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.

A084211 G.f. A(x) defined by: A(x)^11 consists entirely of integer coefficients between 1 and 11 (A084066); A(x) is the unique power series solution with A(0)=1.

Original entry on oeis.org

1, 1, -4, 26, -189, 1479, -12106, 102224, -883031, 7761209, -69153920, 623018880, -5664270185, 51892998965, -478521450110, 4437418074830, -41350439060725, 386983852716405
Offset: 0

Views

Author

Paul D. Hanna, May 20 2003

Keywords

Comments

Limit a(n)/a(n+1) --> r = -0.100000000011 where A(r)=0.

Crossrefs

Programs

  • Mathematica
    kmax = 20;
    A[x_] = Sum[a[k] x^k, {k, 0, kmax}];
    coes = CoefficientList[A[x]^11 + O[x]^(kmax + 1), x];
    r = {a[0] -> 1, a[1] -> 1};
    coes = coes /. r;
    Do[r = Flatten @ Append[r, Reduce[1 <= coes[[k]] <= 11, a[k-1], Integers] // ToRules];
    coes = coes /. r, {k, 3, kmax + 1}];
    Table[a[k], {k, 0, kmax}] /. r (* Jean-François Alcover, Jul 26 2018 *)

A110644 Every 11th term of A084066 such that the self-convolution 11th power is congruent modulo 121 to A084066, which consists entirely of numbers 1 through 11.

Original entry on oeis.org

1, 1, 7, 4, 9, 5, 5, 1, 5, 5, 11, 11, 9, 5, 11, 4, 8, 10, 10, 8, 10, 5, 11, 6, 1, 7, 1, 11, 5, 10, 1, 9, 4, 3, 9, 6, 8, 1, 6, 3, 4, 8, 2, 4, 4, 8, 10, 2, 4, 11, 1, 7, 11, 9, 11, 5, 2, 1, 4, 7, 9, 3, 2, 5, 8, 1, 8, 7, 4, 3, 2, 3, 5, 9, 1, 9, 5, 4, 1, 4, 6, 8, 5, 6, 9, 7, 4, 4, 5, 4, 6, 4, 10, 6, 6, 9, 9, 9, 4
Offset: 0

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Author

Keywords

Comments

Congruent modulo 11 to A084211, where the self-convolution 11th power of A084211 equals A084066.

Crossrefs

Programs

  • PARI
    {a(n)=local(p=11,A,C,X=x+x*O(x^(p*n)));if(n==0,1, A=sum(i=0,n-1,a(i)*x^(p*i))+p*x*((1-x^(p-1))/(1-X))/(1-X^p); for(k=1,p,C=polcoeff((A+k*x^(p*n))^(1/p),p*n); if(denominator(C)==1,return(k);break)))}

Formula

a(n) = A084066(11*n) for n>=0. G.f. satisfies: A(x^11) = G(x) - 11*x*((1-x^10)/(1-x))/(1-x^11), where G(x) is the g.f. of A084066. G.f. satisfies: A(x)^11 = A(x^11) + 11*x*((1-x^10)/(1-x))/(1-x^11) + 121*x^2*H(x) where H(x) is the g.f. of A111585.

A109626 Consider the array T(n,m) where the n-th row is the sequence of integer coefficients of A(x), where 1<=a(n)<=n, such that A(x)^(1/n) consists entirely of integer coefficients and where m is the (m+1)-th coefficient. This is the antidiagonal read from lower left to upper right.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 4, 3, 2, 1, 1, 5, 2, 1, 2, 1, 1, 6, 5, 4, 3, 2, 1, 1, 7, 3, 5, 3, 3, 1, 1, 1, 8, 7, 2, 5, 4, 3, 2, 1, 1, 9, 4, 7, 3, 1, 4, 3, 2, 1, 1, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 11, 5, 3, 2, 7, 6, 5, 1, 3, 1, 1, 1, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 13, 6, 11, 10, 9, 4, 1, 3, 5
Offset: 1

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Author

Keywords

Examples

			Table begins:
\k...0...1...2...3...4...5...6...7...8...9..10..11..12..13
n\
 1|  1   1   1   1   1   1   1   1   1   1   1   1   1   1
 2|  1   2   1   2   2   2   1   2   2   2   1   2   1   2
 3|  1   3   3   1   3   3   3   3   3   3   3   3   1   3
 4|  1   4   2   4   3   4   4   4   1   4   4   4   3   4
 5|  1   5   5   5   5   1   5   5   5   5   4   5   5   5
 6|  1   6   3   2   3   6   6   6   3   4   6   6   6   6
 7|  1   7   7   7   7   7   7   1   7   7   7   7   7   7
 8|  1   8   4   8   2   8   4   8   7   8   8   8   4   8
 9|  1   9   9   3   9   9   3   9   9   1   9   9   6   9
10|  1  10   5  10  10   2   5  10  10  10   3  10   5  10
11|  1  11  11  11  11  11  11  11  11  11  11   1  11  11
12|  1  12   6   4   9  12   4  12  12   8   6  12   6  12
13|  1  13  13  13  13  13  13  13  13  13  13  13  13   1
14|  1  14   7  14   7  14  14   2   7  14  14  14  14  14
15|  1  15  15   5  15   3  10  15  15  10  15  15   5  15
16|  1  16   8  16   4  16   8  16  10  16   8  16  12  16
		

