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 1-10 of 15 results. Next

A107045 Numerators of the triangle of coefficients T(n,k), read by rows, that satisfy: y^x = Sum_{n=0..x} R_n(y)*x^n for all nonnegative integers x, y, where R_n(y) = Sum_{k=0..n} T(n,k)*y^k and T(n,k) = a(n,k)/A107046(n,k).

Original entry on oeis.org

1, -1, 1, 1, -1, 1, -1, 1, -1, 1, -11, 1, 1, -1, 1, -677, -61, 7, 1, -1, 1, 15311, -259, -1, 7, 1, -1, 1, 1170049273, -971891, -54407, 407, 23, 1, -1, 1, 541293087149, 426148171, -15993079, -58573, 829, 17, 1, -1, 1, -15074636799365429, 31108643619709, -23328513449, -138374321, -53429, 1501, 47, 1
Offset: 0

Views

Author

Paul D. Hanna, May 10 2005

Keywords

Examples

			These are the numerators of the triangle that begins:
1;
-1,1;
1/4,-1/2,1/4;
-1/108,1/18,-1/12,1/27;
-11/6912,1/576,1/192,-1/108,1/256;
-677/21600000,-61/360000,7/24000,1/2700,-1/1280,1/3125; ...
which equals the matrix inverse of triangle A079901(n,k)=n^k:
1;
1,1;
1,2,4;
1,3,9,27;
1,4,16,64,256;
1,5,25,125,625,3125; ...
		

Crossrefs

Programs

  • PARI
    a(n,k)=numerator((matrix(n+1,n+1,r,c,if(r>=c,(r-1)^(c-1)))^-1)[n+1,k+1])

Formula

Numerators of the matrix inverse of triangle A079901(n, k) = n^k.

A107046 Denominators of the triangle of coefficients T(n,k), read by rows, that satisfy: y^x = Sum_{n=0..x} R_n(y)*x^n for all nonnegative integers x, y, where R_n(y) = Sum_{k=0..n} T(n,k)*y^k and T(n,k) = A107045(n,k)/a(n,k).

Original entry on oeis.org

1, 1, 1, 4, 2, 4, 108, 18, 12, 27, 6912, 576, 192, 108, 256, 21600000, 360000, 24000, 2700, 1280, 3125, 2332800000, 12960000, 2592000, 291600, 46080, 18750, 46656, 1921161110400000, 1524731040000, 43563744000, 700131600, 15805440, 918750
Offset: 0

Views

Author

Paul D. Hanna, May 10 2005

Keywords

Examples

			These are the denominators of the triangle that begins:
1;
-1,1;
1/4,-1/2,1/4;
-1/108,1/18,-1/12,1/27;
-11/6912,1/576,1/192,-1/108,1/256;
-677/21600000,-61/360000,7/24000,1/2700,-1/1280,1/3125; ...
which equals the matrix inverse of triangle A079901(n,k)=n^k:
1;
1,1;
1,2,4;
1,3,9,27;
1,4,16,64,256;
1,5,25,125,625,3125; ...
		

Crossrefs

Programs

  • PARI
    a(n,k)=denominator((matrix(n+1,n+1,r,c,if(r>=c,(r-1)^(c-1)))^-1)[n+1,k+1])

Formula

Denominators of the matrix inverse of triangle A079901(n, k) = n^k.

A107047 Numerators of coefficients that satisfy: 2^n = Sum_{k=0..n} c(k)*x^k for n>=0, where c(k) = a(k)/A107048(k).

Original entry on oeis.org

1, 1, 1, 7, 77, 32387, 395159, 31824093937, 44855117331581, 1825389561156191099, 1571879809058619206897, 28070265610073576492663157851903, 2782861136717279135850604073374039
Offset: 0

Views

Author

Paul D. Hanna, May 10 2005

Keywords

Comments

Sum_{k>=0} a(k)/A107048(k) = 2.3276417590495914492697647475269004042620542650376396714...

