-1, 4] = (zeta(2*m-7) - 14*zeta(2*m-5) + 49*zeta(2*m-3) - 36*zeta(2*m-1))/72
The first few rs(2*m,p) are (m >= p+2)
  rs(2*m, p=0) = ZG1[2*m-1,1]
  rs(2*m, p=1) = ZG1[2*m-1,1] + ZG1[2*m-1,2]
  rs(2*m, p=2) = ZG1[2*m-1,1] + 3*ZG1[2*m-1,2] + 2*ZG1[2*m-1,3]
  rs(2*m, p=3) = ZG1[2*m-1,1] + 7*ZG1[2*m-1,2] + 12*ZG1[2*m-1,3] + 6*ZG1[2*m-1,4]
The first few rs(2*m,p) are (m >= p+2)
  rs(2*m, p=-1) = zeta(2*m+1)/(1/2)
  rs(2*m, p=0) = zeta(2*m-1)/(1/2)
  rs(2*m, p=1) = (zeta(2*m-1) + zeta(2*m-3))/1
  rs(2*m, p=2) = (zeta(2*m-1) + 4*zeta(2*m-3) + zeta(2*m-5))/3
  rs(2*m, p=3) = (0*zeta(2*m-1) + 13*zeta(2*m-3) + 10*zeta(2*m-5) + zeta(2*m-7))/12
The first few rows of the RSEG2 triangle are:
  [1]
  [0, 1]
  [0, 1, 1]
  [0, 1, 4, 1]
  [0, 0, 13, 10, 1]
  [0, -4, 30, 73, 20, 1]
		

Crossrefs

A000007, A129825, A161742 and A161743 are the first four left hand columns.
A000012, A000292, A107963, A161740 and A161741 are the first five right hand columns.
A010790 equals 2*r(n) and A054977 equals denom(r(n)).
A001710 equals numer(q(n)) and A141044 equals denom(q(n)).
A000142 equals the row sums.
A008955 is a central factorial number triangle.
A028246 is Worpitzky's triangle.

Programs

  • Maple
    nmax:=10; for n from 0 to nmax do A008955(n, 0) := 1 end do: for n from 0 to nmax do A008955(n, n) := (n!)^2 end do: for n from 1 to nmax do for m from 1 to n-1 do A008955(n, m) := A008955(n-1, m-1)*n^2 + A008955(n-1, m) end do: end do: for n from 1 to nmax do A028246(n, 1) := 1 od: for n from 1 to nmax do A028246(n, n) := (n-1)! od: for n from 3 to nmax do for m from 2 to n-1 do A028246(n, m) := m*A028246(n-1, m) + (m-1)*A028246(n-1, m-1) od: od: for i from 0 to nmax-2 do s(i) := ((i+1)!/2)*sum(A028246(i+1, k1+1)*(sum((-1)^(j)*A008955(k1, j)*2*x^(2*nmax-(2*k1+1-2*j)), j=0..k1)/ (k1!*(k1+1)!)), k1=0..i) od: a(0,0) := 1: for n from 1 to nmax-1 do for m from 0 to n do a(n,m) := coeff(s(n-1), x, 2*nmax-1-2*m+2) od: od: seq(seq(a(n, m), m=0..n), n=0..nmax-1); for n from 0 to nmax-1 do seq(a(n, m), m=0..n) od;
    m:=7: row := 2*m; rs(2*m, -2) := 2*eta(2*m+2); for p from -1 to m-2 do q(p+1) := (p+1)!/2 od: for p from -1 to m-2 do rs(2*m, p) := sum(a(p+1, k)*zeta(2*m+1-2*k), k=0..p+1)/q(p+1) od;

Formula

RGEG2(2*m,z) = Sum_{n >= 1} EG2[2*m,n]*z^(n-1) = Integral_{y = 0..oo}((2*y)^(2*m)/(2*m)!)* cosh(y)/(cosh(y)^2 - z)^(3/2) for m >= 0.
EG2[2*m,n] = Sum_{k = 1..n} (-1)^(k+n)* A008955(n-1, k-1)*2*eta(2*m-2*n+2*k)/((n-1)!)^2.
ZG1[2*m-1,p+1] = Sum_{j = 0..p} (-1)^j*A008955(p, j)*zeta(2*m-(2*p+1-2*j))/ r(p) with r(p)= p!*(p+1)!/2 and p >= 0.
rs(2*m,p) = Sum_{k = 0..p} A028246(p+1,k+1)*ZG1[2*m-1,k+1] and p >= 0; p <= m-2.
rs(2*m,p) = Sum_{k = 0..p+1} A161739(p+1,k)*zeta(2*m+1-2*k)/q(p+1) with q(p+1) = (p+1)!/2 and p >= -1; p <= m-2.
From Peter Bala, Mar 19 2022: (Start)
It appears that the k-th row polynomial (with indexing starting at k = 1) is given by R(k,n^2) = (k-1)!*Sum_{i = 0..n} (-1)^(n-i)*(i^k)* binomial(n,i)*binomial(n+i,i)/(n+i) for n >= 1.
For example, for k = 6, Maple's SumTools:-Summation procedure gives 5!*Sum_{i = 0..n} (-1)^(n-i)*(i^6)*binomial(n,i)*binomial(n+i,i)/(n+i) = -4*n^2 + 30*n^4 + 73*n^6 + 20*n^8 + n^10 = R(6,n^2). (End)

Extensions

Minor error corrected and edited by Johannes W. Meijer, Sep 22 2012
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