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 11 results. Next

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

A101913 G.f. satisfies: A(x) = 1/(1 + x*A(x^3)) and also the continued fraction: 1+x*A(x^4) = [1;1/x,1/x^3,1/x^9,1/x^27,...,1/x^(3^(n-1)),...].

Original entry on oeis.org

1, -1, 1, -1, 2, -3, 4, -6, 9, -13, 19, -28, 41, -61, 90, -132, 195, -288, 424, -625, 922, -1359, 2004, -2955, 4356, -6423, 9471, -13963, 20587, -30355, 44755, -65987, 97293, -143449, 211503, -311844, 459785, -677912, 999524, -1473709, 2172854, -3203685, 4723551, -6964461, 10268490, -15139986
Offset: 0

Views

Author

Paul D. Hanna, Dec 20 2004

Keywords

Crossrefs

Cf. A101912, A101914-A101918.

Programs

  • PARI
    {a(n)=local(A);A=1-x;for(i=1,n\3+1, A=1/(1+x*subst(A,x,x^3)+x*O(x^n)));polcoeff(A,n,x)}
  • PARI
    {a(n)=local(M=contfracpnqn(concat(1, vector(ceil(log(n+1)/log(3))+1,n,1/x^(3^(n-1)))))); polcoeff(M[1,1]/M[2,1]+x*O(x^(4*n+1)),4*n+1)}

Formula

From Joerg Arndt, Oct 15 2011: (Start)
For the sequence abs(a(n)) we have
g.f. B(x) 1/(1-x/(1-x^3/(1-x^9/(1-x^27(1- ... ))))) and
B(x) satisfies B(x) = 1 + x*B(x)*B(x^3) (cf. A000621)
(End)
G.f.: T(0), where T(k) = 1 - (-x)^(3^k)/((-x)^(3^k) - 1/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 18 2013
a(0) = 1; a(n) = -Sum_{k=0..floor((n-1)/3)} a(k) * a(n-3*k-1). - Ilya Gutkovskiy, Mar 01 2022

A101917 G.f. satisfies: A(x) = 1/(1 + x*A(x^7)) and also the continued fraction: 1 + x*A(x^8) = [1; 1/x, 1/x^7, 1/x^49, 1/x^343, ..., 1/x^(7^(n-1)), ...].

Original entry on oeis.org

1, -1, 1, -1, 1, -1, 1, -1, 2, -3, 4, -5, 6, -7, 8, -10, 13, -17, 22, -28, 35, -43, 53, -66, 83, -105, 133, -168, 211, -264, 330, -413, 518, -651, 819, -1030, 1294, -1624, 2037, -2555, 3206, -4025, 5055, -6349, 7973, -10010, 12565, -15771, 19796, -24851, 31200, -39173, 49183, -61748, 77519, -97315, 122166
Offset: 0

Views

Author

Paul D. Hanna, Dec 20 2004

Keywords

Crossrefs

Cf. A101912, A101913, A101914, A101915, A101916, A101918.

Programs

  • Maple
    nmax:=57: kmax:=nmax: for k from 0 to kmax do A:= proc(x): add(A101917(n)*x^n, n=0..k) end: f(x):=series(1/(1 + x*A(x^7)),x,k+1); for n from 0 to k do x(n):=coeff(f(x),x,n) od: A101917(k):=x(k): od: seq(A101917(n), n=0..nmax); # Johannes W. Meijer, Aug 08 2011
  • Mathematica
    m = 57; A[] = 0; Do[A[x] = 1/(1 + x A[x^7]) + 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\7+1, A=1/(1+x*subst(A,x,x^7)+x*O(x^n)));polcoeff(A,n,x)
  • PARI
    a(n)=local(M=contfracpnqn(concat(1, vector(ceil(log(n+1)/log(7))+1,n,1/x^(7^(n-1)))))); polcoeff(M[1,1]/M[2,1]+x*O(x^(8*n+1)),8*n+1)

Formula

This was conjectured to have g.f. (1+x^7) / (1+x+x^7) by Ralf Stephan, May 17 2007, but this is wrong. This g.f. produces a sequence which differs at a(57) = -153367. The g.f. gives a(57) = -153366. - Johannes W. Meijer, Aug 08 2011
a(0) = 1; a(n) = -Sum_{k=0..floor((n-1)/7)} a(k) * a(n-7*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

A218031 G.f. A(x) satisfies A(x) = 1 + x / A(x^2).

