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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A006095 Gaussian binomial coefficient [n, 2] for q = 2.

Original entry on oeis.org

0, 0, 1, 7, 35, 155, 651, 2667, 10795, 43435, 174251, 698027, 2794155, 11180715, 44731051, 178940587, 715795115, 2863245995, 11453115051, 45812722347, 183251413675, 733006703275, 2932028910251, 11728119835307, 46912487729835
Offset: 0

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Author

Keywords

Comments

Number of 4-block coverings of an n-set where every element of the set is covered by exactly 3 blocks (if offset is 3), so a(n) = (1/4!)*(4^n-6*2^n+8). - Vladeta Jovovic, Feb 20 2001
Number of non-coprime pairs of polynomials (f,g) with binary coefficients where both f and g have degree n+1 and nonzero constant term. - Luca Mariot and Enrico Formenti, Sep 26 2016
Number of triplets found from the integers 1 to 2^n-1 by converting to binary and performing an XOR operation on the corresponding bits of each pair. Defining addition in this carryless way (0+0=1+1=0, 0+1=1+0=1), each triplet (A,B,C) has the property A+B=C, A+C=B and B+C=A. For example, n=3 gives the 7 triplets (1,2,3), (1,4,5), (1,6,7), (2,4,6), (2,5,7), (3,4,7) and (3,5,6). Each integer appears in the set of triplets 2^(n-1)-1 times, for example 3 for n=3. - Ian Duff, Oct 05 2019
Number of 2-dimensional vector subspaces of (Z_2)^n, so also number of Klein subgroups of the group (C_2)^n. - Robert FERREOL, Jul 28 2021

References

  • J. Goldman and G.-C. Rota, The number of subspaces of a vector space, pp. 75-83 of W. T. Tutte, editor, Recent Progress in Combinatorics. Academic Press, NY, 1969.
  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration. Wiley, NY, 1983, p. 99.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • M. Sved, Gaussians and binomials, Ars. Combinatoria, 17A (1984), 325-351.

Crossrefs

First differences: A006516.
Gaussian binomial coefficient [n, k] for q = 2: A000225 (k = 1), this sequence (k = 2), A006096 (k = 3), A006097 (k = 4), A006110 (k = 5), A022189 - A022195 (k = 6 thru 12).

Programs

  • Maple
    a:= n-> add((4^(n-1-j) - 2^(n-1-j))/2, j=0..n-1):
    seq(a(n), n=0..24); # Zerinvary Lajos, Jan 04 2007
    A006095 := -z^2/(z-1)/(2*z-1)/(4*z-1); # Simon Plouffe in his 1992 dissertation. [adapted to offset 0 by Peter Luschny, Jul 20 2021]
    a := n -> (2^n - 2)*(2^n - 1)/6:
    seq(a(n), n = 0..24); # Peter Luschny, Jul 20 2021
  • Mathematica
    Join[{a=0,b=0},Table[c=6*b-8*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 06 2011 *)
    CoefficientList[Series[x^2/((1-x)(1-2x)(1-4x)),{x,0,30}],x] (* or *) LinearRecurrence[{7,-14,8},{0,0,1},30] (* Harvey P. Dale, Jul 22 2011 *)
    (* Next, using elementary symmetric functions *)
    f[k_] := 2^(k - 1); t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[2, t[n]]
    Table[a[n], {n, 2, 32}]    (* A203235 *)
    Table[a[n]/2, {n, 2, 32}]  (* A006095 *)
    (* Clark Kimberling, Dec 31 2011 *)
    Table[QBinomial[n, 2, 2], {n, 0, 24}] (* Arkadiusz Wesolowski, Nov 12 2015 *)
  • PARI
    a(n) = (2^n - 1)*(2^(n-1) - 1)/3 \\ Charles R Greathouse IV, Jul 25 2011
    
  • PARI
    concat([0, 0], Vec(x^2/((1-x)*(1-2*x)*(1-4*x)) + O(x^50))) \\ Altug Alkan, Nov 12 2015
  • Sage
    [gaussian_binomial(n,2,2) for n in range(0,25)] # Zerinvary Lajos, May 24 2009
    

Formula

G.f.: x^2/((1-x)(1-2x)(1-4x)).
a(n) = (2^n - 1)*(2^(n-1) - 1)/3 = 4^n/6 - 2^(n-1) + 1/3.
Row sums of triangle A130324. - Gary W. Adamson, May 24 2007
a(n) = Stirling2(n+1,3) + Stirling2(n+1,4). - Zerinvary Lajos, Oct 04 2007; corrected by R. J. Mathar, Mar 19 2011
a(n) = A139250(2^(n-1) - 1), n >= 1. - Omar E. Pol, Mar 03 2011
a(n) = 4*a(n-1) + 2^(n-1) - 1, n >= 2. - Vincenzo Librandi, Mar 19 2011
a(0) = 0, a(1) = 0, a(2) = 1, a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3). - Harvey P. Dale, Jul 22 2011
a(n) = Sum_{k=0..n-2} 2^k*C(2*n-k-2, k), n >= 2. - Johannes W. Meijer, Aug 19 2013
a(n) = Sum_{i=0..n-2, j=i..n-2} 2^{i+j} = 2^0 * (2^0 + 2^1 + ... + 2^(n-2)) + 2^1 * (2^1 + 2^2 + ... + 2^(n-2)) + ... + 2^(n-2) * 2^(n-2), n>1. - J. M. Bergot, May 08 2017
a(n) = a(n-1) + A000217(A000225(n-1)), n > 0. - Ivan N. Ianakiev, Dec 11 2017
E.g.f.: (2*exp(x)-3*exp(2*x)+exp(4*x))/6. - Paul Weisenhorn, Aug 22 2021
From Peter Bala, Jul 01 2025: (Start)
G.f. assuming an offset of 0: exp( Sum_{n >= 1} b(3*n)/b(n)*x^n/n ) = 1 + 7*x + 35*x^2 + ..., where b(n) = A000225(n) = 2^n - 1.
The following are examples of telescoping series:
Sum_{n >= 2} 2^n/a(n) = 6, follows from 1 - (1/6)*Sum_{k = 2..n} 2^k/a(k) = 1/(2^n - 1).
Sum_{n >= 2} 2^n/(a(n)*a(n+2)) = 6/49, follows from 1 - (49/6)*Sum_{k = 2..n} 2^k/(a(k)*a(k+2)) = 1/A006096(n+2);
Sum_{n >= 2} 4^n/(a(n)*a(n+2)) = 26/49, follows from 13 - (49/2)*Sum_{k = 2..n} 4^k/(a(k)*a(k+2)) = A086224(n)/A006096(n+2);
Sum_{n >= 2} 8^n/(a(n)*a(n+2)) = 129/49, follows from 43 - (49/3)*Sum_{k = 2..n} 8^k/(a(k)*a(k+2)) = A171479(n+1)/A006096(n+2). (End)

