cp's OEIS Frontend

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

Showing 1-10 of 20 results. Next

A001787 a(n) = n*2^(n-1).

Original entry on oeis.org

0, 1, 4, 12, 32, 80, 192, 448, 1024, 2304, 5120, 11264, 24576, 53248, 114688, 245760, 524288, 1114112, 2359296, 4980736, 10485760, 22020096, 46137344, 96468992, 201326592, 419430400, 872415232, 1811939328, 3758096384, 7784628224, 16106127360, 33285996544
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of edges in an n-dimensional hypercube.
Number of 132-avoiding permutations of [n+2] containing exactly one 123 pattern. - Emeric Deutsch, Jul 13 2001
Number of ways to place n-1 nonattacking kings on a 2 X 2(n-1) chessboard for n >= 2. - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), May 22 2001
Arithmetic derivative of 2^n: a(n) = A003415(A000079(n)). - Reinhard Zumkeller, Feb 26 2002
(-1) times the determinant of matrix A_{i,j} = -|i-j|, 0 <= i,j <= n.
a(n) is the number of ones in binary numbers 1 to 111...1 (n bits). a(n) = A000337(n) - A000337(n-1) for n = 2,3,... . - Emeric Deutsch, May 24 2003
The number of 2 X n 0-1 matrices containing n+1 1's and having no zero row or column. The number of spanning trees of the complete bipartite graph K(2,n). This is the case m = 2 of K(m,n). See A072590. - W. Edwin Clark, May 27 2003
Binomial transform of 0,1,2,3,4,5,... (A001477). Without the initial 0, binomial transform of odd numbers.
With an additional leading zero, [0,0,1,4,...] this is the binomial transform of the integers repeated A004526. Its formula is then (2^n*(n-1) + 0^n)/4. - Paul Barry, May 20 2003
Number of zeros in all different (n+1)-bit integers. - Ralf Stephan, Aug 02 2003
From Lekraj Beedassy, Jun 03 2004: (Start)
Final element of a summation table (as opposed to a difference table) whose first row consists of integers 0 through n (or first n+1 nonnegative integers A001477); illustrating the case n=5:
0 1 2 3 4 5
1 3 5 7 9
4 8 12 16
12 20 28
32 48
80
and the final element is a(5)=80. (End)
This sequence and A001871 arise in counting ordered trees of height at most k where only the rightmost branch at the root actually achieves this height and the count is by the number of edges, with k = 3 for this sequence and k = 4 for A001871.
Let R be a binary relation on the power set P(A) of a set A having n = |A| elements such that for all elements x,y of P(A), xRy if x is a proper subset of y and there are no z in P(A) such that x is a proper subset of z and z is a proper subset of y. Then a(n) = |R|. - Ross La Haye, Sep 21 2004
Number of 2 X n binary matrices avoiding simultaneously the right-angled numbered polyomino patterns (ranpp) (00;1) and (10;1). An occurrence of a ranpp (xy;z) in a matrix A=(a(i,j)) is a triple (a(i1,j1), a(i1,j2), a(i2,j1)) where i1 < i2, j1 < j2 and these elements are in same relative order as those in the triple (x,y,z). - Sergey Kitaev, Nov 11 2004
Number of subsequences 00 in all binary words of length n+1. Example: a(2)=4 because in 000,001,010,011,100,101,110,111 the sequence 00 occurs 4 times. - Emeric Deutsch, Apr 04 2005
If you expand the n-factor expression (a+1)*(b+1)*(c+1)*...*(z+1), there are a(n) variables in the result. For example, the 3-factor expression (a+1)*(b+1)*(c+1) expands to abc+ab+ac+bc+a+b+c+1 with a(3) = 12 variables. - David W. Wilson, May 08 2005
An inverse Chebyshev transform of n^2, where g(x)->(1/sqrt(1-4*x^2))*g(x*c(x^2)), c(x) the g.f. of A000108. - Paul Barry, May 13 2005
Sequences A018215 and A058962 interleaved. - Graeme McRae, Jul 12 2006
The number of never-decreasing positive integer sequences of length n with a maximum value of 2*n. - Ben Paul Thurston, Nov 13 2006
Total size of all the subsets of an n-element set. For example, a 2-element set has 1 subset of size 0, 2 subsets of size 1 and 1 of size 2. - Ross La Haye, Dec 30 2006
Convolution of the natural numbers [A000027] and A045623 beginning [0,1,2,5,...]. - Ross La Haye, Feb 03 2007
If M is the matrix (given by rows) [2,1;0,2] then the sequence gives the (1,2) entry in M^n. - Antonio M. Oller-Marcén, May 21 2007
If X_1,X_2,...,X_n is a partition of a 2n-set X into 2-blocks then, for n > 0, a(n) is equal to the number of (n+1)-subsets of X intersecting each X_i (i=1,2,...,n). - Milan Janjic, Jul 21 2007
Number of n-permutations of 3 objects u,v,w, with repetition allowed, containing exactly one u. Example: a(2)=4 because we have uv, vu, uw and wu. - Zerinvary Lajos, Dec 27 2007
A member of the family of sequences defined by a(n) = n*[c(1)*...*c(r)]^(n-1); c(i) integer. This sequence has c(1)=2, A027471 has c(1)=3. - Ctibor O. Zizka, Feb 23 2008
a(n) is the number of ways to split {1,2,...,n-1} into two (possibly empty) complementary intervals {1,2,...,i} and {i+1,i+2,...,n-1} and then select a subset from each interval. - Geoffrey Critzer, Jan 31 2009
Equals the Jacobsthal sequence A001045 convolved with A003945: (1, 3, 6, 12, ...). - Gary W. Adamson, May 23 2009
Starting with offset 1 = A059570: (1, 2, 6, 14, 34, ...) convolved with (1, 2, 2, 2, ...). - Gary W. Adamson, May 23 2009
Equals the first left hand column of A167591. - Johannes W. Meijer, Nov 12 2009
The number of tatami tilings of an n X n square with n monomers is n*2^(n-1). - Frank Ruskey, Sep 25 2010
Under T. D. Noe's variant of the hypersigma function, this sequence gives hypersigma(2^n): a(n) = A191161(A000079(n)). - Alonso del Arte, Nov 04 2011
Number of Dyck (n+2)-paths with exactly one valley at height 1 and no higher valley. - David Scambler, Nov 07 2011
Equals triangle A059260 * A016777 as a vector, where A016777 = (3n + 1): [1, 4, 7, 10, 13, ...]. - Gary W. Adamson, Mar 06 2012
Main transitions in systems of n particles with spin 1/2 (see A212697 with b=2). - Stanislav Sykora, May 25 2012
Let T(n,k) be the triangle with (first column) T(n,1) = 2*n-1 for n >= 1, otherwise T(n,k) = T(n,k-1) + T(n-1,k-1), then a(n) = T(n,n). - J. M. Bergot, Jan 17 2013
Sum of all parts of all compositions (ordered partitions) of n. The equivalent sequence for partitions is A066186. - Omar E. Pol, Aug 28 2013
Starting with a(1)=1: powers of 2 (A000079) self-convolved. - Bob Selcoe, Aug 05 2015
Coefficients of the series expansion of the normalized Schwarzian derivative -S{p(x)}/6 of the polynomial p(x) = -(x-x1)*(x-x2) with x1 + x2 = 1 (cf. A263646). - Tom Copeland, Nov 02 2015
a(n) is the number of North-East lattice paths from (0,0) to (n+1,n+1) that have exactly one east step below y = x-1 and no east steps above y = x+1. Details can be found in Pan and Remmel's link. - Ran Pan, Feb 03 2016
Also the number of maximal and maximum cliques in the n-hypercube graph for n > 0. - Eric W. Weisstein, Dec 01 2017
Let [n]={1,2,...,n}; then a(n-1) is the total number of elements missing in proper subsets of [n] that contain n to form [n]. For example, for n = 3, a(2) = 4 since the proper subsets of [3] that contain 3 are {3}, {1,3}, {2,3} and the total number of elements missing in these subsets to form [3] is 4: 2 in the first subset, 1 in the second, and 1 in the third. - Enrique Navarrete, Aug 08 2020
Number of 3-permutations of n elements avoiding the patterns 132, 231. See Bonichon and Sun. - Michel Marcus, Aug 19 2022

