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.

A101912 G.f. satisfies: A(x) = 1/(1 + x*A(x^2)) and also the continued fraction: 1 + x*A(x^3) = [1; 1/x, 1/x^2, 1/x^4, 1/x^8, ..., 1/x^(2^(n-1)), ...].

Original entry on oeis.org

1, -1, 1, 0, -1, 1, 0, -2, 3, -1, -3, 6, -4, -4, 12, -10, -5, 23, -25, -2, 43, -57, 12, 74, -124, 56, 120, -258, 172, 170, -516, 454, 187, -989, 1095, 40, -1811, 2487, -604, -3128, 5375, -2567, -4991, 11140, -7704, -6976, 22164, -20062, -7220, 42288, -48020, -36, 76928, -108334, 29476, 131898, -233020, 117166
Offset: 0

Views

Author

Paul D. Hanna, Dec 20 2004

Keywords

Comments

Sequence appears to have a rational g.f. - Ralf Stephan, May 17 2007
The conjecture is wrong. The g.f. is dependent on the number of terms. - Johannes W. Meijer, Aug 08 2011

Crossrefs

Cf. A101913, A101914, A101915, A101916, A101917, A101918.

Programs

  • Maple
    nmax:=57: kmax:=nmax: for k from 0 to kmax do A:= proc(x): add(A101912(n)*x^n, n=0..k) end: f(x):=series(1/(1 + x*A(x^2)),x,k+1); for n from 0 to k do x(n):=coeff(f(x),x,n) od: A101912(k):=x(k): od: seq(A101912(n), n=0..nmax); # Johannes W. Meijer, Aug 08 2011
  • Mathematica
    m = 58; A[] = 0; Do[A[x] = 1/(1 + x A[x^2]) + O[x]^m // Normal, {m}]; CoefficientList[A[x], x] (* Jean-François Alcover, Nov 03 2019 *)
  • PARI
    {a(n)=local(A);A=1-x;for(i=1,n\2+1, A=1/(1+x*subst(A,x,x^2)+x*O(x^n)));polcoeff(A,n,x)}
  • PARI
    {a(n)=local(M=contfracpnqn(concat(1, vector(#binary(n)+1,n,1/x^(2^(n-1)))))); polcoeff(M[1,1]/M[2,1]+x*O(x^(3*n+1)),3*n+1)}

Formula

G.f.: 1/(1 + x/(1 + x^2/(1 + x^4/(1 + x^8/(1 + ...))))) (continued fraction). - Joerg Arndt, Oct 19 2012
G.f. A(x) = 1/B(x) where B(x) is the g.f. of A218031. - Joerg Arndt, Oct 19 2012
a(0) = 1; a(n) = -Sum_{k=0..floor((n-1)/2)} a(k) * a(n-2*k-1). - Ilya Gutkovskiy, Mar 01 2022

A101918 G.f. satisfies: A(x) = 1/(1 + x*A(x^8)) and also the continued fraction: 1 + x*A(x^9) = [1; 1/x, 1/x^8, 1/x^64, 1/x^512, ..., 1/x^(8^(n-1)), ...].

Original entry on oeis.org

1, -1, 1, -1, 1, -1, 1, -1, 1, 0, -1, 2, -3, 4, -5, 6, -7, 7, -6, 4, -1, -3, 8, -14, 21, -28, 34, -38, 39, -36, 28, -14, -7, 35, -69, 107, -146, 182, -210, 224, -217, 182, -113, 6, 140, -322, 532, -756, 973, -1155, 1268, -1274, 1134, -812, 280, 476, -1449, 2604, -3872, 5146, -6280, 7092, -7372, 6896, -5447, 2843, 1029
Offset: 0

Views

Author

Paul D. Hanna, Dec 20 2004

Keywords

Links

Crossrefs

Cf. A101912-A101917.

