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 1-3 of 3 results.

A158985 Coefficients of polynomials (in descending powers of x) P(n,x) := 1 + P(n-1,x)^2, where P(1,x) = x + 1.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 4, 8, 8, 5, 1, 8, 32, 80, 138, 168, 144, 80, 26, 1, 16, 128, 672, 2580, 7664, 18208, 35296, 56472, 74944, 82432, 74624, 54792, 31776, 13888, 4160, 677, 1, 32, 512, 5440, 43048, 269920, 1393728, 6082752, 22860480, 75010560, 217147904
Offset: 1

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Author

Clark Kimberling, Apr 02 2009

Keywords

Examples

			Row 1: 1 1 (from x + 1)
Row 2: 1 2 2 (from x^2 + 2*x + 2)
Row 3: 1 4 8 8 5
Row 4: 1 8 32 80 138 168 144 80 26
		

Crossrefs

Programs

  • PARI
    tabf(nn) = {my(P = x+1); print(Vec(P)); for (n=1, nn, P = 1 + P^2; print(Vec(P)););} \\ Michel Marcus, Jul 01 2015

Formula

From Peter Bala, Jul 01 2015: (Start)
P(n,x) = P(n,-2 - x) for n >= 2.
P(n+1,x)= P(n,(1 + x)^2). Thus if alpha is a zero of P(n,x) then sqrt(alpha) - 1 is a zero of P(n+1,x).
Define a sequence of polynomials Q(n,x) by setting Q(1,x) = 1 + x^2 and Q(n,x) = Q(n-1, 1 + x^2) for n >= 2. Then P(n,x) = Q(n,sqrt(x)).
Q(n,x) = Q(k,Q(n-k,x)) for 1 <= k <= n-1; P(n,x) = P(k,P(n-k,x)^2) for 1 <= k <= n - 1.
n-th row sum = P(n,1) = A003095(n+1);
P(n,1) = P(n+1,0) = P(n+1,-2); P(n,1) = P(n,-3) for n >= 2.
P(n,2) = A062013(n). (End)

A088674 Coefficients of the eigenfunction of a sequence transformation.

Original entry on oeis.org

1, 3, 6, 45, 126, 750, 2796, 19389, 75894, 449562, 2027796, 12211794, 57895596, 332787324, 1677545304, 9766642077, 50378641830, 286825948194, 1529968671492, 8729259097158, 47374697101572, 269062276076868, 1484430536591592
Offset: 0

Views

Author

Michael Somos, Oct 04 2003

Keywords

Comments

G.f. A(x) satisfies A(x^2) = (A(x/2)-1)/x - A(x/2)^2/2.
B(x) := 1/(2*x) - x*A(x^2) satisfies B(x)^2 + 1 = B(2*x^2).
Define f(n, c) := x - Sum_{k>=0} a(k)/(2*x)^(2*k+1) where x = c^(2^n). Then A003095(n+1) = A004019(n) + 1 = f(n, 1.502836801...). Also, A062013(n) = f(n, 1.78050350...). - Michael Somos, Jun 07 2021

Examples

			G.f. = A(x) = 1 + 3*x + 6*x^2 + 45*x^3 + 126*x^4 + 750*x^5 + 2796*x^6 + ...
B(x) = 1/(2*x) - x - 3*x^3 - 6*x^5 - 45*x^7 - 126*x^9 - 750*x^11 - ... - _Michael Somos_, Jul 11 2019
		

Crossrefs

Programs

  • Mathematica
    a[ n_] := If[n < 0, 0, Module[{A = 1 + O[x], m = 2}, While[m < n + 2, m *= 2; A = (Normal[ 1/x - Sqrt[ 1/x^2 - 2/x - 2*(Normal[A] /. x -> x^2) + O[x]^(m - 2)]] /. x -> 2*x) + O[x]^(m - 1) //PowerExpand]; SeriesCoefficient[A, n]]]; (* Michael Somos, Jun 07 2021 *)
  • PARI
    {a(n) = my(A, m); if( n < 0, 0, m=2; A = 1 + O(x); while( m < n+2, m*=2; A = subst(1/x - sqrt(2*(subst((1/2)/x - A, x, x^2) - 1/x)), x, 2*x)); polcoeff(A, n))};

A135378 Main diagonal of "square and add k" array.

Original entry on oeis.org

2, 5, 38, 2707, 21418388, 3000279372337641, 255122481276683701099886061668842
Offset: 0

Views

Author

Jonathan Vos Post, Dec 09 2007

Keywords

Comments

Array of recurrence "start with 2, square and add k" begins:
k..|.A[k,n]=A[k,n-1]^2 + k
-1.|.2..3...8....63.......3968..15745023.247905749270528.............A003096
0..|.2..4..16...256......65536..4294967296.18446744073709551616......A001146
1..|.2..5..26...677.....458330..210066388901.44127887745906175987802.A003095
2..|.2..6..38..1446....2090918.4371938082726...19113842599189892819591078...
3..|.2..7..52..2707....7327852.53697414933907..2883412370584178505178284652.
4..|.2..8..68..4628...21418388.458747344518548.210449126102819371741916028308.
5..|.2..9..86..7401...54774806.3000279372337641.9001676312074749038996905444886.
6..|.2.10.106.11242..126382570.1597255405035792810...
7..|.2.11.128.16391..268664888.72180822044052551...
8..|.2.12.152.23112..534164552.285331768613360712..
9..|.2.13.178.31693.1004446258.1008912285210202573.
10.|.2.14.206.42446.1801662926.3245989298922881486.

Crossrefs

Programs

  • Mathematica
    A[k_,0] = 2; A[k_,n_] := A[k,n] = A[k, n-1]^2 + k; a[n_] := A[n, n]; a /@ Range[0, 6] (* Giovanni Resta, Jun 20 2016 *)

Formula

a(n) = A[n,n] where A[k,n] = n-th term of recurrence A[k,0] = 2, A[k,n] = A[k,n-1]^2 + k.

Extensions

Corrected and edited by Giovanni Resta, Jun 20 2016
Showing 1-3 of 3 results.