Crossrefs

Diagonals: A000027 (main), A111614 (first upper), A111627 (2nd), A111615 (3rd), A111618 (first lower), A111623 (2nd).
Other diagonals: A005408 (T(2*n-1, n)), A111626, A111627, A111628, A111629, A111630.

Programs

  • Mathematica
    f[n_]:= f[n]= Block[{a}, a[0] = 1; a[l_]:= a[l]= Block[{k = 1, s = Sum[ a[i]*x^i, {i,0,l-1}]}, While[ IntegerQ[Last[CoefficientList[Series[(s + k*x^l)^(1/n), {x, 0, l}], x]]] != True, k++ ]; k]; Table[a[j], {j,0,32}]];
    T[n_, m_]:= f[n][[m]];
    Flatten[Table[T[i,n-i], {n,15}, {i,n-1,1,-1}]]
  • PARI
    A109626_row(n, len=40)={my(A=1, m); vector(len, k, if(k>m=1, while(denominator(polcoeff(sqrtn(O(x^k)+A+=x^(k-1), n), k-1))>1, m++); m, 1))} \\ M. F. Hasler, Jan 27 2025

Formula

When m is prime, column m is T(n,m) = n/gcd(m, n) = numerator of n/(n+m). - M. F. Hasler, Jan 27 2025

A084067 Integer coefficients of A(x), where 1<=a(n)<=12, such that A(x)^(1/12) consists entirely of integer coefficients.

Original entry on oeis.org

1, 12, 6, 4, 9, 12, 4, 12, 12, 8, 6, 12, 6, 12, 12, 12, 12, 12, 8, 12, 9, 12, 12, 12, 12, 12, 6, 12, 6, 12, 10, 12, 6, 12, 12, 12, 2, 12, 6, 8, 6, 12, 12, 12, 12, 4, 12, 12, 8, 12, 12, 8, 3, 12, 4, 12, 12, 4, 12, 12, 9, 12, 6, 4, 6, 12, 4, 12, 12, 12, 12, 12, 2, 12, 6, 12, 3, 12, 6, 12, 3, 8
Offset: 0

Views

Author

Paul D. Hanna, May 10 2003

Keywords

Comments

More generally, the sequence: "integer coefficients of A(x), where 1<=a(n)<=m, such that A(x)^(1/m) consists entirely of integer coefficients", appears to have a unique solution for all m>0. Are these sequences ever periodic?

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = Block[{k = 1, s = Sum[a[i]*x^i, {i, 0, n-1}]}, While[ Union[ IntegerQ /@ CoefficientList[ Series[(s+k*x^n)^(1/12), {x, 0, n}], x]] != {True}, k++ ]; k]; Table[ a[n], {n, 0, 81}] (* Robert G. Wilson v *)

Extensions

More terms from Robert G. Wilson v, Jul 26 2005

A111603 Consider the array T(n, m) where the n-th row is the sequence of integer coefficients of A(x), where 1<=a(n)<=n, such that A(x)^(1/n) consists entirely of integer coefficients and where m is the (m+1)-th coefficient. This is the antidiagonal read from upper right to lower left.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 3, 4, 1, 1, 2, 1, 2, 5, 1, 1, 2, 3, 4, 5, 6, 1, 1, 1, 3, 3, 5, 3, 7, 1, 1, 2, 3, 4, 5, 2, 7, 8, 1, 1, 2, 3, 4, 1, 3, 7, 4, 9, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 1, 1, 3, 1, 5, 6, 7, 2, 3, 5, 11, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 1, 1, 3, 4, 5, 3, 1, 4, 9, 10
Offset: 1

