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 1-4 of 4 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

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

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 1-4 of 4 results.

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