Examples

			a(2)=4 since 2314, 2341,3124 and 4123 are the only 132-avoiding permutations of 1234 containing exactly one increasing subsequence of length 3.
x + 4*x^2 + 12*x^3 + 32*x^4 + 80*x^5 + 192*x^6 + 448*x^7 + ...
a(5) = 1*0 + 5*1 + 10*2 + 10*3 + 5*4 + 1*5 = 80, with 1,5,10,10,5,1 the 5th row of Pascal's triangle. - _J. M. Bergot_, Apr 29 2014

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 131.
  • Clifford A. Pickover, The Math Book, From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publ., NY, 2009, page 282.
  • 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).

Links

Crossrefs

Three other versions, essentially identical, are A085750, A097067, A118442.
Partial sums of A001792.
A058922(n+1) = 4*A001787(n).
Equals A090802(n, 1).
Column k=1 of A038207.
Row sums of A003506, A322427, A322428.
Cf. A053109, A001788, A001789, A000337, A130300, A134083, A002064, A027471, A003945, A059670, A167591, A059260, A016777, A212697, A000079, A263646.

Programs

  • Haskell
    a001787 n = n * 2 ^ (n - 1)
a001787_list = zipWith (*) [0..] $ 0 : a000079_list
-- Reinhard Zumkeller, Jul 11 2014
  • Magma
    [n*2^(n-1): n in [0..40]]; // Vincenzo Librandi, Feb 04 2016
  • Maple
    spec := [S, {B=Set(Z, 0 <= card), S=Prod(Z, B, B)}, labeled]: seq(combstruct[count](spec, size=n), n=0..29); # Zerinvary Lajos, Oct 09 2006
A001787:=1/(2*z-1)^2; # Simon Plouffe in his 1992 dissertation, dropping the initial zero
  • Mathematica
    Table[Sum[Binomial[n, i] i, {i, 0, n}], {n, 0, 30}] (* Geoffrey Critzer, Mar 18 2009 *)
f[n_] := n 2^(n - 1); f[Range[0, 40]] (* Vladimir Joseph Stephan Orlovsky, Feb 09 2011 *)
Array[# 2^(# - 1) &, 40, 0] (* Harvey P. Dale, Jul 26 2011 *)
Join[{0}, Table[n 2^(n - 1), {n, 20}]] (* Eric W. Weisstein, Dec 01 2017 *)
Join[{0}, LinearRecurrence[{4, -4}, {1, 4}, 20]] (* Eric W. Weisstein, Dec 01 2017 *)
CoefficientList[Series[x/(-1 + 2 x)^2, {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)
  • PARI
    {a(n) = if( n<0, 0, n * 2^(n-1))}
  • PARI
    concat(0, Vec(x/(1-2*x)^2 + O(x^50))) \\ Altug Alkan, Nov 03 2015
  • Python
    def A001787(n): return n*(1<Chai Wah Wu, Nov 14 2022

