A041025 -id:A041025 - cp's OEIS frontend

A121740 Solutions to the Pell equation x^2 - 17y^2 = 1 (y values).

0, 8, 528, 34840, 2298912, 151693352, 10009462320, 660472819768, 43581196642368, 2875698505576520, 189752520171407952, 12520790632807348312, 826182429245113580640, 54515519539544688973928

Offset: 1

Rick L. Shepherd, Jul 31 2006

After initial term this sequence bisects A041025. See A099370 for corresponding x values. a(n+1)/a(n) apparently converges to (4+sqrt(17))^2.

The first solution to the equation x^2 - 17*y^2 = 1 is (X(1); Y(1)) = (1, 0) and the other solutions are defined by: (X(n), Y(n))= (33*X(n-1) + 136*Y(n-1), 8*X(n-1) + 33*Y(n-1)) with n >= 2. - Mohamed Bouhamida, Jan 16 2020

			A099370(1)^2 - 17*a(1)^2 = 33^2 - 17*8^2 = 1089 - 1088 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Tanya Khovanova, Recursive Sequences.
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for linear recurrences with constant coefficients, signature (66,-1).

Cf. A099370, A041025, A040012.

I:=[0, 8]; [n le 2 select I[n] else 66*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 18 2011

LinearRecurrence[{66,-1},{0,8},30] (* Vincenzo Librandi, Dec 18 2011 *)

makelist(expand(((33+8*sqrt(17))^n - (33-8*sqrt(17))^n) /(4*sqrt(17)/2)), n, 0, 16); /* Vincenzo Librandi, Dec 18 2011 */

\\ Program uses fact that continued fraction for sqrt(17) = [4,8,8,...].
print1("0, "); forstep(n=2,40,2,v=vector(n,i,if(i>1,8,4)); print1(contfracpnqn(v)[2,1],", "))

a(n) = ((33+8*sqrt(17))^(n-1) - (33-8*sqrt(17))^(n-1))/(2*sqrt(17)).
From Mohamed Bouhamida, Feb 07 2007: (Start)
a(n) = 65*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 67*(a(n-1) - a(n-2)) + a(n-3). (End)
From Philippe Deléham, Nov 18 2008: (Start)
a(n) = 66*a(n-1) - a(n-2) for n > 1; a(1)=0, a(2)=8.
G.f.: 8*x^2/(1 - 66*x + x^2). (End)
E.g.f.: (1/17)*exp(33*x)*(33*sqrt(17)*sinh(8*sqrt(17)*x) + 136*(1 - cosh(8*sqrt(17)*x))). - Stefano Spezia, Feb 08 2020

Offset changed from 0 to 1 and g.f. adapted by Vincenzo Librandi, Dec 18 2011

A200069 a(n) = 4a(n-1) + 13a(n-2) for n>2, a(1)=1, a(2)=4.

Original entry on oeis.org

1, 4, 29, 168, 1049, 6380, 39157, 239568, 1467313, 8983636, 55009613, 336825720, 2062427849, 12628445756, 77325345061, 473471175072, 2899114186081, 17751582020260, 108694812500093, 665549816263752, 4075231827556217, 24953074921653644, 152790313444845397

Offset: 1

Sture Sjöstedt, Nov 13 2011

De Moivres formula : a(n)=(r^n-s^n)/(r-s), for r>s gives sequences with integers if r and s are conjugates. With r=2+sqrt(17) and s=2-sqrt(17), a(n+1)/a(n) converges to 2+sqrt(17).

			a(3) = 4*4+13*1 = 29.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,13).

Cf. A041025.

a200069 n = a200069_list !! (n-1)
a200069_list = 1 : 4 : zipWith (+)
   (map (* 4) $ tail a200069_list) (map (* 13) a200069_list)
-- Reinhard Zumkeller, Nov 15 2011

Mathematica
```
LinearRecurrence[{4,13}, {1,4}, 50]
```

a(n) = ((2+sqrt(17))^n-(2-sqrt(17))^n)/(2*sqrt(17)).

G.f.: x/(1-4*x-13*x^2). - Bruno Berselli, Nov 15 2011

More terms from Bruno Berselli, Nov 15 2011

A317028 Triangle read by rows: T(0,0) = 1; T(n,k) = 8 * T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, 8, 64, 1, 512, 16, 4096, 192, 1, 32768, 2048, 24, 262144, 20480, 384, 1, 2097152, 196608, 5120, 32, 16777216, 1835008, 61440, 640, 1, 134217728, 16777216, 688128, 10240, 40, 1073741824, 150994944, 7340032, 143360, 960, 1, 8589934592, 1342177280, 75497472, 1835008, 17920, 48

Offset: 0

Zagros Lalo, Jul 19 2018

The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A013615 ((1+8*x)^n) and along skew diagonals pointing top-right in center-justified triangle given in A038279 ((8+x)^n).
The coefficients in the expansion of 1/(1-8x-x^2) are given by the sequence generated by the row sums.
The row sums are Denominators of continued fraction convergents to sqrt(17), see A041025.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 8.12310562561766054982... (a metallic mean), when n approaches infinity (see A176458: (4+sqrt(17))).

