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 1-4 of 4 results.

A101479 Triangular matrix T, read by rows, where row n equals row (n-1) of T^(n-1) after appending '1' for the main diagonal.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 19, 9, 3, 1, 1, 191, 70, 18, 4, 1, 1, 2646, 795, 170, 30, 5, 1, 1, 46737, 11961, 2220, 335, 45, 6, 1, 1, 1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1, 25330125, 5051866, 758814, 92652, 9730, 924, 84, 8, 1, 1, 735180292, 132523155, 18301950, 2065146, 199692, 17226, 1380, 108, 9, 1, 1
Offset: 0

Views

Author

Paul D. Hanna, Jan 21 2005, Jul 26 2006, May 27 2007

Keywords

Comments

Remarkably, T equals the product of these triangular matrices: T = A107867*A107862^-1 = A107870*A107867^-1 = A107873*A107870^-1; reversing the order of these products yields triangle A107876.

Examples

			Triangle begins:
1;
1, 1;
1, 1, 1;
3, 2, 1, 1;
19, 9, 3, 1, 1;
191, 70, 18, 4, 1, 1;
2646, 795, 170, 30, 5, 1, 1;
46737, 11961, 2220, 335, 45, 6, 1, 1;
1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1;
25330125, 5051866, 758814, 92652, 9730, 924, 84, 8, 1, 1;
735180292, 132523155, 18301950, 2065146, 199692, 17226, 1380, 108, 9, 1, 1; ...
Row 4 starts with row 3 of T^3 which begins:
1;
3, 1;
6, 3, 1;
19, 9, 3, 1; ...
row 5 starts with row 4 of T^4 which begins:
1;
4, 1;
10, 4, 1;
34, 14, 4, 1;
191, 70, 18, 4, 1; ...
An ALTERNATE GENERATING METHOD is illustrated as follows.
For row 4:
Start with a '1' and append 2 zeros,
take partial sums and append 1 zero,
take partial sums thrice more, resulting in:
1, 0, 0;
1, 1, 1, 0;
1, 2, 3, 3;
1, 3, 6, 9;
1, 4,10,19.
Final nonzero terms form row 4: [19,9,3,1,1].
For row 5:
Start with a '1' and append 3 zeros,
take partial sums and append 2 zeros,
take partial sums and append 1 zero,
take partial sums thrice more, resulting in:
1, 0, 0, 0;
1, 1, 1, 1, 0,  0;
1, 2, 3, 4, 4,  4,  0;
1, 3, 6,10,14, 18, 18;
1, 4,10,20,34, 52, 70;
1, 5,15,35,69,121,191;
where the final nonzero terms form row 5: [191,70,18,4,1,1].
Likewise, for row 6:
1, 0, 0, 0,  0;
1, 1, 1, 1,  1,  0,  0,  0;
1, 2, 3, 4,  5,  5,  5,  5,   0,   0;
1, 3, 6,10, 15, 20, 25, 30,  30,  30,   0;
1, 4,10,20, 35, 55, 80,110, 140, 170, 170;
1, 5,15,35, 70,125,205,315, 455, 625, 795;
1, 6,21,56,126,251,456,771,1226,1851,2646;
where the final nonzero terms form row 6: [2646,795,170,30,5,1,1].
Continuing in this way generates all rows of this triangle.

Links

Crossrefs

Columns are A101481, A101482, A101483, row sums form A101484.
Cf. A107876 (dual triangle).
Cf. A304184, A304185, A304186, A304187.

Programs

  • Maple
    b:= proc(n) option remember;
      Matrix(n, (i,j)-> T(i-1,j-1))^(n-1)
    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}], n-1];
T[n_, k_] := T[n, k] = If[n == k, 1, If[k > n, 0, b[n][[n, k+1]]]];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Apr 25 2020, after Alois P. Heinz *)
  • PARI
    {T(n,k) = my(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^(i-2))[i-1,j]);)); A=B); return(A[n+1,k+1])}
for(n=0,10, for(k=0,n, print1(T(n,k),", ")); print(""))
  • PARI
    {T(n,k) = my(A=vector(n+1),p); A[1]=1; for(j=1,n-k-1, p=(n-1)*(n-2)/2-(n-j-1)*(n-j-2)/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(""))

A304184 G.f. A(x) satisfies: 0 = [x^n] (1+x)^(n*(n-1)/2) / A(x) for n>0.

