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.

Previous Showing 11-18 of 18 results.

A078536 Infinite lower triangular matrix, M, that satisfies [M^4](i,j) = M(i+1,j+1) for all i,j>=0 where [M^n](i,j) denotes the element at row i, column j, of the n-th power of matrix M, with M(0,k)=1 and M(k,k)=1 for all k>=0.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 28, 16, 1, 1, 524, 496, 64, 1, 1, 29804, 41136, 8128, 256, 1, 1, 5423660, 10272816, 2755264, 130816, 1024, 1, 1, 3276048300, 8220685104, 2804672704, 178301696, 2096128, 4096, 1, 1, 6744720496300, 21934062166320, 9139625620672, 729250931456, 11442760704, 33550336, 16384, 1
Offset: 0

Views

Author

Paul D. Hanna, Nov 29 2002

Keywords

Comments

M also satisfies: [M^(4k)](i,j) = [M^k](i+1,j+1) for all i,j,k>=0; thus [M^(4^n)](i,j) = M(i+n,j+n) for all n>=0. Conjecture: sum of the n-th row equals the partitions of 4^n into powers of 4.

Examples

			The 4th power of matrix is the same matrix excluding the first row and column:
[1,__0,__0,_0,0]^4=[____1,____0,___0,__0,0]
[1,__1,__0,_0,0]___[____4,____1,___0,__0,0]
[1,__4,__1,_0,0]___[___28,___16,___1,__0,0]
[1,_28,_16,_1,0]___[__524,__496,__64,__1,0]
[1,524,496,64,1]___[29804,41136,8128,256,1]

Crossrefs

Cf. A078121, A078122, A078537.

Programs

  • Mathematica
    dim = 9;
a[, 0] = 1; a[i, i_] = 1; a[i_, j_] /; j > i = 0;
M = Table[a[i, j], {i, 0, dim-1}, {j, 0, dim-1}];
M4 = MatrixPower[M, 4];
sol = Table[M4[[i, j]] == M[[i+1, j+1]], {i, 1, dim-1}, {j, 1, dim-1}] // Flatten // Solve;
Table[a[i, j], {i, 0, dim-1}, {j, 0, i}] /. sol // Flatten (* Jean-François Alcover, Oct 20 2019 *)

Formula

M(n, k) = the coefficient of x^(4^n - 4^(n-k)) in the power series expansion of 1/Product_{j=0..n-k}(1-x^(4^j)) whenever 0<=k0 (conjecture).

Extensions

More terms from Jean-François Alcover, Oct 20 2019

A111825 Triangle P, read by rows, that satisfies [P^6](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(6*m)](n,k) = [P^m](n+1,k+1) for all m, where [P^m](n,k) denotes the element at row n, column k, of the matrix power m of P, with P(0,k)=1 and P(k,k)=1 for all k>=0.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 96, 36, 1, 1, 6306, 3816, 216, 1, 1, 1883076, 1625436, 139536, 1296, 1, 1, 2700393702, 3121837776, 360839016, 5036256, 7776, 1, 1, 19324893252552, 28794284803908, 4200503990976, 78293629296, 181382976, 46656, 1
Offset: 0

Views

Author

Gottfried Helms and Paul D. Hanna, Aug 22 2005

Keywords

Comments

Also P(n,k) = the partitions of (6^n - 6^(n-k)) into powers of 6 <= 6^(n-k).

Examples

			Let q=6; the g.f. of column k of matrix power P^m is:
1 + (m*q^k)*L(x) + (m*q^k)^2/2!*L(x)*L(q*x) +
(m*q^k)^3/3!*L(x)*L(q*x)*L(q^2*x) +
(m*q^k)^4/4!*L(x)*L(q*x)*L(q^2*x)*L(q^3*x) + ...
where L(x) satisfies:
x/(1-x) = L(x) + L(x)*L(q*x)/2! + L(x)*L(q*x)*L(q^2*x)/3! + ...
and L(x) = x - 4/2!*x^2 + 42/3!*x^3 + 7296/4!*x^4 +... (A111829).
Thus the g.f. of column 0 of matrix power P^m is:
1 + m*L(x) + m^2/2!*L(x)*L(6*x) + m^3/3!*L(x)*L(6*x)*L(6^2*x) +
m^4/4!*L(x)*L(6*x)*L(6^2*x)*L(6^3*x) + ...
Triangle P begins:
1;
1,1;
1,6,1;
1,96,36,1;
1,6306,3816,216,1;
1,1883076,1625436,139536,1296,1;
1,2700393702,3121837776,360839016,5036256,7776,1; ...
where P^6 shifts columns left and up one place:
1;
6,1;
96,36,1;
6306,3816,216,1; ...