Formula

a(n) = Sum_{k=1..n} k*binomial(n, k). - Benoit Cloitre, Dec 06 2002
E.g.f.: x*exp(2x). - Paul Barry, Apr 10 2003
G.f.: x/(1-2*x)^2.
G.f.: x / (1 - 4*x / (1 + x / (1 - x))). - Michael Somos, Apr 07 2012
A108666(n) = Sum_{k=0..n} binomial(n, k)^2 * a(n). - Michael Somos, Apr 07 2012
PSumSIGN transform of A053220. PSumSIGN transform is A045883. Binomial transform is A027471(n+1). - Michael Somos, Jul 10 2003
Starting at a(1)=1, INVERT transform is A002450, INVERT transform of A049072, MOBIUS transform of A083413, PSUM transform is A000337, BINOMIAL transform is A081038, BINOMIAL transform of A005408. - Michael Somos, Apr 07 2012
a(n) = 2*a(n-1)+2^(n-1).
a(2*n) = n*4^n, a(2*n+1) = (2*n+1)4^n.
G.f.: x/det(I-x*M) where M=[1,i;i,1], i=sqrt(-1). - Paul Barry, Apr 27 2005
Starting 1, 1, 4, 12, ... this is 0^n + n2^(n-1), the binomial transform of the 'pair-reversed' natural numbers A004442. - Paul Barry, Jul 24 2003
Convolution of [1, 2, 4, 8, ...] with itself. - Jon Perry, Aug 07 2003
The signed version of this sequence, n(-2)^(n-1), is the inverse binomial transform of n(-1)^(n-1) (alternating sign natural numbers). - Paul Barry, Aug 20 2003
a(n-1) = (Sum_{k=0..n} 2^(n-k-1)*C(n-k, k)*C(1,(k+1)/2)*(1-(-1)^k)/2) - 0^n/4. - Paul Barry, Oct 15 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)(n-2k)^2. - Paul Barry, May 13 2005
a(n+2) = A049611(n+2) - A001788(n).
a(n) = n! * Sum_{k=0..n} 1/((k - 1)!(n - k)!). - Paul Barry, Mar 26 2003
a(n+1) = Sum_{k=0..n} 4^k * A109466(n,k). - Philippe Deléham, Nov 13 2006
Row sums of A130300 starting (1, 4, 12, 32, ...). - Gary W. Adamson, May 20 2007
Equals row sums of triangle A134083. Equals A002064(n) + (2^n - 1). - Gary W. Adamson, Oct 07 2007
a(n) = 4*a(n-1) - 4*a(n-2), a(0)=0, a(1)=1. - Philippe Deléham, Nov 16 2008
Sum_{n>0} 1/a(n) = 2*log(2). - Jaume Oliver Lafont, Feb 10 2009
a(n) = A000788(A000225(n)) = A173921(A000225(n)). - Reinhard Zumkeller, Mar 04 2010
a(n) = n * A011782(n). - Omar E. Pol, Aug 28 2013
a(n-1) = Sum_{t_1+2*t_2+...+n*t_n=n} (t_1+t_2+...+t_n-1)*multinomial(t_1+t_2 +...+t_n,t_1,t_2,...,t_n). - Mircea Merca, Dec 06 2013
a(n+1) = Sum_{r=0..n} (2*r+1)*C(n,r). - J. M. Bergot, Apr 07 2014
a(n) = A007283(n)*n/6. - Enxhell Luzhnica, Apr 16 2016
a(n) = (A000225(n) + A000337(n))/2. - Anton Zakharov, Sep 17 2016
Sum_{n>0} (-1)^(n+1)/a(n) = 2*log(3/2) = 2*A016578. - Ilya Gutkovskiy, Sep 17 2016
a(n) = Sum_{k=0..n-1} Sum_{i=0..n-1} (i+1) * C(k,i). - Wesley Ivan Hurt, Sep 21 2017
a(n) = Sum_{i=1..n} Sum_{j=1..n} phi(i)*binomial(n, i*j). - Ridouane Oudra, Feb 17 2024

A002391 Decimal expansion of natural logarithm of 3.

Original entry on oeis.org

1, 0, 9, 8, 6, 1, 2, 2, 8, 8, 6, 6, 8, 1, 0, 9, 6, 9, 1, 3, 9, 5, 2, 4, 5, 2, 3, 6, 9, 2, 2, 5, 2, 5, 7, 0, 4, 6, 4, 7, 4, 9, 0, 5, 5, 7, 8, 2, 2, 7, 4, 9, 4, 5, 1, 7, 3, 4, 6, 9, 4, 3, 3, 3, 6, 3, 7, 4, 9, 4, 2, 9, 3, 2, 1, 8, 6, 0, 8, 9, 6, 6, 8, 7, 3, 6, 1, 5, 7, 5, 4, 8, 1, 3, 7, 3, 2, 0, 8, 8, 7, 8, 7, 9, 7
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Examples

			1.098612288668109691395245236922525704647490557822749451734694333637494...

References

  • Calvin C. Clawson, Mathematical Mysteries: The Beauty and Magic of Numbers, Springer, 2013. See p. 221.
  • W. E. Mansell, Tables of Natural and Common Logarithms. Royal Society Mathematical Tables, Vol. 8, Cambridge Univ. Press, 1964, p. 2.
  • 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).

Links

Crossrefs

Cf. A058962, A154920, A002162, A016731 (continued fraction), A073000, A105531, A254619.