Original entry on oeis.org

1, 0, 0, 1, 9, 117, 1851, 34923, 765933, 19155084, 538051164, 16771165230, 574424285076, 21443516818065, 866521903003641, 37683366660458208, 1754777541925339779, 87115221430910051901, 4592968693335470802627, 256294382115032521083411, 15090698035153332532531074
Offset: 0

Views

Author

Paul D. Hanna, May 08 2018

Keywords

Examples

			G.f.: A(x) = 1 + x^3 + 9*x^4 + 117*x^5 + 1851*x^6 + 34923*x^7 + 765933*x^8 + 19155084*x^9 + 538051164*x^10 + 16771165230*x^11 + 574424285076*x^12 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^(n*(n-1)/2) / A(x) begins:
n=0: [1, 0, 0, -1, -9, -117, -1850, -34905, -765618, ...];
n=1: [1, 0, 0, -1, -9, -117, -1850, -34905, -765618, ...];
n=2: [1, 1, 0, -1, -10, -126, -1967, -36755, -800523, ...];
n=3: [1, 3, 3, 0, -12, -147, -2229, -40815, -876000, ...];
n=4: [1, 6, 15, 19, 0, -180, -2706, -47955, -1005279, ...];
n=5: [1, 10, 45, 119, 191, 0, -3335, -59840, -1214055, ...];
n=6: [1, 15, 105, 454, 1341, 2646, 0, -73965, -1545531, ...];
n=7: [1, 21, 210, 1329, 5955, 19833, 46737, 0, -1913457, ...];
n=8: [1, 28, 378, 3275, 20438, 97533, 364936, 1003150, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that 0 = [x^n] (1+x)^(n*(n-1)/2) / A(x) for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 1, 3, 19, 191, 2646, 46737, 1003150, 25330125,  ...]
yields A101481, column 0 of triangle A101479.
Related triangular matrix T = A101479 begins:
1;
1, 1;
1, 1, 1;
3, 2, 1, 1;
19, 9, 3, 1, 1;
191, 70, 18, 4, 1, 1;
2646, 795, 170, 30, 5, 1, 1;
46737, 11961, 2220, 335, 45, 6, 1, 1;
1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1; ...
in which row n equals row (n-1) of T^(n-1) followed by '1' for n>0.

Crossrefs

Cf. A101481, A304185, A304186, A304187, A304189, A101479.

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

Formula

[x^n] (1+x)^(n*(n+1)/2) / A(x) = A101481(n+1) = A101479(n+1,0) for n>=0.
[x^n] (1+x)^((n+1)*(n+2)/2) / A(x) = Sum_{k=0..n} A101479(n+2,k+1) * A101479(k+1,0) for n>=0.

A304186 G.f. A(x) satisfies: 0 = [x^n] (1+x)^((n+1)*(n+2)/2) / A(x) for n>0.

Original entry on oeis.org

1, 3, 6, 24, 189, 2199, 33495, 625743, 13778469, 348258723, 9916111584, 313642702743, 10901017499472, 412730651160567, 16902257604014685, 744247858158013245, 35058132248539742325, 1758994171367292095805, 93646661943861501833100, 5272500660870261306912750
Offset: 0

Views

Author

Paul D. Hanna, May 08 2018

Keywords

Examples

			G.f.: A(x) = 1 + 3*x + 6*x^2 + 24*x^3 + 189*x^4 + 2199*x^5 + 33495*x^6 + 625743*x^7 + 13778469*x^8 + 348258723*x^9 + 9916111584*x^10 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^((n+1)*(n+2)/2) / A(x) begins:
n=0: [1, -2, 0, -12, -105, -1434, -23877, -473730, -10881882, ...];
n=1: [1, 0, -3, -14, -129, -1656, -26850, -522918, -11853219, ...];
n=2: [1, 3, 0, -22, -180, -2088, -32219, -608565, -13504179, ...];
n=3: [1, 7, 18, 0, -255, -2937, -41739, -750711, -16140285, ...];
n=4: [1, 12, 63, 170, 0, -3996, -58877, -991308, -20341875, ...];
n=5: [1, 18, 150, 748, 2220, 0, -78435, -1401570, -27251715, ...];
n=6: [1, 25, 297, 2211, 11271, 37149, 0, -1843458, -38615364, ...];
n=7: [1, 33, 525, 5343, 38745, 207663, 758814, 0, -50361381, ...];
n=8: [1, 42, 858, 11340, 108630, 797100, 4541805, 18301950, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that 0 = [x^n] (1+x)^((n+1)*(n+2)/2) / A(x) for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 3, 18, 170, 2220, 37149, 758814, 18301950, 508907970, ...]
yields A101483, column 2 of triangle A101479.
Related triangular matrix T = A101479 begins:
1;
1, 1;
1, 1, 1;
3, 2, 1, 1;
19, 9, 3, 1, 1;
191, 70, 18, 4, 1, 1;
2646, 795, 170, 30, 5, 1, 1;
46737, 11961, 2220, 335, 45, 6, 1, 1;
1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1; ...
in which row n equals row (n-1) of T^(n-1) followed by '1' for n>0.

