A304191 G.f. A(x) satisfies: [x^n] (1+x)^(n^2) / A(x) = 0 for n > 0.

1, 1, 3, 35, 611, 14691, 448873, 16606825, 720241161, 35786093321, 2002505540123, 124546575282555, 8520012343770331, 635618668572015451, 51348334729127568273, 4465119223213849398545, 415808496978034659793361, 41283870149540066960271441, 4353184675864365012327673843, 485828603554439779231472806675

Offset: 0

Author

Paul D. Hanna, May 07 2018

Keywords

• nonn

Comments

Note that [x^n] (1+x)^(n*k) / G(x) = 0 for n > 0 holds when G(x) = (1+x)/(1 - (k-1)*x) given some fixed k; this sequence explores the case where k varies with n.

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 35*x^3 + 611*x^4 + 14691*x^5 + 448873*x^6 + 16606825*x^7 + 720241161*x^8 + 35786093321*x^9 + 2002505540123*x^10 + ...
ILLUSTRATION OF DEFINITION.
(EX. 1) The table of coefficients of x^k in (1+x)^(n^2) / A(x) begins:
n=0: [1, -1,   -2,   -30,   -540,  -13380, -416910, -15634290, ...];
n=1: [1,  0,   -3,   -32,   -570,  -13920, -430290, -16051200, ...];
n=2: [1,  3,    0,   -40,   -675,  -15729, -473792, -17384400, ...];
n=3: [1,  8,   25,     0,   -840,  -19488, -559584, -19917600, ...];
n=4: [1, 15,  102,   378,      0,  -24192, -712590, -24272754, ...];
n=5: [1, 24,  273,  1920,   8460,       0, -883740, -31495200, ...];
n=6: [1, 35,  592,  6408,  48885,  252087,       0, -39049296, ...];
n=7: [1, 48, 1125, 17120, 189090, 1583040, 9392890,         0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] (1+x)^(n^2) / A(x) = 0 for n > 0.
RELATED SEQUENCES.
(EX. 2) The secondary diagonal in the above table (EX. 1) that begins
[1, 3, 25, 378, 8460, 252087, 9392890, 420142350, ...]
yields A132617, column 1 of triangle A132615.
Related triangular matrix T = A132615 begins:
         1;
         1,       1;
         1,       1,      1;
         6,       3,      1,     1;
        80,      25,      5,     1,    1;
      1666,     378,     56,     7,    1,   1;
     47232,    8460,   1020,    99,    9,   1,  1;
   1694704,  252087,  26015,  2134,  154,  11,  1, 1;
  73552752, 9392890, 855478, 61919, 3848, 221, 13, 1, 1; ...
in which row n equals row (n-1) of T^(2*n-1) followed by '1' for n > 0.
(EX. 3) The next diagonal in the table (EX. 1) that begins:
[1, 8, 102, 1920, 48885, 1583040, 1583040, 62467314, ...]
yields the first column in the following matrix product.
Let TSL(m) denote the table T = A132615, with the diagonal of 1's truncated, as SHIFTED LEFT m times, so that
TSL(1) begins
  [   1];
  [   3,    1];
  [  25,    5,  1];
  [ 378,   56,  7, 1];
  [8460, 1020, 99, 9, 1]; ...
TSL(2) begins
  [    1];
  [    5,    1];
  [   56,    7,   1];
  [ 1020,   99,   9,  1];
  [26015, 2134, 154, 11, 1]; ...
etc.,
then the matrix product TSL(2)*TSL(1) begins
  [       1];
  [       8,       1];
  [     102,      12,      1];
  [    1920,     200,     16,     1];
  [   48885,    4540,    330,    20,   1];
  [ 1583040,  132810,   8816,   492,  24,  1];
  [62467314, 4790156, 293419, 15148, 686, 28, 1]; ...
in which the first column equals the secondary diagonal in the table of (EX. 1).
The subsequent diagonal in the table of (EX. 1) also equals the first column of matrix product TSL(3)*TSL(2)*TSL(1). This process can be continued to produce all the lower diagonals of the table of (EX. 1).
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..300

Crossrefs

Cf. A132617, A304190, A304192, A304193, A132615.

Programs

PARI

{a(n) = my(A=[1]); for(i=1,n, A=concat(A,0); m=#A; A[m] = Vec( (1+x +x*O(x^m))^((m-1)^2)/Ser(A) )[m] ); A[n+1]}
for(n=0,30, print1(a(n),", "))

Formula

A132617(n+1) = [x^n] (1+x)^((n+1)^2) / A(x) for n >= 0.