Views

Author

Keywords

Examples

			Table begins
k= 0 1 2 3 4 5 6 7 8 9 10 11 12 13
n\
1| 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2| 1 2 1 2 2 2 1 2 2 2 1 2 1 2
3| 1 3 3 1 3 3 3 3 3 3 3 3 1 3
4| 1 4 2 4 3 4 4 4 1 4 4 4 3 4
5| 1 5 5 5 5 1 5 5 5 5 4 5 5 5
6| 1 6 3 2 3 6 6 6 3 4 6 6 6 6
7| 1 7 7 7 7 7 7 1 7 7 7 7 7 7
8| 1 8 4 8 2 8 4 8 7 8 8 8 4 8
9| 1 9 9 3 9 9 3 9 9 1 9 9 6 9
10| 1 10 5 10 10 2 5 10 10 10 3 10 5 10
11| 1 11 11 11 11 11 11 11 11 11 11 1 11 11
12| 1 12 6 4 9 12 4 12 12 8 6 12 6 12
13| 1 13 13 13 13 13 13 13 13 13 13 13 13 1
14| 1 14 7 14 7 14 14 2 7 14 14 14 14 14
15| 1 15 15 5 15 3 10 15 15 10 15 15 5 15
16| 1 16 8 16 4 16 8 16 10 16 8 16 12 16
		

Crossrefs

Programs

  • Mathematica
    f[n_] := f[n] = Block[{a}, a[0] = 1; a[l_] := a[l] = Block[{k = 1, s = Sum[ a[i]*x^i, {i, 0, l - 1}]}, While[ IntegerQ[ Last[ CoefficientList[ Series[(s + k*x^l)^(1/n), {x, 0, l}], x]]] != True, k++ ]; k]; Table[ a[j], {j, 0, 32}]]; g[n_, m_] := f[n][[m]]; Flatten[ Table[ f[i, n - i], {n, 15}, {i, n - 1, 1, -1}]]

A111604 Consider the array T(n, m) where the n-th row is the sequence of integer coefficients of A(x), where 1<=a(n)<=n, such that A(x)^(1/n) consists entirely of integer coefficients and where m is the (m+1)-th coefficient. This is the antidiagonal read zig-zag.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 4, 3, 2, 1, 1, 2, 1, 2, 5, 1, 1, 6, 5, 4, 3, 2, 1, 1, 1, 3, 3, 5, 3, 7, 1, 1, 8, 7, 2, 5, 4, 3, 2, 1, 1, 2, 3, 4, 1, 3, 7, 4, 9, 1, 1, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 1, 3, 1, 5, 6, 7, 2, 3, 5, 11, 1, 1, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 1, 3, 4, 5, 3, 1, 4, 9, 10, 11
Offset: 1

Views

Author

Keywords

Comments

T(n,n)=T(n,n+2)=A111627.

Examples

			Table begins
\k...0...1....2....3....4....5....6....7....8....9...10...11...12...13
n\
1| 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2| 1 2 1 2 2 2 1 2 2 2 1 2 1 2
3| 1 3 3 1 3 3 3 3 3 3 3 3 1 3
4| 1 4 2 4 3 4 4 4 1 4 4 4 3 4
5| 1 5 5 5 5 1 5 5 5 5 4 5 5 5
6| 1 6 3 2 3 6 6 6 3 4 6 6 6 6
7| 1 7 7 7 7 7 7 1 7 7 7 7 7 7
8| 1 8 4 8 2 8 4 8 7 8 8 8 4 8
9| 1 9 9 3 9 9 3 9 9 1 9 9 6 9
10| 1 10 5 10 10 2 5 10 10 10 3 10 5 10
11| 1 11 11 11 11 11 11 11 11 11 11 1 11 11
12| 1 12 6 4 9 12 4 12 12 8 6 12 6 12
13| 1 13 13 13 13 13 13 13 13 13 13 13 13 1
14| 1 14 7 14 7 14 14 2 7 14 14 14 14 14
15| 1 15 15 5 15 3 10 15 15 10 15 15 5 15
16| 1 16 8 16 4 16 8 16 10 16 8 16 12 16
		

Crossrefs

Programs

  • Mathematica
    f[n_] := f[n] = Block[{a}, a[0] = 1; a[l_] := a[l] = Block[{k = 1, s = Sum[ a[i]*x^i, {i, 0, l - 1}]}, While[ IntegerQ[ Last[ CoefficientList[ Series[(s + k*x^l)^(1/n), {x, 0, l}], x]]] != True, k++ ]; k]; Table[a[j], {j, 0, 32}]]; g[n_, m_] := f[n][[m]];

A111585 G.f.: A(x) = ( G(x)^11 - G(x^11) - 11*x*((1-x^10)/(1-x))/(1-x^11) )/(121*x^2) where G(x) is the g.f. of A110644.

Original entry on oeis.org

1, 8, 58, 331, 1786, 8339, 36732, 145822, 547140, 1892180, 6204220, 19012264, 55425015, 152433682, 400145401, 997800881, 2382495077, 5433427757, 11903695647, 25025392965, 50712232586, 99047934945, 187153799666
Offset: 0

Views

Author

Paul D. Hanna, Aug 28 2005

Keywords

Comments

A110644 is formed by selecting every 11th term of A084066; surprisingly, the self-convolution 11th power of A110644 is congruent modulo 121 to A084066, which consists entirely of numbers 1 through 11.

Crossrefs

Showing 1-7 of 7 results.