			Triangle begins:
1;
8;
64, 1;
512, 16;
4096, 192, 1;
32768, 2048, 24;
262144, 20480, 384, 1;
2097152, 196608, 5120, 32;
16777216, 1835008, 61440, 640, 1;
134217728, 16777216, 688128, 10240, 40;
1073741824, 150994944, 7340032, 143360, 960, 1;
8589934592, 1342177280, 75497472, 1835008, 17920, 48;
68719476736, 11811160064, 754974720, 22020096, 286720, 1344, 1;
549755813888, 103079215104, 7381975040, 251658240, 4128768, 28672, 56;
4398046511104, 893353197568, 70866960384, 2768240640, 55050240, 516096, 1792, 1;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, Pages 70, 98

Zagros Lalo, Left-justified triangle
Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (1 + 8x)^n
Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (8 + x)^n

Row sums give A041025.
Cf. A013615, A038279, A176458.
Cf. A001018 (column 0), A053539 (column 1), A081138 (column 2), A140802 (column 3), A172510 (column 4).

t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, 8 t[n - 1, k] + t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 11}, {k, 0, Floor[n/2]}] // Flatten

T(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, 8*T(n-1, k)+T(n-2, k-1)));
tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n, k), ", ")); print); \\ Michel Marcus, Jul 20 2018

A172346 Triangle t(n,k) read by rows: fibonomial ratios c(n)/(c(k)*c(n-k)) where c are partial products of a generalized Fibonacci sequence with multiplier m=8.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 65, 65, 1, 1, 528, 4290, 528, 1, 1, 4289, 283074, 283074, 4289, 1, 1, 34840, 18678595, 151727664, 18678595, 34840, 1, 1, 283009, 1232504195, 81326315267, 81326315267, 1232504195, 283009, 1, 1, 2298912, 81326598276

Offset: 0

Roger L. Bagula, Feb 01 2010

Start from the generalized Fibonacci sequence A041025 and its partial products c(n) = 1, 1, 8, 520, 274560, 1177587840, 41027160345600,... Then t(n,k) = c(n)/(c(k)*c(n-k)).

Row sums are 1, 2, 10, 132, 5348, 574728, 189154536, 165118204944, 441439547818768, 3130197658239760416, 67978275921898969849504,...

			1;
1, 1;
1, 8, 1;
1, 65, 65, 1;
1, 528, 4290, 528, 1;
1, 4289, 283074, 283074, 4289, 1;
1, 34840, 18678595, 151727664, 18678595, 34840, 1;
1, 283009, 1232504195, 81326315267, 81326315267, 1232504195, 283009, 1;
1, 2298912, 81326598276, 43591056675936, 354094776672518, 43591056675936, 81326598276, 2298912, 1;

Cf. A010048 (m=1), A099927 (m=2), A172345 (m=7).

Clear[f, c, a, t];
f[0, a_] := 0; f[1, a_] := 1;
f[n_, a_] := f[n, a] = a*f[n - 1, a] + f[n - 2, a];
c[n_, a_] := If[n == 0, 1, Product[f[i, a], {i, 1, n}]];
t[n_, m_, a_] := c[n, a]/(c[m, a]*c[n - m, a]);
Table[Table[Table[t[n, m, a], {m, 0, n}], {n, 0, 10}], {a, 1, 10}];
Table[Flatten[Table[Table[t[n, m, a], {m, 0, n}], {n, 0, 10}]], {a, 1, 10}]

A305534 Index of the smallest prime in the n-Fibonacci sequence, or the Lucas U(n,-1) sequence.

Original entry on oeis.org

3, 2, 2, 3, 2, 3, 2, 5, 29, 3, 2, 5, 2, 3, 23, 3, 2, 7, 2, 3, 29, 19, 2, 3, 83, 3, 53, 19, 2, 5, 2, 5, 5, 5479, 71, 3, 2, 17, 11, 3, 2, 37, 2, 31, 5, 11, 2, 5