Programs

  • Mathematica
    RealDigits[Log[3],10,120][[1]]  (* Harvey P. Dale, Apr 23 2011 *)
  • PARI
    log(3) \\ Charles R Greathouse IV, Jan 24 2012
  • Python
    # Use some guard digits when computing.
# BBP formula P(1, 4, 2, (1, 0)).
from decimal import Decimal as dec, getcontext
def BBPlog3(n: int) -> dec:
    getcontext().prec = n
    s = dec(0); f = dec(1); g = dec(4)
    for k in range(2 * n):
        s += f / dec(2 * k + 1)
        f /= g
    return s
print(BBPlog3(200))  # Peter Luschny, Nov 03 2023

Formula

log(3) = Sum_{n>=1} (9*n-4)/((3*n-2)*(3*n-1)*3*n). [Jolley, Summation of Series, Dover (1961) eq 74]
log(3) = (1/4)*(1 + Sum_{m>=0} (1/9)^(k+1)*(27/(2*k+1) + 4/(2*k+2) + 1/(2*k+3))) (a BBP-type formula). - Alexander R. Povolotsky, Dec 01 2008
log(3) = 4/5 + (1/5)*Sum_{n>=0} (1/4)^n*(1/(2*n+1) + 1/(2*n+3)). - Alexander R. Povolotsky, Dec 18 2008
log(3) = Sum_{k>=0} (1/9)^(k+1)*(9/(2k+1) + 1/(2k+2)). - Jaume Oliver Lafont, Dec 22 2008
Sum_{i>=1} 1/(9^i*i) + Sum_{i>=0} 1/(9^i*(i+1/2)) = 2*log(3) (Huvent 2001). - Jaume Oliver Lafont, Oct 12 2009
Conjecture: log(3) = Sum_{k>=1} A191907(3,k)/k. - Mats Granvik, Jun 19 2011
log(3) = lim_{n->oo} Sum_{k=3^n..3^(n+1)-1} 1/k. Also see A002162. By analogy to the integral of 1/x, log(m) = lim_{n->oo} Sum_{k=m^n..m^(n+1)-1} 1/k, for any value of m > 1. - Richard R. Forberg, Aug 16 2014
From Peter Bala, Feb 04 2015: (Start)
log(3) = Sum {k >= 0} 1/((2*k + 1)*4^k).
Define a pair of integer sequences A(n) = 4^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} 1/((2*k + 1)*4^k). Both sequences satisfy the same second-order recurrence equation u(n) = (20*n + 6)*u(n-1) - 16*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion log(3) = 1 + 2/(24 - 16*3^2/(46 - 16*5^2/(66 - ... - 16*(2*n - 1)^2/((20*n + 6) - ... )))). Cf. A002162, A073000 and A105531 for similar expansions.
log(3) = 2 * Sum_{k >= 1} (-1)^(k+1)*(4/3)^k/(k*binomial(2*k,k)).
log(3) = (1/4) * Sum_{k >= 1} (-1)^(k+1) (55*k - 23)*(8/9)^k/( 2*k*(2*k - 1)*binomial(3*k,k) ).
log(3) = (1/4) * Sum_{k >= 1} (7*k + 1)*(8/3)^k/( 2*k*(2*k - 1)*binomial(3*k,k) ). (End)
log(3) = -lim_{n->oo} (n+1)th derivative of zeta(n) / n-th derivative of zeta(n). By n = 1000 there is convergence to 25 digits. A related expression: lim_{n->oo} n-th derivative of zeta(n-1) / n-th derivative of zeta(n) = 3. Also see A002581. - Richard R. Forberg, Feb 24 2015
From Peter Bala, Nov 02 2019: (Start)
log(3) = 2*Integral_{x = 0..1} (1 - x^2)/(1 + x^2 + x^4) dx = 2*( 1 - (2/3) + 1/5 + 1/7 - (2/9) + 1/11 + 1/13 - (2/15) + ... ).
log(3) = 16*Sum_{n >= 0} 1/( (6*n + 1)*(6*n + 3)*(6*n + 5) ).
log(3) = 4/5 + 64*Sum_{n >= 0} (18*n + 1)/((6*n - 5)*(6*n - 3)*(6*n - 1)*(6*n + 1)*(6*n + 7)). (End)
From Amiram Eldar, Jul 05 2020: (Start)
Equals 2*arctanh(1/2).
Equals Sum_{k>=1} (2/3)^k/k.
Equals Integral_{x=0..Pi} sin(x)dx/(2 + cos(x)). (End)
log(3) = Integral_{x = 0..1} (x^2 - 1)/log(x) dx. - Peter Bala, Nov 14 2020
From Peter Bala, Oct 28 2023: (Start)
The series representation log(3) = 16*Sum_{n >= 0} 1/((6*n + 1)*(6*n + 3)*(6*n + 5)) given above appears to be the case k = 0 of the following infinite family of series representations for log(3):
log(3) = c(k) + (-1)^k*d(k)*Sum_{n >= 0} 1/((6*n + 1)*(6*n + 3)*...*(6*n + 12*k + 5)), where c(k) is a rational approximation to log(3) and d(k) = 2^(6*k+3)/27^k * (6*k + 2)!.
The first few values of c(k) for k >= 0 are [0, 2996/2673, 89195548/81236115, 23239436137364/21153065697225, 3345533089100222564/3045237239236561677, ...]. Cf A304656. (End)
log(3) = 1 + 2*Sum_{k>=1} 1/((3*k)^3 - 3*k) [Ramanujan]. - Stefano Spezia, Jul 01 2024

Extensions

Editing and more terms from Charles R Greathouse IV, Apr 20 2010

A118416 Triangle read by rows: T(n,k) = (2*k-1)*2^(n-1), 0 < k <= n.

