cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 10 results.

A132616 Column 0 of triangle A132615.

Original entry on oeis.org

1, 1, 1, 6, 80, 1666, 47232, 1694704, 73552752, 3744491970, 218684051648, 14406896813608, 1056681951098592, 85379764462169382, 7534286318509305600, 720884741940337283712, 74330131862002429961712, 8215901579822006354547330, 969069489665924620416715008
Offset: 0

Views

Author

Paul D. Hanna, Aug 24 2007

Keywords

Comments

Triangle T = A132615 is generated by odd matrix powers of itself such that row n+1 of T = row n of T^(2n-1) with appended '1' for n >= 0 with T(0,0) = 1.

Examples

			G.f. = 1 + x + x^2 + 6*x^3 + 80*x^4 + 1666*x^5 + 47232*x^6 + ...

Crossrefs

Cf. A132615 (triangle).
Other columns: A132617, A132618, A132619.

Programs

  • Mathematica
    a[ n_, k_: 1] := a[n, k] = If[ n < 2, Boole[n >= 0], Sum[ a[n - 1, i], {i, k + 2 (n - 2)}]]; (* Michael Somos, Nov 29 2016 *)
  • PARI
    {a(n) = my(A = vector(n+1), p); A[1] = 1; for(j=1, n-1, p = (n-1)*(n-2) - (n-j-1)*(n-j-2); A = Vec((Polrev(A) + x * O(x^p)) / (1-x))); A = Vec((Polrev(A) + x * O(x^p)) / (1-x)); A[p+1]}
  • PARI
    {a(n, k=1) = if( n<2, n>=0, sum(i=1, k + 2*n-4, a(n-1, i)))}; /* Michael Somos, Nov 29 2016 */

Formula

From Benedict W. J. Irwin, Nov 29 2016: (Start)
Conjecture: a(n) is described by a series of nested sums,
a(2) = Sum_{i=1..1} 1,
a(3) = Sum_{i=1..1+2} Sum_{j=1..i} 1,
a(4) = Sum_{i=1..1+4} Sum_{j=1..i+2} Sum_{k=1..j} 1,
a(5) = Sum_{i=1..1+6} Sum_{j=1..i+4} Sum_{k=1..j+2} Sum_{l=1..k} 1,
and so on, where 2, 4, 6,... are the even numbers. (End)

A132617 Column 1 of triangle A132615.

Original entry on oeis.org

1, 1, 3, 25, 378, 8460, 252087, 9392890, 420142350, 21927528948, 1307723670020, 87712176219801, 6534001434836758, 535159878432210000, 47792303301224799288, 4621416976280491118910, 481020078381722064175750
Offset: 0

Views

Author

Paul D. Hanna, Aug 24 2007

Keywords

Comments

Triangle T=A132615 is generated by odd matrix powers of itself such that row n+1 of T = row n of T^(2n-1) with appended '1' for n>=0 with T(0,0)=1.

Crossrefs

Cf. A132615 (triangle); other columns: A132616, A132618; A132619.

Programs

  • PARI
    {a(n)=local(A=vector(n+1), p); A[1]=1; for(j=1, n-1, p=n*(n-1)-(n-j)*(n-j-1); A=Vec((Polrev(A)+x*O(x^p))/(1-x))); A=Vec((Polrev(A)+x*O(x^p))/(1-x)); A[p+1]}

Formula

a(n) is divisible by 2n-1 for n>0; a(n)/(2n-1) = A132619(n).

A132618 Column 2 of triangle A132615.

Original entry on oeis.org

1, 1, 5, 56, 1020, 26015, 855478, 34461260, 1642995124, 90456911140, 5646312067585, 393937815588880, 30374808071994000, 2564601377235725520, 235302361169390146650, 23309583579201438877060
Offset: 0

Views

Author

Paul D. Hanna, Aug 24 2007

Keywords

Comments

Triangle T=A132615 is generated by odd matrix powers of itself such that row n+1 of T = row n of T^(2n-1) with appended '1' for n>=0 with T(0,0)=1.

Crossrefs

Cf. A132615 (triangle); other columns: A132616, A132617; A132619.

Programs

  • PARI
    {a(n)=local(A=vector(n+1), p); A[1]=1; for(j=1, n-1, p=(n+1)*n-(n-j+1)*(n-j); A=Vec((Polrev(A)+x*O(x^p))/(1-x))); A=Vec((Polrev(A)+x*O(x^p))/(1-x)); A[p+1]}

A132619 Column 1 of triangle A132615 divided by twice the row index less 1.