Examples

			2^0 = 1;
2^1 = 1 + 1;
2^2 = 1 + 1*2 + (1/4)*2^2;
2^3 = 1 + 1*3 + (1/4)*3^2 + (7/108)*3^3;
2^4 = 1 + 1*4 + (1/4)*4^2 + (7/108)*4^3 + (77/6912)*4^4.
Initial fractional coefficients are:
A107047/A107048 = {1, 1, 1/4, 7/108, 77/6912, 32387/21600000,
395159/2332800000, 31824093937/1921161110400000,
44855117331581/31476303632793600000, ... }.
		

Crossrefs

Programs

  • PARI
    {a(n)=numerator(sum(k=0,n,2^k*(matrix(n+1,n+1,r,c,if(r>=c,(r-1)^(c-1)))^-1)[n+1,k+1]))}

Formula

a(n)/A107048(n) = Sum_{k=0..n} T(n, k)*2^k where T(n, k) = A107045(n, k)/A107046(n, k) = [A079901^-1](n, k) (matrix inverse of A079901).

A107048 Denominators of coefficients that satisfy: 2^n = Sum_{k=0..n} c(k)*x^k for n>=0, where c(k) = A107047(k)/a(k).

Original entry on oeis.org

1, 1, 4, 108, 6912, 21600000, 2332800000, 1921161110400000, 31476303632793600000, 16727798278915463577600000, 209097478486443294720000000000
Offset: 0

Views

Author

Paul D. Hanna, May 10 2005

Keywords

Examples

			2^0 = 1;
2^1 = 1 + 1;
2^2 = 1 + 1*2 + (1/4)*2^2;
2^3 = 1 + 1*3 + (1/4)*3^2 + (7/108)*3^3;
2^4 = 1 + 1*4 + (1/4)*4^2 + (7/108)*4^3 + (77/6912)*4^4.
Initial fractional coefficients are:
A107047/A107048 = {1, 1, 1/4, 7/108, 77/6912, 32387/21600000,
395159/2332800000, 31824093937/1921161110400000,
44855117331581/31476303632793600000, ... }.
		

Crossrefs

Programs

  • PARI
    {a(n)=denominator(sum(k=0,n,2^k*(matrix(n+1,n+1,r,c,if(r>=c,(r-1)^(c-1)))^-1)[n+1,k+1]))}

Formula

A107047(n)/a(n) = Sum_{k=0..n} T(n, k)*2^k where T(n, k) = A107045(n, k)/A107046(n, k) = [A079901^-1](n, k) (matrix inverse of A079901).

A107049 Numerators of coefficients that satisfy: 3^n = Sum_{k=0..n} c(k)*x^k for n>=0, where c(k) = a(k)/A107050(k).

Original entry on oeis.org

1, 2, 1, 11, 101, 71723, 1462111, 194269981673, 224103520039487, 14876670160046176873, 20871062802926443547323, 606768727432357137728440774281877, 97827345788163051844748893917483101
Offset: 0

Views

Author

Paul D. Hanna, May 10 2005

Keywords

Comments

Sum_{k>=0} a(k)/A107050(k) = 4.5568226185870666883519278484116281050682807568451524897...

Examples

			3^0 = 1;
3^1 = 1 + (2)*1;
3^2 = 1 + (2)*2 + (1)*2^2;
3^3 = 1 + (2)*3 + (1)*3^2 + (11/27)*3^3;
3^4 = 1 + (2)*4 + (1)*4^2 + (11/27)*4^3 + (101/864)*4^4.
Initial coefficients are:
A107049/A107050 = {1, 2, 1, 11/27, 101/864, 71723/2700000,
1462111/291600000, 194269981673/240145138800000,
224103520039487/1967268977049600000, ...}.
		

Crossrefs

Programs

  • PARI
    {a(n)=numerator(sum(k=0,n,3^k*(matrix(n+1,n+1,r,c,if(r>=c,(r-1)^(c-1)))^-1)[n+1,k+1]))}

Formula

a(n)/A107050(n) = Sum_{k=0..n} T(n, k)*3^k where T(n, k) = A107045(n, k)/A107046(n, k) = [A079901^-1](n, k) (matrix inverse of A079901).

