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 1-5 of 5 results.

A113393 Column 1 of triangle A113392, also equals column 0 of A113381^6.

Original entry on oeis.org

1, 6, 48, 605, 11196, 280440, 8981460, 353283128, 16567072675, 905357065354, 56632746126107, 3997082539456084, 314584709388906568, 27340439653453247728, 2602372304420672868499, 269388182085308601450047
Offset: 0

Views

Author

Paul D. Hanna, Nov 14 2005

Keywords

Crossrefs

Cf. A113392, A113389, A113381, A113388 (column 0).

Programs

  • PARI
    a(n)=local(A,B);A=Mat(1);for(m=2,n+1,B=matrix(m,m); for(i=1,m, for(j=1,i,if(i<3 || j==i || j>m-1,B[i,j]=1,if(j==1, B[i,1]=1,B[i,j]=(A^(3*j-2))[i-j+1,1]));));A=B); (matrix(#A,#A,r,c,if(r>=c,(A^(3*c))[r-c+1,1]))^2)[n+1,1]

Formula

A113392 equals the matrix square of A113389, which has the property: Column k of A113389^2 = column 0 of A113381^(3*k+3) for k>=0.

A113370 Triangle P, read by rows, such that P^3 transforms column k of P into column k+1 of P, so that column k of P equals column 0 of P^(3*k+1), where P^3 denotes the matrix cube of P.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 28, 7, 1, 1, 326, 91, 10, 1, 1, 5702, 1722, 190, 13, 1, 1, 136724, 43764, 4945, 325, 16, 1, 1, 4226334, 1415799, 163705, 10751, 496, 19, 1, 1, 161385532, 56096733, 6617605, 437723, 19896, 703, 22, 1, 1, 7378504140, 2644883675
Offset: 0

Views

Author

Paul D. Hanna, Nov 14 2005

Keywords

Comments

Triangle A114150 illustrates the identity: R^2*Q^-1 = Q^3*P^-2.
See also A114152 for the matrix product: R^3*P^-1.

Examples

			Triangle P begins:
1;
1,1;
1,4,1;
1,28,7,1;
1,326,91,10,1;
1,5702,1722,190,13,1;
1,136724,43764,4945,325,16,1;
1,4226334,1415799,163705,10751,496,19,1;
1,161385532,56096733,6617605,437723,19896,703,22,1;
1,7378504140,2644883675,317416204,21179483,960696,33136,946,25,1;
Matrix cube P^3 (A113378) starts:
1;
3,1;
15,12,1;
136,168,21,1;
1998,3190,483,30,1;
41973,80136,13615,960,39,1; ...
where P^3 transforms column k of P into column k+1 of P:
at k=0, [P^3]*[1,1,1,1,1,...] = [1,4,28,326,5702,...];
at k=1, [P^3]*[1,4,28,326,5702,...] = [1,7,91,1722,43764,...].

Crossrefs

Cf. A113371 (column 1), A113372 (column 2), A113373 (column 3).
Cf. A113374 (P^2), A113378 (P^3), A113381 (Q), A113384 (Q^2), A113387 (Q^3), A113389 (R), A113392 (R^2), A113394 (R^3), A114156 (P^-1).
Cf. A114150 (R^2*Q^-1=Q^3*P^-2), A114152 (R^3*P^-1).
Cf. variants: A113340, A113350.

Programs

  • PARI
    P(n,k)=local(A,B);A=Mat(1);for(m=2,n+1,B=matrix(m,m); for(i=1,m, for(j=1,i,if(i<3 || j==i || j>m-1,B[i,j]=1,if(j==1, B[i,1]=1,B[i,j]=(A^(3*j-2))[i-j+1,1]));));A=B);A[n+1,k+1]

