A264871 -id:A264871 - cp's OEIS frontend

A136516 a(n) = (2^n+1)^n.

1, 3, 25, 729, 83521, 39135393, 75418890625, 594467302491009, 19031147999601100801, 2460686496619787545743873, 1280084544196357822418212890625, 2672769719437237714909813214827010049, 22366167213460480200139104627873703828439041

Offset: 0

Paul D. Hanna, Jan 02 2008

nonn

More generally, Sum_{n>=0} m^n * q^(n^2) * exp(b*q^n*x) * x^n / n! = Sum_{n>=0} (m*q^n + b)^n * x^n / n! for all q, m, b.

Main diagonal of A264871. - Omar E. Pol, Nov 27 2015

			A(x) = 1 + 3x + 5^2*x^2/2! + 9^3*x^3/3! + 17^4*x^4/4! +... + (2^n+1)^n*x^n/n! +...
A(x) = exp(x) + 2*exp(2x) + 2^4*exp(4x)*x^2/2! + 2^9*exp(8x)*x^3/3! +...+ 2^(n^2)*exp(2^n*x)*x^n/n! +...
This is a special case of the more general statement:
Sum_{n>=0} m^n * F(q^n*x)^b * log( F(q^n*x) )^n / n! = Sum_{n>=0} x^n * [y^n] F(y)^(m*q^n + b) where F(x) = exp(x), q=2, m=1, b=1.

Vincenzo Librandi, Table of n, a(n) for n = 0..45

Cf. A014050, A055601, A243918, A202989, A326011, A326012.

[(2^n+1)^n: n in [0..45]]; // Vincenzo Librandi, Apr 21 2011

seq((2^n+1)^n, n=0..30); # Robert Israel, Nov 27 2015

Table[(2^n+1)^n,{n,0,16}] (* Vladimir Joseph Stephan Orlovsky, Feb 14 2011*)

a(n)=polcoeff(sum(k=0,n,2^(k^2)*exp(2^k*x)*x^k/k!),n)

{a(n)=polcoeff(sum(k=0, n, 2^(k^2)*x^k/(1-2^k*x +x*O(x^n))^(k+1)), n)} \\ Paul D. Hanna, Sep 15 2009

E.g.f.: A(x) = Sum_{n>=0} 2^(n^2) * exp(2^n*x) * x^n/n!.
O.g.f.: Sum_{n>=0} 2^(n^2)*x^n/(1 - 2^n*x)^(n+1) = Sum_{n>=0} (2^n+1)^n*x^n. [Paul D. Hanna, Sep 15 2009]
a(n) = 2^(n^2) + n 2^(n^2-n) + O(n^2 2^(n^2-2n)). - Robert Israel, Nov 27 2015

A264872 Array read by antidiagonals: T(n,m) = 2^n*(1+2^n)^m; n,m >= 0.

Original entry on oeis.org

1, 2, 2, 4, 6, 4, 8, 18, 20, 8, 16, 54, 100, 72, 16, 32, 162, 500, 648, 272, 32, 64, 486, 2500, 5832, 4624, 1056, 64, 128, 1458, 12500, 52488, 78608, 34848, 4160, 128, 256, 4374, 62500, 472392, 1336336, 1149984, 270400, 16512, 256, 512, 13122, 312500

Offset: 0

R. J. Mathar, Nov 27 2015

Start with an n X m rectangle and cut it vertically along any set of the m-1 separators. There are binomial(m-1,c) ways of doing this with 0 <= c < m cuts. Inside each of these 1+c regions cut vertically, for which there are 2^(n-1) choices. The total number of ways of dissecting the rectangle into rectangles in this way is Sum_{c=0..m-1} binomial(m-1,c) 2^((1+c)(n-1)) = 2^(n-1)*(1+2^(n-1))^(m-1) = T(n-1,m-1).

The symmetrized version of the array is S(n,m) = T(n,m) + T(m,n) - 2^(m+n) <= A116694(n,m), which counts tilings that start with guillotine cuts either horizontally or vertically, avoiding double counting of the tilings where the order of the cuts does not matter. - R. J. Mathar, Nov 29 2015

			   1,    2,     4,       8,       16,         32, ...
   2,    6,    18,      54,      162,        486, ...
   4,   20,   100,     500,     2500,      12500, ...
   8,   72,   648,    5832,    52488,     472392, ...
  16,  272,  4624,   78608,  1336336,   22717712, ...
  32, 1056, 34848, 1149984, 37949472, 1252332576, ...
.
The symmetrized version S(n,m) starts
   1,    2,     4,       8,       16,         32, ...
   2,    8,    30,     110,      402,       1478, ...
   4,   30,   184,    1116,     7060,      47220, ...
   8,  110,  1116,   11600,   130968,    1622120, ...
  16,  402,  7060,  130968,  2672416,   60666672, ...
  32, 1478, 47220, 1622120, 60666672, 2504664128, ...

R. J. Mathar, Counting 2-way monotonic terrace forms over rectangular landscapes, vixra:1511.0225 (2015), Section 6.

Cf. A000079 (row and column 0), A008776 (row 1), A005054 (row 2), A055275 (row 3), A063376 (column 1).

A264872 := proc(n,m)
    2^n*(1+2^n)^m ;
end proc:
seq(seq(A264872(n,d-n),n=0..d),d=0..12) ; # R. J. Mathar, Aug 14 2024

Table[2^(n - m) (1 + 2^(n - m))^m, {n, 9}, {m, 0, n}] // Flatten (* Michael De Vlieger, Nov 27 2015 *)

T(n,m) = 2^n*A264871(n,m).

T(n,m) <= A116694(n+1,m+1).

cp's OEIS Frontend

A136516 a(n) = (2^n+1)^n.

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