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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A104035 Triangle T(n,k), 0 <= k <= n, read by rows, defined by T(0,0) = 1; T(0,k) = 0 if k>0 or if k<0; T(n,k) = k*T(n-1,k-1) + (k+1)*T(n-1,k+1).

Original entry on oeis.org

1, 0, 1, 1, 0, 2, 0, 5, 0, 6, 5, 0, 28, 0, 24, 0, 61, 0, 180, 0, 120, 61, 0, 662, 0, 1320, 0, 720, 0, 1385, 0, 7266, 0, 10920, 0, 5040, 1385, 0, 24568, 0, 83664, 0, 100800, 0, 40320, 0, 50521, 0, 408360, 0, 1023120, 0, 1028160, 0, 362880, 50521, 0, 1326122, 0, 6749040
Offset: 0

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Author

Philippe Deléham, Apr 06 2005

Keywords

Comments

Or, triangle of coefficients (with exponents in increasing order) in polynomials Q_n(u) defined by d^n sec x / dx^n = Q_n(tan x)*sec x.
Interpolates between factorials and Euler (or secant) numbers. Related to Springer numbers.
Companion triangles are A155100 (derivative polynomials of tangent function) and A185896 (derivative polynomials of squared secant function).
A combinatorial interpretation for the polynomial Q_n(u) as the generating function for a sign change statistic on certain types of signed permutation can be found in [Verges]. A signed permutation is a sequence (x_1,x_2,...,x_n) of integers such that {|x_1|,|x_2|,...,|x_n|} = {1,2,...,n}. They form a group, the hyperoctahedral group of order 2^n*n! = A000165(n), isomorphic to the group of symmetries of the n dimensional cube.
Let x_1,...,x_n be a signed permutation. Adjoin x_0 = 0 to the front of the permutation and x_(n+1) = (-1)^n*(n+1) to the end to form x_0,x_1,...,x_n,x_(n+1). Then x_0,x_1,...,x_n,x_(n+1) is a snake of type S(n;0) when x_0 < x_1 > x_2 < ... x_(n+1). For example, 0 3 -1 2 -4 is a snake of type S(3;0).
Let sc be the number of sign changes through a snake ... sc = #{i, 0 <= i <= n, x_i*x_(i+1) < 0}. For example, the snake 0 3 -1 2 -4 has sc = 3. The polynomial Q_n(u) is the generating function for the sign change statistic on snakes of type S(n;0): ... Q_n(u) = sum {snakes in S(n;0)} u^sc. See the example section below for the cases n = 2 and n = 3.
PRODUCTION MATRIX
Let D = subdiag(1,2,3,...) be the array with the indicated sequence on the first subdiagonal and zeros elsewhere and let C = transpose(D). The production matrix for this triangle is C+D: the first row of (C+D)^n is the n-th row of this triangle. D represents the derivative operator d/dx and C represents the operator p(x) -> x*d/dx(x*p(x)) acting on the basis monomials {x^n}n>=0. See Formula (1) below.

Examples

			The polynomials Q_0(u) through Q_6(u) (with exponents in decreasing order) are:
  1
  u
  2*u^2 + 1
  6*u^3 + 5*u
  24*u^4 + 28*u^2 + 5
  120*u^5 + 180*u^3 + 61*u
  720*u^6 + 1320*u^4 + 662*u^2 + 61
Triangle begins:
  1
  0 1
  1 0 2
  0 5 0 6
  5 0 28 0 24
  0 61 0 180 0 120
  61 0 662 0 1320 0 720
  0 1385 0 7266 0 10920 0 5040
  1385 0 24568 0 83664 0 100800 0 40320
  0 50521 0 408360 0 1023120 0 1028160 0 362880
  50521 0 1326122 0 6749040 0 13335840 0 11491200 0 3628800
  0 2702765 0 30974526 0 113760240 0 185280480 0 139708800 0 39916800
  2702765 0 98329108 0 692699304 0 1979524800 0 2739623040 0 1836172800 0 479001600
Examples of sign change statistic sc on snakes of type S(n;0)
= = = = = = = = = = = = = = = = = = = = = =
.....Snakes....# sign changes sc.......u^sc
= = = = = = = = = = = = = = = = = = = = = =
n=2
...0 1 -2 3...........2.................u^2
...0 2  1 3...........0.................1
...0 2 -1 3...........2.................u^2
yields Q_2(u) = 2*u^2 + 1.
n=3
...0 1 -2  3 -4.......3.................u^3
...0 1 -3  2 -4.......3.................u^3
...0 1 -3 -2 -4.......1.................u
...0 2  1  3 -4.......1.................u
...0 2 -1  3 -4.......3.................u^3
...0 2 -3  1 -4.......3.................u^3
...0 2 -3 -2 -4.......1.................u
...0 3  1  2 -4.......1.................u
...0 3 -1  2 -4.......3.................u^3
...0 3 -2  1 -4.......3.................u^3
...0 3 -2 -1 -4.......1.................u
yields Q_3(u) = 6*u^3 + 5*u.
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998, p. 287.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see pp. 445 and 469.

Crossrefs

See A008294 for another version of this triangle.
Setting u=0,1,2,3,4 gives A000364, A001586, A156129, A156131, A156132.
Setting u=sqrt(2) gives A156134 and A156138; u=sqrt(3) gives A002437 and A002439.

