A176824 -id:A176824 - cp's OEIS frontend

A048160 Triangle giving T(n,k) = number of (n,k) labeled rooted Greg trees (n >= 1, 0<=k<=n-1).

1, 2, 1, 9, 10, 3, 64, 113, 70, 15, 625, 1526, 1450, 630, 105, 7776, 24337, 31346, 20650, 6930, 945, 117649, 450066, 733845, 650188, 329175, 90090, 10395, 2097152, 9492289, 18760302, 20925065, 14194180, 5845455, 1351350, 135135, 43046721

Offset: 1

N. J. A. Sloane

An (n,k) rooted Greg tree can be described as a rooted tree with n black nodes and k white nodes where only the black nodes are labeled and the white nodes have at least 2 children. - Christian G. Bower, Nov 15 1999

			Triangle begins:
  1;
  2, 1;
  9, 10, 3;
  64, 113, 70, 15;
  ...

C. Flight, How many stemmata?, Manuscripta, 34 (1990), 122-128.
C. Flight, How many stemmata?, Manuscripta, 34 (1990), 122-128. (Annotated scanned copy)
C. Flight, Letter to N. J. A. Sloane, Nov 1990
D. J. Jeffrey, G. A. Kalugin, N. Murdoch, Lagrange inversion and Lambert W, Preprint 2015.
M. Josuat-Vergès, Derivatives of the tree function, arXiv preprint arXiv:1310.7531 [math.CO], 2013.
Index entries for sequences related to rooted trees
Index entries for sequences related to trees

Row sums give A005264. Cf. A005263, A048159, A052300-A052303. A054589.

t[n_ /; n >= 1, k_ /; k >= 0] /; 0 <= k <= n-1 := t[n, k] = (n+k-2) t[n-1, k-1] + (2n + 2k - 2)*t[n-1, k] + (k+1) t[n-1, k+1]; t[1, 0] = 1; t[, ] = 0; Flatten[Table[t[n, k], {n, 1, 9}, {k, 0, n-1}]] (* Jean-François Alcover, Jul 20 2011, after formula *)

T(n, 0)=n^(n-1), T(n, k)=(n+k-2)*T(n-1, k-1)+(2*n+2*k-2)*T(n-1, k)+(k+1)*T(n-1, k+1).
From Peter Bala, Sep 29 2011: (Start)
E.g.f.: compositional inverse with respect to x of t*(exp(-x)-1) + (1+t)*x*exp(-x) = compositional inverse with respect to x of (x - (2+t)*x^2/2! + (3+2*t)*x^3/3! - (4+3*t)*x^4/4! + ...) = x + (2+t)*x^2/2! + (9+10*t+3*t^2)*x^3/3! + ....
The row generating polynomials R(n,t) satisfy the recurrence R(n+1,t) = (1+t)^2*R'(n,t)+n*(2+t)*R(n,t) with R(1,t) = 1.
The shifted row polynomials R(n,t-1) are the row generating polynomials of A054589. (End)
From Peter Bala, Sep 12 2012: (Start)
It appears that the entries in column k = 1 are given by T(n,1) = (n+1)^n - 2*n^n (checked up to n = 15) - see A176824.
Assuming this, we could then use the recurrence equation to obtain explicit formulas for columns k = 2,3,....
For example, T(n,2) = 1/2*{(n+2)^(n+1) - 4*(n+1)^(n+1) + (4*n+3)*n^n}. (End)

More terms from Larry Reeves (larryr(AT)acm.org), Apr 07 2000

A152260 Triangle T(n, k) = [x^k] p(n, x), where p(n, x) = (1/n)(1-x)^(2n) * Sum_{j >= 0} binomial(n+j-1, j) * j^n * x^(j-1).

