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.

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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.
		

Crossrefs

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

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.
		

Crossrefs

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).
		

Crossrefs

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.

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.

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

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.

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

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.

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

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 1-7 of 7 results.