A018215 -id:A018215 - cp's OEIS frontend

A230540 a(n) = 2n3^(2*n-1).

0, 6, 108, 1458, 17496, 196830, 2125764, 22320522, 229582512, 2324522934, 23245229340, 230127770466, 2259436291848, 22029503845518, 213516729579636, 2058911320946490, 19765548681086304, 189008059262887782, 1801135623563989452, 17110788423857899794

Offset: 0

Bruno Berselli, Oct 23 2013

Arithmetic derivative of 9^n: a(n) = A003415(9^n).

Sum of reciprocals of a(n), for n>0: (3/2)*log(9/8).

Bruno Berselli, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (18,-81).

Cf. A001019, A003415.

Cf. arithmetic derivative of k^n: A001787 (k=2), A027471 (k=3), A018215 (k=4), A053464 (k=5), A212700 (k=6), A027473 (k=7), A230539 (k=8), this sequence, A085708 (k=10), A081127 (k=11).

Magma
```
[2*n*3^(2*n-1): n in [0..20]];
```
Mathematica
```
Table[2 n 3^(2 n - 1), {n, 0, 20}]
```

a(n) = 2*n*3^(2*n-1); \\ Michel Marcus, Oct 23 2013

G.f.: 6*x/(1-9*x)^2.

a(n) = 6*A053540(n), with A053540(0)=0.

A005985 Length of longest trail (i.e., path with all distinct edges) on the edges of an n-cube.

Original entry on oeis.org

0, 1, 4, 9, 32, 65, 192, 385, 1024, 2049, 5120, 10241, 24576, 49153, 114688, 229377, 524288, 1048577, 2359296, 4718593, 10485760, 20971521, 46137344, 92274689, 201326592, 402653185, 872415232, 1744830465, 3758096384, 7516192769, 16106127360, 32212254721

Offset: 0

Colin Mallows

Walk along edges of n-cube without walking along any edge twice; a(n) = number of edges in longest path.
For even n we can traverse all the edges, so a(n) = number of edges = n*2^(n-1). For n odd, every vertex has odd degree, so we need (# vertices)/2 = 2^(n-1) separate paths to cover them all; we will not be able to traverse more than n*2^(n-1) - (2^(n-1)-1) edges before Euler blocks the way. There is a recursive construction (temporarily lost) which achieves this bound.
Suppose n is odd. Delete all but one edge between {0,1}^(n-1) x {0} = A and {0,1}^(n-1) x {1} = B. Starting at the vertex v of A that has an edge to B, do an Euler tour of A coming back to v, then cross over to B and do an Euler tour of B.
This gives you a longest possible trail. - Robert Israel, Jun 02 2015

			For n=3, let the vertices be labeled with Cartesian coordinates (0,0,0), (0,0,1), ..., (1,1,1). An example of a maximal path (of length 9) visits the ten vertices: (0,0,0), (1,0,0), (1,0,1), (1,1,1), (0,1,1), (0,0,1), (0,0,0), (0,1,0), (1,1,0), (1,0,0).

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..300
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (2,5,-10,-4,8).

Cf. A018215 (bisection).

[(2*n*2^n-(1-(-1)^n)*(2^n-2))/4 : n in [0..50]]; // Wesley Ivan Hurt, May 31 2015

A005985:=n->(2*n*2^n-(1-(-1)^n)*(2^n-2))/4: seq(A005985(n), n=0..50); # Wesley Ivan Hurt, May 31 2015

Table[(2*n*2^n - (1 - (-1)^n)(2^n - 2))/4, {n, 0, 20}] (* Giovanni Resta, May 31 2015 *)
LinearRecurrence[{2,5,-10,-4,8},{0,1,4,9,32},40] (* Harvey P. Dale, Jun 11 2015 *)

a(n)=(2*n<Charles R Greathouse IV, Jun 03 2015

G.f.: -x*(1 + 2*x - 4*x^2 + 4*x^3) / ( (x - 1)*(2*x + 1)*(1 + x)*(-1 + 2*x)^2 ). - Simon Plouffe in his 1992 dissertation.
a(n) = (2*n*2^n-(1-(-1)^n)*(2^n-2))/4. - Giovanni Resta, May 31 2015
a(n) = 2*a(n-1)+5*a(n-2)-10*a(n-3)-4*a(n-4)+8*a(n-5), n>5. - Wesley Ivan Hurt, May 31 2015

Revised by Colin Mallows, Jun 13 2005

More terms from Erich Friedman, Aug 08 2005

A362353 Triangle read by rows: T(n,k) = (-1)^(n-k)binomial(n, k)(k+3)^n, for n >= 0, and k = 0,1, ..., n. Coefficients of certain Sidi polynomials.