A000392 Stirling numbers of second kind S(n,3).

Original entry on oeis.org

0, 0, 0, 1, 6, 25, 90, 301, 966, 3025, 9330, 28501, 86526, 261625, 788970, 2375101, 7141686, 21457825, 64439010, 193448101, 580606446, 1742343625, 5228079450, 15686335501, 47063200806, 141197991025, 423610750290, 1270865805301
Offset: 0

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Author

Keywords

Comments

Number of palindromic structures using exactly three different symbols; Mobius transform: A056279. - Marks R. Nester
Number of ways of placing n labeled balls into k=3 indistinguishable boxes. - Thomas Wieder, Nov 30 2004
With two leading zeros, this is the second binomial transform of cosh(x)-1 and the binomial transform of A000225 (with extra leading zero). - Paul Barry, May 13 2003
Let [m] denote the first m positive integers. Then a(n) is the number of functions f from [n] to [n+1] that satisfy (i) f(x) > x for all x, (ii) f(x) = n+1 for exactly 3 elements and (iii) f(f(x)) = n+1 for the remaining n-3 elements of [n]. For example, a(4)=6 since there are exactly 6 functions from {1,2,3,4} to {1,2,3,4,5} such that f(x) > x, f(x) = 5 for 3 elements and f(f(x)) = 5 for the remaining element. The functions are f1 = {(1,5), (2,5), (3,4), (4,5)}, f2 = {(1,5), (2,3), (3,5), (4,5)}, f3 = {(1,5), (2,4), (3,5), (4,5)}, f4 = {(1,2), (2,5), (3,5), (4,5)}, f5 = {(1,3), (2,5), (3,5), (4,5)}, f6 = {(1,4), (2,5), (3,5), (4,5)}. - Dennis P. Walsh, Feb 20 2007
Conjecture. Let S(1)={1} and, for n > 1, let S(n) be the smallest set containing x, 2x and 3x for each element x in S(n-1). Then a(n+2) is the sum of the elements in S(n). (It is easy to prove that the number of elements in S(n) is the n-th triangular number given by A001952.) See A122554 for a sequence defined in this way. - John W. Layman, Nov 21 2007; corrected (a(n) to a(n+2) due to offset change) by Fred Daniel Kline, Oct 02 2014
Let P(A) be the power set of an n-element set A. Then a(n+1) = the number of pairs of elements {x,y} of P(A) for which x and y are disjoint and for which x is not a subset of y and y is not a subset of x. Wieder calls these "disjoint strict 2-combinations". - Ross La Haye, Jan 11 2008; corrected by Ross La Haye, Oct 29 2008
Also, let P(A) be the power set of an n-element set A. Then a(n+2) = the number of pairs of elements {x,y} of P(A) for which either 0) x and y are disjoint and for which either x is a subset of y or y is a subset of x, or 1) x and y are disjoint and for which x is not a subset of y and y is not a subset of x, or 2) x and y are intersecting and for which either x is a proper subset of y or y is a proper subset of x. - Ross La Haye, Jan 11 2008
3 * a(n+1) = p(n+1) where p(x) is the unique degree-n polynomial such that p(k) = a(k+1) for k = 0, 1, ..., n. - Michael Somos, Apr 29 2012
John W. Layman's conjecture that a(n+2) is the sum of elements in S(n) follows from the identification of S(n) with the first n rows of A036561, whose row sums are A001047. - Fred Daniel Kline, Oct 02 2014
From M. Sinan Kul, Sep 08 2016: (Start)
Let m be equal to the product of n-1 distinct primes. Then a(n) is equal to the number of distinct fractions >=1 that may be created by dividing a divisor of m by another divisor. For example for m = 2*3*5 = 30, we would have the following 6 fractions: 6/5, 3/2, 5/3, 5/2, 10/3, 15/2.
Here finding the number of fractions would be equivalent to distributing n-1 balls (distinct primes) to two bins (numerator and denominator) with no empty bins which can be found by Stirling numbers of the second kind. So another definition for a(n) is a(n) = Sum_{i=2..n-1} Stirling2(i,2)*binomial(n-1,i).
Also for n > 0, a(n) = (d(m^2)+1)/2 - d(m) where m is equal to the product of n-1 distinct primes. Example for a(5): m = 2*3*5*7 = 210 (product of four distinct primes) so a(5) = (d(210^2)+1)/2 - d(210) = 41 - 16 = 25. (End)
6*a(n) is the number of ternary strings of length n that contain at least one of each of the 3 symbols on which they are defined. For example, for n=4, the strings are the 12 permutations of 0012, the 12 permutations of 0112, and the 12 permutations of 0122. - Enrique Navarrete, Aug 23 2021
A simpler form of La Haye's first comment is: a(n+1) is the number of ways we can form disjoint unions of two nonempty subsets of [n] (see example below). Cf. A001047 for the requirement that the union contains n. - Enrique Navarrete, Aug 24 2021
As partial sums of the Nicomachus triangle's rows and the differences of the powers of 3 and 2 (A001047), each iteration corresponds to two figurate variations of the Sierpinski triangle (3^n) with cross-correlation to the Nicomachus triangle, see illustrations in links. The Sierpinski half-hexagons of (A001047) stack and conform to the footprint of 2^n - 1 triangular numbers. The 3^n Sierpinski triangle minus its 2^n bottom row, also correlates to the Nicomachus triangle according to each Sierpinski triangular sub-row. - John Elias, Oct 04 2021