Crossrefs

Cf. A101481, A304184, A304185, A304187, A101479.

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

Formula

[x^n] (1+x)^((n+2)*(n+3)/2) / A(x) = A101483(n+1) = A101479(n+3,2) for n>=0.
[x^n] (1+x)^((n+3)*(n+4)/2) / A(x) = Sum_{k=0..n} A101479(n+4,k+3) * A101479(k+3,2) for n>=0.

A304187 G.f. A(x) satisfies: 0 = [x^n] (1+x)^((n+2)*(n+3)/2) / A(x) for n>0.

Original entry on oeis.org

1, 6, 21, 86, 606, 6756, 102316, 1931046, 43250376, 1114876536, 32394654066, 1045240099026, 37027935179016, 1427410628324856, 59449956448178106, 2659215814433980056, 127108810653344820456, 6464722863550156435146, 348541208165221134718986, 19854709880058367829287716, 1191556960000156185148449636
Offset: 0

Views

Author

Paul D. Hanna, May 08 2018

Keywords

Examples

			G.f.: A(x) = 1 + 6*x + 21*x^2 + 86*x^3 + 606*x^4 + 6756*x^5 + 102316*x^6 + 1931046*x^7 + 43250376*x^8 + 1114876536*x^9 + 32394654066*x^10 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^((n+1)*(n+2)/2) / A(x) begins:
n=0: [1, -3, 0, -22, -216, -3180, -56540, -1186170, -28599870, ...];
n=1: [1, 0, -6, -30, -285, -3894, -66750, -1365546, -32331180, ...];
n=2: [1, 4, 0, -50, -440, -5238, -84162, -1657080, -38209725, ...];
n=3: [1, 9, 30, 0, -645, -7917, -115248, -2134920, -47391375, ...];
n=4: [1, 15, 99, 335, 0, -11046, -171920, -2957874, -62097600, ...];
n=5: [1, 22, 225, 1378, 4984, 0, -233730, -4379370, -86791905, ...];
n=6: [1, 30, 429, 3850, 23610, 92652, 0, -5860422, -127938780, ...];
n=7: [1, 39, 735, 8875, 76350, 483684, 2065146, 0, -169402725, ...];
n=8: [1, 49, 1170, 18100, 203065, 1743201, 11567124, 53636520, 0, ...]; ...
in which the main diagonal is all zeros after the initial term, illustrating that 0 = [x^n] (1+x)^((n+2)*(n+3)/2) / A(x) for n>0.
RELATED SEQUENCES.
The secondary diagonal in the above table that begins
[1, 4, 30, 335, 4984, 92652, 2065146, 53636520, 1589752230, ...]
yields column 3 of triangle A101479.
Related triangular matrix T = A101479 begins:
1;
1, 1;
1, 1, 1;
3, 2, 1, 1;
19, 9, 3, 1, 1;
191, 70, 18, 4, 1, 1;
2646, 795, 170, 30, 5, 1, 1;
46737, 11961, 2220, 335, 45, 6, 1, 1;
1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1; ...
in which row n equals row (n-1) of T^(n-1) followed by '1' for n>0.

Crossrefs

Cf. A101481, A304184, A304185, A304186, A101479.

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

Formula

A101479(n+4,3) = [x^n] (1+x)^((n+3)*(n+4)/2) / A(x) for n>=0.
Showing 1-4 of 4 results.

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