Original entry on oeis.org

1, 1, 2, 1, 10, 9, 1, 32, 113, 64, 1, 86, 786, 1526, 625, 1, 212, 4182, 18932, 24337, 7776, 1, 498, 19167, 170332, 477807, 450066, 117649, 1, 1136, 80103, 1266400, 6584615, 12910704, 9492289, 2097152, 1, 2542, 314928, 8313394, 72899230, 254556594, 375886768, 225159022, 43046721

Offset: 1

Roger L. Bagula, Dec 01 2008

A generalization of for n=m: q(x,n,m) = (1-x)^(n+m) * Sum_{j >= 0} binomial(m+j-1, j) * j^n * x^(j-1).

			Polynomials, p(n, x), begin as:
  p(1, x) = 1;
  p(2, x) = 1 +   2*x;
  p(3, x) = 1 +  10*x +    9*x^2;
  p(4, x) = 1 +  32*x +  113*x^2 +    64*x^3;
  p(5, x) = 1 +  86*x +  786*x^2 +  1526*x^3 +   625*x^4;
  p(6, x) = 1 + 212*x + 4182*x^2 + 18932*x^3 + 24337*x^4 + 7776*x^5;
Triangle, T(n, k), begins as:
  1;
  1,    2;
  1,   10,     9;
  1,   32,   113,      64;
  1,   86,   786,    1526,     625;
  1,  212,  4182,   18932,   24337,     7776;
  1,  498, 19167,  170332,  477807,   450066,  117649;
  1, 1136, 80103, 1266400, 6584615, 12910704, 9492289, 2097152;

Douglas C. Montgomery and Lynwood A. Johnson, Forecasting and Time Series Analysis, MaGraw-Hill, New York, 1976, page 91

G. C. Greubel, Table of n, a(n) for n = 1..1275

Cf. A000169, A006963, A123125, A176824.

A152260:= func< n,k | (&+[(-1)^(k+j)*Binomial(2*n,k-j)*Binomial(n+j-1,j)*j^n: j in [1..k]])/n >;
[A152260(n,k): k in [1..n], n in [1..12]]; // G. C. Greubel, May 23 2023

p[x_, n_]:= ((1-x)^(2*n)/n)*Sum[Binomial[k+n-1,k]*k^n*x^(k-1), {k, 0, Infinity}];
Table[CoefficientList[p[x, n], x], {n,12}]//Flatten
(* Second program *)
T[n_, k_]:= (1/n)*Sum[Binomial[2*n,k-j]*Binomial[n+j-1,j]*(-1)^(k+j) *j^n, {j,k}];
Table[T[n,k], {n,12}, {k,n}]//Flatten (* G. C. Greubel, May 23 2023 *)

def A152260(n,k): return (1/n)*sum( (-1)^(k+j)*binomial(2*n,k-j)*binomial(n+j-1,j)*j^n for j in range(1,k+1) )
flatten([[A152260(n,k) for k in range(1,n+1)] for n in range(1,13)]) # G. C. Greubel, May 23 2023

T(n, k) = [x^k] p(n, x), where p(n, x) = (1/n)*(1-x)^(2*n) * Sum_{j >= 0} binomial(n+j-1, j) * j^n * x^(j-1).
Sum_{k=1..n} T(n, k) = A006963(n).
T(n, m) = Sum_{k=1..m} binomial(n+k,k)*binomial(n-k,m-k)*k!*(-1)^(m-k) * Stirling2(n,k)*1/(n+k) (conjectured). - Michael D. Weiner, Jul 01 2020
From G. C. Greubel, May 23 2023: (Start)
T(n, k) = (1/n)*Sum_{j=0..k-1} (-1)^(k+j+1)*binomial(2*n, k-j-1) * binomial(n+j, j+1) * (j+1)^n.
T(n, n) = A000169(n).
T(n, n-1) = A176824(n). (End)

A176825 Primes of the form (k+1)^k mod k^k.

Original entry on oeis.org

113, 24337, 9492289

Offset: 1

Vladimir Joseph Stephan Orlovsky, Apr 26 2010

The next term is too large to include.