Examples

			a(4) = 6. Let denote Z[i] the i-th labeled element = "ball". Then one has for n=4 six different ways to fill sets = "boxes" with the labeled elements:
Set(Set(Z[3], Z[4]), Set(Z[1]), Set(Z[2])), Set(Set(Z[3], Z[1]), Set(Z[4]), Set(Z[2])), Set(Set(Z[4], Z[1]), Set(Z[3]), Set(Z[2])), Set(Set(Z[4]), Set(Z[1]), Set(Z[3], Z[2])), Set(Set(Z[3]), Set(Z[1], Z[2]), Set(Z[4])), Set(Set(Z[3]), Set(Z[1]), Set(Z[4], Z[2])).
G.f. = x^3 + 6*x^4 + 25*x^5 + 90*x^6 + 301*x^7 + 966*x^8 + 3025*x^9 + ...
For example, for n=3, a(4)=6 since the disjoint unions are: {1}U{2}, {1}U{3}, {1}U{2,3}, {2}U{3}, {2}U{1,3}, and {1,2}U{3}. - _Enrique Navarrete_, Aug 24 2021
		

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. 835.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 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).

Crossrefs

Programs

  • GAP
    List([0..400], n->Stirling2(n,3)); # Muniru A Asiru, Feb 04 2018
  • Maple
    A000392 := n -> 9/2*3^n-4*2^n+1/2;  [ seq(9/2*3^n-4*2^n+1/2,n=0..30) ]; # Thomas Wieder
    A000392:=-1/(z-1)/(3*z-1)/(2*z-1); # Simon Plouffe in his 1992 dissertation
  • Mathematica
    StirlingS2[Range[0,30],3] (* Harvey P. Dale, Dec 29 2011 *)
  • PARI
    {a(n) = 3^(n-1) / 2 - 2^(n-1) + 1/2};
    
  • Sage
    [stirling_number2(i,3) for i in (0..40)] # Zerinvary Lajos, Jun 26 2008
    

Formula

G.f.: x^3/((1-x)*(1-2*x)*(1-3*x)).
E.g.f.: ((exp(x) - 1)^3) / 3!.
Recurrence: a(n+3) = 6*a(n+2) - 11*a(n+1) + 6*a(n), a(3) = 1, a(4) = 6, a(5) = 25. - Thomas Wieder, Nov 30 2004
With offset 0, this is 9*3^n/2 - 4*2^n + 1/2, the partial sums of 3*3^n - 2*2^n = A001047(n+1). - Paul Barry, Jun 26 2003
a(n) = (1 + 3^(n-1) - 2^n)/2, n > 0. - Dennis P. Walsh, Feb 20 2007
For n >= 3, a(n) = 3*a(n-1) + 2^(n-2) - 1. - Geoffrey Critzer, Mar 03 2009
a(n) = 5*a(n-1) - 6*a(n-2) + 1, for n > 3. - Vincenzo Librandi Nov 25 2010
a(n) = det(|s(i+3,j+2)|, 1 <= i,j <= n-3), where s(n,k) are Stirling numbers of the first kind. - Mircea Merca, Apr 06 2013
G.f.: x^3 + 12*x^4/(G(0)-12*x), where G(k) = x + 1 + 9*(3*x+1)*3^k - 8*(2*x+1)*2^k - x*(9*3^k+1-8*2^k)*(81*3^k+1-32*2^k)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Feb 01 2014
a(n + 2) = (1 - 2^(2 + n) + 3^(1 + n))/2 for n > 0. - Fred Daniel Kline, Oct 02 2014
For n > 0, a(n) = (1/2) * Sum_{k=1..n-1} Sum_{i=1..n-1} C(n-k-1,i) * C(n-1,k). - Wesley Ivan Hurt, Sep 22 2017
a(n) = Sum_{k=0..n-3} 2^(k-1)*(3^(n-2-k) - 1). - J. M. Bergot, Feb 05 2018

Extensions

Offset changed by N. J. A. Sloane, Feb 08 2008

A052548 a(n) = 2^n + 2.

Original entry on oeis.org

3, 4, 6, 10, 18, 34, 66, 130, 258, 514, 1026, 2050, 4098, 8194, 16386, 32770, 65538, 131074, 262146, 524290, 1048578, 2097154, 4194306, 8388610, 16777218, 33554434, 67108866, 134217730, 268435458, 536870914, 1073741826, 2147483650
Offset: 0

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Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

The most "compact" sequence that satisfies Bertrand's Postulate. Begin with a(1) = 3 = n, then 2n - 2 = 4 = n_1, 2n_1 - 2 = 6 = n_2, 2n_2 - 2 = 10, etc. = a(n), hence there is guaranteed to be at least one prime between successive members of the sequence. - Andrew S. Plewe, Dec 11 2007
Number of 2-sided prudent polygons of area n, for n>0, see Beaton, p. 5. - Jonathan Vos Post, Nov 30 2010

Crossrefs

Programs

  • Haskell
    a052548 = (+ 2) . a000079
    a052548_list = iterate ((subtract 2) . (* 2)) 3
    -- Reinhard Zumkeller, Sep 05 2015
  • Magma
    [2^n + 2: n in [0..35]]; // Vincenzo Librandi, Apr 29 2011
    
  • Maple
    spec := [S,{S=Union(Sequence(Union(Z,Z)),Sequence(Z),Sequence(Z))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    2^Range[0,40]+2 (* Harvey P. Dale, Jun 26 2012 *)
  • PARI
    a(n)=1<Charles R Greathouse IV, Nov 20 2011
    