Original entry on oeis.org

1, -3, 4, 9, -32, 25, -27, 192, -375, 216, 81, -1024, 3750, -5184, 2401, -243, 5120, -31250, 77760, -84035, 32768, 729, -24576, 234375, -933120, 1764735, -1572864, 531441, -2187, 114688, -1640625, 9797760, -28824005, 44040192, -33480783, 10000000, 6561, -524288, 10937500, -94058496, 403536070, -939524096, 1205308188, -800000000, 214358881

Offset: 0

Harlan J. Brothers and Wolfdieter Lang, Apr 27 2023

This is the member N = 2 of a family of signed triangles with row sums n! = A000142(n): T(N; n, k) = (-1)^(n-k)*binomial(n, k)*(k + N + 1)^n, for integer N, n >= 0 and k = 0, 1, ..., n. The row polynomials PS(N; n, z) = Sum_{k=0..n} T(N; n, k)*z^k = ((-1)^n/z^N)*D_{n,N+1,n}(z) in [Sidi 1980].
For N = -1, 0 and 1 see A258773(n, k), A075513(n+1, k) and (-1)^(n-k) * A154715(n, k), respectively.
The column sequences, for k = 0, 1, ..., 6 and n >= k, are A141413(n+2), (-1)^(n+1)*A018215(n) = 4*(-1)^(n+1)*A002697(n), 5^2*(-1)^n*A081135(n), (-1)^(n+1)*A128964(n-1) = 6^3*(-1)^(n+1)*A081144(n), 7^4*(-1)^n*A139641(n-4), 2^15*(-1)^(n+1)*A173155(n-5), 3^12*(-1)^n*A173191(n-6), respectively.
The e.g.f. of the triangle (see below) needs the exponential convolution (LambertW(-z)/(-z))^2 = Sum_{n>=0} c(2; n)*z^n/n!, where c(2; n) = Sum_{m=0..n} |A137352(n+1, m)|*2^m = A007334(n+2).
The row sums give n! = A000142(n).

			The triangle T begins:
n\k    0       1        2         3         4          5          6         7
0:     1
1:    -3       4
2:     9     -32       25
3:   -27     192     -375       216
4:    81   -1024     3750     -5184      2401
5:  -243    5120   -31250     77760    -84035      32768
6:   729  -24576   234375   -933120   1764735   -1572864     531441
7: -2187  114688 -1640625   9797760 -28824005   44040192  -33480783  10000000
...
n = 8:  6561 -524288 10937500 -94058496 403536070 -939524096 1205308188 -800000000 2143588,
n = 9: -19683 2359296 -70312500 846526464 -5084554482 16911433728 -32543321076 36000000000 -21221529219 5159780352.

Paolo Xausa, Table of n, a(n) for n = 0..5049 (rows 0..100 of the triangle, flattened)
Wolfdieter Lang, On a Certain Family of Sidi Polynomials, May 2023.
Avram Sidi, Numerical Quadrature and Nonlinear Sequence Transformations; Unified Rules for Efficient Computation of Integrals with Algebraic and Logarithmic Endpoint Singularities, Math. Comp., 35 (1980), 851-874.

Cf. A000142 (row sums), A075513, A154715, A258773.

Columns k = 0..6 involve (see above): A002697, A007334, A018215, A081135, A081144, A128964, A137352, A139641, A141413, A173155, A173191.

