A002697 -id:A002697 - cp's OEIS frontend

A002802 a(n) = (2n+3)!/(6n!*(n+1)!).

1, 10, 70, 420, 2310, 12012, 60060, 291720, 1385670, 6466460, 29745716, 135207800, 608435100, 2714556600, 12021607800, 52895074320, 231415950150, 1007340018300, 4365140079300, 18839025605400, 81007810103220, 347176329013800, 1483389769422600

Offset: 0

N. J. A. Sloane

For n >= 1 a(n) is also the number of rooted bicolored unicellular maps of genus 1 on n+2 edges. - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 20 2001
a(n) is half the number of (n+2) X 2 Young tableaux with a three horizontal walls between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021], A000984 for one horizontal wall, and A002457 for two. - Michael Wallner, Jan 31 2022
From Robert Coquereaux, Feb 12 2024: (Start)
Call B(p,g) the number of genus g partitions of a set with p elements (genus-dependent Bell number). Up to an appropriate shift the given sequence counts the genus 1 partitions of a set: we have a(n) = B(n+4,1), with a(0)= B(4,1)=1.
When shifted with an offset 4 (i.e., defining b(p)=a(p-4), which starts with 0,0,0,1,10,70, etc., and b(4)=1), the given sequence reads b(p) = (1/( 2^4 3 )) * (1/( (2 p - 1) (2 p - 3))) * (1/(p - 4)!) * (2p)!/p!. In this form it appears as a generalization of Catalan numbers (that indeed count the genus 0 partitions).
Call C[p, [alpha], g] the number of partitions of a set with p elements, of cyclic type [alpha], and of genus g (genus g Faa di Bruno coefficients of type [alpha]). Up to an appropriate shift the given sequence also counts the genus 1 partitions of p=2k into k parts of length 2, which is then called C[2k, [2^k], 1], and we have a(n) = C[2k, [2^k], 1] for k=n+2.
The two previous interpretations of this sequence, leading to a(n) = B(n+4, 1) and to a(n) = C[2(n+2), [2^(n+2)], 1] are not related in any obvious way. (End)

			G.f. = 1 + 10*x + 70*x^2 + 420*x^3 + 2310*x^4 + 12012*x^5 + 60060*x^6 + ...

C. Jordan, Calculus of Finite Differences. Röttig and Romwalter, Budapest, 1939; Chelsea, NY, 1965, p. 449.
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).

T. D. Noe, Table of n, a(n) for n = 0..200
Cyril Banderier and Michael Wallner, Young Tableaux with Periodic Walls: Counting with the Density Method, Séminaire Lotharingien de Combinatoire, 85B (2021), Art. 47, 12 pp.
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See pp. 4, 12.
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus: a compendium of results, Journal of Integer Sequences, Vol. 27 (2024), Article 24.2.6. See p. 9. See also arXiv:2305.01100, 2023. See pp. 9, 19.
R. Cori and G. Hetyei, Counting genus one partitions and permutations, arXiv preprint arXiv:1306.4628 [math.CO], 2013.
R. Cori and G. Hetyei, How to count genus one partitions, FPSAC 2014, Chicago, Discrete Mathematics and Theoretical Computer Science (DMTCS), Nancy, France, 2014, 333-344.
Alain Goupil and Gilles Schaeffer, Factoring N-Cycles and Counting Maps of Given Genus, Europ. J. Combinatorics (1998) 19 819-834.
T. R. S. Walsh and A. B. Lehman, Counting rooted maps by genus. I, J. Comb. Theory B 13 (1972), 192-218 (Tab.1).
Liang Zhao and Fengyao Yan, Note on Total Positivity for a Class of Recursive Matrices, Journal of Integer Sequences, Vol. 19 (2016), Article 16.6.5.
Notes.

Cf. A035309, A000108 (for genus 0 maps), A046521 (third column).
Cf. A000984, A000911, A002457, A002697, A051133.
Column g=1 of A370235.

