A046717 -id:A046717 - cp's OEIS frontend

A247540 a(n) = 2a(n-1) - 3a(n-1)^2 / a(n-2), with a(0) = a(1) = 1.

1, 1, -1, -5, 65, 2665, -322465, -117699725, 128645799425, 422086867913425, -4153756867136015425, -122639671502190855423125, 10862563623963550637392450625, 2886411268723218638918559372525625, -2300934493386669693418957707961899750625

Offset: 0

Michael Somos, Sep 18 2014

sign

Reinhard Zumkeller, Table of n, a(n) for n = 0..60

Cf. A046717.

a247540 n = a247540_list !! n
a247540_list = 1 : 1 : zipWith (-)
   (map (* 2) xs) (zipWith div (map ((* 3) . (^ 2)) xs) a247540_list)
   where xs = tail a247540_list
-- Reinhard Zumkeller, Sep 20 2014

I:=[1, 1]; [n le 2 select I[n] else 2*Self(n-1) - 3*Self(n-1)^2/Self(n-2): n in [1..30]]; // G. C. Greubel, Aug 04 2018

RecurrenceTable[{a[n] == 2*a[n-1] - 3*a[n-1]^2/a[n-2], a[0]==1, a[1]==1}, a, {n,0,30}] (* G. C. Greubel, Aug 04 2018 *)

{a(n) = if( n<0, 1 / prod(k=1, -n, (1 + (-3)^-k) / 2), prod(k=0, n-1, (1 + (-3)^k) / 2))};

0 = a(n)*(-2*a(n+1) + a(n+2)) + a(n+1)*(+3*a(n+1)) for all n in Z.
a(n+1) = a(n) * (-1)^n * A046717(n) for all n in Z.
a(1-n) = (-3)^(n*(n-1)/2) / a(n) for all n in Z.

A247584 a(n) = 5a(n-1) - 10a(n-2) + 10a(n-3) - 5a(n-4) + 3*a(n-5) with a(0) = a(1) = a(2) = a(3) = a(4) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 13, 43, 113, 253, 509, 969, 1849, 3719, 8009, 18027, 40897, 91257, 198697, 423777, 894081, 1886011, 4007301, 8594411, 18560081, 40181493, 86872293, 187197193, 402060793, 861827743, 1846685729, 3960390059, 8504658049, 18283290609, 39325827729

Offset: 0

Alexander Samokrutov, Sep 20 2014

a(n)/a(n-1) tends to 2.1486... = 1 + 2^(1/5), the real root of the polynomial x^5 - 5*x^4 + 10*x^3 - 10*x^2 + 5*x - 3.
If x^5 = 2 and n >= 0, then there are unique integers a, b, c, d, g  such that (1 + x)^n = a + b*x + c*x^2 + d*x^3 + g*x^4. The coefficient a is a(n) (from A052102). - Alexander Samokrutov, Jul 11 2015
If x=a(n), y=a(n+1), z=a(n+2), s=a(n+3), t=a(n+4) then x, y, z, s, t satisfies Diophantine equation (see link). - Alexander Samokrutov, Jul 11 2015

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..48 from Alexander Samokrutov)
Alexander Samokrutov, Diophantine equation
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,3).

The following sequences belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A052101, A052102, A052103, A063727, A083098, A083099, A083100, A084057, A093406, A247344.

Cf. A005531.

[n le 5 select 1 else 5*Self(n-1) -10*Self(n-2) +10*Self(n-3) -5*Self(n-4) +3*Self(n-5): n in [1..40]]; // Vincenzo Librandi, Jul 11 2015

m:=50; S:=series( (1-x)^4/(1 -5*x +10*x^2 -10*x^3 +5*x^4 -3*x^5), x, m+1):
seq(coeff(S, x, j), j=0..m); # G. C. Greubel, Apr 15 2021

LinearRecurrence[{5,-10,10,-5,3}, {1,1,1,1,1}, 50] (* Vincenzo Librandi, Jul 11 2015 *)

makelist(sum(2^k*binomial(n,5*k), k, 0, floor(n/5)), n, 0, 50); /* Alexander Samokrutov, Jul 11 2015 */

Vec((1-x)^4/(1-5*x+10*x^2-10*x^3+5*x^4-3*x^5) + O(x^100)) \\ Colin Barker, Sep 22 2014

[sum(2^j*binomial(n, 5*j) for j in (0..n//5)) for n in (0..50)] # G. C. Greubel, Apr 15 2021

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + 3*a(n-5).
a(n) = Sum_{k=0...floor(n/5)} (2^k*binomial(n,5*k)). - Alexander Samokrutov, Jul 11 2015
G.f.: (1-x)^4/(1 -5*x +10*x^2 -10*x^3 +5*x^4 -3*x^5). - Colin Barker, Sep 22 2014

A292848 a(n) is the smallest prime of form (1/2)((1 + sqrt(2n))^k + (1 - sqrt(2*n))^k).