The corresponding values of k are 4, 6, 8, 132, ... - Amiram Eldar, Jul 18 2019

			5^4 mod 4^4 = 113 is a prime.

Amiram Eldar, Table of n, a(n) for n = 1..4

Cf. A048160, A176824.

Select[Table[Mod[(n+1)^n,n^n],{n,140}],PrimeQ[ # ]&]

A305559 [0, -1, -1] together with A000290.

Original entry on oeis.org

0, -1, -1, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500

Offset: 0

Paul Curtz, Jun 21 2018

Squares leading to an autosequence of the first kind.
The third sequence of the array
A060576(n+1)=         0,  1, 1,  1,  1,   1,   1,   1,   1,  1, ...
A289207(n)=      0,   0,  0, 1,  2,  3,   4,   5,   6,   7, ...
a(n)=       0,  -1,  -1,  0, 1,  4,  9,  16,  25,  36, ...
      0,   10,  10,   5,  0, 1,  8, 27,  64, 125, ...
0, -113, -113, -68, -23,  0, 1, 16, 81, 256, ... .
The first full vertical is (-1)^n*A033312(n).
From 0, the first two nonzero antidiagonals are 0, -1, 10, -113, 1526, ... = (-1)^n* A176824(n+1).
See OEIS Wiki, Autosequence.
a(n) difference table:
0, -1, -1, 0, 1, 4, 9, 16, 25, ...
-1, 0,  1, 1, 3, 5, 7,  9, 11, ...
1,  1,  0, 2, 2, 2, 2,  2,  2, ...
0, -1,  2, 0, 0, 0, 0,  0,  0, ...

Paolo Xausa, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A003992, A033312, A060576, A176824, A289207, A000290, A113679.

Join[{0,-1,-1},Range[0,100]^2] (* Paolo Xausa, Nov 13 2023 *)

From Stefano Spezia, May 28 2025: (Start)
G.f.: x*(1 - 2*x + x^3 - 2*x^4)/(1 - x)^3.
E.g.f.: 9 + 5*x + x^2 - exp(x)*(9 - 5*x + x^2). (End)

A225190 a(n) = (n+2)^(n+2) mod n^n.

Original entry on oeis.org

0, 0, 0, 20, 64, 1668, 27712, 355279, 779264, 170190707, 6100448256, 159424073982, 2545667031040, 239361355053790, 812743283245056, 58702956893404802, 17949710147773530112, 488189490082385976772, 38768887410023899070464, 313775221076492698014434, 11531764219557396646723584

Offset: 0

Alex Ratushnyak, May 01 2013

The sequence of indices of odd terms b(n) begins: 7, 9, 25, 27, 39, 43, 45, 49, 53, 57, 59, 65, 67, 71, ...
a(n) may be odd only if n is odd. For even n's, both (n+2)^(n+2) and n^n are even, therefore a(n) is even.
c(n) = (b(n)-1)/2 begins: 3, 4, 12, 13, 19, 21, 22, 24, 26, 28, 29, 32, 33, 35, 36, 37, 38, 41, ...

			a(3) = 5^5 mod 3^3 = 3125 mod 27 = 20.

Cf. A066611, A176823, A176824.

for i in range(21):  print(((i+2)**(i+2) % (i**i)), end=', ')

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A048160 Triangle giving T(n,k) = number of (n,k) labeled rooted Greg trees (n >= 1, 0<=k<=n-1).

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A152260 Triangle T(n, k) = [x^k] p(n, x), where p(n, x) = (1/n)*(1-x)^(2*n) * Sum_{j >= 0} binomial(n+j-1, j) * j^n * x^(j-1).

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A176825 Primes of the form (k+1)^k mod k^k.

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A305559 [0, -1, -1] together with A000290.

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A225190 a(n) = (n+2)^(n+2) mod n^n.

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A152260 Triangle T(n, k) = [x^k] p(n, x), where p(n, x) = (1/n)(1-x)^(2n) * Sum_{j >= 0} binomial(n+j-1, j) * j^n * x^(j-1).