F:=Factorial;; List([0..25], n-> F(2*n+3)/(6*F(n)*F(n+1)) ); # G. C. Greubel, Jul 20 2019

F:=Factorial; [F(2*n+3)/(6*F(n)*F(n+1)): n in [0..25]]; // G. C. Greubel, Jul 20 2019

seq(simplify(4^n*hypergeom([-n,-3/2], [1], 1)),n=0..25); # Peter Luschny, Apr 26 2016

Table[(2*n+3)!/(6*n!*(n+1)!), {n, 0, 25}] (* Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *)

{a(n) = if( n<0, 0, (2*n + 3)! / (6 * n! * (n+1)!))}; /* Michael Somos, Sep 16 2013 */

{a(n) = 2^(n+3) * polcoeff( pollegendre(n+4), n) / 3}; /* Michael Somos, Sep 16 2013 */

f=factorial; [f(2*n+3)/(6*f(n)*f(n+1)) for n in (0..25)] # G. C. Greubel, Jul 20 2019

G.f.: (1 - 4*x)^(-5/2) = 1F0(5/2;;4x).
Asymptotic expression for a(n) is a(n) ~ (n+2)^(3/2) * 4^(n+2) / (sqrt(Pi) * 48).
a(n) = Sum_{a+b+c+d+e=n} f(a)*f(b)*f(c)*f(d)*f(e) with f(n) = binomial(2n, n) = A000984(n). - Philippe Deléham, Jan 22 2004
a(n-1) = (1/4)*Sum_{k=1..n} k*(k+1)*binomial(2*k, k). - Benoit Cloitre, Mar 20 2004
a(n) = A051133(n+1)/3 = A000911(n)/6. - Zerinvary Lajos, Jun 02 2007
From Rui Duarte, Oct 08 2011: (Start)
Also convolution of A000984 with A002697, also convolution of A000302 with A002457.
a(n) = ((2n+3)(2n+1)/(3*1)) * binomial(2n, n).
a(n) = binomial(2n+4, 4) * binomial(2n, n) / binomial(n+2, 2).
a(n) = binomial(n+2, 2) * binomial(2n+4, n+2) / binomial(4, 2).
a(n) = binomial(2n+4, n+2) * (n+2)*(n+1) / 12. (End)
D-finite with recurrence: n*a(n) - 2*(2*n+3)*a(n-1) = 0. - R. J. Mathar, Jan 31 2014
a(n) = 4^n*hypergeom([-n,-3/2], [1], 1). - Peter Luschny, Apr 26 2016
Boas-Buck recurrence: a(n) = (10/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+2, 2). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = (-4)^n*binomial(-5/2, n). - Peter Luschny, Oct 23 2018
Sum_{n>=0} 1/a(n) = 12 - 2*sqrt(3)*Pi. - Amiram Eldar, Oct 13 2020
E.g.f.: (1/12) exp(2 x) x^2 BesselI[2, 2 x]. - Robert Coquereaux, Feb 12 2024

A027465 Cube of lower triangular normalized binomial matrix.

Original entry on oeis.org

1, 3, 1, 9, 6, 1, 27, 27, 9, 1, 81, 108, 54, 12, 1, 243, 405, 270, 90, 15, 1, 729, 1458, 1215, 540, 135, 18, 1, 2187, 5103, 5103, 2835, 945, 189, 21, 1, 6561, 17496, 20412, 13608, 5670, 1512, 252, 24, 1, 19683, 59049, 78732, 61236, 30618, 10206, 2268