Programs

Formula

T(n, n) = n!; T(n, 0) = 0 if n = 2m + 1; T(n, 0) = A000364(m) if n = 2m.
Sum_{k>=0} T(m, k)*T(n, k) = T(m+n, 0).
Sum_{k>=0} T(n, k) = A001586(n): Springer numbers.
G.f.: Sum_{n >= 0} Q_n(u)*t^n/n! = 1/(cos t - u sin t).
From Peter Bala: (Start)
RECURRENCE RELATION
For n>=0,
(1)... Q_(n+1)(u) = d/du Q_n(u) + u*d/du(u*Q_n(u))
... = (1+u^2)*d/du Q_n(u) + u*Q_n(u),
with starting condition Q_0(u) = 1. Compare with Formula (4) of A186492.
RELATION WITH TYPE B EULERIAN NUMBERS
(2)... Q_n(u) = ((u+i)/2)^n*B(n,(u-i)/(u+i)), where i = sqrt(-1) and
[B(n,u)]n>=0 = [1,1+u,1+6*u+u^2,1+23*u+23*u^2+u^3,...] is the sequence of type B Eulerian polynomials (with a factor of u removed) - see A060187.
(End)
T(n,0) = abs(A122045(n)). - Reinhard Zumkeller, Apr 27 2014

Extensions

Entry revised by N. J. A. Sloane, Nov 06 2009

A097592 Triangle read by rows: T(n,k) is the number of permutations of [n] with exactly k increasing runs of even length.

Original entry on oeis.org

1, 1, 1, 1, 2, 4, 7, 12, 5, 25, 52, 43, 102, 299, 258, 61, 531, 1750, 1853, 906, 3141, 11195, 15634, 8965, 1385, 20218, 83074, 133697, 94398, 31493, 146215, 675304, 1207256, 1088575, 460929, 50521, 1174889, 5880354, 11974457, 12625694, 6632158
Offset: 0

Views

Author

Emeric Deutsch, Aug 29 2004

Keywords

Comments

Row n has 1+floor(n/2) entries.

Examples

			Triangle starts:
    1;
    1;
    1,   1;
    2,   4;
    7,  12,   5;
   25,  52,  43;
  102, 299, 258, 61;
Example: T(4,2) = 5 because we have 13/24, 14/23, 23/14, 24/13 and 34/12.
		

Crossrefs

Row sums give A000142.
T(n,floor(n/2)) gives A317139.
T(2n,n) gives A000364.
T(2n+1,n) gives A317140.

Programs

  • Maple
    G:=2*(t-1)*u/(-2*u+(2-t+t*u)*exp((-1+u)*x/2)+(t-2+t*u)*exp(-(1+u)*x/2)): u:=sqrt(5-4*t): Gser:=simplify(series(G,x=0,12)): P[0]:=1: for n from 1 to 11 do P[n]:=sort(n!*coeff(Gser,x^n)) od: seq(seq(coeff(t*P[n],t^k),k=1..1+floor(n/2)),n=0..11);
    # second Maple program:
    b:= proc(u, o, t) option remember; `if`(u+o=0, x^t, expand(
          add(b(u+j-1, o-j, irem(t+1, 2)), j=1..o)+
          add(b(u-j, o+j-1, 0)*x^t, j=1..u)))
        end:
    T:= n->(p->seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0$2)):
    seq(T(n), n=0..12);  # Alois P. Heinz, Nov 19 2013
  • Mathematica
    b[u_, o_, t_] := b[u, o, t] = If[u+o == 0, x^t, Expand[Sum[b[u+j-1, o-j, Mod[t+1, 2]], {j, 1, o}] + Sum[b[u-j, o+j-1, 0]*x^t, {j, 1, u}]]]; T[n_] := Function[ {p}, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, 0, 0]]; Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Apr 29 2015, after Alois P. Heinz *)

Formula

E.g.f.: 2(t-1)u/[ -2u+(2-t+tu)exp((-1+u)x/2)+(t-2+tu)exp(-(1+u)x/2)], where u=sqrt(5-4t).
Sum_{k=1..floor(n/2)} k * T(n,k) = A097593(n). - Alois P. Heinz, Jul 04 2019

A000281 Expansion of cos(x)/cos(2x).

Original entry on oeis.org

1, 3, 57, 2763, 250737, 36581523, 7828053417, 2309644635483, 898621108880097, 445777636063460643, 274613643571568682777, 205676334188681975553003, 184053312545818735778213457, 193944394596325636374396208563
Offset: 0

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Author

Keywords

Comments

a(n) is (2n)! times the coefficient of x^(2n) in the Taylor series for cos(x)/cos(2x).

Examples

			cos x / cos 2*x = 1 + 3*x^2/2 + 19*x^4/8 + 307*x^6/80 + ...
		