Original entry on oeis.org

3, 5, 7, 113, 11, 13, 43, 17, 19, 61, 23, 73, 79, 29, 31, 97, 103, 37, 1241463763, 41, 43, 664973, 47, 2593, 151, 53, 163, 14972833, 59, 61, 4217, 193, 67, 23801, 71, 73, 223, 229, 79, 241, 83, 7561, 61068909859, 89, 271, 277, 283, 97, 10193, 101, 103, 313

Offset: 1

XU Pingya, Sep 24 2017

nonn

When 2n + 1 = p is prime, a(n) = p.
From Robert Israel, Sep 26 2017: (Start)
a(n) is also the first prime in the sequence defined by the recursion x(k+2)=2*x(k+1)+(2*n-1)*x(k) with x(0)=x(1)=1.
a(307), if it exists, has more than 10000 digits.
It appears that x(n*k) is divisible by x(k) if n is odd.  Thus a(n) (if it exists) must be x(k) where k is either a power of 2 or a prime. (End)

			For k = {1, 2, 3, 4}, (1/2)((1 + sqrt(8))^k + (1 - sqrt(8))^k) = {1, 9, 25, 113}. 113 is prime, so a(4) = 113.

Robert Israel, Table of n, a(n) for n = 1..306

Cf. A001333, A026150, A046717, A084057, A002533, A083098, A083100, A003665, A002535, A133294, A090042, A125816, A133343, A133345, A120612, A133356, A125818.

f:= proc(n) local a,b,t;
  a:= 1; b:= 1;
  do
    t:= a; a:= 2*a + (2*n-1)*b;
    if isprime(a) then return a fi;
    b:= t;
  od
end proc:
map(f, [$1..100]); # Robert Israel, Sep 26 2017

f[n_, k_] := ((1 + Sqrt[n])^k + (1 - Sqrt[n])^k)/2;
Table[k = 1; While[! PrimeQ[Expand@f[2n, k]], k++]; Expand@f[2n, k], {n, 52}]

A373547 Triangle read by rows: T(n,k) = 4^k*binomial(n+k, n-k) with 0 <= k <= n.

Original entry on oeis.org

1, 1, 4, 1, 12, 16, 1, 24, 80, 64, 1, 40, 240, 448, 256, 1, 60, 560, 1792, 2304, 1024, 1, 84, 1120, 5376, 11520, 11264, 4096, 1, 112, 2016, 13440, 42240, 67584, 53248, 16384, 1, 144, 3360, 29568, 126720, 292864, 372736, 245760, 65536, 1, 180, 5280, 59136, 329472, 1025024, 1863680, 1966080, 1114112, 262144

Offset: 0

Stefano Spezia, Jun 09 2024

T(n,k) is the number of occurrences of the periodic substring (01)^k in the periodic string (0011)^n (see Proposition 4.3 at page 6 in Fang).
The word (w_1, w_2, ..., w_r)^m is defined as the word obtained by concatenating (w_1, w_2, ..., w_r) m times.
A word w' = (w'1, w'_2, ..., w'_s) is said be a subword of a given word w = (w_1, w_2, ..., w_r), if there is some set P = {p_1 < ... < p_s} of integers from 1 to r satisfying w{p_j} = w'_j for all 1 <= j <= s, and we call the set P an occurrence of w' in w (see Preliminaries section at pp. 2-3 in Fang).

			The triangle begins as:
  1;
  1,  4;
  1, 12,   16;
  1, 24,   80,   64;
  1, 40,  240,  448,   256;
  1, 60,  560, 1792,  2304,  1024;
  1, 84, 1120, 5376, 11520, 11264, 4096;
  ...
T(2,1) = 12 since there are 12 occurrences of (01)^1 = 01 in (0011)^2 = 00110011: {1, 3}, {1, 4}, {1, 7}, {1, 8}, {2, 3}, {2, 4}, {2, 7}, {2, 8}, {5, 7}, {5, 8}, {6, 7}, {6, 8}.

Wenjie Fang, Maximal number of subword occurrences in a word, arXiv:2406.02971 [math.CO], 2024.

Cf. A000012 (k=0), A000302 (diagonal), A001653 (row sums), A046092 (k=1), A046717, A085478, A130810, A130812, A373628.

T[n_,k_]:=4^k Binomial[n+k,n-k]; Table[T[n,k],{n,0,9},{k,0,n}]//Flatten (* or *)
T[n_,k_]:=SeriesCoefficient[(1-x)/((1-x)^2-4x y),{x,0,n},{y,0,k}]; Table[T[n,k],{n,0,9},{k,0,n}]//Flatten

G.f.: (1 - x)/((1 - x)^2 - 4*x*y).
T(n,k) = A000302(k)*A085478(n,k).
Sum_{k=0..n} T(n-k,k) = A046717(n).
T(n,2) = A130810(n+2).
T(n,3) = A130812(n+3).