Original entry on oeis.org

1, 1, 5, 54, 940, 22917, 722530, 28009490, 1289854644, 68827561580, 4176770296181, 284087018905946, 21406395137288400, 1770085307452770344, 159359206078637624790, 15516776721991034328250
Offset: 0

Views

Author

Paul D. Hanna, Aug 24 2007

Keywords

Comments

Triangle T=A132615 is generated by odd matrix powers of itself such that row n+1 of T = row n of T^(2n-1) with appended '1' for n>=0 with T(0,0)=1.

Crossrefs

Cf. A132615 (triangle); columns: A132616, A132617, A132618.

Programs

  • PARI
    {a(n)=local(A=vector(n+1), p); A[1]=1; for(j=1, n-1, p=n*(n-1)-(n-j)*(n-j-1); A=Vec((Polrev(A)+x*O(x^p))/(1-x))); A=Vec((Polrev(A)+x*O(x^p))/(1-x)); A[p+1]/(2*n-1)}

Formula

a(n) = A132615(n+1,1)/(2n-1) = A132617(n)/(2n-1) for n>=1.

A132610 Triangle T, read by rows, where row n+1 of T = row n of matrix power T^(2n) with appended '1' for n>=0 with T(0,0)=1.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 14, 4, 1, 1, 194, 39, 6, 1, 1, 4114, 648, 76, 8, 1, 1, 118042, 15465, 1510, 125, 10, 1, 1, 4274612, 483240, 41121, 2908, 186, 12, 1, 1, 186932958, 18685905, 1424178, 89670, 4970, 259, 14, 1, 1, 9577713250, 861282832, 59857416, 3437248, 171700, 7824, 344, 16, 1, 1
Offset: 0

Views

Author

Paul D. Hanna, Aug 23 2007

Keywords

Examples

			Triangle begins:
1;
1, 1;
2, 1, 1;
14, 4, 1, 1;
194, 39, 6, 1, 1;
4114, 648, 76, 8, 1, 1;
118042, 15465, 1510, 125, 10, 1, 1;
4274612, 483240, 41121, 2908, 186, 12, 1, 1;
186932958, 18685905, 1424178, 89670, 4970, 259, 14, 1, 1; ...
GENERATE T FROM EVEN MATRIX POWERS OF T.
Matrix square T^2 begins:
1;
2, 1; <-- row 2 of T
5, 2, 1;
34, 9, 2, 1;
453, 88, 13, 2, 1; ...
where row 2 of T = row 1 of T^2 with appended '1'.
Matrix fourth powers T^4 begins:
1;
4, 1;
14, 4, 1; <-- row 3 of T
96, 22, 4, 1;
1215, 220, 30, 4, 1; ...
where row 3 of T = row 2 of T^4 with appended '1'.
Matrix sixth power T^6 begins:
1;
6, 1;
27, 6, 1;
194, 39, 6, 1; <-- row 4 of T
2394, 404, 51, 6, 1; ...
where row 4 of T = row 3 of T^6 with appended '1'.
ALTERNATE GENERATING METHOD.
Start with [1,0,0,0]; take partial sums and append 1 zero;
take partial sums twice more:
(1), 0, 0, 0;
1, 1, 1, (1), 0;
1, 2, 3, 4, (4);
1, 3, 6, 10, (14);
the final nonzero terms form row 3: [14,4,1,1].
Start with [1,0,0,0,0,0]; take partial sums and append 3 zeros;
take partial sums and append 1 zero; take partial sums twice more:
(1), 0, 0, 0, 0, 0;
1, 1, 1, 1, 1, (1), 0, 0, 0;
1, 2, 3, 4, 5, 6, 6, 6, (6), 0;
1, 3, 6, 10, 15, 21, 27, 33, 39, (39);
1, 4, 10, 20, 35, 56, 83, 116, 155, (194);
the final nonzero terms form row 4: [194,39,6,1,1].
Continuing in this way produces all the rows of this triangle.

Links

Crossrefs

Cf. columns: A132611, A132612, A132613; A132614; variants: A132615, A101479.
Cf. A304189, A304192, A304188.