Crossrefs

Cf. A111826 (column 1), A111827 (row sums), A111828 (matrix log); triangles: A110503 (q=-1), A078121 (q=2), A078122 (q=3), A078536 (q=4), A111820 (q=5), A111830 (q=7), A111835 (q=8).

Programs

  • PARI
    P(n,k,q=6)=local(A=Mat(1),B);if(n

Formula

Let q=6; the g.f. of column k of P^m (ignoring leading zeros) equals: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} L(q^j*x) where L(x) satisfies: x/(1-x) = Sum_{n>=1} Product_{j=0..n-1} L(q^j*x)/(j+1) and L(x) equals the g.f. of column 0 of the matrix log of P (A111829).

A111820 Triangle P, read by rows, that satisfies [P^5](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(5*m)](n,k) = [P^m](n+1,k+1) for all m, where [P^m](n,k) denotes the element at row n, column k, of the matrix power m of P, with P(0,k)=1 and P(k,k)=1 for all k>=0.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 55, 25, 1, 1, 2055, 1525, 125, 1, 1, 291430, 311525, 38875, 625, 1, 1, 165397680, 239305275, 40338875, 975625, 3125, 1, 1, 390075741430, 735920617775, 157056792000, 5077475625, 24409375, 15625, 1
Offset: 0

Views

Author

Gottfried Helms and Paul D. Hanna, Aug 22 2005

Keywords

Comments

Also P(n,k) = the partitions of (5^n - 5^(n-k)) into powers of 5 <= 5^(n-k).

Examples

			Let q=5; the g.f. of column k of matrix power P^m is:
1 + (m*q^k)*L(x) + (m*q^k)^2/2!*L(x)*L(q*x) +
(m*q^k)^3/3!*L(x)*L(q*x)*L(q^2*x) +
(m*q^k)^4/4!*L(x)*L(q*x)*L(q^2*x)*L(q^3*x) + ...
where L(x) satisfies:
x/(1-x) = L(x) + L(x)*L(q*x)/2! + L(x)*L(q*x)*L(q^2*x)/3! + ...
and L(x) = x - 3/2!*x^2 + 16/3!*x^3 + 2814/4!*x^4 +... (A111824).
Thus the g.f. of column 0 of matrix power P^m is:
1 + m*L(x) + m^2/2!*L(x)*L(5*x) + m^3/3!*L(x)*L(5*x)*L(5^2*x) +
m^4/4!*L(x)*L(5*x)*L(5^2*x)*L(5^3*x) + ...
Triangle P begins:
1;
1,1;
1,5,1;
1,55,25,1;
1,2055,1525,125,1;
1,291430,311525,38875,625,1;
1,165397680,239305275,40338875,975625,3125,1; ...
where P^5 shifts columns left and up one place:
1;
5,1;
55,25,1;
2055,1525,125,1;
291430,311525,38875,625,1; ...

Crossrefs

Cf. A111821 (column 1), A111822 (row sums), A111823 (matrix log); triangles: A110503 (q=-1), A078121 (q=2), A078122 (q=3), A078536 (q=4), A111825 (q=6), A111830 (q=7), A111835 (q=8).

Programs

  • PARI
    P(n,k,q=5)=local(A=Mat(1),B);if(n

Formula

Let q=5; the g.f. of column k of P^m (ignoring leading zeros) equals: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} L(q^j*x) where L(x) satisfies: x/(1-x) = Sum_{n>=1} Product_{j=0..n-1} L(q^j*x)/(j+1) and L(x) equals the g.f. of column 0 of the matrix log of P (A111824).

A111830 Triangle P, read by rows, that satisfies [P^7](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(7*m)](n,k) = [P^m](n+1,k+1) for all m, where [P^m](n,k) denotes the element at row n, column k, of the matrix power m of P, with P(0,k)=1 and P(k,k)=1 for all k>=0.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 154, 49, 1, 1, 16275, 8281, 343, 1, 1, 9106461, 6558209, 410914, 2401, 1, 1, 28543862991, 27307109501, 2298650515, 20170801, 16807, 1, 1, 521136519414483, 636922972420469, 67522139062441, 790856748801, 988621354
Offset: 0

Views

Author

Gottfried Helms and Paul D. Hanna, Aug 22 2005

Keywords

Comments

Also P(n,k) = partitions of (7^n - 7^(n-k)) into powers of 7 <= 7^(n-k).