Offset: 0

Olivier Gérard, N. J. A. Sloane

Rows of A013610 reversed. - Michael Somos, Feb 14 2002
Row sums are powers of 4 (A000302), antidiagonal sums are A006190 (a(n) = 3*a(n-1) + a(n-2)). - Gerald McGarvey, May 17 2005
Triangle of coefficients in expansion of (3+x)^n.
Also: Pure Galton board of scheme (3,1). Also: Multiplicity (number) of pairs of n-dimensional binary vectors with dot product (overlap) k. There are 2^n = A000079(n) binary vectors of length n and 2^(2n) = 4^n = A000302(n) different pairs to form dot products k = Sum_{i=1..n} v[i]*u[i] between these, 0 <= k <= n. (Since dot products are symmetric, there are only 2^n*(2^n-1)/2 different non-ordered pairs, actually.) - R. J. Mathar, Mar 17 2006
Mirror image of A013610. - Zerinvary Lajos, Nov 25 2007
T(i,j) is the number of i-permutations of 4 objects a,b,c,d, with repetition allowed, containing j a's. - Zerinvary Lajos, Dec 21 2007
The antidiagonals of the sequence formatted as a square array (see Examples section) and summed with alternating signs gives a bisection of Fibonacci sequence, A001906. Example: 81-(27-1)=55. Similar rule applied to rows gives A000079. - Mark Dols, Sep 01 2009
Triangle T(n,k), read by rows, given by (3,0,0,0,0,0,0,0,...)DELTA (1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 09 2011
T(n,k) = binomial(n,k)*3^(n-k), the number of subsets of [2n] with exactly k symmetric pairs, where elements i and j of [2n] form a symmetric pair if i+j=2n+1. Equivalently, if n couples attend a (ticketed) event that offers door prizes, then the number of possible prize distributions that have exactly k couples as dual winners is T(n,k). - Dennis P. Walsh, Feb 02 2012
T(n,k) is the number of ordered pairs (A,B) of subsets of {1,2,...,n} such that the intersection of A and B contains exactly k elements. For example, T(2,1) = 6 because we have ({1},{1}); ({1},{1,2}); ({2},{2}); ({2},{1,2}); ({1,2},{1}); ({1,2},{2}). Sum_{k=0..n} T(n,k)*k = A002697(n) (see comment there by Ross La Haye). - Geoffrey Critzer, Sep 04 2013
Also the convolution triangle of A000244. - Peter Luschny, Oct 09 2022

			Example: n = 3 offers 2^3 = 8 different binary vectors (0,0,0), (0,0,1), ..., (1,1,0), (1,1,1). a(3,2) = 9 of the 2^4 = 64 pairs have overlap k = 2: (0,1,1)*(0,1,1) = (1,0,1)*(1,0,1) = (1,1,0)*(1,1,0) = (1,1,1)*(1,1,0) = (1,1,1)*(1,0,1) = (1,1,1)*(0,1,1) = (0,1,1)*(1,1,1) = (1,0,1)*(1,1,1) = (1,1,0)*(1,1,1) = 2.
For example, T(2,1)=6 since there are 6 subsets of {1,2,3,4} that have exactly 1 symmetric pair, namely, {1,4}, {2,3}, {1,2,3}, {1,2,4}, {1,3,4}, and {2,3,4}.
The present sequence formatted as a triangular array:
     1
     3     1
     9     6     1
    27    27     9     1
    81   108    54    12    1
   243   405   270    90   15    1
   729  1458  1215   540  135   18   1
  2187  5103  5103  2835  945  189  21  1
  6561 17496 20412 13608 5670 1512 252 24 1
  ...
A013610 formatted as a triangular array:
  1
  1  3
  1  6   9
  1  9  27   27
  1 12  54  108   81
  1 15  90  270  405   243
  1 18 135  540 1215  1458   729
  1 21 189  945 2835  5103  5103  2187
  1 24 252 1512 5670 13608 20412 17496 6561
   ...
A099097 formatted as a square array:
      1     0     0    0   0 0 0 0 0 0 0 ...
      3     1     0    0   0 0 0 0 0 0 ...
      9     6     1    0   0 0 0 0 0 ...
     27    27     9    1   0 0 0 0 ...
     81   108    54   12   1 0 0 ...
    243   405   270   90  15 1 ...
    729  1458  1215  540 135 ...
   2187  5103  5103 2835 ...
   6561 17496 20412 ...
  19683 59049 ...
  59049 ...

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
B. N. Cyvin et al., Isomer enumeration of unbranched catacondensed polygonal systems with pentagons and heptagons, Match, No. 34 (Oct 1996), pp. 109-121.
Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 No. 1-3, 33-51 (2001).