References

  • J. W. L. Glaisher, "On the coefficients in the expansions of cos x / cos 2x and sin x / cos 2x", Quart. J. Pure and Applied Math., 45 (1914), 187-222.
  • I. J. Schwatt, Intro. to Operations with Series, Chelsea, p. 278.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    a := n -> (-1)^n*2^(6*n+1)*(Zeta(0,-2*n,1/8)-Zeta(0,-2*n,5/8)):
    seq(a(n), n=0..13); # Peter Luschny, Mar 11 2015
  • Mathematica
    With[{nn=30},Take[CoefficientList[Series[Cos[x]/Cos[2x],{x,0,nn}],x] Range[0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Oct 06 2011 *)
  • PARI
    {a(n) = if( n<0, 0, n*=2; n! * polcoeff( cos(x + x * O(x^n)) / cos(2*x + x * O(x^n)), n))}; /* Michael Somos, Feb 09 2006 */

Formula

a(n) = Sum_{k=0..n} (-1)^k*binomial(2n, 2k)*A000364(n-k)*4^(n-k). - Philippe Deléham, Jan 26 2004
E.g.f.: Sum_{k>=0} a(k)x^(2k)/(2k)! = cos(x)/cos(2x).
a(n-1) is approximately 2^(4*n-3)*(2*n-1)!*sqrt(2)/((Pi^(2*n-1))*(2*n-1)). The approximation is quite good a(250) is of the order of 10^1181 and this formula is accurate to 238 digits. - Simon Plouffe, Jan 31 2007
G.f.: 1 / (1 - 1*3*x / (1 - 4*4*x / (1 - 5*7*x / (1 - 8*8*x / (1 - 9*11*x / ... ))))). - Michael Somos, May 12 2012
G.f.: 1/E(0) where E(k) = 1 - 3*x - 16*x*k*(2*k+1) - 16*x^2*(k+1)^2*(4*k+1)*(4*k+3)/E(k+1) (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 17 2012
G.f.: T(0)/(1-3*x), where T(k) = 1 - 16*x^2*(4*k+1)*(4*k+3)*(k+1)^2/( 16*x^2*(4*k+1)*(4*k+3)*(k+1)^2 - (32*x*k^2+16*x*k+3*x-1 )*(32*x*k^2+80*x*k+51*x -1)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 11 2013
From Peter Bala, Mar 09 2015: (Start)
a(n) = (-1)^n*4^(2*n)*E(2*n,1/4), where E(n,x) denotes the n-th Euler polynomial.
O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 + x*(4*k + 1)^2) = 1 + 3*x + 57*x^2 + 2763*x^3 + ....
We appear to have the asymptotic expansion Pi/(2*sqrt(2)) - Sum {k = 0..n - 1} (-1)^floor(k/2)/(2*k + 1) ~ 1/(2*n) - 3/(2*n)^3 + 57/(2*n)^5 - 2763/(2*n)^7 + .... See A093954.
Bisection of A001586. See also A188458 and A212435. Second row of A235605 (read as a square array).
The expansion of exp( Sum_{n >= 1} a(n)*x^n/n ) appears to have integer coefficients. See A255883. (End)
From Peter Luschny, Mar 11 2015: (Start)
a(n) = ((-64)^n/((n+1/2)))*(B(2*n+1,7/8)-B(2*n+1,3/8)), B(n,x) Bernoulli polynomials.
a(n) = 2*(-16)^n*LerchPhi(-1, -2*n, 1/4).
a(n) = (-1)^n*Sum_{0..2*n} 2^k*C(2*n,k)*E(k), E(n) the Euler secant numbers A122045.
a(n) = (-4)^n*SKP(2*n,1/2) where SKP are the Swiss-Knife polynomials A153641.
a(n) = (-1)^n*2^(6*n+1)*(Zeta(-2*n,1/8) - Zeta(-2*n,5/8)), where Zeta(a,z) is the generalized Riemann zeta function. (End)
From Peter Bala, May 13 2017: (Start)
G.f.: 1/(1 + x - 4*x/(1 - 12*x/(1 + x - 40*x/(1 - 56*x/(1 + x - ... - 4*n(4*n - 3)*x/(1 - 4*n(4*n - 1)*x/(1 + x - ...
G.f.: 1/(1 + 9*x - 12*x/(1 - 4*x/(1 + 9*x - 56*x/(1 - 40*x/(1 + 9*x - ... - 4*n(4*n - 1)*x/(1 - 4*n(4*n - 3)*x/(1 + 9*x - .... (End)
From Peter Bala, Nov 08 2019: (Start)
a(n) = sqrt(2)*4^n*Integral_{x = 0..inf} x^(2*n)*cosh(Pi*x/2)/cosh(Pi*x) dx. Cf. A002437.
The L-series 1 + 1/3^(2*n+1) - 1/5^(2*n+1) - 1/7^(2*n+1) + + - - ... = sqrt(2)*(Pi/4)^(2*n+1)*a(n)/(2*n)! (see Shanks), which gives a(n) ~ (1/sqrt(2))*(2*n)!*(4/Pi)^(2*n+1). (End)

A094665 Another version of triangular array in A083061: triangle T(n,k), 0<=k<=n, read by rows; given by [0, 1, 3, 6, 10, 15, 21, 28, ...] DELTA [1, 2, 3, 4, 5, 6, 7, 8, ...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 4, 15, 15, 0, 34, 147, 210, 105, 0, 496, 2370, 4095, 3150, 945, 0, 11056, 56958, 111705, 107415, 51975, 10395, 0, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135, 0, 14873104, 85389132, 197722980, 244909665, 178378200, 77567490, 18918900, 2027025
Offset: 0