Original entry on oeis.org

1, 2, 6, 4, 12, 20, 8, 24, 40, 56, 16, 48, 80, 112, 144, 32, 96, 160, 224, 288, 352, 64, 192, 320, 448, 576, 704, 832, 128, 384, 640, 896, 1152, 1408, 1664, 1920, 256, 768, 1280, 1792, 2304, 2816, 3328, 3840, 4352, 512, 1536, 2560, 3584, 4608, 5632, 6656, 7680
Offset: 1

Views

Author

Reinhard Zumkeller, Apr 27 2006

Keywords

Comments

Row sums give A014477: Sum_{k=1..n} T(n,k) = A014477(n-1);
central terms give A118415; T(2*k-1,k) = A058962(k-1);
T(n,1) = A000079(n-1);
T(n,2) = A007283(n-1) for n > 1;
T(n,3) = A020714(n-1) for n > 2;
T(n,4) = A005009(n-1) for n > 3;
T(n,5) = A005010(n-1) for n > 4;
T(n,n-1) = A118417(n-1) for n > 1;
T(n,n) = A014480(n-1) = A118413(n,n);
A001511(T(n,k)) = A002024(n,k);
A003602(T(n,k)) = A002260(n,k).
The alternating row sums, Sum_{k=1..n} (-1)^(k+1)*T(n,k), are: (a) in odd rows, the central term, T(n,(n+1)/2) = A058962((n-1)/2); (b) in even rows, the negation of the average of the two central terms, -(T(2n,n) + T(2n,+1))/2 = -A018215(m/2). The absolute values of the alternating row sums give the plain row means, Sum_{k=1..n} T(n,k)/n; the alternating sign row means are (-2)^(n-1). - Gregory Gerard Wojnar, Feb 10 2024

Examples

			Triangle begins:
   1;
   2,   6;
   4,  12,  20;
   8,  24,  40,  56;
  16,  48,  80, 112, 144;
  32,  96, 160, 224, 288, 352;
  64, 192, 320, 448, 576, 704, 832;

Links

Crossrefs

Cf. A118413, A117303, A054582.

Programs

  • Haskell
    a118416 n k = a118416_tabl !! (n-1) !! (k-1)
a118416_row 1 = [1]
a118416_row n = (map (* 2) $ a118416_row (n-1)) ++ [a014480 (n-1)]
a118416_tabl = map a118416_row [1..]
-- Reinhard Zumkeller, Jan 22 2012
  • Maple
    A118416 := proc(n,k) 2^(n-1)*(2*k-1) ; end proc: # R. J. Mathar, Sep 04 2011
  • Mathematica
    Flatten[Table[(2k-1)2^(n-1),{n,10},{k,n}]] (* Harvey P. Dale, Aug 26 2014 *)
  • Python
    from math import isqrt
def A118416(n): return (a:=(m:=isqrt(k:=n<<1))+(k>m*(m+1)))*(1-a)+(n<<1)-1<Chai Wah Wu, Jun 20 2025

Formula

T(n,k) = 2*T(n-1,k), 1 <= k < n; T(n,n) = A014480(n-1).

A160487 The Lambda triangle.

Original entry on oeis.org

1, -107, 10, 59845, -7497, 210, -6059823, 854396, -35574, 420, 5508149745, -827924889, 41094790, -765534, 4620, -8781562891079, 1373931797082, -75405128227, 1738417252, -17219202, 60060
Offset: 2

Views

Author

Johannes W. Meijer, May 24 2009, Sep 18 2012

Keywords

Comments

The coefficients of the LS1 matrix are defined by LS1[2*m,n] = int(y^(2*m)/(sinh(y))^(2*n-1),y=0..infinity)/factorial(2*m) for m = 1, 2, 3, .. and n = 1, 2, 3, .. under the condition that n <= m.
This definition leads to LS1[2*m,n=1] = 2*lambda(2*m+1), for m = 1, 2, .. , and the recurrence relation LS1[2*m,n] = ((2*n-3)/(2*n-2))*(LS1[2*m-2,n-1]/(2*n-3)^2- LS1[2*m,n-1]). As usual lambda(m) = (1-2^(-m))*zeta(m) with zeta(m) the Riemann zeta function.
These two formulas enable us to determine the values of the LS1[2*m,n] coefficients, for all integers m and all positive integers n, but not for all n. If we choose, somewhat but not entirely arbitrarily, LS1[m=0,n=1] = gamma, with gamma the Euler-Mascheroni constant, we can determine them all.
The coefficients in the columns of the LS1 matrix, for m = 0, 1, 2, .. , and n = 2, 3, 4 .. , can be generated with the GL(z;n) polynomials for which we found the following general expression GL(z;n) = (h(n)*CFN2(z;n)*GL(z;n=1) + LAMBDA(z;n))/p(n).
The CFN2(z;n) polynomials depend on the central factorial numbers A008956.
The LAMBDA(z;n) are the Lambda polynomials which lead to the Lambda triangle.
The zero patterns of the Lambda polynomials resemble a UFO. These patterns resemble those of the Eta, Zeta and Beta polynomials, see A160464, A160474 and A160480.
The first Maple algorithm generates the coefficients of the Lambda triangle. The second Maple algorithm generates the LS1[2*m,n] coefficients for m= -1, -2, -3, .. .
Some of our results are conjectures based on numerical evidence.