Programs

  • Maple
    b:= proc(n) option remember;
      Matrix(n, (i,j)-> T(i-1,j-1))^(2*n-2)
    end:
T:= proc(n,k) option remember;
     `if`(n=k, 1, `if`(k>n, 0, b(n)[n,k+1]))
    end:
seq(seq(T(n,k), k=0..n), n=0..10);  # Alois P. Heinz, Apr 13 2020
  • Mathematica
    b[n_] := b[n] = MatrixPower[Table[T[i-1, j-1], {i, n}, {j, n}], 2n-2];
T[n_, k_] := T[n, k] = If[n == k, 1, If[k > n, 0, b[n][[n, k+1]]]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 27 2020, after Alois P. Heinz *)
  • PARI
    {T(n, k)=local(A=vector(n+1), p); A[1]=1; for(j=1, n-k-1, p=(n-1)^2-(n-j-1)^2; A=Vec((Polrev(A)+x*O(x^p))/(1-x))); A=Vec((Polrev(A)+x*O(x^p))/(1-x)); A[p+1]}
for(n=0,10, for(k=0,n, print1(T(n,k),", "));print(""))

Formula

T(n+1,1) is divisible by n for n>=1.

A304193 G.f. A(x) satisfies: [x^n] (1+x)^((n+1)^2) / A(x) = 0 for n>0.

Original entry on oeis.org

1, 4, 16, 144, 2346, 55236, 1688084, 63040736, 2770165274, 139623836116, 7925496107656, 499719554537584, 34625595715906866, 2613946666882042164, 213475621178226876156, 18748792440158256161216, 1761875767691411063734514, 176383456081424163875684516, 18739798321516251204837796864, 2105891800817103192582808107856
Offset: 0

Views

Author

Paul D. Hanna, May 07 2018

Keywords

Comments

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

Examples

			G.f.: A(x) = 1 + 4*x + 16*x^2 + 144*x^3 + 2346*x^4 + 55236*x^5 + 1688084*x^6 + 63040736*x^7 + 2770165274*x^8 + 139623836116*x^9 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^((n+1)^2) / A(x) begins:
n=0: [1, -3, -4, -80, -1530, -40222, -1316104, -51439572, ...];
n=1: [1, 0, -10, -100, -1785, -45056, -1441440, -55510080, ...];
n=2: [1, 5, 0, -140, -2380, -55080, -1685620, -63186200, ...];
n=3: [1, 12, 56, 0, -3150, -74484, -2125948, -76230384, ...];
n=4: [1, 21, 200, 1020, 0, -96492, -2901052, -98301840, ...];
n=5: [1, 32, 486, 4540, 26015, 0, -3718000, -135081440, ...];
n=6: [1, 45, 980, 13640, 132810, 855478, 0, -172046940, ...];
n=7: [1, 60, 1760, 33520, 462150, 4790156, 34461260, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] (1+x)^((n+1)^2) / A(x) = 0 for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 5, 56, 1020, 26015, 855478, 34461260, 1642995124, ...]
yields A132618, column 2 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.

Links

Crossrefs

Cf. A132618, A304190, A304191, A304192, 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^2)/Ser(A) )[m] ); A[n+1]}
for(n=0,30, print1(a(n),", "))

Formula

A132618(n+1) = [x^n] (1+x)^((n+2)^2) / A(x) for n>=0.

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

Original entry on oeis.org

1, 1, 3, 35, 611, 14691, 448873, 16606825, 720241161, 35786093321, 2002505540123, 124546575282555, 8520012343770331, 635618668572015451, 51348334729127568273, 4465119223213849398545, 415808496978034659793361, 41283870149540066960271441, 4353184675864365012327673843, 485828603554439779231472806675
Offset: 0

Views

Author

Paul D. Hanna, May 07 2018

Keywords

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

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.

A132625 Triangle T, read by rows, where row n+1 of T = row n of T^(2^n) with appended '1' for n>=0 with T(0,0)=1.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 14, 4, 1, 1, 336, 60, 8, 1, 1, 25836, 2960, 248, 16, 1, 1, 6251504, 454072, 24800, 1008, 32, 1, 1, 4838830976, 216266368, 7603952, 202944, 4064, 64, 1, 1, 12344615283200, 328381917376, 7190266752, 124427232, 1641856, 16320, 128, 1, 1
Offset: 0

Views

Author

Paul D. Hanna, Aug 25 2007, Jan 07 2008

Keywords

Comments

Let R_{n} equal row n of square array A136555, where A136555(n,k) = C(2^k + n-1, k); this triangle transforms rows of A136555: T * R_{n} = R_{n+1} for n>=0.