Programs

  • Maple
    nmax:=66: kmax:=nmax: for k from 0 to kmax do A:= proc(x): add(A101918(n)*x^n, n=0..k) end: f(x):=series(1/(1 + x*A(x^8)),x,k+1); for n from 0 to k do x(n):=coeff(f(x),x,n) od: A101918(k):=x(k): od: seq(A101918(n), n=0..nmax); # Johannes W. Meijer, Aug 08 2011
  • Mathematica
    nmax = 66; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x] - (1/(1 + x A[x^8])) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 03 2019 *)
  • PARI
    {a(n)=local(A);A=1-x;for(i=1,n\8+1, A=1/(1+x*subst(A,x,x^8)+x*O(x^n)));polcoeff(A,n,x)}
for(n=0,120,print1(a(n),", "))
  • PARI
    {a(n)=local(M=contfracpnqn(concat(1, vector(ceil(log(n+1)/log(8))+1,n,1/x^(8^(n-1)))))); polcoeff(M[1,1]/M[2,1]+x*O(x^(9*n+1)),9*n+1)}
for(n=0,120,print1(a(n),", "))

Formula

The conjecture that this has g.f. (1+x^8) / (1+x+x^8) by Ralf Stephan, May 17 2007 is wrong. The first difference occurs at a(73) = -42106. The g.f. gives a(73) = -42105. - Johannes W. Meijer, Aug 08 2011
a(0) = 1; a(n) = -Sum_{k=0..floor((n-1)/8)} a(k) * a(n-8*k-1). - Ilya Gutkovskiy, Mar 01 2022

A101914 G.f. satisfies: A(x) = 1/(1 + x*A(x^4)) and also the continued fraction: 1 + x*A(x^5) = [1; 1/x, 1/x^4, 1/x^16, 1/x^64, ..., 1/x^(4^(n-1)), ...].

Original entry on oeis.org

1, -1, 1, -1, 1, 0, -1, 2, -3, 3, -2, 0, 3, -6, 8, -8, 5, 1, -9, 17, -22, 20, -10, -8, 31, -51, 60, -50, 16, 38, -100, 150, -163, 119, -11, -147, 315, -432, 433, -268, -70, 522, -964, 1222, -1118, 542, 484, -1756, 2887, -3385, 2793, -879, -2176, 5678, -8472, 9186, -6672, 542, 8372, -17816, 24384, -24350, 14952
Offset: 0

Views

Author

Paul D. Hanna, Dec 20 2004

Keywords

Comments

Sequence appears to have a rational g.f. - Ralf Stephan, May 17 2007
The conjecture is wrong. The g.f. is dependent on the number of terms. With twenty terms the g.f. is (1 + x^4)/(1 + x + x^4). - Johannes W. Meijer, Aug 08 2011

Crossrefs

Cf. A101912, A101913, A101915, A101916, A101917, A101918.

Programs

  • Maple
    nmax:=62: kmax:=nmax: for k from 0 to kmax do A:= proc(x): add(A101914(n)*x^n, n=0..k) end: f(x):=series(1/(1 + x*A(x^4)),x,k+1); for n from 0 to k do x(n):=coeff(f(x),x,n) od: A101914(k):=x(k): od: seq(A101914(n), n=0..nmax); # Johannes W. Meijer, Aug 08 2011
  • Mathematica
    m = 63; A[] = 0; Do[A[x] = 1/(1 + x A[x^4]) + O[x]^m // Normal, {m}]; CoefficientList[A[x], x] (* Jean-François Alcover, Nov 03 2019 *)
  • PARI
    a(n)=local(A);A=1-x;for(i=1,n\4+1, A=1/(1+x*subst(A,x,x^4)+x*O(x^n)));polcoeff(A,n,x)
  • PARI
    a(n)=local(M=contfracpnqn(concat(1, vector(ceil(log(n+1)/log(4))+1,n,1/x^(4^(n-1)))))); polcoeff(M[1,1]/M[2,1]+x*O(x^(5*n+1)),5*n+1)

Formula

a(0) = 1; a(n) = -Sum_{k=0..floor((n-1)/4)} a(k) * a(n-4*k-1). - Ilya Gutkovskiy, Mar 01 2022

A101916 G.f. satisfies: A(x) = 1/(1 + x*A(x^6)) and also the continued fraction: 1+x*A(x^7) = [1;1/x,1/x^6,1/x^36,1/x^216,...,1/x^(6^(n-1)),...].