Formula

Let [P^m]_k denote column k of matrix power P^m,
so that triangular matrix P may be defined by
[P]_k = [P^(3*k+1)]_0, k>=0.
Define the triangular matrix Q = A113381 by
[Q]_k = [P^(3*k+2)]_0, k>=0.
Define the triangular matrix R = A113389 by
[R]_k = [P^(3*k+3)]_0, k>=0.
Then P, Q and R are related by:
Q^2 = R*P = R*Q*(R^-2)*Q*R = P*Q*(P^-2)*Q*P,
P^2 = Q*(R^-2)*Q^3, R^2 = Q^3*(P^-2)*Q.
Amazingly, columns in powers of P, Q, R, obey:
[P^(3*j+1)]_k = [P^(3*k+1)]_j,
[Q^(3*j+1)]_k = [P^(3*k+2)]_j,
[R^(3*j+1)]_k = [P^(3*k+3)]_j,
[Q^(3*j+2)]_k = [Q^(3*k+2)]_j,
[R^(3*j+2)]_k = [Q^(3*k+3)]_j,
[R^(3*j+3)]_k = [R^(3*k+3)]_j,
for all j>=0, k>=0.
Also, we have the column transformations:
P^3 * [P]k = [P]{k+1},
P^3 * [Q]k = [Q]{k+1},
P^3 * [R]k = [R]{k+1},
Q^3 * [P^2]k = [P^2]{k+1},
Q^3 * [Q^2]k = [Q^2]{k+1},
Q^3 * [R^2]k = [R^2]{k+1},
R^3 * [P^3]k = [P^3]{k+1},
R^3 * [Q^3]k = [Q^3]{k+1},
R^3 * [R^3]k = [R^3]{k+1},
for all k>=0.

A113381 Triangle Q, read by rows, such that Q^3 transforms column k of Q^2 into column k+1 of Q^2, so that column k of Q^2 equals column 0 of Q^(3*k+2), where Q^3 denotes the matrix cube of Q.

Original entry on oeis.org

1, 2, 1, 6, 5, 1, 37, 45, 8, 1, 429, 635, 120, 11, 1, 7629, 12815, 2556, 231, 14, 1, 185776, 343815, 71548, 6556, 378, 17, 1, 5817106, 11651427, 2508528, 233706, 13391, 561, 20, 1, 224558216, 480718723, 106427700, 10069521, 579047, 23817, 780, 23, 1
Offset: 0

Views

Author

Paul D. Hanna, Nov 14 2005

Keywords

Comments

Related matrix products are: R^3*Q^-2 (A114154), Q^-2*P^3 (A114155).

Examples

			Triangle Q begins:
1;
2,1;
6,5,1;
37,45,8,1;
429,635,120,11,1;
7629,12815,2556,231,14,1;
185776,343815,71548,6556,378,17,1;
5817106,11651427,2508528,233706,13391,561,20,1;
224558216,480718723,106427700,10069521,579047,23817,780,23,1;
Matrix square Q^2 (A113384) starts:
1;
4,1;
22,10,1;
212,130,16,1;
3255,2365,328,22,1;
70777,57695,8640,616,28,1; ...
Matrix cube Q^3 (A113387) starts:
1;
6,1;
48,15,1;
605,255,24,1;
11196,5630,624,33,1;
280440,159210,19484,1155,42,1; ...
where Q^3 transforms column k of Q^2 into column k+1:
at k=0, [Q^3]*[1,4,22,212,3255,...] = [1,10,130,2365,...];
at k=1, [Q^3]*[1,10,130,2365,...] = [1,16,328,8640,...].

Crossrefs

Cf. A113375 (column 0), A113382 (column 1), A113383 (column 2).
Cf. A113370 (P), A113374 (P^2), A113378 (P^3), A113384 (Q^2), A113387 (Q^3), A113389 (R), A113392 (R^2), A113394 (R^3).
Cf. A114154 (R^3*Q^-2), A114155 (Q^-2*P^3).
Cf. variants: A113340, A113350.

