A006131 -id:A006131 - cp's OEIS frontend

A231775 Triangular array read by rows: row n shows the coefficients of the polynomial u(n) = c(0) + c(1)x + ... + c(n)x^(n) which is the denominator of the n-th convergent of the continued fraction [k, k, k, ... ], where k = (x + 1)/(x + 2).

2, 1, 2, 3, 1, 10, 17, 10, 2, 18, 47, 45, 19, 3, 58, 173, 210, 129, 40, 5, 130, 491, 769, 642, 302, 76, 8, 362, 1545, 2850, 2940, 1830, 687, 144, 13, 882, 4391, 9565, 11925, 9315, 4671, 1469, 265, 21, 2330, 12901, 31898, 46195, 43170, 26994, 11294, 3049, 482

Offset: 1

Clark Kimberling, Nov 13 2013

Sum of numbers in row n: 3*A002534(n). Left edge: 2*A006131. Right edge: A000045 (Fibonacci numbers).

			First 3 rows:
2 .... 1
2 .... 3 .... 1
10 ... 17 ... 10 ... 2
First 3 polynomials:  2 + x, 2 + 3*x + x^2, 10 + 17*x + 10*x^2 + 2*x^3.

Cf. A230000, A231774, A000045.

t[n_] := t[n] = Table[(x + 1)/(x + 2), {k, 0, n}];
b = Table[Factor[Convergents[t[n]]], {n, 0, 10}];
p[x_, n_] := p[x, n] = Last[Expand[Denominator[b]]][[n]];
u = Table[p[x, n], {n, 1, 10}]
v = CoefficientList[u, x]; Flatten[v]

A242763 a(n) = 1 for n <= 7; a(n) = a(n-5) + a(n-7) for n>7.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 7, 7, 8, 9, 9, 12, 12, 15, 16, 17, 21, 21, 27, 28, 32, 37, 38, 48, 49, 59, 65, 70, 85, 87, 107, 114, 129, 150, 157, 192, 201, 236, 264, 286, 342, 358, 428, 465, 522, 606, 644, 770, 823, 950, 1071, 1166, 1376

Offset: 1

Vicente Jesús Maniega Granado, Oct 03 2016

Generalized Fibonacci growth sequence using i = 2 as maturity period, j = 5 as conception period, and k = 2 as growth factor.

Maturity period is the number of periods that a Fibonacci tree node needs for being able to start developing branches. Conception period is the number of periods in a Fibonacci tree node needed to develop new branches since its maturity. Growth factor is the number of additional branches developed by a Fibonacci tree node, plus 1, and equals the base of the exponential series related to the given tree if maturity factor would be zero. Standard Fibonacci would use 1 as maturity period, 1 as conception period, and 2 as growth factor as the series becomes equal to 2^n with a maturity period of 0. Related to Lucas sequences.

			For n = 13 the a(13) = a(8) + a(6) = 2 + 1 = 3.

Colin Barker, Table of n, a(n) for n = 1..1000
Julia Collins, Fibonacci Tree
Fractal Foundation, Fibonacci Fractals
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1,0,1).