Cf. A000244, A007318, A013610, A013610, A099097, A027471, A027472, A036216, A036217, A036219, A036220, A036221, A036222, A036223.

a027465 n k = a027465_tabl !! n !! k
a027465_row n = a027465_tabl !! n
a027465_tabl = iterate (\row ->
   zipWith (+) (map (* 3) (row ++ [0])) (map (* 1) ([0] ++ row))) [1]
-- Reinhard Zumkeller, May 26 2013

for i from 0 to 12 do seq(binomial(i, j)*3^(i-j), j = 0 .. i) od; # Zerinvary Lajos, Nov 25 2007
# Uses function PMatrix from A357368. Adds column 1, 0, 0, ... to the left.
PMatrix(10, n -> 3^(n-1)); # Peter Luschny, Oct 09 2022

t[n_, k_] := Binomial[n, k]*3^(n-k); Table[t[n, n-k], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Sep 19 2012 *)

{T(n, k) = polcoeff( (3 + x)^n, k)}; /* Michael Somos, Feb 14 2002 */

Numerators of lower triangle of (b^2)[ i, j ] where b[ i, j ] = binomial(i-1, j-1)/2^(i-1) if j <= i, 0 if j > i.
Triangle whose (i, j)-th entry is binomial(i, j)*3^(i-j).
a(n, m) = 4^(n-1)*Sum_{j=m..n} b(n, j)*b(j, m) = 3^(n-m)*binomial(n-1, m-1), n >= m >= 1; a(n, m) := 0, n < m. G.f. for m-th column: (x/(1-3*x))^m (m-fold convolution of A000244, powers of 3). - Wolfdieter Lang, Feb 2006
G.f.: 1 / (1 - x(3+y)).
a(n,k) = 3*a(n-1,k) + a(n-1,k-1) - R. J. Mathar, Mar 17 2006
From the formalism of A133314, the e.g.f. for the row polynomials of A027465 is exp(x*t)*exp(3x). The e.g.f. for the row polynomials of the inverse matrix is exp(x*t)*exp(-3x). p iterates of the matrix give the matrix with e.g.f. exp(x*t)*exp(p*3x). The results generalize for 3 replaced by any number. - Tom Copeland, Aug 18 2008
T(n,k) = A164942(n,k)*(-1)^k. - Philippe Deléham, Oct 09 2011
Let P and P^T be the Pascal matrix and its transpose and H = P^3 = A027465. Then from the formalism of A132440 and A218272,
  exp[x*z/(1-3z)]/(1-3z) = exp(3z D_z z) e^(x*z)= exp(3D_x x D_x) e^(z*x)
  = (1 z z^2 z^3 ...) H (1 x x^2/2! x^3/3! ...)^T
  = (1 x x^2/2! x^3/3! ...) H^T (1 z z^2 z^3 ...)^T = Sum_{n>=0} (3z)^n L_n(-x/3), where D is the derivative operator and L_n(x) are the regular (not normalized) Laguerre polynomials. - Tom Copeland, Oct 26 2012
E.g.f. for column k: x^k/k! * exp(3x). - Geoffrey Critzer, Sep 04 2013

A038845 3-fold convolution of A000302 (powers of 4).

Original entry on oeis.org

1, 12, 96, 640, 3840, 21504, 114688, 589824, 2949120, 14417920, 69206016, 327155712, 1526726656, 7046430720, 32212254720, 146028888064, 657129996288, 2937757630464, 13056700579840, 57724360458240, 253987186016256, 1112705767309312, 4855443348258816

Offset: 0

Wolfdieter Lang

Also convolution of A002802 with A000984 (central binomial coefficients).
With a different offset, number of n-permutations of 5 objects u, v, w, z, x with repetition allowed, containing exactly two u's. - Zerinvary Lajos, Dec 29 2007
Also convolution of A000302 with A002697, also convolution of A002457 with itself. - Rui Duarte, Oct 08 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Adam Ehrenberg, Joseph T. Iosue, Abhinav Deshpande, Dominik Hangleiter, and Alexey V. Gorshkov, The Second Moment of Hafnians in Gaussian Boson Sampling, arXiv:2403.13878 [quant-ph], 2024. See p. 30.
Index entries for linear recurrences with constant coefficients, signature (12,-48,64).

Cf. A000302, A000984, A002457, A002697, A002802, A038231, A052780.

Sequences similar to the form q^(n-2)*binomial(n, 2): A000217 (q=1), A001788 (q=2), A027472 (q=3), this sequence (q=4), A081135 (q=5), A081136 (q=6), A027474 (q=7), A081138 (q=8), A081139 (q=9), A081140 (q=10), A081141 (q=11), A081142 (q=12), A027476 (q=15).