Original entry on oeis.org

1, -1, 1, -1, 1, -1, 1, 0, -1, 2, -3, 4, -5, 5, -4, 2, 1, -5, 10, -15, 19, -21, 20, -15, 5, 10, -29, 50, -70, 85, -90, 80, -51, 1, 69, -154, 244, -324, 375, -376, 307, -153, -91, 414, -788, 1163, -1469, 1621, -1529, 1115, -328, -833, 2299, -3916, 5440, -6550, 6874, -6039, 3741, 170, -5600, 12135, -18990, 25008
Offset: 0

Views

Author

Paul D. Hanna, Dec 20 2004

Keywords

Comments

This sequence resembles the series expansion of B(x) = (1+x^6)/(1+x+x^6). The first difference occurs at a(43) = 415 versus a(43) = 414. - Johannes W. Meijer, Aug 08 2011

Crossrefs

Cf. A101912-A101915, A101917-A101918.

Programs

  • Maple
    nmax:=63: kmax:=nmax: for k from 0 to kmax do A:= proc(x): add(a(n)*x^n, n=0..k) end: f(x):=series(1/(1 + x*A(x^6)),x,k+1); for n from 0 to k do x(n):=coeff(f(x),x,n) od: a(k):=x(k): od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Aug 08 2011
  • PARI
    a(n)=local(A);A=1-x;for(i=1,n\6+1, A=1/(1+x*subst(A,x,x^6)+x*O(x^n)));polcoeff(A,n,x)
  • PARI
    a(n)=local(M=contfracpnqn(concat(1, vector(ceil(log(n+1)/log(6))+1,n,1/x^(6^(n-1)))))); polcoeff(M[1,1]/M[2,1]+x*O(x^(7*n+1)),7*n+1)

Formula

a(0) = 1; a(n) = -Sum_{k=0..floor((n-1)/6)} a(k) * a(n-6*k-1). - Ilya Gutkovskiy, Mar 01 2022

A101915 G.f. satisfies: A(x) = 1/(1 + x*A(x^5)) and also the continued fraction: 1+x*A(x^6) = [1;1/x,1/x^5,1/x^25,1/x^125,...,1/x^(5^(n-1)),...].

Original entry on oeis.org

1, -1, 1, -1, 1, -1, 2, -3, 4, -5, 6, -8, 11, -15, 20, -26, 34, -45, 60, -80, 106, -140, 185, -245, 325, -431, 571, -756, 1001, -1326, 1757, -2329, 3086, -4088, 5415, -7173, 9504, -12593, 16685, -22105, 29284, -38796, 51400, -68100, 90225, -119535, 158365, -209810, 277970, -368275, 487916, -646421, 856416
Offset: 0

Views

Author

Paul D. Hanna, Dec 20 2004

Keywords

Crossrefs

Cf. A101912, A101913, A101914, A101916, A101917, A101918.

Programs

  • PARI
    a(n)=local(A);A=1-x;for(i=1,n\5+1, A=1/(1+x*subst(A,x,x^5)+x*O(x^n)));polcoeff(A,n,x)
  • PARI
    a(n)=local(M=contfracpnqn(concat(1, vector(ceil(log(n+1)/log(5))+1,n,1/x^(5^(n-1)))))); polcoeff(M[1,1]/M[2,1]+x*O(x^(6*n+1)),6*n+1)

Formula

a(0) = 1; a(n) = -Sum_{k=0..floor((n-1)/5)} a(k) * a(n-5*k-1). - Ilya Gutkovskiy, Mar 01 2022
Showing 1-5 of 5 results.

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