Examples

			Let q=7; the g.f. of column k of matrix power P^m is:
1 + (m*q^k)*L(x) + (m*q^k)^2/2!*L(x)*L(q*x) +
(m*q^k)^3/3!*L(x)*L(q*x)*L(q^2*x) +
(m*q^k)^4/4!*L(x)*L(q*x)*L(q^2*x)*L(q^3*x) + ...
where L(x) satisfies:
x/(1-x) = L(x) + L(x)*L(q*x)/2! + L(x)*L(q*x)*L(q^2*x)/3! + ...
and L(x) = x - 5/2!*x^2 + 83/3!*x^3 + 16110/4!*x^4 +... (A111834).
Thus the g.f. of column 0 of matrix power P^m is:
1 + m*L(x) + m^2/2!*L(x)*L(7*x) + m^3/3!*L(x)*L(7*x)*L(7^2*x) +
m^4/4!*L(x)*L(7*x)*L(7^2*x)*L(7^3*x) + ...
Triangle P begins:
1;
1,1;
1,7,1;
1,154,49,1;
1,16275,8281,343,1;
1,9106461,6558209,410914,2401,1;
1,28543862991,27307109501,2298650515,20170801,16807,1; ...
where P^7 shifts columns left and up one place:
1;
7,1;
154,49,1;
16275,8281,343,1; ...

Crossrefs

Cf. A111831 (column 1), A111832 (row sums), A111833 (matrix log); triangles: A110503 (q=-1), A078121 (q=2), A078122 (q=3), A078536 (q=4), A111820 (q=5), A111825 (q=6), A111835 (q=8).

Programs

  • PARI
    P(n,k,q=7)=local(A=Mat(1),B);if(n

Formula

Let q=7; the g.f. of column k of P^m (ignoring leading zeros) equals: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} L(q^j*x) where L(x) satisfies: x/(1-x) = Sum_{n>=1} Product_{j=0..n-1} L(q^j*x)/(j+1) and L(x) equals the g.f. of column 0 of the matrix log of P (A111834).

A111835 Triangle P, read by rows, that satisfies [P^8](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(8*m)](n,k) = [P^m](n+1,k+1) for all m, where [P^m](n,k) denotes the element at row n, column k, of the matrix power m of P, with P(0,k)=1 and P(k,k)=1 for all k>=0.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 232, 64, 1, 1, 36968, 16192, 512, 1, 1, 35593832, 21928768, 1047040, 4096, 1, 1, 219379963496, 178379459392, 11424946688, 67096576, 32768, 1, 1, 9003699178010216, 9288403489672000, 748093366229504, 5862250172416
Offset: 0

Views

Author

Gottfried Helms and Paul D. Hanna, Aug 22 2005

Keywords

Comments

Also P(n,k) = partitions of (8^n - 8^(n-k)) into powers of 8 <= 8^(n-k).

Examples

			Let q=8; the g.f. of column k of matrix power P^m is:
1 + (m*q^k)*L(x) + (m*q^k)^2/2!*L(x)*L(q*x) +
(m*q^k)^3/3!*L(x)*L(q*x)*L(q^2*x) +
(m*q^k)^4/4!*L(x)*L(q*x)*L(q^2*x)*L(q^3*x) + ...
where L(x) satisfies:
x/(1-x) = L(x) + L(x)*L(q*x)/2! + L(x)*L(q*x)*L(q^2*x)/3! + ...
and L(x) = x - 6/2!*x^2 + 142/3!*x^3 + 31800/4!*x^4 +... (A111839).
Thus the g.f. of column 0 of matrix power P^m is:
1 + m*L(x) + m^2/2!*L(x)*L(8*x) + m^3/3!*L(x)*L(8*x)*L(8^2*x) + m^4/4!*L(x)*L(8*x)*L(8^2*x)*L(8^3*x) + ...
Triangle P begins:
1;
1,1;
1,8,1;
1,232,64,1;
1,36968,16192,512,1;
1,35593832,21928768,1047040,4096,1;
1,219379963496,178379459392,11424946688,67096576,32768,1; ...
where P^8 shifts columns left and up one place:
1;
8,1;
232,64,1;
36968,16192,512,1; ...

Crossrefs

Cf. A111836 (column 1), A111837 (row sums), A111838 (matrix log); triangles: A110503 (q=-1), A078121 (q=2), A078122 (q=3), A078536 (q=4), A111820 (q=5), A111825 (q=6), A111830 (q=7).

Programs

  • PARI
    P(n,k,q=8)=local(A=Mat(1),B);if(n

Formula

Let q=8; the g.f. of column k of P^m (ignoring leading zeros) equals: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} L(q^j*x) where L(x) satisfies: x/(1-x) = Sum_{n>=1} Product_{j=0..n-1} L(q^j*x)/(j+1) and L(x) equals the g.f. of column 0 of the matrix log of P (A111839).

A111840 Triangle P, read by rows, that satisfies [P^3](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(3*m)](n,k) = [P^m](n+1,k+1) for all m, where [P^m](n,k) denotes the element at row n, column k, of the matrix power m of P, with P(k,k)=1 and P(k+1,1)=P(k+1,0) for k>=0.