Original entry on oeis.org

1, 1, 0, -1, 0, 1, 0, 0, 0, -1, 0, 1, 0, 0, 0, -2, 0, 3, 0, -1, 0, -3, 0, 6, 0, -4, 0, -4, 0, 12, 0, -10, 0, -5, 0, 23, 0, -25, 0, -2, 0, 43, 0, -57, 0, 12, 0, 74, 0, -124, 0, 56, 0, 120, 0, -258, 0, 172, 0, 170, 0, -516, 0, 454, 0, 187, 0, -989, 0, 1095, 0, 40, 0, -1811, 0, 2487
Offset: 0

Views

Author

Joerg Arndt, Oct 18 2012

Keywords

Links

Crossrefs

Cf. A101912, A238429.

Programs

  • Maple
    P:= 1+x: d:= 1:
while d < 127 do
  P:= convert(series(1+x/subs(x=x^2,P),x,2+2*d),polynom);
  d:= 1+2*d;
od:
seq(coeff(P,x,i),i=0..d); # Robert Israel, Mar 13 2018
  • Mathematica
    nmax = 75; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x] - (1 + x/A[x^2]) + 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 01 2019 *)
  • PARI
    N=166;  R=O('x^N);  x='x+R;
A= 1; for (k=1,N+1, A = 1 + x / subst(A,'x,'x^2) + R; );
Vec(A)

Formula

G.f. A(x) = 1/B(x) = 1 + x*B(x^2) where B(x) is the g.f. of A101912.
G.f.: 1+x/(1+x^2/(1+x^4/(1+x^8/(1+ ...)))) (continued fraction).
(A(x) + 1) / (A(x) - 1) = 1 + 2*A(x^2) / x. [Joerg Arndt, Feb 28 2014]
A(x^3) = F(x) - x where F(x) is the g.f. of A238429. [Joerg Arndt, Feb 28 2014]

A352009 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} (-2)^k * a(k) * a(n-2*k-1).

Original entry on oeis.org

1, 1, 1, -1, -3, -1, 5, 23, 29, -33, -139, -217, 13, 943, 1765, -1545, -8963, -11265, 6229, 73671, 126701, -65713, -567611, -793449, 415197, 4231583, 7471669, -4933529, -37928499, -52823313, 28920485, 346647351, 610476733, -316142785, -2913394667, -4922323705
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 28 2022

Keywords

Crossrefs

Cf. A015097, A101912, A352010, A352011.

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = Sum[(-2)^k a[k] a[n - 2 k - 1], {k, 0, Floor[(n - 1)/2]}]; Table[a[n], {n, 0, 35}]
nmax = 35; A[] = 0; Do[A[x] = 1/(1 - x A[-2 x^2]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 - x * A(-2*x^2)).

A352010 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} (-3)^k * a(k) * a(n-2*k-1).

Original entry on oeis.org

1, 1, 1, -2, -5, 1, 16, 94, 127, -317, -1103, -2258, 160, 18922, 39664, -163064, -524489, -187577, 1711591, 12353800, 18926119, -53449661, -207918068, -15689834, 960812728, 3295064650, 3870965368, -33248209286, -96830084768, 17122290256, 505711364524
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 28 2022

Keywords

Crossrefs

Cf. A015098, A101912, A352009, A352011.

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = Sum[(-3)^k a[k] a[n - 2 k - 1], {k, 0, Floor[(n - 1)/2]}]; Table[a[n], {n, 0, 30}]
nmax = 30; A[] = 0; Do[A[x] = 1/(1 - x A[-3 x^2]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 - x * A(-3*x^2)).

A352011 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} (-4)^k * a(k) * a(n-2*k-1).

Original entry on oeis.org

1, 1, 1, -3, -7, 5, 33, 269, 393, -1451, -4815, -14115, -2791, 171685, 398145, -3887699, -10399319, 6567925, 63031889, 558518141, 853157689, -4400392635, -14954126751, 29904043597, 151457170889, 344861133205, 170895616881, -12627954103779, -30049168949927
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 28 2022

Keywords

Crossrefs

Cf. A015099, A101912, A352009, A352010.

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = Sum[(-4)^k a[k] a[n - 2 k - 1], {k, 0, Floor[(n - 1)/2]}]; Table[a[n], {n, 0, 28}]
nmax = 28; A[] = 0; Do[A[x] = 1/(1 - x A[-4 x^2]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 - x * A(-4*x^2)).
Showing 1-10 of 11 results. Next

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