A107050 Denominators of coefficients that satisfy: 3^n = Sum_{k=0..n} c(k)*x^k for n>=0, where c(k) = A107049(k)/a(k).

Original entry on oeis.org

1, 1, 1, 27, 864, 2700000, 291600000, 240145138800000, 1967268977049600000, 1045487392432216473600000, 13068592405402705920000000000, 3728621931719673008255139717120000000000
Offset: 0

Views

Author

Paul D. Hanna, May 10 2005

Keywords

Examples

			3^0 = 1;
3^1 = 1 + (2)*1;
3^2 = 1 + (2)*2 + (1)*2^2;
3^3 = 1 + (2)*3 + (1)*3^2 + (11/27)*3^3;
3^4 = 1 + (2)*4 + (1)*4^2 + (11/27)*4^3 + (101/864)*4^4.
Initial coefficients are:
A107049/A107050 = {1, 2, 1, 11/27, 101/864, 71723/2700000,
1462111/291600000, 194269981673/240145138800000,
224103520039487/1967268977049600000, ...}.
		

Crossrefs

Programs

  • PARI
    {a(n)=denominator(sum(k=0,n,3^k*(matrix(n+1,n+1,r,c,if(r>=c,(r-1)^(c-1)))^-1)[n+1,k+1]))}

Formula

A107049(n)/a(n) = Sum_{k=0..n} T(n, k)*3^k where T(n, k) = A107045(n, k)/A107046(n, k) = [A079901^-1](n, k) (matrix inverse of A079901).

A107051 Numerators of coefficients that satisfy: 4^n = Sum_{k=0..n} c(k)*x^k for n>=0, where c(k) = a(k)/A107052(k).

Original entry on oeis.org

1, 3, 9, 5, 127, 124273, 385829, 70009765747, 220026935042111, 59574747365570286907, 113453152114585319883313, 4471148647570383262775217527741887
Offset: 0

Views

Author

Paul D. Hanna, May 10 2005

Keywords

Comments

Sum_{k>=0} a(k)/A107052(k) = 8.2025187671791748426202820386803825244610468145759213023...

Examples

			4^0 = 1;
4^1 = 1 + (3)*1;
4^2 = 1 + (3)*2 + (9/4)*2^2;
4^3 = 1 + (3)*3 + (9/4)*3^2 + (5/4)*3^3;
4^4 = 1 + (3)*4 + (9/4)*4^2 + (5/4)*4^3 + (127/256)*4^4.
Initial coefficients are:
A107051/A107052 = {1, 3, 9/4, 5/4, 127/256, 124273/800000,
385829/9600000, 70009765747/7906012800000,
220026935042111/129532113715200000, ...}.
		

Crossrefs

Programs

  • PARI
    {a(n)=numerator(sum(k=0,n,4^k*(matrix(n+1,n+1,r,c,if(r>=c,(r-1)^(c-1)))^-1)[n+1,k+1]))}

Formula

a(n)/A107052(n) = Sum_{k=0..n} T(n, k)*4^k where T(n, k) = A107045(n, k)/A107046(n, k) = [A079901^-1](n, k) (matrix inverse of A079901).

A107052 Denominators of coefficients that satisfy: 4^n = Sum_{k=0..n} c(k)*x^k for n>=0, where c(k) = A107051(k)/a(k).

Original entry on oeis.org

1, 1, 4, 4, 256, 800000, 9600000, 7906012800000, 129532113715200000, 206516028134758809600000, 2581450351684485120000000000, 736517912438453927556570808320000000000
Offset: 0

Views

Author

Paul D. Hanna, May 10 2005

Keywords

Examples

			4^0 = 1;
4^1 = 1 + (3)*1;
4^2 = 1 + (3)*2 + (9/4)*2^2;
4^3 = 1 + (3)*3 + (9/4)*3^2 + (5/4)*3^3;
4^4 = 1 + (3)*4 + (9/4)*4^2 + (5/4)*4^3 + (127/256)*4^4.
Initial coefficients are:
A107051/A107052 = {1, 3, 9/4, 5/4, 127/256, 124273/800000,
385829/9600000, 70009765747/7906012800000,
220026935042111/129532113715200000, ...}.
		