A362353row[n_]:=Table[(-1)^(n-k)Binomial[n,k](k+3)^n,{k,0,n}];Array[A362353row,10,0] (* Paolo Xausa, Jul 30 2023 *)

T(n, k) = (-1)^(n-k)*binomial(n, k)*(k + 3)^n, for n >= 0, k = 0, 1, ..., n.
O.g.f. of column k: (x*(k + 3))^k/(1 - (k + 3)*x)^(k+1), for k >= 0.
E.g.f. of column k: exp(-(k + 3)*x)*((k + 3)*x)^k/k!, for k >= 0.
E.g.f. of the triangle, that is, the e.g.f. of its row polynomials {PS(2;n,y)}_{n>=0}): ES(2;y,x) = exp(-3*x)*(1/3)*(d/dz)(W(-z)/(-z))^2, after replacing z by x*y*exp(-x), where W is the Lambert W-function for the principal branch. This becomes ES(2;y,x) = exp(-3*x)*exp(3*(-W(-z)))/(1 - (-W(-z)), with z = x*y*exp(-x).

a(41)-a(44) from Paolo Xausa, Jul 31 2023

A134574 Array, a(n,k) = total size of all n-length words on a k-letter alphabet, read by antidiagonals.

Original entry on oeis.org

1, 2, 2, 3, 8, 3, 4, 24, 18, 4, 5, 64, 81, 32, 5, 6, 160, 324, 192, 50, 6, 7, 384, 1215, 1024, 375, 72, 7, 8, 896, 4374, 5120, 2500, 648, 98, 8, 9, 2048, 15309, 24576, 15625, 5184, 1029, 128, 9, 10, 4608, 52488, 114688, 93750, 38880, 9604, 1536, 162, 10

Offset: 1

Ross La Haye, Jan 22 2008

			a(2,2) = 8 because there are 2^2 = 4 2-length words on a 2 letter alphabet, each of size 2 and 2*4 = 8.
Array begins:
==================================================================
n\k|  1     2       3        4         5         6          7  ...
---|--------------------------------------------------------------
1  |  1     2       3        4         5         6          7  ...
2  |  2     8      18       32        50        72         98  ...
3  |  3    24      81      192       375       648       1029  ...
4  |  4    64     324     1024      2500      5184       9604  ...
5  |  5   160    1215     5120     15625     38880      84035  ...
6  |  6   384    4374    24576     93750    279936     705894  ...
7  |  7   896   15309   114688    546875   1959552    5764801  ...
8  |  8  2048   52488   524288   3125000  13436928   46118408  ...
9  |  9  4608  177147  2359296  17578125  90699264  363182463  ...
... - _Franck Maminirina Ramaharo_, Aug 07 2018

Cf. a(n, 1) = a(1, k) = A000027(n); a(n, 2) = A036289(n); a(n, 3) = A036290(n); a(n, 4) = A018215(n); a(n, 5) = A036291(n); a(n, 6) = A036292(n); a(n, 7) = A036293(n); a(n, 8) = A036294(n); a(2, k) = A001105(k); a(3, k) = A117642(k); a(n, n) = A007778(n); a(n, n+1) = A066274(n+1): sum[a(i-1, n-i+1), {i, 1, n}] = A062807(n).

t[n_, k_] := Sum[k^n, {j, n}]; Table[ t[n - k + 1, k], {n, 10}, {k, n}] // Flatten (* Robert G. Wilson v, Aug 07 2018 *)

a(n,k) = n*k^n.
O.g.f. (by columns): (k*x)/(-1+k*x)^2.
E.g.f. (by columns): k*x*exp(k*x).
a(n,k) = Sum[k^n,{j,1,n}] = n*Sum[C(n,m)*(k-1)^m,{m,0,n}]. - Ross La Haye, Jan 26 2008

cp's OEIS Frontend

A230540 a(n) = 2*n*3^(2*n-1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A005985 Length of longest trail (i.e., path with all distinct edges) on the edges of an n-cube.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

A362353 Triangle read by rows: T(n,k) = (-1)^(n-k)*binomial(n, k)*(k+3)^n, for n >= 0, and k = 0,1, ..., n. Coefficients of certain Sidi polynomials.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A134574 Array, a(n,k) = total size of all n-length words on a k-letter alphabet, read by antidiagonals.

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Formula

A230540 a(n) = 2n3^(2*n-1).

A362353 Triangle read by rows: T(n,k) = (-1)^(n-k)binomial(n, k)(k+3)^n, for n >= 0, and k = 0,1, ..., n. Coefficients of certain Sidi polynomials.