Examples

			The first few rows of the triangle LAMBDA(n,m) with n=2,3,.. and m=1,2,.. are
  [1]
  [ -107, 10]
  [59845, -7497, 210]
  [ -6059823, 854396, -35574, 420]
The first few LAMBDA(z;n) polynomials are
  LAMBDA (z;n=2) = 1
  LAMBDA (z;n=3) = -107 +10*z^2
  LAMBDA (z;n=4) = 59845-7497*z^2+210*z^4
The first few CFN2(z;n) polynomials are
  CFN2(z;n=2) = (z^2-1)
  CFN2(z;n=3) = (z^4-10*z^2+9)
  CFN2(z;n=4) = (z^6- 35*z^4+259*z^2-225)
The first few generating functions GL(z;n) are:
  GL(z;n=2) = (6*(z^2-1)*GL(z,n=1) + (1)) /12
  GL(z;n=3) = (60*(z^4-10*z^2+9)*GL(z,n=1)+ (-107+10*z^2)) / 1440
  GL(z;n=4) = (1260*( z^6- 35*z^4+259*z^2-225)*GL(z,n=1) + (59845-7497*z^2+ 210*z^4))/907200

References

  • Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.

Links

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, Chapter 23, pp. 811-812.
  • Johannes W. Meijer, The zeros of the Eta, Zeta, Beta and Lambda polynomials, jpg and pdf, Mar 03 2013.

Crossrefs

A160488 equals the first left hand column.
A160476 equals the first right hand column and 6*h(n).
A160489 equals the rows sums.
A160490 equals the p(n) sequence.
A160479 equals the ZL(n) sequence.
A001620 is the Euler-Mascheroni constant gamma.
The LS1[ -2, n] coefficients lead to A002197, A002198 and A058962.
The LS1[ -2*m, 1] coefficients equal (-1)^(m+1)*A036282/A036283.
The CFN2(z, n) and the cfn2(n, k) lead to A008956.
Cf. The Eta, Zeta and Beta triangles A160464, A160474 and A160480.
Cf. A162448 (LG1 matrix)

Programs

  • Maple
    nmax:=7; for n from 0 to nmax do cfn2(n, 0) := 1: cfn2(n, n) := (doublefactorial(2*n-1))^2 od: for n from 1 to nmax do for k from 1 to n-1 do cfn2(n, k) := (2*n-1)^2*cfn2(n-1, k-1) + cfn2(n-1, k) od: od: for n from 1 to nmax do Delta(n-1) := sum((1-2^(2*k1-1))* (-1)^(n+1)*(-bernoulli(2*k1)/(2*k1))*(-1)^(k1+n)*cfn2(n-1,n-k1, n), k1=1..n) / (2*4^(n-1)*(2*n-1)!); LAMBDA(-2, n) := sum(2*(1-2^(2*k1-1))*(-bernoulli(2*k1) / (2*k1))*(-1)^(k1+n)* cfn2(n-1,n-k1), k1=1..n)/ factorial(2*n-2) end do: Lcgz(2) := 1/12: f(2) := 1/12: for n from 3 to nmax do Lcgz(n) := LAMBDA(-2, n-1)/((2*n-2)*(2*n-3)): f(n) := Lcgz(n)-((2*n-3)/(2*n-2))*f(n-1) end do: for n from 1 to nmax do b(n) := denom(Lcgz(n+1)) end do: for n from 1 to nmax do b(n) := 2*n*denom(Delta(n-1))/2^(2*n) end do: p(2) := b(1): for n from 2 to nmax do p(n+1) := lcm(p(n)*(2*n)*(2*n-1), b(n)) end do: for n from 2 to nmax do LAMBDA(n, 1) := p(n)*f(n) end do: mmax:=nmax: for n from 2 to nmax do LAMBDA(n, n) := 0 end do: for n from 1 to nmax do b(n) := (2*n)*(2*n-1)*denom(Delta(n-1))/ (2^(2*n)*(2*n-1)) end do: c(1) := b(1): for n from 1 to nmax-1 do c(n+1) := lcm(c(n)*(2*n+2)* (2*n+1), b(n+1)) end do: for n from 1 to nmax do cm(n) := c(n)/(6*(2*n)!) end do: for n from 1 to nmax-1 do ZL(n+2) := cm(n+1)/cm(n) end do: for m from 2 to mmax do for n from m+1 to nmax do LAMBDA(n, m) := ZL(n)*(LAMBDA(n-1, m-1)-(2*n-3)^2*LAMBDA(n-1, m)) end do end do; seq(seq(LAMBDA(n,m), m=1..n-1), n=2..nmax);
# End first program.
nmax1:=10; m:=1; LS1row:=-2*m; for n from 0 to nmax1 do cfn2(n, 0) := 1: cfn2(n, n) := (doublefactorial(2*n-1))^2 od: for n from 1 to nmax1 do for k from 1 to n-1 do cfn2(n, k) := (2*n-1)^2*cfn2(n-1, k-1) + cfn2(n-1, k) od: od: mmax1:=nmax1: for m1 from 1 to mmax1 do LS1[-2*m1, 1] := 2*(1-2^(-(-2*m1+1)))*(-bernoulli(2*m1)/(2*m1)) od: for n from 2 to nmax1 do for m1 from 1 to mmax1-n+1 do LS1[ -2*m1, n] := sum((-1)^(k1+1)*cfn2(n-1,k1-1)* LS1[2*k1-2*n-2*m1, 1], k1=1..n)/(2*n-2)! od: od: seq(LS1[ -2*m, n], n=1..nmax1-m+1);
# End second program.

