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 1-7 of 7 results.

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])}

A135890 Column 1 of triangle A135888, which equals the matrix cube of triangle A135880; also equals column 1 of triangle A135895.

Original entry on oeis.org

1, 6, 39, 300, 2741, 29380, 363922, 5135894, 81557270, 1441771540, 28114817877, 600012111858, 13919315033624, 348932593149877, 9403371859278914, 271183690566871863, 8335374900994682248, 272083236017290793444
Offset: 0

Views

Author

Paul D. Hanna, Dec 15 2007

Keywords

Crossrefs

Cf. A135888, A135889; A135880, A135895.

Programs

  • PARI
    {a(n)=local(P=Mat(1),R,PShR);if(n==0,1,for(i=0,n+1, 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+2,2])}

A135880 Triangle P, read by rows, 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 one place left, with P(0,0)=1.

Original entry on oeis.org

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, 8390, 16220, 8057, 2171, 400, 57, 7, 1, 86796, 182714, 93627, 25628, 4740, 666, 77, 8, 1, 1049546, 2378780, 1252752, 348050, 64805, 9080, 1029, 100, 9, 1, 14563135
Offset: 0

Views

Author

Paul D. Hanna, Dec 15 2007

Keywords

Comments

Amazingly, column 0 (A135881) also equals column 0 of tables A135878 and A135879, both of which have unusual recurrences seemingly unrelated to this triangle.

Examples

			Triangle P 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;
8390, 16220, 8057, 2171, 400, 57, 7, 1;
86796, 182714, 93627, 25628, 4740, 666, 77, 8, 1;
1049546, 2378780, 1252752, 348050, 64805, 9080, 1029, 100, 9,
1;
14563135, 35219202, 19003467, 5352788, 1004176, 140908, 15855,
1504, 126, 10, 1;
where column k of P equals column 0 of R^(k+1) where R =
A135894.
Triangle Q = P^2 = A135885 begins:
1;
2, 1;
6, 4, 1;
25, 20, 6, 1;
138, 126, 42, 8, 1;
970, 980, 351, 72, 10, 1;
8390, 9186, 3470, 748, 110, 12, 1;
86796, 101492, 39968, 8936, 1365, 156, 14, 1;
1049546, 1296934, 528306, 121532, 19090, 2250, 210, 16, 1; ...
where column k of Q equals column 0 of Q^(k+1) for k>=0;
thus column k of P^2 equals column 0 of P^(2k+2).
Triangle 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;
8390, 34380, 20340, 5733, 1026, 132, 13, 1;
86796, 399463, 247066, 72030, 13305, 1771, 182, 15, 1; ...
where column k of R equals column 0 of P^(2k+1) for k>=0.
Surprisingly, column 0 is also found in triangle A135879:
1;
1,1;
2,2,1,1;
6,6,4,4,2,2,1;
25,25,19,19,13,13,9,5,5,3,1,1;
138,138,113,113,88,88,69,50,50,37,24,24,15,10,5,5,2,1; ...
and is generated by a process that seems completely unrelated.

Crossrefs

Cf. columns: A135881, A135882, A135883, A135884.
Cf. related tables: A135885 (Q=P^2), A135894 (R).
Cf. A135888 (P^3), A135891 (P^4), A135892 (P^5), A135893 (P^6).
Cf. A135898 (P^-1*R), A135899 (P*R^-1*P), A135900 (R^-1*Q).
Cf. A135878, A135879.

Programs

  • PARI
    {T(n,k)=local(P=Mat(1),R,PShR);if(n>0,for(n=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[n+1,k+1]}

Formula

Denote this triangle by P and define as follows.
Let [P^m]_k denote column k of matrix power P^m,
so that triangular matrix Q = A135885 may be defined by
[Q]_k = [P^(2k+2)]_0, for k>=0, such that
(1) Q = P^2 and (2) [Q]_0 = [P]_0 shifted left.
Define the dual triangular matrix R = A135894 by
[R]_k = [P^(2k+1)]_0, for k>=0.
Then columns of P may be formed from powers of R:
[P]_k = [R^(k+1)]_0, for k>=0.
Further, columns of powers of P, Q and R satisfy:
[R^(j+1)]_k = [P^(2k+1)]_j,
[Q^(j+1)]_k = [P^(2k+2)]_j,
[Q^(j+1)]_k = [Q^(k+1)]_j,
[P^(2j+2)]_k = [P^(2k+2)]_j, for all j>=0, k>=0.
Also, we have the column transformations:
R * [P]k = [P]{k+1},
Q * [Q]k = [Q]{k+1},
Q * [R]k = [R]{k+1},
P^2 * [Q]k = [Q]{k+1},
P^2 * [R]k = [R]{k+1}, for all k>=0.
Other identities include the matrix products:
P^-1*R (A135898) = P shifted right one column;
P*R^-1*P (A135899) = Q shifted down one row;
R^-1*Q (A135900) = R shifted down one row.