List([0..30], n-> 4^n*Binomial(n+2,n) ); # G. C. Greubel, Jul 20 2019

[4^n*Binomial(n+2, 2): n in [0..30]]; // Vincenzo Librandi, Oct 15 2011

seq((n+2)*(n+1)*4^n/2, n=0..30); # Zerinvary Lajos, Apr 25 2007

Table[4^n*Binomial[n+2,n], {n,0,30}] (* G. C. Greubel, Jul 20 2019 *)

a(n)=(n+2)*(n+1)<<(2*n-1) \\ Charles R Greathouse IV, Aug 21 2015

[4^(n-2)*binomial(n,2) for n in range(2, 30)] # Zerinvary Lajos, Mar 11 2009

a(n) = (n+2)*(n+1)*2^(2*n-1).
G.f.: 1/(1-4*x)^3.
a(n) = Sum_{u+v+w+x+y+z=n} f(u)*f(v)*f(w)*f(x)*f(y)*f(z) with f(n)=A000984(n). - Philippe Deléham, Jan 22 2004
a(n) = binomial(n+2,n) * 4^n. - Rui Duarte, Oct 08 2011
E.g.f.: (1 + 8*x + 8*x^2)*exp(4*x). - G. C. Greubel, Jul 20 2019
From Amiram Eldar, Jan 05 2022: (Start)
Sum_{n>=0} 1/a(n) = 8 - 24*log(4/3).
Sum_{n>=0} (-1)^n/a(n) = 40*log(5/4) - 8. (End)

A038846 4-fold convolution of A000302 (powers of 4); expansion of g.f. 1/(1-4*x)^4.

Original entry on oeis.org

1, 16, 160, 1280, 8960, 57344, 344064, 1966080, 10813440, 57671680, 299892736, 1526726656, 7633633280, 37580963840, 182536110080, 876173328384, 4161823309824, 19585050869760, 91396904058880, 423311976693760, 1947235092791296, 8901646138474496, 40462027902156800

Offset: 0

Wolfdieter Lang

Also minimal 3-covers of a labeled n-set that cover 3 points of that set uniquely (if offset is 3). Cf. A057524 for unlabeled case. - Vladeta Jovovic, Sep 02 2000
Also convolution of A020918 with A000984 (central binomial coefficients).
Let M=[1,0,0,i;0,1,i,0;0,i,1,0;i,0,0,1], i=sqrt(-1). Then 1/det(I-xM) = 1/(1-4x)^4. - Paul Barry, Apr 27 2005
With a different offset, number of n-permutations (n=4) of 5 objects u, v, w, z, x with repetition allowed, containing exactly three u's. Example: a(1)=16 because we have uuuv, uuvu, uvuu, vuuu, uuuw, uuwu, uwuu, wuuu, uuuz, uuzu, uzuu, zuuu, uuux, uuxu, uxuu and xuuu. - Zerinvary Lajos, May 19 2008
From A152818. a(n) = A006044/6. - Paul Curtz, Jan 07 2009
Also convolution of A000302 with A038845, also convolution of A002457 with A002802, also convolution of A002697. - Rui Duarte, Oct 08 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (16,-96,256,-256).

Cf. A000302, A000984, A002457, A002697, A002802, A006044, A020918, A038231, A038845, A057524, A152818.

List([0..30], n-> 4^n*Binomial(n+3,3) ) # G. C. Greubel, Jul 20 2019

[4^n*Binomial(n+3, 3): n in [0..30]]; // Vincenzo Librandi, Oct 15 2011

seq(seq(binomial(i, j)*4^(i-3), j =i-3), i=3..33); # Zerinvary Lajos, Dec 03 2007
seq(binomial(n+3,3)*4^n,n=0..30); # Zerinvary Lajos, May 19 2008

Table[4^n*Binomial[n+3,3], {n,0,30}] (* G. C. Greubel, Jul 20 2019 *)