Formula

We discovered a remarkable relation between the Lambda triangle coefficients Lambda(n,m) = ZL(n)*(Lambda(n-1,m-1)-(2*n-3)^2*Lambda(n-1,m)) for n = 3, 4, .. and m = 2, 3, .. . See A160488 for LAMBDA(n,m=1) and furthermore LAMBDA(n,n) = 0 for n = 2, 3, .. .
We observe that the ZL(n) = A160479(n) sequence also rules the Zeta triangle A160474.
The generating functions GL(z;n) of the coefficients in the matrix columns are defined by
GL(z;n) = sum(LS1[2*m-2,n]*z^(2*m-2), m=1..infinity), with n = 1, 2, 3, .. .
This definition, and our choice of LS1[m=0,n=1] = gamma, leads to GL(z;n=1) = -2*Psi(1-z)+Psi(1-(z/2))-(Pi/2)*tan(Pi*z/2) with Psi(z) the digamma-function. Furthermore we discovered that GL(z;n) =GL(z;n-1)*(z^2/((2*n-2)*(2*n-3)) -(2*n-3)/((2*n-2)))+LS1[ -2,n-1]/((2*n-2)*(2*n-3)) for n = 2, 3 , .. . with LS1[ -2,n] = (-1)^(n-1)*4*A058962(n-1)*A002197(n-1)/A002198(n-1) for n = 1, 2, .. , with A058962(n-1) = 2^(2*n-2)*(2*n-1).
We found the following general expression for the GL(z;n) polynomials, for n = 2, 3, ..
GL(z;n) = (h(n)*CFN2(z;n)*GL(z;n=1) + LAMBDA(z;n))/p(n) with
h(n) = 6*A160476(n) and p(n) = A160490(n).

A155988 a(n) = (2*n + 1)*9^n.

Original entry on oeis.org

1, 27, 405, 5103, 59049, 649539, 6908733, 71744535, 731794257, 7360989291, 73222472421, 721764371007, 7060738412025, 68630377364883, 663426981193869, 6382625094934119, 61149666232110753, 583701359488329915, 5553501505988967477, 52683216989246691471, 498464283821334080841
Offset: 0

Views

Author

Jaume Oliver Lafont, Feb 01 2009

Keywords

Links

Crossrefs

Cf. A058962 for the similar (2n+1)4^n.
Cf. A001019, A005408, A060851, A096949, A096950, A154920, A164985, A165132.

Programs

  • Magma
    [(2*n+1)*9^n: n in [0..20]]; // Vincenzo Librandi, Jun 08 2011
  • Maxima
    makelist((2*n+1)*9^n, n, 0, 20); /* Martin Ettl, Nov 11 2012 */
  • PARI
    a(n)=(2*n+1)*9^n;

Formula

G.f.: (1 + 9*x)/(1 - 9*x)^2.
a(n) = 18*a(n-1) - 81*a(n-2) for n>=2.
Sum_{n>=0} 1/a(n) = (3/2)*log(2).
a(n) = A005408(n) * A001019(n).
a(n) = (2*n - 1)*3^(2*n-1)/3 = A060851(n)/3.
Sum_{n>=0} (-1)^n/a(n) = 3*arctan(1/3). - Amiram Eldar, Feb 26 2022
E.g.f.: exp(9*x)*(1 + 18*x). - Stefano Spezia, May 07 2023

A154920 Denominators of a ternary BBP-type formula for log(3).

Original entry on oeis.org

1, 18, 27, 324, 405, 4374, 5103, 52488, 59049, 590490, 649539, 6377292, 6908733, 66961566, 71744535, 688747536, 731794257, 6973568802, 7360989291, 69735688020, 73222472421, 690383311398, 721764371007, 6778308875544
Offset: 0

Views

Author

Jaume Oliver Lafont, Jan 17 2009, Jan 18 2009

Keywords

Comments

log(3) = Sum_{k>=0} (9/(2k+1)+1/(2k+2))/9^(k+1).
log(3) = 1 + Sum_{k>=0} (1/(2k+2)+1/(2k+3))/9^(k+1).

Links

Crossrefs

Cf. A002391, A058962.
Cf. A164985, A165132. - Jaume Oliver Lafont, Sep 25 2009

Programs

Formula

a(n) = (n+1)*9^[(n+1)/2] = 18*a(n-2) - 81*a(n-4).
Sum_{n>=0} 1/a(n) = log(3).
G.f.: (1+18*x+9*x^2)/(1-9*x^2)^2. - Jaume Oliver Lafont, Jan 29 2009
a(n) = (2-(-1)^n)*(n+1)*3^n. - Jaume Oliver Lafont, Sep 27 2009
Sum_{n>=0} (-1)^n/a(n) = log(8/3). - Amiram Eldar, Feb 26 2022

A171220 a(n) = (2n + 1)*5^n.

Original entry on oeis.org

1, 15, 125, 875, 5625, 34375, 203125, 1171875, 6640625, 37109375, 205078125, 1123046875, 6103515625, 32958984375, 177001953125, 946044921875, 5035400390625, 26702880859375, 141143798828125, 743865966796875, 3910064697265625, 20503997802734375, 107288360595703125
Offset: 0

Views

Author

Jaume Oliver Lafont, Dec 05 2009

Keywords

Comments

Inserting x=1/sqrt(b) into the power series expansion of arctanh(x) yields the general BBP-type formula log((sqrt(b)+1)/(sqrt(b)-1))*sqrt(b)/2 = Sum_{k>=0} 1/((2k+1)b^k).
This sequence illustrates case b=5, with
Sum_{k>=0} 1/a(k) = sqrt(5)*log((1+sqrt(5))/2).