Formula

G.f.: (3-5*x)/((1-2*x)*(1-x)) = (3-5*x)/(1 - 3*x + 2*x^2) = 2/(1-x) + 1/(1-2*x).
a(0)=3, a(1)=4, a(n) = 3*a(n-1) - 2*a(n-2).
a(n) = A058896(n)/A000918(n), for n>0. - Reinhard Zumkeller, Feb 14 2009
a(n) = A173786(n,1), for n>0. - Reinhard Zumkeller, Feb 28 2010
a(n)*A000918(n) = A028399(2*n), for n>0. - Reinhard Zumkeller, Feb 28 2010
a(0)=3, a(n) = 2*a(n-1) - 2. - Vincenzo Librandi, Aug 06 2010
E.g.f.: (2 + exp(x))*exp(x). - Ilya Gutkovskiy, Aug 16 2016

Extensions

More terms from James Sellers, Jun 06 2000

A016123 a(n) = (11^(n+1) - 1)/10.

Original entry on oeis.org

1, 12, 133, 1464, 16105, 177156, 1948717, 21435888, 235794769, 2593742460, 28531167061, 313842837672, 3452271214393, 37974983358324, 417724816941565, 4594972986357216, 50544702849929377, 555991731349223148
Offset: 0

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Author

Keywords

Comments

11^a(n) is highest power of 11 dividing (11^(n+1))!.
Partial sums of powers of 11 (A001020).
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=11, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=det(A). - Milan Janjic, Feb 21 2010
Let A be the Hessenberg matrix of the order n, defined by: A[1,j]=1, A[i,i]:=12, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=2, a(n-1)=(-1)^n*charpoly(A,1). - Milan Janjic, Feb 21 2010

Crossrefs

For analogs with primes 2, 3, 5, 7, 13 and 17 see: A000225, A003462, A003463, A023000, A091030 and A091045, respectively.

Programs

Formula

a(n) = Sum_{k=0..n} 11^k = (11^(n+1) - 1)/10.
G.f.: (1/(1-11*x) - 1/(1-x))/(10*x) = 1/((1-11*x)*(1-x)).
a(0)=1, a(n) = 11*a(n-1) + 1. - Vincenzo Librandi, Feb 05 2011
a(0)=0, a(1)=1, a(n) = 12*a(n-1) - 11*a(n-2). - Harvey P. Dale, Apr 05 2012
E.g.f.: exp(x)*(11*exp(10*x) - 1)/10. - Stefano Spezia, Mar 11 2023

Extensions

Title edited by Daniel Forgues, Jul 08 2011

A106566 Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, 1, 1, 1, 1, 1, 1, ... ] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ... ] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 5, 5, 3, 1, 0, 14, 14, 9, 4, 1, 0, 42, 42, 28, 14, 5, 1, 0, 132, 132, 90, 48, 20, 6, 1, 0, 429, 429, 297, 165, 75, 27, 7, 1, 0, 1430, 1430, 1001, 572, 275, 110, 35, 8, 1, 0, 4862, 4862, 3432, 2002, 1001, 429, 154, 44, 9, 1
Offset: 0

Views

Author

Philippe Deléham, May 30 2005

Keywords

Comments

Catalan convolution triangle; g.f. for column k: (x*c(x))^k with c(x) g.f. for A000108 (Catalan numbers).
Riordan array (1, xc(x)), where c(x) the g.f. of A000108; inverse of Riordan array (1, x*(1-x)) (see A109466).
Diagonal sums give A132364. - Philippe Deléham, Nov 11 2007

Examples

			Triangle begins:
  1;
  0,   1;
  0,   1,   1;
  0,   2,   2,  1;
  0,   5,   5,  3,  1;
  0,  14,  14,  9,  4,  1;
  0,  42,  42, 28, 14,  5, 1;
  0, 132, 132, 90, 48, 20, 6, 1;
From _Paul Barry_, Sep 28 2009: (Start)
Production array is
  0, 1,
  0, 1, 1,
  0, 1, 1, 1,
  0, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1, 1, 1 (End)
		

Crossrefs

The three triangles A059365, A106566 and A099039 are the same except for signs and the leading term.
See also A009766, A033184, A059365 for other versions.
The following are all versions of (essentially) the same Catalan triangle: A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.

Programs

  • Magma
    A106566:= func< n,k | n eq 0 select 1 else (k/n)*Binomial(2*n-k-1, n-k) >;
    [A106566(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 06 2021
    
  • Maple
    A106566 := proc(n,k)
        if n = 0 then
            1;
        elif k < 0 or k > n then
            0;
        else
            binomial(2*n-k-1,n-k)*k/n ;
        end if;
    end proc: # R. J. Mathar, Mar 01 2015
  • Mathematica
    T[n_, k_] := Binomial[2n-k-1, n-k]*k/n; T[0, 0] = 1; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 18 2017 *)
    (* The function RiordanArray is defined in A256893. *)
    RiordanArray[1&, #(1-Sqrt[1-4#])/(2#)&, 11] // Flatten (* Jean-François Alcover, Jul 16 2019 *)
  • PARI
    {T(n, k) = if( k<=0 || k>n, n==0 && k==0, binomial(2*n - k, n) * k/(2*n - k))}; /* Michael Somos, Oct 01 2022 */
  • Sage
    def A106566(n, k): return 1 if (n==0) else (k/n)*binomial(2*n-k-1, n-k)
    flatten([[A106566(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 06 2021
    