Vec(1/(1-4*x)^4+O(x^30)) \\ Charles R Greathouse IV, Oct 03 2016

[lucas_number2(n, 4, 0)*binomial(n,3)/2^6 for n in range(3, 33)] # Zerinvary Lajos, Mar 11 2009

a(n) = binomial(n+3, 3)*4^n.
G.f.: 1/(1-4*x)^4.
a(n) = Sum_{a+b+c+d+e+f+g+h=n} f(a)*f(b)*f(c)*f(d)*f(e)*f(f)*f(g)*f(h) with f(n)=A000984(n). - Philippe Deléham, Jan 22 2004
From Amiram Eldar, Jan 05 2022: (Start)
Sum_{n>=0} 1/a(n) = 108*log(4/3) - 30.
Sum_{n>=0} (-1)^n/a(n) = 300*log(5/4) - 66. (End)
E.g.f.: exp(4*x)*(3 + 36*x + 72*x^2 + 32*x^3)/3. - Stefano Spezia, Jan 01 2023

A020918 Expansion of 1/(1-4*x)^(7/2).

Original entry on oeis.org

1, 14, 126, 924, 6006, 36036, 204204, 1108536, 5819814, 29745716, 148728580, 730122120, 3528923580, 16830250920, 79342611480, 370265520240, 1712478031110, 7857252142740, 35794148650260

Offset: 0

N. J. A. Sloane

Also convolution of A000984 with A038845, also convolution of A000302 with A002802, also convolution of A002457 with A002697. - Rui Duarte, Oct 08 2011

5*a(n) is the number of (n+3) X 2 Young tableaux with a four horizontal walls between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021], A000984 for one horizontal wall, A002457 for two, and A002802 for three. - Michael Wallner, Mar 09 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Cyril Banderier and Michael Wallner, Young Tableaux with Periodic Walls: Counting with the Density Method, Séminaire Lotharingien de Combinatoire, 85B (2021), Art. 47, 12 pp.

Cf. A000302, A000984, A001622, A002457, A002697, A002802, A000292, A038845, A046521 (fourth column).

List([0..30], n-> Binomial(2*n+6, n+3)*Binomial(n+3, 3)/20); # G. C. Greubel, Jul 20 2019

[Binomial(2*n+6, n+3)*Binomial(n+3, 3)/20: n in [0..30]]; // G. C. Greubel, Jul 20 2019

seq(binomial(2*n,n)*binomial(n,(n-3))/20, n=2..21); # Zerinvary Lajos, May 05 2007
seq(simplify(4^n*hypergeom([-n,-5/2], [1], 1)),n=0..18); # Peter Luschny, Apr 26 2016

CoefficientList[Series[1/(1-4x)^(7/2), {x, 0, 30}], x] (* Vincenzo Librandi, Jul 04 2013 *)

vector(30, n, n--; binomial(2*n+6, n+3)*binomial(n+3, 3)/20 ) \\ G. C. Greubel, Jul 20 2019

[binomial(2*n+6, n+3)*binomial(n+3, 3)/20 for n in (0..30)] # G. C. Greubel, Jul 20 2019

a(n) = binomial(n+3, 3)*A000984(n+3)/A000984(3), where A000984 are the central binomial coefficients. - Wolfdieter Lang
a(n) ~ 8/15*Pi^(-1/2)*n^(5/2)*2^(2*n)*{1 + 35/8*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 22 2001
a(n) = Sum_{a+b+c+d+e+f+g=n} f(a)*f(b)*f(c)*f(d)*f(e)*f(f)*f(g) with f(n)=A000984(n). - Philippe Deléham, Jan 22 2004
a(n) = A000292(n)*A000984(n+2)/20. - Zerinvary Lajos, May 05 2007
From Rui Duarte, Oct 08 2011: (Start)
a(n) = ((2n+5)(2n+3)(2n+1)/(5*3*1)) * binomial(2n, n).
a(n) = binomial(2n+6, 6) * binomial(2n, n) / binomial(n+3, 3).
a(n) = binomial(n+3, 3) * binomial(2n+6, n+3) / binomial(6, 3). (End)
a(n) = 4^n*hypergeom([-n,-5/2], [1], 1). - Peter Luschny, Apr 26 2016
Boas-Buck recurrence: a(n) = (14/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+3, 3). See a comment there. - Wolfdieter Lang, Aug 10 2017
From Amiram Eldar, Apr 07 2022: (Start)
Sum_{n>=0} 1/a(n) = 10*sqrt(3)*Pi - 160/3.
Sum_{n>=0} (-1)^n/a(n) = 10*sqrt(5)*log(phi) - 320/3, where phi is the golden ratio (A001622). (End)
D-finite with recurrence n*a(n) +2*(-2*n-5)*a(n-1)=0. - R. J. Mathar, Aug 01 2022

A053464 a(n) = n*5^(n-1).