Views

Author

Philippe Deléham, Jun 07 2004, Jun 12 2007

Keywords

Comments

Define polynomials P(n,x) = x(2x+1)P(n-1,x+1) - 2x^2P(n-1,x), P(0,x) = 1. Sequence gives triangle read by rows, defined by P(n,x) = Sum_{k = 0..n} T(n,k)*x^k. - Philippe Deléham, Jun 20 2004
From Johannes W. Meijer, May 24 2009: (Start)
In A160464 we defined the coefficients of the ES1 matrix by ES1[2*m-1,n=1] = 2*eta(2*m-1) and the recurrence relation ES1[2*m-1,n] = ((2*n-2)/(2*n-1))*(ES1[2*m-1,n-1] - ES1[2*m-3,n-1]/(n-1)^2) for m the positive and negative integers and n >= 1. As usual eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. It is well-known that ES1[1-2*m,n=1] = (4^m-1)*(-bernoulli(2*m))/m for m >= 1. and together with the recurrence relation this leads to ES1[-1,n] = 0.5 for n >= 1.
We discovered that the n-th term of the row coefficients ES1[1-2*m,n] for m >= 1, can be generated with the rather simple polynomials RES1(1-2*m,n) = (-1)^(m+1)*ECGP(1-2*m, n)/2^m. This discovery was enabled by the recurrence relation for the RES1(1-2*m,n) which we derived from the recurrence relation for the ES1[2*m-1,n] coefficients and the fact that RES1(-1,n) = 0.5. The coefficients of the ECGP(1-2*m,n) polynomials led to this triangle and subsequently to triangle A083061. (End)
From David Callan, Jan 03 2011: (Start)
T(n,k) is the number of increasing 0-2 trees (A002105) on 2n edges in which the minimal path from the root has length k.
Proof. The number a(n,k) of such trees satisfies the recurrence a(0,0)=1, a(1,1)=1 and, counting by size of the subtree rooted at the smaller child of the root,
a(n,k) = Sum_{j=1..n-1} C(2n-1,j)*a(j,k-1)*a(n-1-j)
for 2<=k<=n, where a(n) = Sum_{k>=0} a(n,k) is the reduced tangent number A002105 (indexed from 0). The recurrence translates into the differential equation
F_x(x,y) = y*F(x,y)*G(x)
for the GF F(x,y) = Sum_{n,k>=0} a(n,k)x^(2n)/(2n)!*y^k, where G(x):=Sum_{n>=0} a(n)x^(2n+1)/(2n+1)! is known to be sqrt(2)*tan(x/sqrt(2)). The differential equation has solution F(x,y) = sec(x/sqrt(2))^(2y). (End)

Examples

			Triangle begins:
.1;
.0, 1;
.0, 1, 3;
.0, 4, 15, 15;
.0, 34, 147, 210, 105;
.0, 496, 2370, 4095, 3150, 945;
.0, 11056, 56958, 111705, 107415, 51975, 10395;
.0, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135;
From _Johannes W. Meijer_, May 24 2009: (Start)
The first few ECGP(1-2*m,n) polynomials are: ECGP(-1,n) = 1; ECGP(-3,n) = n; ECGP(-5,n) = n + 3*n^2; ECGP(-7,n) = 4*n + 15*n^2+ 15*n^3 .
The first few RES1(1-2*m,n) are: RES1(-1,n) = (1/2)*(1); RES1(-3,n) = (-1/4)*(n); RES1(-5,n) = (1/8)*(n+3*n^2); RES1(-7,n) = (-1/16)*(4*n+15*n^2+15*n^3).
(End)
		

Crossrefs

From Johannes W. Meijer, May 24 2009 and Jun 27 2009: (Start)
A001147, A001880, A160470, A160471 and A160472 are the first five right hand columns.
Appears in A162005, A162006 and A162007.
(End)

Programs

  • Maple
    nmax:=7; imax := nmax: T1(0, x) := 1: T1(0, x+1) := 1: for i from 1 to imax do T1(i, x) := expand((2*x+1) * (x+1) * T1(i-1, x+1) - 2 * x^2 * T1(i-1, x)): dx:=degree(T1(i, x)): for k from 0 to dx do c(k) := coeff(T1(i, x), x, k) od: T1(i, x+1) := sum(c(j1)*(x+1)^(j1), j1=0..dx) od: for i from 0 to imax do for j from 0 to i do A083061(i, j) := coeff(T1(i, x), x, j) od: od: for n from 0 to nmax do for k from 0 to n do T(n+1, k+1) := A083061(n, k) od: od: T(0, 0):=1: for n from 1 to nmax do T(n, 0):=0 od: seq(seq(T(n, k), k=0..n), n=0..nmax);
    # Johannes W. Meijer, Jun 27 2009, revised Sep 23 2012
  • Mathematica
    nmax = 8;
    T[n_, k_] := SeriesCoefficient[Sec[x/Sqrt[2]]^(2y), {x, 0, 2n}, {y, 0, k}]* (2n)!;
    Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 10 2018 *)

Formula

Sum_{k = 0..n} T(n, k) = A002105(n+1).
Sum_{k = 0..n} T(n, k)*2^(n-k) = A000364(n); Euler numbers.
Sum_{k = 0..n} T(n, k)*(-2)^(n-k) = 1.
RES1(1-2*m,n) = n^2*RES1(3-2*m,n)-n*(2*n+1)*RES1(3-2*m,n+1)/2 for m >= 2, with RES1(-1,n) = 0.5 for n >= 1. - Johannes W. Meijer, May 24 2009
G.f.: Sum_{n,k>=0} T(n,k)x^n/n!*y^k = sec(x/sqrt(2))^(2y).