Programs

  • PARI
    Q(n,k)=local(A,B);A=Mat(1);for(m=2,n+1,B=matrix(m,m); for(i=1,m, for(j=1,i,if(i<3 || j==i || j>m-1,B[i,j]=1,if(j==1, B[i,1]=1,B[i,j]=(A^(3*j-2))[i-j+1,1]));));A=B);(A^(3*k+2))[n-k+1,1]

Formula

Let [Q^m]_k denote column k of matrix power Q^m,
so that triangular matrix Q may be defined by
[Q]_k = [P^(3*k+2)]_0, k>=0,
where the triangular matrix P = A113370 satisfies:
[P]_k = [P^(3*k+1)]_0, k>=0.
Define the triangular matrix R = A113389 by
[R]_k = [P^(3*k+3)]_0, k>=0.
Then P, Q and R are related by:
Q^2 = R*P = R*Q*(R^-2)*Q*R = P*Q*(P^-2)*Q*P,
P^2 = Q*(R^-2)*Q^3, R^2 = Q^3*(P^-2)*Q.
Amazingly, columns in powers of P, Q, R, obey:
[P^(3*j+1)]_k = [P^(3*k+1)]_j,
[Q^(3*j+1)]_k = [P^(3*k+2)]_j,
[R^(3*j+1)]_k = [P^(3*k+3)]_j,
[Q^(3*j+2)]_k = [Q^(3*k+2)]_j,
[R^(3*j+2)]_k = [Q^(3*k+3)]_j,
[R^(3*j+3)]_k = [R^(3*k+3)]_j,
for all j>=0, k>=0.
Also, we have the column transformations:
P^3 * [P]k = [P]{k+1},
P^3 * [Q]k = [Q]{k+1},
P^3 * [R]k = [R]{k+1},
Q^3 * [P^2]k = [P^2]{k+1},
Q^3 * [Q^2]k = [Q^2]{k+1},
Q^3 * [R^2]k = [R^2]{k+1},
R^3 * [P^3]k = [P^3]{k+1},
R^3 * [Q^3]k = [Q^3]{k+1},
R^3 * [R^3]k = [R^3]{k+1},
for all k>=0.

A113389 Triangle R, read by rows, such that R^3 transforms column k of R^3 into column k+1 of R^3, so that column k of R^3 equals column 0 of R^(3*k+3), where R^3 denotes the matrix cube of R.

Original entry on oeis.org

1, 3, 1, 15, 6, 1, 136, 66, 9, 1, 1998, 1091, 153, 12, 1, 41973, 24891, 3621, 276, 15, 1, 1166263, 737061, 110637, 8482, 435, 18, 1, 40747561, 27110418, 4176549, 323874, 16430, 630, 21, 1, 1726907675, 1199197442, 188802141, 14813844, 751920, 28221
Offset: 0

Views

Author

Paul D. Hanna, Nov 14 2005

Keywords

Comments

Related matrix products: identity R^-2*Q^3 = Q^-1*P^2 (A114151) and R^-1*P^3 (A114153).

Examples

			Triangle R begins:
1;
3,1;
15,6,1;
136,66,9,1;
1998,1091,153,12,1;
41973,24891,3621,276,15,1;
1166263,737061,110637,8482,435,18,1;
40747561,27110418,4176549,323874,16430,630,21,1;
1726907675,1199197442,188802141,14813844,751920,28221,861,24,1;
Matrix cube R^3 (A113394) starts:
1;
9,1;
99,18,1;
1569,360,27,1;
34344,9051,783,36,1;
980487,284148,26820,1368,45,1; ...
where R^3 transforms column k of R^3 into column k+1:
at k=0, [R^3]*[1,9,99,1569,...] = [1,18,360,9051,...];
at k=1, [R^3]*[1,18,360,9051,..] = [1,27,783,26820,..].

Crossrefs

Cf. A113379 (column 0), A113390 (column 1), A113391 (column 2).
Cf. A113370 (P), A113374 (P^2), A113378 (P^3), A113381 (Q), A113384 (Q^2), A113387 (Q^3), A113392 (R^2), A113394 (R^3).
Cf. A114151 (R^-2*Q^3 = Q^-1*P^2), A114153 (R^-1*P^3).
Cf. variants: A113340, A113350.