Crossrefs

Programs

  • PARI
    {a(n)=denominator(sum(k=0,n,4^k*(matrix(n+1,n+1,r,c,if(r>=c,(r-1)^(c-1)))^-1)[n+1,k+1]))}

Formula

A107051(n)/a(n) = Sum_{k=0..n} T(n, k)*4^k where T(n, k) = A107045(n, k)/A107046(n, k) = [A079901^-1](n, k) (matrix inverse of A079901).

A107053 Numerators of coefficients that satisfy: 5^n = Sum_{k=0..n} c(k)*x^k for n>=0, where c(k) = a(k)/A107054(k).

Original entry on oeis.org

1, 4, 4, 76, 307, 380989, 13464073, 3084163593839, 6109976845914041, 694491088545589897439, 1664245369537759004769053, 82473629015170976645702130970352147
Offset: 0

Views

Author

Paul D. Hanna, May 10 2005

Keywords

Comments

Sum_{k>=0} a(k)/A107054(k) = 14.052297927432224441845709796250699506418496460894575328...

Examples

			5^0 = 1;
5^1 = 1 + (4)*1;
5^2 = 1 + (4)*2 + (4)*2^2;
5^3 = 1 + (4)*3 + (4)*3^2 + (76/27)*3^3;
5^4 = 1 + (4)*4 + (4)*4^2 + (76/27)*4^3 + (307/216)*4^4.
Initial coefficients are:
A107053/A107054 = {1, 4, 4, 76/27, 307/216, 380989/675000,
13464073/72900000, 3084163593839/60036284700000,
6109976845914041/491817244262400000, ...}
		

Crossrefs

Programs

  • PARI
    {a(n)=numerator(sum(k=0,n,5^k*(matrix(n+1,n+1,r,c,if(r>=c,(r-1)^(c-1)))^-1)[n+1,k+1]))}

Formula

a(n)/A107054(n) = Sum_{k=0..n} T(n, k)*5^k where T(n, k) = A107045(n, k)/A107046(n, k) = [A079901^-1](n, k) (matrix inverse of A079901).

A107054 Denominators of coefficients that satisfy: 5^n = Sum_{k=0..n} c(k)*x^k for n>=0, where c(k) = A107053(k)/a(k).

Original entry on oeis.org

1, 1, 1, 27, 216, 675000, 72900000, 60036284700000, 491817244262400000, 261371848108054118400000, 3267148101350676480000000000, 932155482929918252063784929280000000000
Offset: 0

Views

Author

Paul D. Hanna, May 10 2005

Keywords

Examples

			5^0 = 1;
5^1 = 1 + (4)*1;
5^2 = 1 + (4)*2 + (4)*2^2;
5^3 = 1 + (4)*3 + (4)*3^2 + (76/27)*3^3;
5^4 = 1 + (4)*4 + (4)*4^2 + (76/27)*4^3 + (307/216)*4^4.
Initial coefficients are:
A107053/A107054 = {1, 4, 4, 76/27, 307/216, 380989/675000,
13464073/72900000, 3084163593839/60036284700000,
6109976845914041/491817244262400000, ...}
		

Crossrefs

Programs

  • PARI
    {a(n)=denominator(sum(k=0,n,5^k*(matrix(n+1,n+1,r,c,if(r>=c,(r-1)^(c-1)))^-1)[n+1,k+1]))}

Formula

A107053(n)/a(n) = Sum_{k=0..n} T(n, k)*5^k where T(n, k) = A107045(n, k)/A107046(n, k) = [A079901^-1](n, k) (matrix inverse of A079901).
Showing 1-10 of 15 results. Next