Examples

			Triangle begins:
1;
1, 1;
2, 1, 1;
14, 4, 1, 1;
336, 60, 8, 1, 1;
25836, 2960, 248, 16, 1, 1;
6251504, 454072, 24800, 1008, 32, 1, 1;
4838830976, 216266368, 7603952, 202944, 4064, 64, 1, 1;
12344615283200, 328381917376, 7190266752, 124427232, 1641856, 16320, 128, 1, 1; ...
GENERATE T FROM MATRIX POWERS OF T.
Matrix power T^4 begins:
1;
4, 1;
14, 4, 1; <-- row 3 of T
96, 22, 4, 1;
1941, 316, 38, 4, 1;
129206, 14185, 1140, 70, 4, 1; ...
where row 3 of T = row 2 of T^(2^2) with appended '1'.
Matrix power T^8 begins:
1;
8, 1;
44, 8, 1;
336, 60, 8, 1; <-- row 4 of T
6062, 872, 92, 8, 1;
345596, 35734, 2712, 156, 8, 1; ...
where row 4 of T = row 3 of T^(2^3) with appended '1'.
Matrix power T^16 begins:
1;
16, 1;
152, 16, 1;
1504, 184, 16, 1;
25836, 2960, 248, 16, 1; <-- row 5 of T
1197304, 109500, 7408, 376, 16, 1; ...
where row 5 of T = row 4 of T^(2^4) with appended '1'.
Alternate generating method:
RoW 3: start with '1' followed by (2^2 - 1) zeros;
take partial sums and append (2^1 - 1) zero;
take partial sums twice more:
(1), 0, 0, 0;
1, 1, 1, (1), 0;
1, 2, 3, 4, (4);
1, 3, 6, 10, (14);
the final nonzero terms form row 3: [14, 4, 1, 1].
Row 4: start with '1' followed by (2^3 - 1) zeros;
take partial sums and append (2^2 - 1) zeros;
take partial sums and append (2^1 - 1) zero;
take partial sums twice more:
(1), 0, 0, 0, 0, 0, 0, 0;
1, 1, 1, 1, 1, 1, 1, (1), 0, 0, 0;
1, 2, 3, 4, 5, 6, 7, 8, 8, 8, (8), 0;
1, 3, 6, 10, 15, 21, 28, 36, 44, 52, 60, (60);
1, 4, 10, 20, 35, 56, 84, 120, 164, 216, 276, (336);
the final nonzero terms form row 4: [336, 60, 8, 1, 1].
Continuing in this way produces all the rows of this triangle.

Crossrefs

Cf. variants: A101479, A132610, A132615; columns: A132626, A132627.
Cf. A136555.

Programs

  • PARI
    T(n, k)=local(A=Mat(1), B); for(m=1, n+1, B=matrix(m, m); for(i=1, m, for(j=1, i, if(j==i, B[i, j]=1, B[i, j]=(A^(2^(i-2)))[i-1, j]); )); A=B); return( ((A)[n+1, k+1]))
  • PARI
    /* Generate using partial sums method (faster) */ T(n, k)=local(A=vector(n+1), p); A[1]=1; for(j=1, n-k, p=2^n-2^(n-j)-j; A=Vec((Polrev(A)+x*O(x^p))/(1-x))); A[p+1]
  • PARI
    /* As Row Transformation of Square Array A136555(n,k) = C(2^k + n-1, k): */ T(n,k)=local(M=matrix(n+2,n+2,r,c,binomial(2^(c-1)+r-2,c-1)), N=matrix(n+1,n+1,r,c,M[r,c]),P=matrix(n+1,n+1,r,c,M[r+1,c]),R=P~*N~^-1); R[n+1,k+1]

A136170 Triangle T, read by rows, where row n of T = row n-1 of T^fibonacci(n) with appended '1' for n>=1 starting with a single '1' in row 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 19, 9, 3, 1, 1, 310, 105, 25, 5, 1, 1, 10978, 2702, 480, 68, 8, 1, 1, 868140, 154609, 20657, 2184, 182, 13, 1, 1, 149688297, 19092682, 1906051, 152579, 9562, 483, 21, 1, 1, 57339888914, 5161046609, 378639419, 22799907
Offset: 0