Original entry on oeis.org

1, 1, 1, 3, 3, 1, 18, 18, 9, 1, 216, 216, 135, 27, 1, 5589, 5589, 4050, 1134, 81, 1, 336555, 336555, 269730, 95256, 9963, 243, 1, 49768101, 49768101, 42724503, 17926839, 2450898, 88938, 729, 1, 18707873562, 18707873562, 16835895603, 8074043145
Offset: 0

Views

Author

Paul D. Hanna, Aug 22 2005

Keywords

Comments

Column 0 and column 1 are equal for n>0.

Examples

			Let q=3; the g.f. of column k of matrix power P^m is:
1 + (m*q^k)*L(x) + (m*q^k)^2/2!*L(x)*L(q*x) +
(m*q^k)^3/3!*L(x)*L(q*x)*L(q^2*x) +
(m*q^k)^4/4!*L(x)*L(q*x)*L(q^2*x)*L(q^3*x) + ...
where L(x) satisfies:
x = L(x) - L(x)*L(q*x)/2! + L(x)*L(q*x)*L(q^2*x)/3! +- ...
and L(x) = x + 3/2!*x^2 + 27/3!*x^3 + 486/4!*x^4 + ... (A111844).
Thus the g.f. of column 0 of matrix power P^m is:
1 + m*L(x) + m^2/2!*L(x)*L(3*x) + m^3/3!*L(x)*L(3*x)*L(3^2*x) +
m^4/4!*L(x)*L(3*x)*L(3^2*x)*L(3^3*x) + ...
Triangle P begins:
1;
1,1;
3,3,1;
18,18,9,1;
216,216,135,27,1;
5589,5589,4050,1134,81,1;
336555,336555,269730,95256,9963,243,1; ...
where P^3 shifts columns left and up one place:
1;
3,1;
18,9,1;
216,135,27,1;
5589,4050,1134,81,1; ...

Crossrefs

Cf. A111841 (column 0), A111842 (row sums), A111843 (matrix log), A078122 (variant).

Programs

  • PARI
    {P(n,k,q=3) = my(A=Mat(1),B);if(nPaul D. Hanna, Jul 11 2025):
for(n=0,10, for(k=0,n, print1(P(n,k),", ")); print(""))

Formula

Let q=3; the g.f. of column k of P^m (ignoring leading zeros) equals: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} L(q^j*x) where L(x) satisfies: x = -Sum_{n>=1} Product_{j=0..n-1} -L(q^j*x)/(j+1); L(x) equals the g.f. of column 0 of the matrix log of P (A111844).

A125804 Main diagonal of table A125800.

Original entry on oeis.org

1, 2, 12, 238, 15200, 3013980, 1828979530, 3373190565626, 18837339867421686, 317817051628161116674, 16176220447967300610844988, 2481251352301850541661479580329, 1146112129196402690505198891390847384
Offset: 0

Views

Author

Paul D. Hanna, Dec 10 2006

Keywords

Comments

Table A125800 is related to partitions into powers of 3; column k of A125800 equals row sums of matrix power A078122^k, where triangle A078122 shifts left one column under matrix cube.

Crossrefs

Cf. A125800, A078122; columns: A078125, A078124, A125801, A125802, A125803; A125805 (antidiagonal sums).

Programs

  • PARI
    a(n)=local(q=3,A=Mat(1), B); for(m=1, n+1, B=matrix(m, m); for(i=1, m, for(j=1, i, if(j==i || j==1, B[i, j]=1, B[i, j]=(A^q)[i-1, j-1]); )); A=B); return(sum(c=0,n,(A^n)[n+1,c+1]))

A125805 Antidiagonal sums of table A125800.

Original entry on oeis.org

1, 2, 4, 10, 41, 361, 7741, 417212, 57581062, 20688363559, 19625079296963, 49742424992663959, 340292157995636104240, 6337196928437059669994069, 323627960380394115802942263514, 45610724032832026072070666274435391
Offset: 0

Views

Author

Paul D. Hanna, Dec 10 2006

Keywords

Comments

Table A125800 is related to partitions into powers of 3; column k of A125800 equals row sums of matrix power A078122^k, where triangle A078122 shifts left one column under matrix cube.

Crossrefs

Cf. A125800, A078122; columns: A078125, A078124, A125801, A125802, A125803; A125804 (diagonal).

Programs

  • PARI
    a(n)=local(q=3,A=Mat(1), B); for(m=1, n+1, B=matrix(m, m); for(i=1, m, for(j=1, i, if(j==i || j==1, B[i, j]=1, B[i, j]=(A^q)[i-1, j-1]); )); A=B); return(sum(c=0,n,(A^(c+1))[n-c+1,1]))
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