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.

A132611 Column 0 of triangle A132610.

Original entry on oeis.org

1, 1, 2, 14, 194, 4114, 118042, 4274612, 186932958, 9577713250, 562450162646, 37232881004442, 2742420824107648, 222414345991567630, 19691735781407563460, 1889658596054736522248, 195353325211864635176182, 21643625444562045727620930
Offset: 0

Views

Author

Paul D. Hanna, Aug 23 2007

Keywords

Comments

Triangle T=A132610 is generated by even matrix powers of itself such that row n+1 of T = row n of T^(2n) with appended '1' for n>=0 with T(0,0)=1.

Crossrefs

Cf. A132610 (triangle); other columns: A132612, A132613; A132614.

Programs

  • PARI
    {a(n)=local(k=0,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=0,25,print1(a(n),", "))
  • PARI
    {a(n)=local(A=[1]);for(i=1,n,A=Vec(Ser(A)/(1-x)^(2*(#A)-3));A=concat(A,A[#A]));A[#A]}
for(n=0,25,print1(a(n),", "))

A132612 Column 1 of triangle A132610.

Original entry on oeis.org

1, 1, 4, 39, 648, 15465, 483240, 18685905, 861282832, 46085893011, 2807152825020, 191731595897600, 14510053796849640, 1205013817282706730, 108941005329522201360, 10650027832902977866245, 1119401271751383414197280, 125879457463215695125460535
Offset: 0

Views

Author

Paul D. Hanna, Aug 23 2007

Keywords

Comments

Triangle T=A132610 is generated by even matrix powers of itself such that row n+1 of T = row n of T^(2n) with appended '1' for n>=0 with T(0,0)=1.

Crossrefs

Cf. A132610 (triangle); other columns: A132611, A132613; A132614.
Cf. A304192.

Programs

  • PARI
    {a(n)=local(A=vector(n+2), p); A[1]=1; for(j=1, n-1, p=n^2-(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]}
for(n=0,25,print1(a(n),", "))
  • PARI
    {a(n)=local(A=[1]);for(i=1,n,A=Vec(Ser(A)/(1-x)^(2*(#A)-1));A=concat(A,A[#A]));A[#A]}
for(n=0,25,print1(a(n),", "))

Formula

a(n) is divisible by n for n>0; a(n)/n = A132614(n).
a(n) = [x^(n-1)] (1+x)^(n*(n+1)) / F(x) for n>0, where F(x) is the g.f. of A304192.

A132613 Column 2 of triangle A132610.

Original entry on oeis.org

1, 1, 6, 76, 1510, 41121, 1424178, 59857416, 2957282370, 167852629965, 10757980606208, 768190770422700, 60461731639747100, 5199414726620992073, 484974399630368105130, 48763257278485285019472
Offset: 0

Views

Author

Paul D. Hanna, Aug 23 2007

Keywords

Comments

Triangle T=A132610 is generated by even matrix powers of itself such that row n+1 of T = row n of T^(2n) with appended '1' for n>=0 with T(0,0)=1.

Crossrefs

Cf. A132610 (triangle); other columns: A132611, A132612; A132614.

Programs

  • PARI
    {a(n)=local(A=vector(n+3), p); A[1]=1; for(j=1, n-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]}

A132614 Column 1 of triangle A132610 divided by row index less one.

Original entry on oeis.org

1, 2, 13, 162, 3093, 80540, 2669415, 107660354, 5120654779, 280715282502, 17430145081600, 1209171149737470, 92693370560208210, 7781500380680157240, 710001855526865191083, 69962579484461463387330
Offset: 1

Views

Author

Paul D. Hanna, Aug 23 2007

Keywords

Comments

Triangle T=A132610 is generated by even matrix powers of itself such that row n+1 of T = row n of T^(2n) with appended '1' for n>=0 with T(0,0)=1.

Crossrefs

Cf. A132610 (triangle); columns: A132611, A132612, A132613.

Programs

  • PARI
    {a(n)=local(A=vector(n+2), p); A[1]=1; if(n<1,0,for(j=1, n-1, p=n^2-(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]/n)}

Formula

a(n) = A132610(n+1,1)/n = A132612(n)/n for n>=1.

A132615 Triangle T, read by rows, where row n+1 of T = row n of T^(2n-1) with appended '1' for n>=0 with T(0,0)=1.

