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.

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

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

A367800 G.f. A(x) satisfies A(x) = 1 / (1 - x * A(x^8)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 14, 18, 23, 29, 36, 44, 53, 64, 78, 96, 119, 148, 184, 228, 281, 345, 423, 519, 638, 786, 970, 1198, 1479, 1824, 2247, 2766, 3404, 4190, 5160, 6358, 7837, 9661, 11908, 14674, 18078, 22268, 27428, 33786, 41623
Offset: 0

Views

Author

Seiichi Manyama, Dec 01 2023

Keywords

Comments

a(n) = A005710(n-1) up to n=72, but then the two sequences start to differ. - R. J. Mathar, Dec 04 2023

Links

Crossrefs

Cf. A000621, A367794, A367637.
Cf. A005710, A101918.

Programs

  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=0, (i-1)\8, v[j+1]*v[i-8*j])); v;

Formula

a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/8)} a(k) * a(n-1-8*k).
Showing 1-7 of 7 results.

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