Formula

T(n, k) = binomial(2n-k-1, n-k)*k/n for 0 <= k <= n with n > 0; T(0, 0) = 1; T(0, k) = 0 if k > 0.
T(0, 0) = 1; T(n, 0) = 0 if n > 0; T(0, k) = 0 if k > 0; for k > 0 and n > 0: T(n, k) = Sum_{j>=0} T(n-1, k-1+j).
Sum_{j>=0} T(n+j, 2j) = binomial(2n-1, n), n > 0.
Sum_{j>=0} T(n+j, 2j+1) = binomial(2n-2, n-1), n > 0.
Sum_{k>=0} (-1)^(n+k)*T(n, k) = A064310(n). T(n, k) = (-1)^(n+k)*A099039(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000007(n), A000108(n), A000984(n), A007854(n), A076035(n), A076036(n), A127628(n), A126694(n), A115970(n) for x = 0,1,2,3,4,5,6,7,8 respectively.
Sum_{k>=0} T(n, k)*x^(n-k) = C(x, n); C(x, n) are the generalized Catalan numbers.
Sum_{j=0..n-k} T(n+k,2*k+j) = A039599(n,k).
Sum_{j>=0} T(n,j)*binomial(j,k) = A039599(n,k).
Sum_{k=0..n} T(n,k)*A000108(k) = A127632(n).
Sum_{k=0..n} T(n,k)*(x+1)^k*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x= 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Aug 25 2007
Sum_{k=0..n} T(n,k)*A000108(k-1) = A121988(n), with A000108(-1)=0. - Philippe Deléham, Aug 27 2007
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Oct 27 2007
T(n,k)*2^(n-k) = A110510(n,k); T(n,k)*3^(n-k) = A110518(n,k). - Philippe Deléham, Nov 11 2007
Sum_{k=0..n} T(n,k)*A000045(k) = A109262(n), A000045: Fibonacci numbers. - Philippe Deléham, Oct 28 2008
Sum_{k=0..n} T(n,k)*A000129(k) = A143464(n), A000129: Pell numbers. - Philippe Deléham, Oct 28 2008
Sum_{k=0..n} T(n,k)*A100335(k) = A002450(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A100334(k) = A001906(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A099322(k) = A015565(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A106233(k) = A003462(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A151821(k+1) = A100320(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A082505(k+1) = A144706(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A000045(2k+2) = A026671(n). - Philippe Deléham, Feb 11 2009
Sum_{k=0..n} T(n,k)*A122367(k) = A026726(n). - Philippe Deléham, Feb 11 2009
Sum_{k=0..n} T(n,k)*A008619(k) = A000958(n+1). - Philippe Deléham, Nov 15 2009
Sum_{k=0..n} T(n,k)*A027941(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
G.f.: Sum_{n>=0, k>=0} T(n, k)*x^k*z^n = 1/(1 - x*z*c(z)) where c(z) the g.f. of A000108. - Michael Somos, Oct 01 2022

Extensions

Formula corrected by Philippe Deléham, Oct 31 2008
Corrected by Philippe Deléham, Sep 17 2009
Corrected by Alois P. Heinz, Aug 02 2012

A048473 a(0)=1, a(n) = 3*a(n-1) + 2; a(n) = 2*3^n - 1.

Original entry on oeis.org

1, 5, 17, 53, 161, 485, 1457, 4373, 13121, 39365, 118097, 354293, 1062881, 3188645, 9565937, 28697813, 86093441, 258280325, 774840977, 2324522933, 6973568801, 20920706405, 62762119217, 188286357653, 564859072961, 1694577218885, 5083731656657, 15251194969973, 45753584909921
Offset: 0

Views

Author

Keywords

Comments

The number of triangles (of all sizes, including holes) in Sierpiński's triangle after n inscriptions. - Lee Reeves, May 10 2004
The sequence is not only related to Sierpiński's triangle, but also to "Floret's cube" and the quaternion factor space Q X Q / {(1,1), (-1,-1)}. It can be written as a_n = ves((A+1)x)^n) as described at the Math Forum Discussions link. - Creighton Dement, Jul 28 2004
Relation to C(n) = Collatz function iteration using only odd steps: If we look for record subsequences where C(n) > n, this subsequence starts at 2^n - 1 and stops at the local maximum of 2*3^n - 1. Examples: [3,5], [7,11,17], [15,23,35,53], ..., [127,191,287,431,647,971,1457]. - Lambert Klasen, Mar 11 2005
Group the natural numbers so that the (2n-1)-th group sum is a multiple of the (2n)-th group containing one term. (1,2),(3),(4,5,6,7,8,9,10,11),(12),(13,14,15,16,17,18,19,...,38),(39),(40,41,...,118,119),(120), (121,122,123,...) ... a(n) = {the sum of the terms of (2n-1)-th group}/{the term of (2n)th group}. The first term of the odd numbered group is given by A003462. The only term of even numbered group is given by A029858. - Amarnath Murthy, Aug 01 2005
a(n)+1 = A008776(n); it appears that this gives the number of terms in the (n+1)-th "gap" of numbers missing in A171884. - M. F. Hasler, May 09 2013
Sum of n-th row of triangle of powers of 3: 1; 1 3 1; 1 3 9 3 1; 1 3 9 27 9 3 1; ... - Philippe Deléham, Feb 23 2014
For n >= 3, also the number of dominating sets in the n-helm graph. - Eric W. Weisstein, May 28 2017
The number of elements of length <= n in the free group on two generators. - Anton Mellit, Aug 10 2017
In general, a first order inhomogeneous recurrence of the form s(0) = a, s(n) = m*s(n-1) + k, n>0, will have a closed form of a*m^n + ((m^n-1)/(m-1))*k. - Gary Detlefs, Jun 07 2024

Examples

			a(0) = 1;
a(1) = 1 + 3 + 1 = 5;
a(2) = 1 + 3 + 9 + 3 + 1 = 17;
a(3) = 1 + 3 + 9 + 27 + 9 + 3 + 1 = 53; etc. - _Philippe Deléham_, Feb 23 2014
		

References

  • Theoni Pappas, Math Stuff, Wide World Publ/Tetra, San Carlos CA, page 15, 2002.

Crossrefs

a(n)=T(2,n), array T given by A048471.
Cf. A003462, A029858. A column of A119725.