Links

Crossrefs

Cf. A014480 ((2n+1)*2^n), A124647 ((2n+1)*3^n), A058962 ((2n+1)*4^n), A155988 ((2n+1)*9^n), A165283 ((2n+1)*16^n), A166725 ((2n+1)*25^n).

Programs

  • Magma
    [(2*n+1)*5^n: n in [0..25]]; // Vincenzo Librandi, Jun 08 2011
  • PARI
    a(n)=(2*n+1)*5^n

Formula

a(n) = 10*a(n-1) - 25*a(n-2).
O.g.f: (1+5*x)/(1-5*x)^2.
Sum_{n>=0} (-1)^n/a(n) = sqrt(5)*arctan(1/sqrt(5)). - Amiram Eldar, Feb 26 2022
E.g.f.: exp(5*x)*(1 + 10*x). - Stefano Spezia, May 09 2023

A199299 a(n) = (2*n + 1)*6^n.

Original entry on oeis.org

1, 18, 180, 1512, 11664, 85536, 606528, 4199040, 28553472, 191476224, 1269789696, 8344332288, 54419558400, 352638738432, 2272560758784, 14575734521856, 93096626946048, 592433080565760, 3757718396731392, 23765029860409344, 149902496042582016, 943288877536247808
Offset: 0

Views

Author

Philippe Deléham, Nov 04 2011

Keywords

Links

Crossrefs

Cf. A014480, A058962, A124647, A155988, A171220, A199300, A199301.

Programs

  • Magma
    [(2*n+1)*6^n: n in [0..30]]; // Vincenzo Librandi, Nov 05 2011
  • Mathematica
    a[n_] := (2*n + 1)*6^n; Array[a, 25, 0] (* Amiram Eldar, Dec 10 2022 *)
  • PARI
    a(n) = (2*n+1)*6^n \\ Amiram Eldar, Dec 10 2022

Formula

a(n) = 12*a(n-1) - 36*a(n-2).
G.f.: (1+6*x)/(1-6*x)^2.
a(n) = 6*a(n-1) + 2*6^n. - Vincenzo Librandi, Nov 05 2011
From Amiram Eldar, Dec 10 2022: (Start)
Sum_{n>=0} 1/a(n) = sqrt(6)*arccoth(sqrt(6)).
Sum_{n>=0} (-1)^n/a(n) = sqrt(6)*arccot(sqrt(6)). (End)
E.g.f.: exp(6*x)*(1 + 12*x). - Stefano Spezia, May 07 2023

A118415 a(n) = (4*n - 3) * 2^(n - 1).

Original entry on oeis.org

1, 10, 36, 104, 272, 672, 1600, 3712, 8448, 18944, 41984, 92160, 200704, 434176, 933888, 1998848, 4259840, 9043968, 19136512, 40370176, 84934656, 178257920, 373293056, 780140544, 1627389952, 3388997632, 7046430720, 14629732352, 30333206528, 62813896704
Offset: 1

Views

Author

Reinhard Zumkeller, Apr 27 2006

Keywords

Comments

Central terms of the triangle in A118413.

Links

Crossrefs

Cf. A058962.

Programs

  • Magma
    [(4*n-3)*2^(n-1): n in [1..40]]; // Vincenzo Librandi, Dec 26 2010
  • Mathematica
    CoefficientList[Series[(1 + 6 x)/(-1 + 2 x)^2, {x, 0, 40}], x] (* Vincenzo Librandi, May 21 2014 *)
LinearRecurrence[{4,-4},{1,10},30] (* Harvey P. Dale, Sep 16 2022 *)

Formula

a(n) = A016813(n-1)*A000079(n-1).
O.g.f.: x*(1+6*x)/(-1+2*x)^2 . - R. J. Mathar, Feb 26 2008

Extensions

More terms from R. J. Mathar, Feb 26 2008

A199300 a(n) = (2*n + 1)*7^n.

Original entry on oeis.org

1, 21, 245, 2401, 21609, 184877, 1529437, 12353145, 98001617, 766718533, 5931980229, 45478515089, 346032180025, 2616003280989, 19668469112621, 147174406808233, 1096686708796833, 8142067989552245, 60251303122686613, 444556912229552577, 3271482918202092041
Offset: 0

Views

Author

Philippe Deléham, Nov 04 2011

Keywords

Links

Crossrefs

Cf. A014480, A058962, A124647, A155988, A171220, A199299, A199301.

Programs

  • Magma
    [(2*n+1)*7^n: n in [0..30]]; // Vincenzo Librandi, Nov 05 2011
  • Mathematica
    a[n_] := (2*n + 1)*7^n; Array[a, 25, 0] (* Amiram Eldar, Dec 10 2022 *)
LinearRecurrence[{14,-49},{1,21},30] (* Harvey P. Dale, Mar 26 2025 *)
  • PARI
    a(n) = (2*n+1)*7^n \\ Amiram Eldar, Dec 10 2022

Formula

a(n) = 14*a(n-1) - 49*a(n-2).
G.f.: (1+7*x)/(1-7*x)^2.
a(n) = 7*a(n-1) + 2*7^n. - Vincenzo Librandi, Nov 05 2011
From Amiram Eldar, Dec 10 2022: (Start)
Sum_{n>=0} 1/a(n) = sqrt(7)*arccoth(sqrt(7)).
Sum_{n>=0} (-1)^n/a(n) = sqrt(7)*arccot(sqrt(7)). (End)
E.g.f.: exp(7*x)*(1 + 14*x). - Stefano Spezia, May 09 2023

Extensions

a(15) corrected by Vincenzo Librandi, Nov 05 2011
Showing 1-10 of 20 results. Next

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.