Extensions

Term corrected by Johannes W. Meijer, Sep 23 2012

A235605 Shanks's array c_{a,n} (a >= 1, n >= 0) that generalizes Euler and class numbers, read by antidiagonals upwards.

Original entry on oeis.org

1, 1, 1, 1, 3, 5, 1, 8, 57, 61, 2, 16, 352, 2763, 1385, 2, 30, 1280, 38528, 250737, 50521, 1, 46, 3522, 249856, 7869952, 36581523, 2702765, 2, 64, 7970, 1066590, 90767360, 2583554048, 7828053417, 199360981, 2, 96, 15872, 3487246, 604935042, 52975108096
Offset: 1

Views

Author

N. J. A. Sloane, Jan 22 2014

Keywords

Examples

			The array begins:
A000364: 1, 1,    5,     61,       1385,         50521,          2702765,..
A000281: 1, 3,   57,   2763,     250737,      36581523,       7828053417,..
A000436: 1, 8,  352,  38528,    7869952,    2583554048,    1243925143552,..
A000490: 1,16, 1280, 249856,   90767360,   52975108096,   45344872202240,..
A000187: 2,30, 3522,1066590,  604935042,  551609685150,  737740947722562,..
A000192: 2,46, 7970,3487246, 2849229890, 3741386059246, 7205584123783010,..
A064068: 1,64,15872,9493504,10562158592,18878667833344,49488442978598912,..
...
		

Crossrefs

Columns: A000003 (class numbers), A000233, A000362, A000508, ...
Cf. A235606.

Programs

  • Mathematica
    amax = 10; nmax = amax-1; km0 = 10; Clear[cc]; L[a_, s_, km_] := Sum[ JacobiSymbol[-a, 2k+1]/(2k+1)^s, {k, 0, km}]; c[1, n_, km_] := 2(2n)! L[1, 2n+1, km] (2/Pi)^(2n+1) // Round; c[a_ /; a>1, n_, km_] := (2n)! L[a, 2n+1, km] (2a/Pi)^(2n+1)/Sqrt[a] // Round; cc[km_] := cc[km] = Table[ c[a, n, km], {a, 1, amax}, {n, 0, nmax}]; cc[km0]; cc[km = 2km0]; While[ cc[km] != cc[km/2, km = 2km]]; A235605[a_, n_] := cc[km][[a, n+1 ]]; Table[ A235605[ a-n, n], {a, 1, amax}, {n, 0, a-1}] // Flatten (* Jean-François Alcover, Feb 05 2016 *)
    ccs[b_, nm_] := With[{ns = Range[0, nm]}, (-1)^ns If[Mod[b, 4] == 3, Sum[JacobiSymbol[k, b] (b - 4 k)^(2 ns), {k, 1, (b - 1)/2}], Sum[JacobiSymbol[-b, 2 k + 1] (b - (2 k + 1))^(2 ns), {k, 0, (b - 2)/2}]]];
    csfs[1, nm_] := csfs[1, nm] = (2 Range[0, nm])! CoefficientList[Series[Sec[x], {x, 0, 2 nm}], x^2];
    csfs[b_, nm_] := csfs[b, nm] = Fold[Function[{cs, cc}, Append[cs, cc - Sum[cs[[-i]] (-b^2)^i Binomial[2 Length[cs], 2 i], {i, Length[cs]}]]], {}, ccs[b, nm]];
    rowA235605[a_, nm_] := With[{facs = FactorInteger[a], ns = Range[0, nm]}, With[{b = Times @@ (#^Mod[#2, 2] &) @@@ facs}, If[a == b, csfs[b, nm], If[b == 1, 1/2, 1] csfs[b, nm] Sqrt[a/b]^(4 ns + 1) Times @@ Cases[facs, {p_, e_} /; p > 2 && e > 1 :> 1 - JacobiSymbol[-b, p]/p^(2 ns + 1)]]]];
    arr = Table[rowA235605[a, 10], {a, 10}];
    Flatten[Table[arr[[r - n + 1, n + 1]], {r, 0, Length[arr] - 1}, {n, 0, r}]] (* Matthew House, Sep 07 2024 *)

Formula

Shanks gives recurrences.

Extensions

a(27) removed, a(29)-a(42) added, and typo in name corrected by Lars Blomberg, Sep 10 2015
Offset corrected by Andrew Howroyd, Oct 25 2024

A002084 Sinh(x) / cos(x) = Sum_{n>=0} a(n)*x^(2n+1)/(2n+1)!.