Programs

  • Magma
    [2*3^n - 1: n in [0..30]]; // Vincenzo Librandi, Sep 23 2011
    
  • Maple
    g:= ((1+x)/(1-3*x)/(1-x)): gser:=series(g, x=0, 43): seq(coeff(gser, x, n), n=0..30); # Zerinvary Lajos, Jan 11 2009; typo fixed by Marko Mihaily, Mar 07 2009
  • Mathematica
    NestList[3 # + 2 &, 1, 30] (* Harvey P. Dale, Mar 06 2012 *)
    LinearRecurrence[{4, -3}, {1, 5}, 30] (* Harvey P. Dale, Mar 06 2012 *)
    Table[2 3^n - 1, {n, 20}] (* Eric W. Weisstein, May 28 2017 *)
    2 3^Range[20] - 1 (* Eric W. Weisstein, May 28 2017 *)
  • PARI
    first(m)=vector(m,n,n--;2*3^n - 1) \\ Anders Hellström, Dec 11 2015

Formula

n-th difference of a(n), a(n-1), ..., a(0) is 2^(n+1) for n=1, 2, 3, ...
a(0)=1, a(n) = a(n-1) + 3^n + 3^(n-1). - Lee Reeves, May 10 2004
a(n) = (3^n + 3^(n+1) - 2)/2. - Creighton Dement, Jul 31 2004
(1, 5, 17, 53, 161, ...) = Ternary (1, 12, 122, 1222, 12222, ...). - Gary W. Adamson, May 02 2005
Row sums of triangle A134347. Also, binomial transform of A046055: (1, 4, 8, 16, 32, 64, ...); and double binomial transform of A010684: (1, 3, 1, 3, 1, 3, ...). - Gary W. Adamson, Oct 21 2007
G.f.: (1+x)/((1-3*x)*(1-x)). - Zerinvary Lajos, Jan 11 2009; corrected by R. J. Mathar, Jan 21 2009
a(0)=1, a(1)=5, a(n) = 4*a(n-1) - 3*a(n-2). - Harvey P. Dale, Mar 06 2012
a(n) = Sum_{k=0..n} A112468(n,k)*4^k. - Philippe Deléham, Feb 23 2014
E.g.f.: exp(x)*(2*exp(2*x) - 1). - Elmo R. Oliveira, Mar 08 2025

Extensions

Better description from Amarnath Murthy, May 27 2001

A109466 Riordan array (1, x(1-x)).

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 0, -2, 1, 0, 0, 1, -3, 1, 0, 0, 0, 3, -4, 1, 0, 0, 0, -1, 6, -5, 1, 0, 0, 0, 0, -4, 10, -6, 1, 0, 0, 0, 0, 1, -10, 15, -7, 1, 0, 0, 0, 0, 0, 5, -20, 21, -8, 1, 0, 0, 0, 0, 0, -1, 15, -35, 28, -9, 1, 0, 0, 0, 0, 0, 0, -6, 35, -56, 36, -10, 1, 0, 0, 0, 0, 0, 0, 1, -21, 70, -84, 45, -11, 1, 0, 0, 0, 0
Offset: 0

Views

Author

Philippe Deléham, Aug 28 2005

Keywords

Comments

Inverse is Riordan array (1, xc(x)) (A106566).
Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, -1, 1, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.
Modulo 2, this sequence gives A106344. - Philippe Deléham, Dec 18 2008
Coefficient array of the polynomials Chebyshev_U(n, sqrt(x)/2)*(sqrt(x))^n. - Paul Barry, Sep 28 2009

Examples

			Rows begin:
  1;
  0,  1;
  0, -1,  1;
  0,  0, -2,  1;
  0,  0,  1, -3,  1;
  0,  0,  0,  3, -4,   1;
  0,  0,  0, -1,  6,  -5,   1;
  0,  0,  0,  0, -4,  10,  -6,   1;
  0,  0,  0,  0,  1, -10,  15,  -7,  1;
  0,  0,  0,  0,  0,   5, -20,  21, -8,  1;
  0,  0,  0,  0,  0,  -1,  15, -35, 28, -9, 1;
From _Paul Barry_, Sep 28 2009: (Start)
Production array is
  0,    1,
  0,   -1,    1,
  0,   -1,   -1,   1,
  0,   -2,   -1,  -1,   1,
  0,   -5,   -2,  -1,  -1,  1,
  0,  -14,   -5,  -2,  -1, -1,  1,
  0,  -42,  -14,  -5,  -2, -1, -1,  1,
  0, -132,  -42, -14,  -5, -2, -1, -1,  1,
  0, -429, -132, -42, -14, -5, -2, -1, -1, 1 (End)
		

Crossrefs

Cf. A026729 (unsigned version), A000108, A030528, A124644.

Programs

  • Magma
    /* As triangle */ [[(-1)^(n-k)*Binomial(k, n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jan 14 2016
  • Mathematica
    (* The function RiordanArray is defined in A256893. *)
    RiordanArray[1&, #(1-#)&, 13] // Flatten (* Jean-François Alcover, Jul 16 2019 *)