A135894 Triangle R, read by rows, where column k of R equals column 0 of P^(2k+1) where P=A135880.

Original entry on oeis.org

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, 8390, 34380, 20340, 5733, 1026, 132, 13, 1, 86796, 399463, 247066, 72030, 13305, 1771, 182, 15, 1, 1049546, 5344770, 3430936, 1028076, 194646, 26565, 2808
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 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;
8390, 34380, 20340, 5733, 1026, 132, 13, 1;
86796, 399463, 247066, 72030, 13305, 1771, 182, 15, 1;
1049546, 5344770, 3430936, 1028076, 194646, 26565, 2808, 240, 17, 1;
14563135, 81097517, 53741404, 16477041, 3182778, 442948, 47801, 4185, 306, 19, 1; ...
where column k of R equals column 0 of P^(2k+1) for k>=0,
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).
The matrix product P^-1*R = A135898 = P (shifted right one column);
the matrix product R^-1*P^2 = A135900 = R (shifted down one row).

Crossrefs

Cf. A135881 (column 0), A135889 (column 1); A135880 (P), A135885 (Q=P^2), A135895 (R^2), A135896 (R^3), A135897 (R^4); A135888 (P^3) A135892 (P^5); A135898 (P^-1*R), A135899 (P*R^-1*P), A135900 (R^-1*Q).

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[n+1,k+1]}

Formula

Column k of R = column 0 of P^(2k+1) for k>=0 where triangle P = A135880; column 0 of R = column 0 of P; column 1 of R = column 0 of P^3; column 2 of R = column 0 of P^5. See more formulas relating triangles P, Q and R, in entry A135880.

Extensions

Typo in formula corrected by Paul D. Hanna, Mar 26 2010

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.

A135896 Triangle, read by rows, equal to R^3, the matrix cube of R = A135894.

Original entry on oeis.org

1, 3, 1, 15, 9, 1, 99, 81, 15, 1, 814, 816, 195, 21, 1, 8057, 9366, 2625, 357, 27, 1, 93627, 122148, 38270, 6006, 567, 33, 1, 1252752, 1795481, 611525, 105910, 11439, 825, 39, 1, 19003467, 29478724, 10721093, 1996988, 236430, 19404, 1131, 45, 1, 322722064
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^3 begins:
1;
3, 1;
15, 9, 1;
99, 81, 15, 1;
814, 816, 195, 21, 1;
8057, 9366, 2625, 357, 27, 1;
93627, 122148, 38270, 6006, 567, 33, 1;
1252752, 1795481, 611525, 105910, 11439, 825, 39, 1;
19003467, 29478724, 10721093, 1996988, 236430, 19404, 1131, 45, 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. A135883 (column 0); 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^3)[n+1,k+1]}

Formula

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

A135897 Triangle, read by rows, equal to R^4, the matrix 4th power of R = A135894.

Original entry on oeis.org

1, 4, 1, 26, 12, 1, 216, 138, 20, 1, 2171, 1716, 330, 28, 1, 25628, 23647, 5440, 602, 36, 1, 348050, 362116, 94515, 12348, 954, 44, 1, 5352788, 6138746, 1761940, 258391, 23400, 1386, 52, 1, 92056223, 114543428, 35429974, 5662412, 572331, 39556, 1898
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^4 begins:
1;
4, 1;
26, 12, 1;
216, 138, 20, 1;
2171, 1716, 330, 28, 1;
25628, 23647, 5440, 602, 36, 1;
348050, 362116, 94515, 12348, 954, 44, 1;
5352788, 6138746, 1761940, 258391, 23400, 1386, 52, 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. A135884 (column 0); 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^4)[n+1,k+1]}

Formula

Column k of R^4 = column 3 of P^(2k+1) for k>=0 where triangle P = A135880; column 0 of R^4 = column 3 of P; column 1 of R^4 = column 3 of P^3; column 2 of R^4 = column 3 of P^5.
Showing 1-7 of 7 results.

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