Original entry on oeis.org

1, 4, 36, 624, 18256, 814144, 51475776, 4381112064, 482962852096, 66942218896384, 11394877025289216, 2336793875186479104, 568240131312188379136, 161669933656307658932224, 53204153193639888357113856, 20053432927718528320240287744
Offset: 0

Views

Author

Keywords

Comments

Gandhi proves that a(n) == 1 (mod 2n+1) if 2n+1 is prime, that a(2n+1) == 4 (mod 10), and that a(2n+2) == 6 (mod 10). - Charles R Greathouse IV, Oct 16 2012

Examples

			x + 2/3*x^3 + 3/10*x^5 + 13/105*x^7 + 163/3240*x^9 + ...
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A002085.

Programs

  • Mathematica
    With[{nn=30},Take[CoefficientList[Series[Sinh[x]/Cos[x],{x,0,nn}],x] Range[0,nn-1]!,{2,-1,2}]] (* Harvey P. Dale, Jul 17 2012 *)
  • PARI
    a(n)=n++;my(v=Vec(1/cos(x+O(x^(2*n+1)))));v=vector(n,i,v[2*i-1]*(2*i-2)!);sum(g=1,n,binomial(2*n-1,2*g-2)*v[g]) \\ Charles R Greathouse IV, Oct 16 2012
    
  • PARI
    list(n)=n++;my(v=Vec(1/cos(x+O(x^(2*n+1)))));v=vector(n,i,v[2*i-1]*(2*i-2)!);vector(n,k,sum(g=1,k,binomial(2*k-1,2*g-2)*v[g])) \\ Charles R Greathouse IV, Oct 16 2012
  • Sage
    # Generalized algorithm of L. Seidel (1877)
    def A002084_list(n) :
        R = []; A = {-1:0, 0:0}
        k = 0; e = 1
        for i in range(2*n) :
            Am = 1 if e == -1 else 0
            A[k + e] = 0
            e = -e
            for j in (0..i) :
                Am += A[k]
                A[k] = Am
                k += e
            if e == 1 : R.append(A[i//2])
        return R
    A002084_list(10) # Peter Luschny, Jun 02 2012
    

Formula

E.g.f.: sinh(x)/cos(x) = Sum_{n>=0} a(n)*x^(2n+1)/(2n+1)!.
a(n) = Sum_{k=0..n} binomial(2n+1, 2k+1)*A000364(n-k) = Sum_{k=0..n} A103327(n, k)*A000324(n-k) = Sum_{k=0..n} (-1)^(n-k)*A104033(n, k). - Philippe Deléham, Aug 27 2005
a(n) ~ sinh(Pi/2) * 2^(2*n + 3) * (2*n + 1)! / Pi^(2*n+2). - Vaclav Kotesovec, Jul 05 2020

Extensions

a(13)-a(15) from Andrew Howroyd, Feb 05 2018

A065547 Triangle of Salie numbers.

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 3, -3, 1, 0, -17, 17, -6, 1, 0, 155, -155, 55, -10, 1, 0, -2073, 2073, -736, 135, -15, 1, 0, 38227, -38227, 13573, -2492, 280, -21, 1, 0, -929569, 929569, -330058, 60605, -6818, 518, -28, 1, 0, 28820619, -28820619, 10233219, -1879038, 211419, -16086, 882, -36, 1, 0, -1109652905
Offset: 0

Views

Author

Wouter Meeussen, Dec 02 2001

Keywords

Comments

Coefficients of polynomials H(n,x) related to Euler polynomials through H(n,x(x-1)) = E(2n,x).

Examples

			Triangle begins:
 1;
 0,   1;
 0,  -1,    1;
 0,   3,   -3,  1;
 0, -17,   17, -6,   1;
 0, 155, -155, 55, -10, 1;
 ...
		

Crossrefs

Sum_{k>=0} (-1)^(n+k)*2^(n-k)*T(n, k) = A005647(n). Sum_{k>=0} (-1)^(n+k)*2^(2n-k)*T(n, k) = A000795(n). Sum_{k>=0} (-1)^(n+k)*T(n, k) = A006846(n), where A006846 = Hammersley's polynomial p_n(1). - Philippe Deléham, Feb 26 2004.
Column sequences (without leading zeros) give, for k=1..10: A065547 (twice), A095652-9.
See A085707 for unsigned and transposed version.
See A098435 for negative values of n, k.

Programs

  • Mathematica
    h[n_, x_] := Sum[c[k]*x^k, {k, 0, n}]; eq[n_] := SolveAlways[h[n, x*(x - 1)] == EulerE[2*n, x], x]; row[n_] := Table[c[k], {k, 0, n}] /. eq[n] // First; Table[row[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Oct 02 2013 *)
  • PARI
    { S2(n, k) = (1/k!)*sum(i=0,k,(-1)^(k-i)*binomial(k,i)*i^n) }{ Eu(n) = sum(m=0,n,(-1)^m*m!*S2(n+1,m+1)*(-1)^floor(m/4)*2^-floor(m/2)*((m+1)%4!=0)) } T(n,k)=if(nRalf Stephan