Formula

Number triangle T(n, k) = (-1)^(n-k)*binomial(k, n-k).
T(n, k)*2^(n-k) = A110509(n, k); T(n, k)*3^(n-k) = A110517(n, k).
Sum_{k=0..n} T(n,k)*A000108(k)=1. - Philippe Deléham, Jun 11 2007
From Philippe Deléham, Oct 30 2008: (Start)
Sum_{k=0..n} T(n,k)*A144706(k) = A082505(n+1).
Sum_{k=0..n} T(n,k)*A002450(k) = A100335(n).
Sum_{k=0..n} T(n,k)*A001906(k) = A100334(n).
Sum_{k=0..n} T(n,k)*A015565(k) = A099322(n).
Sum_{k=0..n} T(n,k)*A003462(k) = A106233(n). (End)
Sum_{k=0..n} T(n,k)*x^(n-k) = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1), A000012(n), A010892(n), A107920(n+1), A106852(n), A106853(n), A106854(n), A145934(n), A145976(n), A145978(n), A146078(n), A146080(n), A146083(n), A146084(n) for x = -12,-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12 respectively. - Philippe Deléham, Oct 27 2008
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A010892(n), A099087(n), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n) for x = 0,1,2,3,4,5,6,7,8,9,10 respectively. - Philippe Deléham, Oct 28 2008
G.f.: 1/(1-y*x+y*x^2). - Philippe Deléham, Dec 15 2011
T(n,k) = T(n-1,k-1) - T(n-2,k-1), T(n,0) = 0^n. - Philippe Deléham, Feb 15 2012
Sum_{k=0..n} T(n,k)*x^(n-k) = F(n+1,-x) where F(n,x)is the n-th Fibonacci polynomial in x defined in A011973. - Philippe Deléham, Feb 22 2013
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Feb 26 2013
Sum_{k=0..n} T(n,k)*T(n+1,k) = -A110320(n). - Philippe Deléham, Feb 26 2013
For T(0,0) = 0, the signed triangle below has the o.g.f. G(x,t) = [t*x(1-x)]/[1-t*x(1-x)] = L[t*Cinv(x)] where L(x) = x/(1-x) and Cinv(x)=x(1-x) with the inverses Linv(x) = x/(1+x) and C(x)= [1-sqrt(1-4*x)]/2, an o.g.f. for the shifted Catalan numbers A000108, so the inverse o.g.f. is Ginv(x,t) = C[Linv(x)/t] = [1-sqrt[1-4*x/(t(1+x))]]/2 (cf. A124644 and A030528). - Tom Copeland, Jan 19 2016

A131865 Partial sums of powers of 16.

Original entry on oeis.org

1, 17, 273, 4369, 69905, 1118481, 17895697, 286331153, 4581298449, 73300775185, 1172812402961, 18764998447377, 300239975158033, 4803839602528529, 76861433640456465, 1229782938247303441, 19676527011956855057, 314824432191309680913, 5037190915060954894609
Offset: 0

Views

Author

Reinhard Zumkeller, Jul 22 2007

Keywords

Comments

16 = 2^4 is the growth measure for the Jacobsthal spiral (compare with phi^4 for the Fibonacci spiral). - Paul Barry, Mar 07 2008
Second quadrisection of A115451. - Paul Curtz, May 21 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=16, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n-1) = det(A). - Milan Janjic, Feb 21 2010
Partial sums are in A014899. Also, the sequence is related to A014931 by A014931(n+1) = (n+1)*a(n) - Sum_{i=0..n-1} a(i) for n>0. - Bruno Berselli, Nov 07 2012
a(n) is the total number of holes in a certain box fractal (start with 16 boxes, 1 hole) after n iterations. See illustration in links. - Kival Ngaokrajang, Jan 28 2015
Except for 1 and 17, all terms are Brazilian repunits numbers in base 16, and so belong to A125134. All terms >= 273 are composite because a(n) = ((4^(n+1) + 1) * (4^(n+1) - 1))/15. - Bernard Schott, Jun 06 2017
The sequence in binary is 1, 10001, 100010001, 1000100010001, 10001000100010001, ... cf. Plouffe link, A330135. - Frank Ellermann, Mar 05 2020

Examples

			a(3) = 1 + 16 + 256 + 4096 = 4369 = in binary: 1000100010001.
a(4) = (16^5 - 1)/15 = (4^5 + 1) * (4^5 - 1)/15 = 1025 * 1023/15 = 205 * 341 = 69905 = 11111_16. - _Bernard Schott_, Jun 06 2017
		

Crossrefs

Programs

Formula

a(n) = if n=0 then 1 else a(n-1) + A001025(n).
for n > 0: A131851(a(n)) = n and abs(A131851(m)) < n for m < a(n).
a(n) = A098704(n+2)/2.
a(n) = (16^(n+1) - 1)/15. - Bernard Schott, Jun 06 2017
a(n) = (A001025(n+1) - 1)/15.
a(n) = 16*a(n-1) + 1. - Paul Curtz, May 20 2008
G.f.: 1 / ( (16*x-1)*(x-1) ). - R. J. Mathar, Feb 06 2011
E.g.f.: exp(x)*(16*exp(15*x) - 1)/15. - Stefano Spezia, Mar 06 2020

A015441 Generalized Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 7, 13, 55, 133, 463, 1261, 4039, 11605, 35839, 105469, 320503, 953317, 2876335, 8596237, 25854247, 77431669, 232557151, 697147165, 2092490071, 6275373061, 18830313487, 56482551853, 169464432775, 508359743893, 1525146340543, 4575304803901, 13726182847159
Offset: 0

Views

Author

Keywords

Comments

a(n) is the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x,y) = xF(n-1)(x,y) + yF(n-2)(x,y), F(0)(x,y)=0, F(1)(x,y)=1, when y=6x^2. - Mario Catalani (mario.catalani(AT)unito.it), Dec 06 2002
For n>=1: number of length-(n-1) words with letters {0,1,2,3,4,5,6,7} where no two consecutive letters are nonzero, see fxtbook link below. - Joerg Arndt, Apr 08 2011
Starting with offset 1 and convolved with (1, 3, 3, 3, ...) = A003462: (1, 4, 13, 40, ...). - Gary W. Adamson, May 28 2009
a(n) is identical to its inverse binomial transform signed. Differences: A102901. - Paul Curtz, Feb 23 2010
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=2, 7*a(n-2) equals the number of 7-colored compositions of n with all parts >=2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
Pisano period lengths: 1, 1, 1, 2, 20, 1, 6, 2, 3, 20, 5, 2, 12, 6, 20, 4, 16, 3, 18, 20, ... - R. J. Mathar, Aug 10 2012
A015441 and A015518 are the only integer sequences (from the family of homogeneous linear recurrence relation of order 2 with positive integer coefficients with initial values a(0)=0 and a(1)=1) whose ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 14 2014
This is an autosequence of the first kind: the array of successive differences shows a main diagonal of zeros and the inverse binomial transform is identical to the sequence (with alternating signs). - Pointed out by Paul Curtz, Dec 05 2016
First two upper diagonals: A000400(n).
This is a variation on the Starhex honeycomb configuration A332243, see illustration in links. It is an alternating pattern of the 2nd iteration of the centered hexagonal numbers A003215 and centered 12-gonal 'Star' numbers A003154. - John Elias, Oct 06 2021

Examples

			G.f. = x + x^2 + 7*x^3 + 13*x^4 + 55*x^5 + 133*x^6 + 463*x^7 + 1261*x^8 + ...
		