Original entry on oeis.org

0, 1, 10, 75, 500, 3125, 18750, 109375, 625000, 3515625, 19531250, 107421875, 585937500, 3173828125, 17089843750, 91552734375, 488281250000, 2593994140625, 13732910156250, 72479248046875, 381469726562500

Offset: 0

Barry E. Williams, Jan 13 2000

Arithmetic derivative of 5^n: a(n) = A003415(5^n). - Darrell Minor, Jul 21 2025

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Frank Ellermann, Illustration of binomial transforms
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 756.
Index entries for linear recurrences with constant coefficients, signature (10,-25).

Cf. A002697, A027471, A001787.

[n*(5^(n-1)): n in [0..30]]; // Vincenzo Librandi, Jun 09 2011

Join[{a=0,b=1},Table[c=10*b-25*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 27 2011 *)
Table[n*5^(n-1),{n,0,20}] (* or *) LinearRecurrence[{10,-25},{0,1},30] (* Harvey P. Dale, Jul 22 2014 *)

{a(n) = n*5^(n-1)}; /* Michael Somos, Sep 12 2005 */

[lucas_number1(n,10,25) for n in range(0, 21)] # Zerinvary Lajos, Apr 26 2009

a(n) = Sum_{k=0..n} 5^(n-k)*binomial(n-k+1, k)*binomial(1, (k+1)/2)*(1-(-1)^k)/2. - Paul Barry, Oct 15 2004
a(n) = 10*a(n-1) - 25*a(n-2); n>1; a(0)=0, a(1)=1.
Fourth binomial transform of n (starting 0, 1, 10...) Convolution of powers of 5.
G.f.: x/(1-5*x)^2; E.g.f.: x*exp(5*x). - Paul Barry, Jul 22 2003
a(n) = - 25^n * a(-n) for all n in Z. - Michael Somos, Jun 26 2017
From Amiram Eldar, Oct 28 2020: (Start)
Sum_{n>=1} 1/a(n) = 5*log(5/4).
Sum_{n>=1} (-1)^(n+1)/a(n) = 5*log(6/5). (End)

More terms from James Sellers, Feb 02 2000

A040075 5-fold convolution of A000302 (powers of 4); expansion of 1/(1-4*x)^5.

Original entry on oeis.org

1, 20, 240, 2240, 17920, 129024, 860160, 5406720, 32440320, 187432960, 1049624576, 5725224960, 30534533120, 159719096320, 821412495360, 4161823309824, 20809116549120, 102821517066240, 502682972323840, 2434043865989120, 11683410556747776, 55635288365465600

Offset: 0

Wolfdieter Lang

Also convolution of A020920 with A000984 (central binomial coefficients).
With a different offset, number of n-permutations (n=5) of 5 objects u, v, w, z, x with repetition allowed, containing exactly four (4)u's. Example: a(1)=20 because we have uuuuv, uuuvu, uuvuu, uvuuu, vuuuu, uuuuw, uuuwu, uuwuu, uwuuu, wuuuu, uuuuz, uuuzu, uuzuu, uzuuu, zuuuu, uuuux, uuuxu, uuxuu, uxuuu and xuuuu. - Zerinvary Lajos, May 19 2008
Also convolution of A000302 with A038846, also convolution of A002457 with A020918, also convolution of A002697 with A038845, also convolution of A002802 with A002802. [Rui Duarte, Oct 08 2011]

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Eric Weisstein's World of Mathematics, Idempotent Number.
Index entries for linear recurrences with constant coefficients, signature (20, -160, 640, -1280, 1024).

Cf. A000302, A020920, A000984.

Cf. A038231.