Formula

E.g.f.: Sum_{n, k=0..oo} T(n, k) t^k x^(2n)/(2n)! = cosh(sqrt(1+4t) x/2) / cosh(x/2).
T(k, n) = Sum_{i=0..n-k} A028296(i)/4^(n-k)*C(2n, 2i)*C(n-i, n-k-i), or 0 if n
Polynomial recurrences: x^n = Sum_{0<=2i<=n} C(n, 2i)*H(n-i, x); (1/4+x)^n = Sum_{m=0..n} C(2n, 2m)*(1/4)^(n-m)*H(m, x).
Dumont/Zeng give a continued fraction and other formulas.
Triangle T(n, k) read by rows; given by [0, -1, -2, -4, -6, -9, -12, -16, ...] DELTA A000035, where DELTA is Deléham's operator defined in A084938.
Sum_{k=0..n} (-4)^(n-k)*T(n,k) = A000364(n) (Euler numbers). - Philippe Deléham, Oct 25 2006

Extensions

Edited by Ralf Stephan, Sep 08 2004

A158690 Expansion of the basic hypergeometric series 1 + (1 - exp(-t)) + (1 - exp(-t))*(1 - exp(-3*t)) + (1 - exp(-t))*(1 - exp(-3*t))*(1 - exp(-5*t)) + ... as a series in t.

Original entry on oeis.org

1, 1, 5, 55, 1073, 32671, 1431665, 85363615, 6646603073, 654896692351, 79656194515025, 11722538113191775, 2052949879753739873, 421931472111868912831, 100568330857984368195185
Offset: 0

Author

Peter Bala, Mar 24 2009

Keywords

Comments

We appear to get the same sequence by expanding 1 - (1 - exp(t)) + (1 - exp(t))*(1 - exp(2*t)) - (1 - exp(t))*(1 - exp(2*t))*(1 - exp(3*t)) + ... as a series in t. Compare with A079144. For other sequences with generating functions of a similar type see A000364, A000464, A002105 and A002439.
From Peter Bala, Mar 13 2017: (Start)
It appears that the g.f. has two other forms: either F(exp(-t)) where F(q) = Sum_{n >= 0} q^(n+1)*Product_{k = 1..n} 1 - q^(2*k) = q + q^2 + q^3 - q^7 - q^8 - q^10 - q^11 - ... is a g.f. for A003475 or 1/2*G(exp(t)) where G(q) = 1 + Sum_{n >= 0} (-1)^n*q^(n+1)*Product_{k = 1..n} 1 - q^k = 1 + q - q^2 + 2*q^3 - 2*q^4 + q^5 + q^7 - 2*q^8 + ... is a g.f. for A003406. See Zagier, Example 1. (End)
From Peter Bala, Dec 18 2021: (Start)
Conjectures:
1) Taking the sequence modulo an integer k gives an eventually periodic sequence with period dividing phi(k). For example, the sequence taken modulo 16 begins [1, 1, 5, 7, 1, 15, 1, 15, 1, 15, 1, 15, ...] with an apparent pre-period of length 4 and a period of length 2.
2) Let i >= 0 and define a_i(n) = a(n+i). Then for each i the Gauss congruences a_i(n*p^k) == a_i(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.
If true, then for each i the expansion of exp( Sum_{n >= 1} a_i(n)*x^n/n ) has integer coefficients. For example, the expansion of exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 3*x^2 + 21*x^3 + 291*x^4 + 6861*x^5 + 246171*x^6 + 12458901*x^7 + 843915891*x^8 + 73640674461*x^9 + 8041227405771*x^10 + ... appears to have integer coefficients. (End)

Examples

			G.f. A(x) = 1 + x + 5*x^2 + 55*x^3 + 1073*x^4 + 32671*x^5 + 1431665*x^6 + ...
		

Programs

  • Mathematica
    max = 14; se = Series[1 + Sum[ Product[1 - E^(-(2*k - 1)*t), {k, 1, n}], {n, 1, max}], {t, 0, max}]; CoefficientList[se, t]*Range[0, max]! (* Jean-François Alcover, Mar 06 2013 *)
  • PARI
    {a(n)=n!*polcoeff(sum(m=0, n, prod(k=1, m, 1-exp(-(2*k-1)*x+x*O(x^n)))), n)} \\ Paul D. Hanna, Aug 01 2012
    
  • PARI
    {a(n)=n!*polcoeff(sum(m=0, n, prod(k=1, m, exp(k*x+x*O(x^n))-1)), n)} \\ Paul D. Hanna, Aug 01 2012

Formula

Basic hypergeometric generating function: 1 + Sum_{n >= 0} Product_{k = 1..n} (1 - exp(2*k-1)*t) = 1 + t + 5*t^2/2! + 55*t^3/3! + ....
a(n) ~ 6*sqrt(2) * 12^n * (n!)^2 / Pi^(2*n+2). - Vaclav Kotesovec, May 04 2014
Conjectural g.f.: G(exp(t)) as a formal power series in t, where G(q) := Sum_{n >= 0} q^(2*n+1) * Product_{k = 1..2*n} (1 - q^k). - Peter Bala, May 16 2017
E.g.f.: Sum_{n>=0} exp(n*(n+1)/2*x) / Product_{k=0..n} (1 + exp(k*x)). - Paul D. Hanna, Oct 14 2020

A002115 Generalized Euler numbers.