Crossrefs

Programs

  • Magma
    I:=[0,1]; [n le 2 select I[n] else Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018
  • Maple
    A015441:=n->(1/5)*((3^n)-((-2)^n)); seq(A015441(n), n=0..30); # Wesley Ivan Hurt, Mar 14 2014
  • Mathematica
    a[n_]:=(MatrixPower[{{1,4},{1,-2}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]], {n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
    LinearRecurrence[{1,6},{0,1},30] (* Harvey P. Dale, Apr 26 2011 *)
    CoefficientList[Series[x/((1 + 2 x) (1 - 3 x)), {x, 0, 29}], x] (* Michael De Vlieger, Dec 05 2016 *)
  • PARI
    {a(n) = (3^n - (-2)^n) / 5};
    
  • Sage
    [lucas_number1(n,1,-6) for n in range(0, 27)] # Zerinvary Lajos, Apr 22 2009
    

Formula

G.f.: x/((1+2*x)*(1-3*x)).
a(n) = a(n-1) + 6*a(n-2).
a(n) = (1/5)*((3^n)-((-2)^n)). - henryk.wicke(AT)stud.uni-hannover.de
E.g.f.: (exp(3*x) - exp(-2*x))/5. - Paul Barry, Apr 20 2003
a(n+1) = Sum_{k=0..ceiling(n/2)} 6^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004
a(n) = (A000244(n) - A001045(n+1)(-1)^n - A001045(n)(-1)^n)/5. - Paul Barry, Apr 27 2004
The binomial transform of [1,1,7,13,55,133,463,...] is A122117. - Philippe Deléham, Oct 19 2006
a(n+1) = Sum_{k=0..n} A109466(n,k)*(-6)^(n-k). - Philippe Deléham, Oct 26 2008
a(n) = 3a(n-1) + (-1)^(n+1)*A000079(n-1). - Paul Curtz, Feb 23 2010
G.f.: Q(0) -1, where Q(k) = 1 + 6*x^2 + (k+2)*x - x*(k+1 + 6*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013
a(n) = (Sum_{1<=k<=n, k odd} binomial(n,k)*5^(k-1))/2^(n-1). - Vladimir Shevelev, Feb 05 2014
a(-n) = -(-1)^n * a(n) / 6^n for all n in Z. - Michael Somos, Mar 18 2014
From Peter Bala, Apr 01 2015: (Start)
Sum_{n >= 0} a(n+1)*x^n = exp( Sum_{n >= 1} A087451(n)*x^n/n ).
For k = 0, 1, 2, ... and for n >= 1, (5^k)*a(n) | a((5^k)*n).
The expansion of exp( Sum_{n >= 1} a(5*n)/(5*a(n))*x^n/n ) has integral coefficients. Cf. A001656. (End)
From Peter Bala, Jun 27 2025: (Start)
Sum_{n >= 1} (-6)^n/(a(n)*a(n+1)) = -2, since (-6)^n/(a(n)*a(n+1)) = (-2)^n/a(n) - (-2)^(n+1)/a(n+1) for n >= 1.
The following are examples of telescoping infinite products:
Product_{n >= 0} (1 + 6^n/a(2*n+2)) = 6, since (1 + 6^(2*n-1)/a(4*n))*(1 + 6^(2*n)/a(4*n+2)) = (6 - 4^(n+1)/b(n)) / (6 - 4^n/b(n-1)), where b(n) = (2*4^n + 3*9^n)/5 = A096951(n). Similarly,
Product_{n >= 1} (1 - 6^n/a(2*n+2)) = 3/13.
Product_{n >= 0} (1 + (-6)^n/a(2*n+2)) = 6/5.
Product_{n >= 1} (1 - (-6)^n/a(2*n+2)) = 15/13.
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n. (End)

A091030 Partial sums of powers of 13 (A001022).

Original entry on oeis.org

1, 14, 183, 2380, 30941, 402234, 5229043, 67977560, 883708281, 11488207654, 149346699503, 1941507093540, 25239592216021, 328114698808274, 4265491084507563, 55451384098598320, 720867993281778161
Offset: 1

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Author

Wolfdieter Lang, Jan 23 2004

Keywords

Comments

13^a(n) is highest power of 13 dividing (13^n)!.
For analogs with primes 2, 3, 5, 7 and 11 see A000225, A003462, A003463, A023000 and A016123 respectively.
Let A be the Hessenberg matrix of the order n, defined by: A[1,j]=1,A[i,i]:=13, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=14, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=(-1)^(n)*charpoly(A,1). - Milan Janjic, Feb 21 2010

Examples

			For n=6, a(6) = 1*6 + 12*15 + 144*20 + 1728*15 + 20736*6 + 248832*1 = 402234. - _Bruno Berselli_, Nov 12 2015
		

Crossrefs

Programs

Formula

G.f.: x/((1-13*x)*(1-x)) = (1/(1-13*x) - 1/(1-x))/12.
a(n) = Sum_{k=0..n-1} 13^k = (13^n-1)/12.
a(n) = 13*a(n-1)+1 for n>1, a(1)=1. - Vincenzo Librandi, Feb 05 2011
a(n) = Sum_{k=0...n-1} 12^k*binomial(n,n-1-k). - Bruno Berselli, Nov 12 2015
E.g.f.: exp(x)*(exp(12*x) - 1)/12. - Stefano Spezia, Mar 11 2023
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