Cf. A000079 (i = 0, j = 1, k = 2), A000244 (i = 0, j = 1, k = 3), A000302 (i = 0, j = 1, k = 4), A000351 (i = 0, j = 1, k = 5), A000400 (i = 0, j = 1, k = 6), A000420 (i = 0, j = 1, k = 7), A001018 (i = 0, j = 1, k = 8), A001019 (i = 0, j = 1, k = 9), A011557 (i = 0, j = 1, k = 10), A001020 (i = 0, j = 1, k = 11), A001021 (i = 0, j = 1, k = 12), A016116 (i = 0, j = 2, k = 2), A108411 (i = 0, j = 2, k = 3), A213173 (i = 0, j = 2, k = 4), A074872 (i = 0, j = 2, k = 5), A173862 (i = 0, j = 3, k = 2), A127975 (i = 0, j = 3, k = 3), A200675 (i = 0, j = 4, k = 2), A111575 (i = 0, j = 4, k = 3), A000045 (i = 1, j = 1, k = 2), A001045 (i = 1, j = 1, k = 3), A006130 (i = 1, j = 1, k = 4), A006131 (i = 1, j = 1, k = 5), A015440 (i = 1, j = 1, k = 6), A015441 (i = 1, j = 1, k = 7), A015442 (i = 1, j = 1, k = 8), A015443 (i = 1, j = 1, k = 9), A015445 (i = 1, j = 1, k = 10), A015446 (i = 1, j = 1, k = 11), A015447 (i = 1, j = 1, k = 12), A000931 (i = 1, j = 2, k = 2), A159284 (i = 1, j = 2, k = 3), A238389 (i = 1, j = 2, k = 4), A097041 (i = 1, j = 2, k = 10), A079398 (i = 1, j = 3, k = 2), A103372 (i = 1, j = 4, k = 2), A103373 (i = 1, j = 5, k = 2), A103374 (i = 1, j = 6, k = 2), A000930 (i = 2, j = 1, k = 2), A077949 (i = 2, j = 1, k = 3), A084386 (i = 2, j = 1, k = 4), A089977 (i = 2, j = 1, k = 5), A178205 (i = 2, j = 1, k = 11), A103609 (i = 2, j = 2, k = 2), A077953 (i = 2, j = 2, k = 3), A226503 (i = 2, j = 3, k = 2), A122521 (i = 2, j = 6, k = 2), A003269 (i = 3, j = 1, k = 2), A052942 (i = 3, j = 1, k = 3), A005686 (i = 3, j = 2, k = 2), A237714 (i = 3, j = 2, k = 3), A238391 (i = 3, j = 2, k = 4), A247049 (i = 3, j = 3, k = 2), A077886 (i = 3, j = 3, k = 3), A003520 (i = 4, j = 1, k = 2), A108104 (i = 4, j = 2, k = 2), A005708 (i = 5, j = 1, k = 2), A237716 (i = 5, j = 2, k = 3), A005709 (i = 6, j = 1, k = 2), A122522 (i = 6, j = 2, k = 2), A005710 (i = 7, j = 1, k = 2), A237718 (i = 7, j = 2, k = 3), A017903 (i = 8, j = 1, k = 2).

[n le 7 select 1 else Self(n-5)+Self(n-7): n in [1..70]]; // Vincenzo Librandi, Nov 30 2016

LinearRecurrence[{0, 0, 0, 0, 1, 0, 1}, {1, 1, 1, 1, 1, 1, 1}, 70] (*  or *)
CoefficientList[ Series[(1+x+x^2+x^3+x^4)/(1-x^5-x^7), {x, 0, 70}], x] (* Robert G. Wilson v, Nov 25 2016 *)
nxt[{a_,b_,c_,d_,e_,f_,g_}]:={b,c,d,e,f,g,a+c}; NestList[nxt,{1,1,1,1,1,1,1},70][[;;,1]] (* Harvey P. Dale, Oct 22 2024 *)

Vec(x*(1+x+x^2+x^3+x^4)/((1-x+x^2)*(1+x-x^3-x^4-x^5)) + O(x^100)) \\ Colin Barker, Oct 27 2016

@CachedFunction # a = A242763
def a(n): return 1 if n<8 else a(n-5) +a(n-7)
[a(n) for n in range(1,76)] # G. C. Greubel, Oct 23 2024

Generic a(n) = 1 for n <= i+j; a(n) = a(n-j) + (k-1)*a(n-(i+j)) for n>i+j where i = maturity period, j = conception period, k = growth factor.
G.f.: x*(1+x+x^2+x^3+x^4) / ((1-x+x^2)*(1+x-x^3-x^4-x^5)). - Colin Barker, Oct 09 2016
Generic g.f.: x*(Sum_{l=0..j-1} x^l) / (1-x^j-(k-1)*x^(i+j)), with i > 0, j > 0 and k > 1.

A305834 Triangle read by rows: T(0,0)= 1; T(n,k)= T(n-1,k) + 4*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, 1, 1, 4, 1, 8, 1, 12, 16, 1, 16, 48, 1, 20, 96, 64, 1, 24, 160, 256, 1, 28, 240, 640, 256, 1, 32, 336, 1280, 1280, 1, 36, 448, 2240, 3840, 1024, 1, 40, 576, 3584, 8960, 6144, 1, 44, 720, 5376, 17920, 21504, 4096

Offset: 0

Shara Lalo, Jun 11 2018

The numbers in rows of the triangle are along skew diagonals pointing top-right in center-justified triangle given in A013611 ((1+4*x)^n).
The coefficients in the expansion of 1/(1-x-4*x^2) are given by the sequence generated by the row sums.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 2.5615528128...: A222132 (sqrt(4 + sqrt(4 + sqrt(4 + sqrt(4 + ... ))))), when n approaches infinity.