Views

Author

Paul D. Hanna, Dec 17 2007

Keywords

Examples

			Triangle T begins:
1;
1, 1;
1, 1, 1;
3, 2, 1, 1;
19, 9, 3, 1, 1;
310, 105, 25, 5, 1, 1;
10978, 2702, 480, 68, 8, 1, 1;
868140, 154609, 20657, 2184, 182, 13, 1, 1;
149688297, 19092682, 1906051, 152579, 9562, 483, 21, 1, 1;
57339888914, 5161046609, 378639419, 22799907, 1090125, 41480, 1275, 34, 1, 1; ...
GENERATE T FROM MATRIX POWERS OF T.
Row n of T = row n-1 of T^fibonacci(n) with appended '1'.
Examples.
Row 5 of T is given by row 4 of matrix power T^fibonacci(5) = T^5:
1;
5, 1;
15, 5, 1;
55, 20, 5, 1;
310, 105, 25, 5, 1; <== row 5 of T
3796, 1070, 215, 35, 5, 1; ...
Row 6 of T is given by row 5 of matrix power T^fibonacci(6) = T^8:
1;
8, 1;
36, 8, 1;
164, 44, 8, 1;
978, 268, 52, 8, 1;
10978, 2702, 480, 68, 8, 1; <== row 6 of T
262838, 53648, 8082, 964, 92, 8, 1; ...
ALTERNATE GENERATING METHOD.
To obtain row n, start with a '1' repeated fibonacci(n) times,
and build a table where row k+1 equals the partial sums of row k
but with the last term appearing fibonacci(n-k) times, for k=1..n-1;
listing the final terms in each row forms row n of this triangle.
Example.
To obtain row 5, start with a '1' repeated fibonacci(5)=5 times:
(1,1,1,1,1);
take partial sums, writing the last term fibonacci(4)=3 times:
1,2,3,4, (5,5,5);
take partial sums, writing the last term fibonacci(3)=2 times:
1,3,6,10,15,20, (25,25);
take partial sums, writing the last term fibonacci(2)=1 times:
1,4,10,20,35,55,80, (105);
take partial sums, writing the last term fibonacci(1)=1 times:
1,5,15,35,70,125,205, (310).
Final terms in the above partial sums forms row 5: [310,105,25,5,1];
repeating this process will generate all the rows of this triangle.

Crossrefs

Cf. columns: A136171, A136172, A136173; variants: A101479, A132610, A132615.

Programs

  • PARI
    /* Generate using matrix power method: */ T(n,k)=local(A=Mat(1), B); for(m=1, n+1, B=matrix(m, m); for(i=1, m, for(j=1, i, if(j==i, B[i, j]=1, B[i, j]=(A^(fibonacci(i-1)))[i-1, j]); )); A=B); return( ((A)[n+1, k+1]))
  • PARI
    /* Generate using partial sums method (faster) */ T(n, k)=local(A=vector(n+1), p); A[1]=1; for(j=1, n-k, p=fibonacci(n+2)-fibonacci(n-j+2)-j; A=Vec((Polrev(A)+x*O(x^p))/(1-x))); A[p+1]

Formula

See example section for two different methods of generating this triangle.

A304190 G.f. A(x) satisfies: [x^n] (1+x)^((n-1)^2) / A(x) = 0 for n>0.

Original entry on oeis.org

1, 0, 0, 4, 90, 2448, 79716, 3058740, 135637242, 6835557984, 386119895256, 24170805494868, 1661105052140226, 124342746871407984, 10070793262851698412, 877493877654988612836, 81848857574562663295026, 8137513480199793111630528, 859067817713438540813133744, 95973644392465888508242272804, 11312379843382901418721437545706
Offset: 0

Views

Author

Paul D. Hanna, May 08 2018

Keywords

Examples

			G.f.: A(x) = 1 + 4*x^3 + 90*x^4 + 2448*x^5 + 79716*x^6 + 3058740*x^7 + 135637242*x^8 + 6835557984*x^9 + 386119895256*x^10 + 24170805494868*x^11 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^((n-1)^2) / A(x) begins:
n=0: [1, 1, 0, -4, -94, -2538, -82148, -3137720, ...];
n=1: [1, 0, 0, -4, -90, -2448, -79700, -3058020, ...];
n=2: [1, 1, 0, -4, -94, -2538, -82148, -3137720, ...];
n=3: [1, 4, 6, 0, -105, -2832, -90048, -3391872, ...];
n=4: [1, 9, 36, 80, 0, -3276, -105224, -3871476, ...];
n=5: [1, 16, 120, 556, 1666, 0, -123900, -4673220, ...];
n=6: [1, 25, 300, 2296, 12460, 47232, 0, -5561820, ...];
n=7: [1, 36, 630, 7136, 58671, 368784, 1694704, 0,  ...];
n=8: [1, 49, 1176, 18420, 211590, 1895322, 13604628, 73552752, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] (1+x)^((n-1)^2) / A(x) = 0 for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 1, 6, 80, 1666, 47232, 1694704, 73552752, ...]
yields A132616, column 0 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.

Crossrefs

Cf. A132616, A304191, 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-2)^2)/Ser(A) )[m] ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))

Formula

A132616(n+1) = [x^n] (1+x)^(n^2) / A(x) for n>=0.
Showing 1-10 of 10 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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