List([0..30], n-> 4^n*Binomial(n+4, 4)); # G. C. Greubel, Jul 20 2019

[4^n*Binomial(n+4, 4): n in [0..30]]; // Vincenzo Librandi, Oct 15 2011

seq(seq(binomial(i, j)*4^(i-4), j =i-4), i=4..22); # Zerinvary Lajos, Dec 03 2007
seq(binomial(n+4,4)*4^n,n=0..30); # Zerinvary Lajos, May 19 2008
spec := [S, {B=Set(Z, 0 <= card), S=Prod(Z, Z, Z, Z, B, B, B, B)}, labeled]: seq(combstruct[count](spec, size=n)/24, n=4..34); # Zerinvary Lajos, Apr 05 2009

Table[Binomial[n+4,4]*4^n, {n,0,30}] (* Michael De Vlieger, Aug 21 2015 *)

vector(30, n, n--; 4^n*binomial(n+4, 4)) \\ G. C. Greubel, Jul 20 2019

[lucas_number2(n, 4, 0)*binomial(n,4)/2^8 for n in range(4, 34)] # Zerinvary Lajos, Mar 11 2009

a(n) = binomial(n+4, 4)*4^n.
G.f.: 1/(1-4*x)^5.
a(n) = Sum_{ i_1+i_2+i_3+i_4+i_5+i_6+i_7+i_8+i_9+i_10 = n } f(i_1)*f(i_2) *f(i_3)*f(i_4)*f(i_5)*f(i_6)*f(i_7)*f(i_8)*f(i_9)*f(i_10) with f(k)=A000984(k). - Rui Duarte, Oct 08 2011
E.g.f.: (3 + 48*x + 144*x^2 + 128*x^3 + 32*x^4)*exp(4*x)/3. - G. C. Greubel, Jul 20 2019
From Amiram Eldar, Mar 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 376/3 - 432*log(4/3).
Sum_{n>=0} (-1)^n/a(n) = 2000*log(5/4) - 1336/3. (End)

A061594 Number of ways to place 3n nonattacking kings on a 6 X 2n chessboard.

Original entry on oeis.org

1, 32, 408, 3600, 26040, 166368, 976640, 5392704, 28432288, 144605184, 714611200, 3449705600, 16333065216, 76081271168, 349524164224, 1586790140800, 7130144209024, 31752978219904, 140298397039232, 615604372260736

Offset: 0

Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), May 22 2001

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. E. Knuth, Nonattacking kings on a chessboard, 1994.
Vaclav Kotesovec, Number of ways of placing non-attacking queens and kings on boards of various sizes [_Vaclav Kotesovec_, Feb 06 2010]
H. S. Wilf, The problem of the kings, Elec. J. Combin. 2, 1995.
Index entries for linear recurrences with constant coefficients, signature (19, -148, 604, -1364, 1644, -928, 192).

Column k=3 of A350819.
Cf. A001787, A061593.
Equals 231*A002697(n+1) - 2608*A000302(n) - 384*A000244(n) + 1103*A007070(n-1) + 780*A006012(n+1) + (n+1)*(17*A048580(n) + 12*A007070(n+1)).

a(n)=polcoeff((1+13*x-52*x^2-20*x^3+60*x^4-20*x^5)/((1-3*x)*(1-4*x)^2*(1-4*x+2*x^2)^2)+x*O(x^n),n)

G.f.: (1+13x-52x^2-20x^3+60x^4-20x^5)/((1-3x)(1-4x)^2(1-4x+2x^2)^2).

Explicit formula: (231n-2377)*4^n - 384*3^n + (1953*sqrt(2)/2+1381+(35*sqrt(2)+99/2)*n)*(2+sqrt(2))^n + (1381-1953*sqrt(2)/2+(99/2-35*sqrt(2))*n)*(2-sqrt(2))^n. - Vaclav Kotesovec, Feb 06 2010

Corrected data by Vincenzo Librandi, Oct 12 2011

cp's OEIS Frontend

A372492 G.f. satisfies A(A(A(A(x)))) = F(x), where F(x) is the g.f. for A002697(n) = n*4^(n-1).

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A083672 Binomial transform of 1,8,48,256,1280,6144,... (cf. A002697).

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A002802 a(n) = (2*n+3)!/(6*n!*(n+1)!).

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A027465 Cube of lower triangular normalized binomial matrix.

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A038845 3-fold convolution of A000302 (powers of 4).

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A038846 4-fold convolution of A000302 (powers of 4); expansion of g.f. 1/(1-4*x)^4.

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A002802 a(n) = (2n+3)!/(6n!*(n+1)!).