Original entry on oeis.org

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, 3744491970, 420142350, 34461260, 2257413, 125760, 6290, 300, 15, 1, 1
Offset: 0

Views

Author

Paul D. Hanna, Aug 24 2007

Keywords

Examples

			Triangle 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; ...
GENERATE T FROM ODD MATRIX POWERS OF T.
Matrix cube, T^3, begins:
  1;
  3, 1;
  6, 3, 1; <-- row 3 of T
  31, 12, 3, 1;
  357, 100, 18, 3, 1;
  6786, 1455, 205, 24, 3, 1; ...
where row 3 of T = row 2 of T^3 with appended '1'.
Matrix fifth power, T^5, begins:
  1;
  5, 1;
  15, 5, 1;
  80, 25, 5, 1; <-- row 4 of T
  855, 215, 35, 5, 1;
  15171, 3065, 410, 45, 5, 1; ...
where row 4 of T = row 3 of T^5 with appended '1'.
Matrix seventh power, T^7, begins:
  1;
  7, 1;
  28, 7, 1;
  161, 42, 7, 1;
  1666, 378, 56, 7, 1; <-- row 5 of T
  28119, 5348, 679, 70, 7, 1; ...
where row 5 of T = row 4 of T^7 with appended '1'.
ALTERNATE GENERATING METHOD.
Row 4: start with a '1' followed by 4 zeros;
take partial sums and append 2 zeros; then
take partial sums thrice more:
  (1), 0, 0, 0, 0;
  1, 1, 1, 1, (1), 0, 0;
  1, 2, 3, 4, 5, 5, (5);
  1, 3, 6, 10, 15, 20, (25);
  1, 4, 10, 20, 35, 55, (80);
the final nonzero terms form row 4: [80, 25, 5, 1, 1].
Row 5: start with a '1' followed by 6 zeros;
take partial sums and append 4 zeros;
take partial sums and append 2 zeros; then
take partial sums thrice more:
  (1), 0, 0, 0, 0, 0, 0;
  1, 1, 1, 1, 1, 1, (1), 0, 0, 0, 0;
  1, 2, 3, 4, 5, 6, 7, 7, 7, 7, (7), 0, 0;
  1, 3, 6, 10, 15, 21, 28, 35, 42, 49, 56, 56, (56);
  1, 4, 10, 20, 35, 56, 84, 119, 161, 210, 266, 322, (378);
  1, 5, 15, 35, 70, 126, 210, 329, 490, 700, 966, 1288, (1666);
the final nonzero terms form row 5: [1666, 378, 56, 7, 1, 1].
Continuing in this way produces all the rows of this triangle.

Links

Crossrefs

Cf. columns: A132616, A132617, A132618; A132619; variants: A132610, A101479.
Cf. A304190, A304191, A304193.

Programs

  • Maple
    b:= proc(n) option remember;
      Matrix(n, (i,j)-> T(i-1,j-1))^(2*n-3)
    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-3];
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)*(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]}

Formula

T(n+1,1) is divisible by 2n-1 for n>=1.

A304192 G.f. A(x) satisfies: [x^n] (1+x)^(n*(n+1)) / A(x) = 0 for n>0.

Original entry on oeis.org

1, 2, 7, 72, 1224, 29184, 892074, 33144288, 1445847756, 72291575784, 4070550314292, 254674699992768, 17518238545282080, 1313558965998605568, 106608039857256267192, 9309469431887521270848, 870250987085629018699728, 86703492688056304091302944, 9171254392641669833788501488, 1026466161170552167031522911104
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 + 2*x + 7*x^2 + 72*x^3 + 1224*x^4 + 29184*x^5 + 892074*x^6 + 33144288*x^7 + 1445847756*x^8 + 72291575784*x^9 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^(n*(n+1)) / A(x) begins:
n=0: [1, -2, -3, -52, -955, -24246, -771113, -29428232, ...];
n=1: [1, 0, -6, -60, -1062, -26208, -820560, -30994704, ...];
n=2: [1, 4, 0, -80, -1337, -30840, -932010, -34438500, ...];
n=3: [1, 10, 39, 0, -1722, -39996, -1138680, -40521096, ...];
n=4: [1, 18, 147, 648, 0, -50832, -1503546, -50844384, ...];
n=5: [1, 28, 372, 3048, 15465, 0, -1898490, -67990260, ...];
n=6: [1, 40, 774, 9580, 83248, 483240, 0, -85539792, ...];
n=7: [1, 54, 1425, 24420, 303363, 2844270, 18685905, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] (1+x)^(n*(n+1)) / A(x) = 0 for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 4, 39, 648, 15465, 483240, 18685905, 861282832, 46085893011, ...]
yields A132612, column 1 of triangle A132610.
Related triangular matrix T = A132610 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; ...
in which  row n+1 of T = row n of matrix power T^(2*n) with appended '1' for n>=0.

Links

Crossrefs

Cf. A132612, A304189, A304188, A304191, A304193, A132610.

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*(m-1))/Ser(A) )[m] ); A[n+1]}
for(n=0,30, print1(a(n),", "))

Formula

A132612(n+1) = [x^n] (1+x)^((n+1)*(n+2)) / A(x) for n>0.