Original entry on oeis.org

1, 1, 19, 1513, 315523, 136085041, 105261234643, 132705221399353, 254604707462013571, 705927677520644167681, 2716778010767155313771539, 14050650308943101316593590153, 95096065132610734223282520762883, 823813936407337360148622860507620561
Offset: 0

Keywords

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    b:= proc(u, o, t) option remember; `if`(u+o=0, 1, `if`(t=0,
           add(b(u-j, o+j-1, irem(t+1, 3)), j=1..u),
           add(b(u+j-1, o-j, irem(t+1, 3)), j=1..o)))
        end:
    a:= n-> b(3*n, 0$2):
    seq(a(n), n=0..17);  # Alois P. Heinz, Aug 12 2019
    # Alternative:
    h := 1 / hypergeom([], [1/3, 2/3], (-x/3)^3): ser := series(h, x, 40):
    seq((3*n)! * coeff(ser, x, 3*n), n = 0..13); # Peter Luschny, Mar 13 2023
  • Mathematica
    max = 12; f[x_] := 1/(1/3*Exp[-x^(1/3)] + 2/3*Exp[1/2*x^(1/3)]*Cos[1/2*3^(1/2)* x^(1/3)]); CoefficientList[Series[f[x], {x, 0, max}], x]*(3 Range[0, max])! (* Jean-François Alcover, Sep 16 2013, after Vladeta Jovovic *)

Formula

E.g.f.: Sum_{n >= 0} a(n)*x^n/(3*n)! = 1/((1/3)*exp(-x^(1/3)) + (2/3)*exp((1/2)*x^(1/3))*cos((1/2)*3^(1/2)*x^(1/3))). - Vladeta Jovovic, Feb 13 2005
E.g.f.: 1/U(0) where U(k) = 1 - x/(6*(6*k+1)*(3*k+1)*(2*k+1) - 6*x*(6*k+1)*(3*k+1)*(2*k+1)/(x - 12*(6*k+5)*(3*k+2)*(k+1)/U(k+1))); (continued fraction). - Sergei N. Gladkovskii, Oct 04 2012
Alternating row sums of A278073. - Peter Luschny, Sep 07 2017
a(n) = A178963(3n). - Alois P. Heinz, Aug 12 2019
a(0) = 1; a(n) = Sum_{k=1..n} (-1)^(k+1) * binomial(3*n,3*k) * a(n-k). - Ilya Gutkovskiy, Jan 27 2020
a(n) = (3*n)! * [x^(3*n)] hypergeom([], [1/3, 2/3], (-x/3)^3)^(-1). - Peter Luschny, Mar 13 2023

Extensions

More terms from Vladeta Jovovic, Feb 13 2005

A083884 a(n) = (3^(2*n) + 1) / 2.

Original entry on oeis.org

1, 5, 41, 365, 3281, 29525, 265721, 2391485, 21523361, 193710245, 1743392201, 15690529805, 141214768241, 1270932914165, 11438396227481, 102945566047325, 926510094425921, 8338590849833285, 75047317648499561, 675425858836496045, 6078832729528464401
Offset: 0

Author

Paul Barry, May 09 2003

Keywords

Comments

Number of compositions of even natural numbers into n parts <= 8. - Adi Dani, May 28 2011
a(n) for n >= 1 gives the number of line segments in the n-th iteration of the Peano curve given by plotting (A163528, A163529) or by (Siromoney 1982) when parallel line segments that are connected end-to-end are counted as a single line segment. - Jason V. Morgan, Oct 08 2021

Examples

			From _Adi Dani_, May 28 2011: (Start)
a(2)=41: there are 41 compositions of even natural numbers into 2 parts <=8:
(0,0);
(0,2),(2,0),(1,1);
(0,4),(4,0),(1,3),(3,1),(2,2);
(0,6),(6,0),(1,5),(5,1),(2,4),(4,2),(3,3);
(0,8),(8,0),(1,7),(7,1),(2,6),(6,2),(3,5),(5,3),(4,4);
(2,8),(8,2),(3,7),(7,3),(4,6),(6,4),(5,5);
(4,8),(8,4),(5,7),(7,5),(6,6);
(6,8),(8,6),(7,7);
(8,8).  (End)
		

References

  • Siromoney, R., & Subramanian, K.G. (1982). Space-filling curves and infinite graphs. Graph-Grammars and Their Application to Computer Science.

Crossrefs

Programs

Formula

a(0) = 1, a(n) = 9*a(n-1) - 4.
a(n) = Sum_{k=0..n} binomial(2*n, 2*k)*4^k.
a(n) = A002438(n) / A000364(n); A000364(n) : Euler numbers.
G.f.: (1-5*x)/((1-x)*(1-9*x)).
a(n) = (3^n + 1^n + (-1)^n + (-3)^n)/4.
E.g.f.: exp(3*x) + exp(x) + exp(-x) + exp(-3*x).
Each term expresses a Pythagorean relationship, along with (a(n)-1) and a power of 3, n>0, such that sqrt((a(n))^2 - (a(n)-1)^2) = 3^n. E.g., 365^2 - 364^2 - 3^3 = 27 (the Pythagorean triangle (365, 364, 27)). - Gary W. Adamson, Jun 25 2006
a(n) = 10*a(n-1) - 9*a(n-2). - Wesley Ivan Hurt, Apr 21 2021

Extensions

Additional comments from Philippe Deléham, Jul 10 2005
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