A067763 Square array read by antidiagonals of base n numbers written as 122...222 with k 2's (and a suitable interpretation for n=0, 1 or 2).

Original entry on oeis.org

1, 2, 1, 2, 3, 1, 2, 5, 4, 1, 2, 7, 10, 5, 1, 2, 9, 22, 17, 6, 1, 2, 11, 46, 53, 26, 7, 1, 2, 13, 94, 161, 106, 37, 8, 1, 2, 15, 190, 485, 426, 187, 50, 9, 1, 2, 17, 382, 1457, 1706, 937, 302, 65, 10, 1, 2, 19, 766, 4373, 6826, 4687, 1814, 457, 82, 11, 1, 2, 21, 1534, 13121

Offset: 0

Henry Bottomley, Feb 06 2002

Start with a node; step one is to connect that node to n+1 new nodes so that it is of degree n+1; further steps are to connect each existing node of degree 1 to n new nodes so that it is of degree n+1; T(n,k) is the total number of nodes after k steps.

			Rows start: 1,2,2,2,2,2,...; 1,3,5,7,9,11,...; 1,4,10,22,46,94,...; 1,5,17,53,161,485,... T(3,2) =122 base 3 =17.

Rows include A040000, A005408, A033484, A048473, A020989, A057651, A061801 etc. For negative n (not shown) absolute values of rows would effectively include A000012, A014113, A046717.

T(n, k) =((n+1)*n^k-2)/(n-1) [with T(1, k)=2k+1] =n*T(n, k-1)+2 =(n+1)*T(n, k-1)-n*T(n, k-2) =T(n, k-1)+(1+1/n)*n^k =A055129(k, n)+A055129(k-1, n). Coefficient of x^k in expansion of (1+x)/((1-x)(1-nx)).

A111009 Starting with the fraction 1/1, the prime numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times bottom to get the new top.

Original entry on oeis.org

5, 13, 41, 1093, 797161, 21523361, 926510094425921, 1716841910146256242328924544641, 3754733257489862401973357979128773, 6957596529882152968992225251835887181478451547013

Offset: 1

Cino Hilliard, Oct 02 2005

Or, A046717(n) is prime.

Is this sequence infinite?

Prime Obsession, John Derbyshire, Joseph Henry Press, April 2004, p 16.

Cf. A088553. [From R. J. Mathar, Aug 18 2008]

Select[NestList[(Numerator[#]+4*Denominator[#])/(Numerator[#]+Denominator[#])&,1/1,200]//Numerator,PrimeQ] (* Harvey P. Dale, Jan 04 2024 *)

primenum(n,k,typ) = \ k=mult,typ=1 num,2 denom. ouyput prime num or denom. { local(a,b,x,tmp,v); a=1;b=1; for(x=1,n, tmp=b; b=a+b; a=k*tmp+a; if(typ==1,v=a,v=b); if(isprime(v),print1(v","); ) ); print(); print(a/b+.) }

Given c(0)=1, b(0)=1 then for i=1, 2, .. c(i)/b(i) = (c(i-1)+4*b(i-1)) /(c(i-1) + b(i-1)).

A046717 INTERSECT A000040. [From R. J. Mathar, Aug 18 2008]

Edited by N. J. A. Sloane, Aug 23 2008

cp's OEIS Frontend

A247540 a(n) = 2*a(n-1) - 3*a(n-1)^2 / a(n-2), with a(0) = a(1) = 1.

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A247584 a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + 3*a(n-5) with a(0) = a(1) = a(2) = a(3) = a(4) = 1.

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A292848 a(n) is the smallest prime of form (1/2)*((1 + sqrt(2*n))^k + (1 - sqrt(2*n))^k).

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A373547 Triangle read by rows: T(n,k) = 4^k*binomial(n+k, n-k) with 0 <= k <= n.

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A067763 Square array read by antidiagonals of base n numbers written as 122...222 with k 2's (and a suitable interpretation for n=0, 1 or 2).

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A111009 Starting with the fraction 1/1, the prime numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times bottom to get the new top.

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A247540 a(n) = 2a(n-1) - 3a(n-1)^2 / a(n-2), with a(0) = a(1) = 1.

A247584 a(n) = 5a(n-1) - 10a(n-2) + 10a(n-3) - 5a(n-4) + 3*a(n-5) with a(0) = a(1) = a(2) = a(3) = a(4) = 1.

A292848 a(n) is the smallest prime of form (1/2)((1 + sqrt(2n))^k + (1 - sqrt(2*n))^k).