			Triangle begins:
1;
1;
1,  4;
1,  8;
1, 12,   16;
1, 16,   48;
1, 20,   96,    64;
1, 24,  160,   256;
1, 28,  240,   640,    256;
1, 32,  336,  1280,   1280;
1, 36,  448,  2240,   3840,   1024;
1, 40,  576,  3584,   8960,   6144;
1, 44,  720,  5376,  17920,  21504,    4096;
1, 48,  880,  7680,  32256,  57344,   28672;
1, 52, 1056, 10560,  53760, 129024,  114688,   16384;
1, 56, 1248, 14080,  84480, 258048,  344064,  131072;
1, 60, 1456, 18304, 126720, 473088,  860160,  589824,  65536;
1, 64, 1680, 23296, 183040, 811008, 1892352, 1966080, 589824;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 371, 372.

Shara Lalo, Right justified triangle
Shara Lalo, Skew diagonals in triangle A013611

Row sums give A006131.
Cf. A000012 (column 0), A008586 (column 1), A035008 (column 2), A141478 (column 3), A120054 (column 4).
Cf. A013611.
Cf. A222132.

t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, t[n - 1, k] + 4 t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}] // Flatten

G.f.: 1/(1 - t*x - 4*t^2).

Column k is binomial (n + k - 1, k) * 4^k.

A330795 Evaluation of the polynomials given by the Riordan square of the Fibonacci sequence with a(0) = 1 (A193737) at 1/2 and normalized with 2^n.

Original entry on oeis.org

1, 3, 9, 39, 153, 615, 2457, 9831, 39321, 157287, 629145, 2516583, 10066329, 40265319, 161061273, 644245095, 2576980377, 10307921511, 41231686041, 164926744167, 659706976665, 2638827906663, 10555311626649, 42221246506599, 168884986026393, 675539944105575

Offset: 0

Peter Luschny, Jan 10 2020

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,4).

Cf. A006131, A015521, A193737, A321620, A324969 (Fibonacci with a(0)=1).

[1] cat [3*(4^n -(-1)^n)/5: n in [1..30]]; // G. C. Greubel, Sep 14 2023

gf := (4*x^2 - 1)/(x*(4*x + 3) - 1): ser := series(gf, x, 32):
seq(coeff(ser, x, n), n=0.. 25);
# Alternative:
gf:= (3/5)*exp(-x)*(exp(5*x) - 1) + 1: ser := series(gf, x, 32):
seq(n!*coeff(ser, x, n), n=0.. 25);
# Or:
a := proc(n) option remember; if n < 3 then return [1, 3, 9][n + 1] fi;
4*a(n-2) + 3*a(n-1) end: seq(a(n), n=0..25);

LinearRecurrence[{3,4}, {1,3,9}, 31] (* G. C. Greubel, Sep 14 2023 *)

[3*(4^n -(-1)^n)//5 + int(n==0) for n in range(31)] # G. C. Greubel, Sep 14 2023

a(n) = 2^n*Sum_{k=0..n} A193737(n,k)/2^k.
a(n) = [x^n] (1 - 4*x^2)/(1 - x*(3 + 4*x)).
a(n) = n! [x^n] (3/5)*exp(-x)*(exp(5*x) - 1) + 1.
a(n) = 4*a(n-2) + 3*a(n-1).
a(n) = 3*A015521(n), n>0. - R. J. Mathar, Aug 19 2022

A101617 The trinomial transform (A027907) gives powers of 3, while the trinomial transform of this sequence shift one place left gives powers of 5.

Original entry on oeis.org

1, 1, 1, 3, -3, 19, -43, 139, -355, 995, -2587, 6907, -17939, 46931, -121419, 314603, -811203, 2091459, -5379963, 13833179, -35527795, 91210035, -234020267, 600258507, -1539135779, 3945762211, -10113490139, 25918908603, -66417608403, 170182721299, -436032111883, 1117120911019

Offset: 0

Paul D. Hanna, Dec 09 2004

sign

			3^3 = 1*(1) + 3*(1) + 6*(1) + 7*(3) + 6*(-3) + 3*(19) + 1*(-43).
5^3 = 1*(1) + 3*(1) + 6*(3) + 7*(-3) + 6*(19) + 3*(-43) + 1*(139).
In general, a sequence A with the property that the
trinomial transform of A gives powers of P, while the
trinomial transform of LSHIFT(A) gives powers of Q
has the g.f.: N(x)/D(x) where
N(x)=(1+3*x-(Q-3)*x^2-(P+Q-2)*x^3) and
D(x)=(1+2*x-(P+Q-3)*x^2-(P+Q-2)*x^3+(P-1)*(Q-1)*x^4).