Programs

  • PARI
    R(n,k)=local(A,B);A=Mat(1);for(m=2,n+1,B=matrix(m,m); for(i=1,m, for(j=1,i,if(i<3 || j==i || j>m-1,B[i,j]=1,if(j==1, B[i,1]=1,B[i,j]=(A^(3*j-2))[i-j+1,1]));));A=B);(A^(3*k+3))[n-k+1,1]

Formula

Let [R^m]_k denote column k of matrix power R^m,
so that triangular matrix R may be defined by
[R]_k = [P^(3*k+3)]_0, k>=0,
where the triangular matrix P = A113370 satisfies:
[P]_k = [P^(3*k+1)]_0, k>=0.
Define the triangular matrix Q = A113381 by
[Q]_k = [P^(3*k+2)]_0, k>=0.
Then P, Q and R are related by:
Q^2 = R*P = R*Q*(R^-2)*Q*R = P*Q*(P^-2)*Q*P,
P^2 = Q*(R^-2)*Q^3, R^2 = Q^3*(P^-2)*Q.
Amazingly, columns in powers of P, Q, R, obey:
[P^(3*j+1)]_k = [P^(3*k+1)]_j,
[Q^(3*j+1)]_k = [P^(3*k+2)]_j,
[R^(3*j+1)]_k = [P^(3*k+3)]_j,
[Q^(3*j+2)]_k = [Q^(3*k+2)]_j,
[R^(3*j+2)]_k = [Q^(3*k+3)]_j,
[R^(3*j+3)]_k = [R^(3*k+3)]_j,
for all j>=0, k>=0.
Also, we have the column transformations:
P^3 * [P]k = [P]{k+1},
P^3 * [Q]k = [Q]{k+1},
P^3 * [R]k = [R]{k+1},
Q^3 * [P^2]k = [P^2]{k+1},
Q^3 * [Q^2]k = [Q^2]{k+1},
Q^3 * [R^2]k = [R^2]{k+1},
R^3 * [P^3]k = [P^3]{k+1},
R^3 * [Q^3]k = [Q^3]{k+1},
R^3 * [R^3]k = [R^3]{k+1},
for all k>=0.

A114153 Triangle, read by rows, given by the product R^-1*P^3 using triangular matrices P=A113370, R=A113389.

Original entry on oeis.org

1, 0, 1, 0, 6, 1, 0, 48, 12, 1, 0, 605, 186, 18, 1, 0, 11196, 3892, 414, 24, 1, 0, 280440, 106089, 12021, 732, 30, 1, 0, 8981460, 3620379, 429345, 27152, 1140, 36, 1, 0, 353283128, 149740555, 18386361, 1196910, 51445, 1638, 42, 1
Offset: 0

Views

Author

Paul D. Hanna, Nov 15 2005

Keywords

Comments

Complementary to A114152, which gives R^3*P^-1.

Examples

			Triangle R^-1*P^3 begins:
1;
0,1;
0,6,1;
0,48,12,1;
0,605,186,18,1;
0,11196,3892,414,24,1;
0,280440,106089,12021,732,30,1; ...
Compare to R^2 (A113392):
1;
6,1;
48,12,1;
605,186,18,1;
11196,3892,414,24,1;
280440,106089,12021,732,30,1; ...
Thus R^-1*P^3 equals R^2 shift right one column.

Crossrefs

Cf. A113392 (R^2), A113370 (P), A113381 (Q), A113389 (R); A114150 (R^2*Q^-1=Q^3*P^-2), A114151 (R^-2*Q^3=Q^-1*P^2), A114152 (R^3*P^-1), A114154 (R^3*Q^-2), A114155 (Q^-2*P^3); A114156 (P^-1), A114158 (Q^-1), A114159 (R^-1).

Programs

  • PARI
    T(n,k)=local(P,Q,R,W);P=Mat(1);for(m=2,n+1,W=matrix(m,m); for(i=1,m, for(j=1,i,if(i<3 || j==i || j>m-1,W[i,j]=1,if(j==1, W[i,1]=1,W[i,j]=(P^(3*j-2))[i-j+1,1]));));P=W); Q=matrix(#P,#P,r,c,if(r>=c,(P^(3*c-1))[r-c+1,1])); R=matrix(#P,#P,r,c,if(r>=c,(P^(3*c))[r-c+1,1])); (R^-1*P^3)[n+1,k+1]
Showing 1-5 of 5 results.

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