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 1-3 of 3 results.

A135888 Triangle, read by rows, equal to the matrix cube of triangle P = A135880.

Original entry on oeis.org

1, 3, 1, 12, 6, 1, 63, 39, 9, 1, 421, 300, 81, 12, 1, 3472, 2741, 816, 138, 15, 1, 34380, 29380, 9366, 1716, 210, 18, 1, 399463, 363922, 122148, 23647, 3105, 297, 21, 1, 5344770, 5135894, 1795481, 362116, 49880, 5088, 399, 24, 1, 81097517, 81557270
Offset: 0

Views

Author

Paul D. Hanna, Dec 15 2007

Keywords

Comments

Matrix square equals triangle A135893.

Examples

			Triangle P^3 begins:
1;
3, 1;
12, 6, 1;
63, 39, 9, 1;
421, 300, 81, 12, 1;
3472, 2741, 816, 138, 15, 1;
34380, 29380, 9366, 1716, 210, 18, 1;
399463, 363922, 122148, 23647, 3105, 297, 21, 1;
5344770, 5135894, 1795481, 362116, 49880, 5088, 399, 24, 1;
81097517, 81557270, 29478724, 6138746, 875935, 93306, 7770, 516, 27, 1;
where P = A135880 begins:
1;
1, 1;
2, 2, 1;
6, 7, 3, 1;
25, 34, 15, 4, 1;
138, 215, 99, 26, 5, 1;
970, 1698, 814, 216, 40, 6, 1; ...
where column k of P^2 equals column 0 of P^(2k+2)
such that column 0 of P^2 equals column 0 of P shift left.

Crossrefs

Cf. columns: A135889, A135890; related tables: A135880 (P), A135894 (R), A135893 (P^6).

Programs

  • PARI
    {T(n,k)=local(P=Mat(1),R,PShR);if(n>0,for(i=0,n, PShR=matrix(#P,#P, r,c, if(r>=c,if(r==c,1,if(c==1,0,P[r-1,c-1]))));R=P*PShR; R=matrix(#P+1, #P+1, r,c, if(r>=c, if(r<#P+1,R[r,c], if(c==1,(P^2)[ #P,1],(P^(2*c-1))[r-c+1,1])))); P=matrix(#R, #R, r,c, if(r>=c, if(r<#R,P[r,c], (R^c)[r-c+1,1])))));(P^3)[n+1,k+1]}

A135889 Column 0 of triangle A135888, which equals the matrix cube of triangle A135880; also equals column 1 of triangle A135894.

Original entry on oeis.org

1, 3, 12, 63, 421, 3472, 34380, 399463, 5344770, 81097517, 1377986373, 25947738574, 536726987593, 12104879913657, 295754724799758, 7784814503249896, 219682110287448760, 6617691928179590112
Offset: 0

Views

Author

Paul D. Hanna, Dec 15 2007

Keywords

Crossrefs

Cf. A135888, A135890; A135880, A135894.

Programs

  • PARI
    {a(n)=local(P=Mat(1),R,PShR);if(n==0,1,for(i=0,n, PShR=matrix(#P,#P, r,c, if(r>=c,if(r==c,1,if(c==1,0,P[r-1,c-1]))));R=P*PShR; R=matrix(#P+1, #P+1, r,c, if(r>=c, if(r<#P+1,R[r,c], if(c==1,(P^2)[ #P,1],(P^(2*c-1))[r-c+1,1])))); P=matrix(#R, #R, r,c, if(r>=c, if(r<#R,P[r,c], (R^c)[r-c+1,1]))));(P^3)[n+1,1])}

A135895 Triangle, read by rows, equal to R^2, the matrix square of R = A135894.

Original entry on oeis.org

1, 2, 1, 7, 6, 1, 34, 39, 10, 1, 215, 300, 95, 14, 1, 1698, 2741, 990, 175, 18, 1, 16220, 29380, 11635, 2296, 279, 22, 1, 182714, 363922, 154450, 32865, 4410, 407, 26, 1, 2378780, 5135894, 2302142, 517916, 74319, 7524, 559, 30, 1, 35219202, 81557270
Offset: 0

Views

Author

Paul D. Hanna, Dec 15 2007

Keywords

Comments

Triangle P = A135880 is defined by: column k of P^2 equals column 0 of P^(2k+2) such that column 0 of P^2 equals column 0 of P shift left.

Examples

			Triangle R^2 begins:
1;
2, 1;
7, 6, 1;
34, 39, 10, 1;
215, 300, 95, 14, 1;
1698, 2741, 990, 175, 18, 1;
16220, 29380, 11635, 2296, 279, 22, 1;
182714, 363922, 154450, 32865, 4410, 407, 26, 1;
2378780, 5135894, 2302142, 517916, 74319, 7524, 559, 30, 1;
35219202, 81557270, 38229214, 8980944, 1353522, 145805, 11830, 735, 34, 1;
where R = A135894 begins:
1;
1, 1;
2, 3, 1;
6, 12, 5, 1;
25, 63, 30, 7, 1;
138, 421, 220, 56, 9, 1;
970, 3472, 1945, 525, 90, 11, 1; ...
where column k of R = column 0 of P^(2k+1)
and P = A135880 begins:
1;
1, 1;
2, 2, 1;
6, 7, 3, 1;
25, 34, 15, 4, 1;
138, 215, 99, 26, 5, 1;
970, 1698, 814, 216, 40, 6, 1; ...
where column k of P equals column 0 of R^(k+1).

Crossrefs

Cf. A135882 (column 0), A135890 (column 1); A135894 (R), A135880 (P), A135888 (P^3), A135892 (P^5).

Programs

  • PARI
    {T(n,k)=local(P=Mat(1),R=Mat(1),PShR);if(n>0,for(i=0,n, PShR=matrix(#P,#P, r,c, if(r>=c,if(r==c,1,if(c==1,0,P[r-1,c-1]))));R=P*PShR; R=matrix(#P+1, #P+1, r,c, if(r>=c, if(r<#P+1,R[r,c], if(c==1,(P^2)[ #P,1],(P^(2*c-1))[r-c+1,1])))); P=matrix(#R, #R, r,c, if(r>=c, if(r<#R,P[r,c], (R^c)[r-c+1,1])))));(R^2)[n+1,k+1]}

Formula

Column k of R^2 = column 1 of P^(2k+1) for k>=0 where triangle P = A135880; column 0 of R^2 = column 1 of P; column 1 of R^2 = column 1 of P^3; column 2 of R^2 = column 1 of P^5.
Showing 1-3 of 3 results.

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