Cf. A027907, A100321.

nn:=31; CoefficientList[Series[(1 + 3*x - 2*x^2 - 6*x^3)/(1 + 2*x - 5*x
^2 - 6*x^3 + 8*x^4),{x,0,nn}],x] (* Georg Fischer, Apr 17 2020 *)

{a(n)=local(P=3,Q=5,V=[1,1]);if(n>1, for(m=1,n, V=concat(V,P^m-sum(k=0,2*m-1,polcoeff((1+x+x^2)^m+x*O(x^k),k)*V[k+1])); V=concat(V,Q^m-sum(k=0,2*m-1,polcoeff((1+x+x^2)^m+x*O(x^k),k)*V[k+2])); ));V[n+1]}
for(n=0,30,print1(a(n),", "))

G.f.: A(x) = (1 + 3*x - 2*x^2 - 6*x^3)/(1 + 2*x - 5*x^2 - 6*x^3 + 8*x^4). [corrected by Georg Fischer, Apr 17 2020]
3^n = Sum_{k=0..2*n} A027907(n, k)*a(k) for n>=0 and
5^n = Sum_{k=0..2*n} A027907(n, k)*a(k+1) for n>=0.
a(n) = (-1)^n*A006131(n-1) + (1/3)[(-2)^n + 2]. - Ralf Stephan, May 16 2007

A358945 Decimal expansion of the positive root of 4*x^2 + x - 1.

Original entry on oeis.org

3, 9, 0, 3, 8, 8, 2, 0, 3, 2, 0, 2, 2, 0, 7, 5, 6, 8, 7, 2, 7, 6, 7, 6, 2, 3, 1, 9, 9, 6, 7, 5, 9, 6, 2, 8, 1, 4, 3, 3, 9, 9, 9, 0, 3, 1, 7, 1, 7, 0, 2, 5, 5, 4, 2, 9, 9, 8, 2, 9, 1, 9, 6, 6, 3, 6, 8, 6, 9, 2, 9, 3, 2, 9, 2, 2

Offset: 0

Wolfdieter Lang, Jan 20 2023

The negative root is -(A189038 - 1) = -0.6403882032... .

c^n = A052923(-n) + A006131(-(n+1))*phi17, for n >= 0, with phi17 = A222132 = (1 + sqrt(17))/2, A052923(-n) = -(-2*i)^(-n)*S(-(n+2), i/2) = (i/2)^n*S(n, i/2), with i = sqrt(-1), and A006131(-(n+1)) = A052923(-n+1)/4 = -(i/2)^(n+1)*S(n-1, i/2), with the S-Chebyshev polynomials (see A049310), and S(-n, x) = -S(n-2, x), for n >= 1. - Wolfdieter Lang, Jan 04 2024

			c = 0.39038820320220756872767623199675962814339990317170255429982919663...

Index entries for sequences related to Chebyshev polynomials.

Cf. A006131, A010473, A049310, A052923, A174930, A189038, A222132.

RealDigits[(Sqrt[17] - 1)/8, 10, 120][[1]] (* Amiram Eldar, Jan 20 2023 *)
RealDigits[Root[4x^2+x-1,2],10,120][[1]] (* Harvey P. Dale, Jan 15 2024 *)

c = (-1 + sqrt(17))/8 = A189038 - 5/4 = A174930 - 5/8.

c = 1/phi17 = (-1 + phi17)/4, with phi17 = A222132. - Wolfdieter Lang, Jan 05 2024

cp's OEIS Frontend

A231775 Triangular array read by rows: row n shows the coefficients of the polynomial u(n) = c(0) + c(1)*x + ... + c(n)*x^(n) which is the denominator of the n-th convergent of the continued fraction [k, k, k, ... ], where k = (x + 1)/(x + 2).

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A242763 a(n) = 1 for n <= 7; a(n) = a(n-5) + a(n-7) for n>7.

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A305834 Triangle read by rows: T(0,0)= 1; T(n,k)= T(n-1,k) + 4*T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

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A330795 Evaluation of the polynomials given by the Riordan square of the Fibonacci sequence with a(0) = 1 (A193737) at 1/2 and normalized with 2^n.

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A101617 The trinomial transform (A027907) gives powers of 3, while the trinomial transform of this sequence shift one place left gives powers of 5.

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A358945 Decimal expansion of the positive root of 4*x^2 + x - 1.

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A231775 Triangular array read by rows: row n shows the coefficients of the polynomial u(n) = c(0) + c(1)x + ... + c(n)x^(n) which is the denominator of the n-th convergent of the continued fraction [k, k, k, ... ], where k = (x + 1)/(x + 2).