A304189 G.f. A(x) satisfies: [x^n] (1+x)^(n*(n-1)) / A(x) = 0 for n>0.

Original entry on oeis.org

1, 0, 1, 14, 262, 6512, 202194, 7540004, 328229124, 16332497152, 914162756076, 56834335366552, 3885119345623448, 289588265286519808, 23372826192097312232, 2030600572225893011568, 188934550189205698385072, 18743556336897311790277824, 1974977055586233987489048976, 220268077592251409442788164320, 25923441737544899398961718119392
Offset: 0

Views

Author

Paul D. Hanna, May 08 2018

Keywords

Examples

			G.f.: A(x) = 1 + x^2 + 14*x^3 + 262*x^4 + 6512*x^5 + 202194*x^6 + 7540004*x^7 + 328229124*x^8 + 16332497152*x^9 + 914162756076*x^10 + 56834335366552*x^11 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^(n*(n-1)) / A(x) begins:
  n=0: [1, 0, -1, -14, -261, -6484, -201475, -7519686, ...];
  n=1: [1, 0, -1, -14, -261, -6484, -201475, -7519686, ...];
  n=2: [1, 2, 0, -16, -290, -7020, -214704, -7929120, ...];
  n=3: [1, 6, 14, 0, -345, -8274, -244588, -8831232, ...];
  n=4: [1, 12, 65, 194, 0, -9968, -299160, -10429680, ...];
  n=5: [1, 20, 189, 1106, 4114, 0, -362790, -13084500, ...];
  n=6: [1, 30, 434, 4016, 26289, 118042, 0, -15934512, ...];
  n=7: [1, 42, 860, 11424, 110220, 809688, 4274612, 0, ...];
  n=8: [1, 56, 1539, 27650, 364705, 3749436, 30746547, 186932958, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] (1+x)^(n*(n-1)) / A(x) = 0 for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 2, 14, 194, 4114, 118042, 4274612, 186932958, 9577713250, ...]
yields A132611, column 0 of triangle A132610.
Related triangular matrix T = A132610 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; ...
in which row n+1 of T = row n of matrix power T^(2*n) with appended '1' for n>=0.

Crossrefs

Cf. A132611, A304192, A304188, A304184, A132610.

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)*(m-2))/Ser(A) )[m] ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))

Formula

A132611(n+1) = [x^n] (1+x)^(n*(n+1)) / 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.

A304188 G.f. A(x) satisfies: [x^n] (1+x)^((n+1)*(n+2)) / A(x) = 0 for n>0.

Original entry on oeis.org

1, 6, 30, 264, 4179, 97758, 3000084, 113020056, 5018695542, 255724146876, 14671199172480, 934467807541824, 65366076594301044, 4978197982191048600, 409875168025688997456, 36268233577292228677728, 3431775207222740657912472, 345742547371677388835049744, 36948141363745699171977916032, 4174429749114285739841190548928
Offset: 0

Views

Author

Paul D. Hanna, May 09 2018

Keywords

Examples

			G.f.: A(x) = 1 + 6*x + 30*x^2 + 264*x^3 + 4179*x^4 + 97758*x^5 + 3000084*x^6 + 113020056*x^7 + 5018695542*x^8 + 255724146876*x^9 + 14671199172480*x^10 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^((n+1)*(n+2)) / A(x) begins:
n=0: [1, -4, -5, -114, -2289, -62568, -2113983, -84889290, ...];
n=1: [1, 0, -15, -154, -2790, -72432, -2378450, -93729900, ...];
n=2: [1, 6, 0, -224, -3924, -91776, -2858196, -109145280, ...];
n=3: [1, 14, 76, 0, -5310, -128964, -3714456, -134815824, ...];
n=4: [1, 24, 261, 1510, 0, -169752, -5223348, -178378752, ...];
n=5: [1, 36, 615, 6446, 41121, 0, -6779045, -251285430, ...];
n=6: [1, 50, 1210, 18696, 201435, 1424178, 0, -323428800, ...];
n=7: [1, 66, 2130, 44616, 675591, 7663626, 59857416, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that [x^n] (1+x)^((n+1)*(n+2)) / A(x) = 0 for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 6, 76, 1510, 41121, 1424178, 59857416, 2957282370, ...]
yields A132613, column 2 of triangle A132610.
Related triangular matrix T = A132610 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; ...
in which  row n+1 of T = row n of matrix power T^(2*n) with appended '1' for n>=0.

Crossrefs

Cf. A132613, A304189, A304192, A304193, A132610.

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*(m+1))/Ser(A) )[m] ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))

Formula

A132613(n+1) = [x^n] (1+x)^((n+2)*(n+3)) / A(x) for n>0.
Showing 1-10 of 10 results.

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