Offset: 1

Eric Chen, Jun 04 2018

Smallest k such that the k-th Fibonacci polynomial evaluated at x=n is prime. (The first few Fibonacci polynomials are 1, x, x^2 + 1, x^3 + 2*x, x^4 + 3*x^2 + 1, x^5 + 4*x^3 + 3*x, ...)
All terms are primes, since if a divides b, then the a-th term of the n-Fibonacci sequence also divides the b-th term of the n-Fibonacci sequence.
Corresponding primes are 2, 2, 3, 17, 5, 37, 7, 4289, 726120289954448054047428229, 101, 11, 21169, 13, 197, 82088569942721142820383601, 257, 17, 34539049, 19, 401, ...
a(n) = 2 if and only if n is prime.
a(n) = 3 if and only if n^2 + 1 is prime (A005574), except n=2 (since 2 is the only prime p such that p^2 + 1 is also prime).
a(34) > 1024, does a(n) exist for all n >= 1? (However, 17 is the only prime in the first 1024 terms of the 4-Fibonacci sequence, and it seems that 17 is the only prime in the 4-Fibonacci sequence.)
a(35)..a(48) = 71, 3, 2, 17, 11, 3, 2, 37, 2, 31, 5, 11, 2, 5, a(50)..a(54) = 11, 11, 23, 2, 3, a(56) = 3, a(58)..a(75) = 5, 2, 47, 2, 5, 311, 13, 233, 3, 2, 5, 11, 5, 2, 7, 2, 3, 5. Unknown terms a(34), a(49), a(55), a(57), exceed 1024, if they exist.
a(49) > 20000, if it exists. - Giovanni Resta, Jun 06 2018

Cf. A001605, A096650, A209493, which are the indices of the primes in the n-Fibonacci sequence for n = 1, 2, 3.
Cf. A005478, A086383, A201001, which are the primes in the n-Fibonacci sequence for n = 1, 2, 3.
Cf. A000045, A000129, A006190, A001076, A052918, A005668, A054413, A041025, A099371, A041041, A049666, A041061 (the n-Fibonacci sequence for n = 1 to 12).
Cf. A302990 (for n-step Fibonacci sequence instead of n-Fibonacci sequence).

b(n,k)=([n,1;1,0]^k)[1,2]
a(n)=for(k=1,2^12,if(ispseudoprime(b(n,k)),return(k)))

a(34)-a(48) from Giovanni Resta, Jun 06 2018

A322129 Digital roots of A057084.

Original entry on oeis.org

1, 8, 2, 6, 5, 1, 4, 6, 7, 8, 8, 9, 8, 1, 7, 3, 4, 8, 5, 3, 2, 1, 1, 9, 1, 8, 2, 6, 5, 1, 4, 6, 7, 8, 8, 9, 8, 1, 7, 3, 4, 8, 5, 3, 2, 1, 1, 9, 1, 8, 2, 6, 5, 1, 4, 6, 7, 8, 8, 9, 8, 1, 7, 3, 4, 8, 5, 3, 2, 1, 1

Offset: 1

Ondrej Janicko, Nov 27 2018

Periodic with period 24. The cycle is in reverse order to that of the digital roots of the Fibonacci numbers (A030132).

Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

Cf. A010888 (digital root), A057084, A030132 (order of cycle digits reversed), A000045, A015454, A041025, A114479, A164546.

A057084:=[1,8];; for n in [3..80] do A057084[n]:=8*(A057084[n-1]-A057084[n-2]);; od; a:=List(A057084,i->1+(i-1) mod 9); # Muniru A Asiru, Nov 29 2018

digRoot[n_]:=FixedPoint[Total[IntegerDigits[#, 10]] &, n] ; digRoot/@LinearRecurrence[{8, -8}, {1, 8}, 100]  (* Amiram Eldar, Nov 29 2018 *)

a(n) = A010888(A041025(n)) for n > 0.
a(n) = A010888(A057084(n)) for n > 0.
a(n) = A010888(A015454(n+3)) for n > 0.
a(n) = A010888(A114479(n+5)) for n > 0.
a(n) = A010888(A164546(n+3)) for n > 0.
a(n) = A030132(24 - (n mod 24)). - Filip Zaludek, Dec 09 2018

cp's OEIS Frontend

A121740 Solutions to the Pell equation x^2 - 17y^2 = 1 (y values).

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A200069 a(n) = 4*a(n-1) + 13*a(n-2) for n>2, a(1)=1, a(2)=4.

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A317028 Triangle read by rows: T(0,0) = 1; T(n,k) = 8 * T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

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A172346 Triangle t(n,k) read by rows: fibonomial ratios c(n)/(c(k)*c(n-k)) where c are partial products of a generalized Fibonacci sequence with multiplier m=8.

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A305534 Index of the smallest prime in the n-Fibonacci sequence, or the Lucas U(n,-1) sequence.

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PARI

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A322129 Digital roots of A057084.

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A200069 a(n) = 4a(n-1) + 13a(n-2) for n>2, a(1)=1, a(2)=4.