A304191 - OEIS

%I #25 Nov 10 2019 20:26:08

%S 1,1,3,35,611,14691,448873,16606825,720241161,35786093321,

%T 2002505540123,124546575282555,8520012343770331,635618668572015451,

%U 51348334729127568273,4465119223213849398545,415808496978034659793361,41283870149540066960271441,4353184675864365012327673843,485828603554439779231472806675

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

%C 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.

%H Paul D. Hanna, <a href="/A304191/b304191.txt">Table of n, a(n) for n = 0..300</a>

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

%e 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 + ...

%e ILLUSTRATION OF DEFINITION.

%e (EX. 1) The table of coefficients of x^k in (1+x)^(n^2) / A(x) begins:

%e n=0: [1, -1,   -2,   -30,   -540,  -13380, -416910, -15634290, ...];

%e n=1: [1,  0,   -3,   -32,   -570,  -13920, -430290, -16051200, ...];

%e n=2: [1,  3,    0,   -40,   -675,  -15729, -473792, -17384400, ...];

%e n=3: [1,  8,   25,     0,   -840,  -19488, -559584, -19917600, ...];

%e n=4: [1, 15,  102,   378,      0,  -24192, -712590, -24272754, ...];

%e n=5: [1, 24,  273,  1920,   8460,       0, -883740, -31495200, ...];

%e n=6: [1, 35,  592,  6408,  48885,  252087,       0, -39049296, ...];

%e n=7: [1, 48, 1125, 17120, 189090, 1583040, 9392890,         0, ...]; ...

%e 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.

%e RELATED SEQUENCES.

%e (EX. 2) The secondary diagonal in the above table (EX. 1) that begins

%e [1, 3, 25, 378, 8460, 252087, 9392890, 420142350, ...]

%e yields A132617, column 1 of triangle A132615.

%e Related triangular matrix T = A132615 begins:

%e          1;

%e          1,       1;

%e          1,       1,      1;

%e          6,       3,      1,     1;

%e         80,      25,      5,     1,    1;

%e       1666,     378,     56,     7,    1,   1;

%e      47232,    8460,   1020,    99,    9,   1,  1;

%e    1694704,  252087,  26015,  2134,  154,  11,  1, 1;

%e   73552752, 9392890, 855478, 61919, 3848, 221, 13, 1, 1; ...

%e in which row n equals row (n-1) of T^(2*n-1) followed by '1' for n > 0.

%e (EX. 3) The next diagonal in the table (EX. 1) that begins:

%e [1, 8, 102, 1920, 48885, 1583040, 1583040, 62467314, ...]

%e yields the first column in the following matrix product.

%e Let TSL(m) denote the table T = A132615, with the diagonal of 1's truncated, as SHIFTED LEFT m times, so that

%e TSL(1) begins

%e   [   1];

%e   [   3,    1];

%e   [  25,    5,  1];

%e   [ 378,   56,  7, 1];

%e   [8460, 1020, 99, 9, 1]; ...

%e TSL(2) begins

%e   [    1];

%e   [    5,    1];

%e   [   56,    7,   1];

%e   [ 1020,   99,   9,  1];

%e   [26015, 2134, 154, 11, 1]; ...

%e etc.,

%e then the matrix product TSL(2)*TSL(1) begins

%e   [       1];

%e   [       8,       1];

%e   [     102,      12,      1];

%e   [    1920,     200,     16,     1];

%e   [   48885,    4540,    330,    20,   1];

%e   [ 1583040,  132810,   8816,   492,  24,  1];

%e   [62467314, 4790156, 293419, 15148, 686, 28, 1]; ...

%e in which the first column equals the secondary diagonal in the table of (EX. 1).

%e 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).

%o (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]}

%o for(n=0,30, print1(a(n),", "))

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

%K nonn

%O 0,3

%A _Paul D. Hanna_, May 07 2018