A015440 -id:A015440 - cp's OEIS frontend

A172350 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=5.

1, 1, 1, 1, 1, 1, 1, 6, 6, 1, 1, 11, 66, 11, 1, 1, 41, 451, 451, 41, 1, 1, 96, 3936, 7216, 3936, 96, 1, 1, 301, 28896, 197456, 197456, 28896, 301, 1, 1, 781, 235081, 3761296, 14019376, 3761296, 235081, 781, 1, 1, 2286, 1785366, 89565861, 781665696

Offset: 0

Roger L. Bagula and Gary W. Adamson, Feb 01 2010

Start from the generalized Fibonacci sequence A015440 and its partial products c(n) = 1, 1, 1, 6, 66, 2706, 259776, 78192576, 61068401856, 139602366642816... Then t(n,k) = c(n)/(c(k)*c(n-k)).

Row sums are 1, 2, 3, 14, 90, 986, 15282, 453308, 22013694, 1746038420, 222562828116,...

			1;
1, 1;
1, 1, 1;
1, 6, 6, 1;
1, 11, 66, 11, 1;
1, 41, 451, 451, 41, 1;
1, 96, 3936, 7216, 3936, 96, 1;
1, 301, 28896, 197456, 197456, 28896, 301, 1;
1, 781, 235081, 3761296, 14019376, 3761296, 235081, 781, 1;
1, 2286, 1785366, 89565861, 781665696, 781665696, 89565861, 1785366, 2286, 1;
1, 6191, 14152626, 1842200151, 50409295041, 118031520096, 50409295041, 1842200151, 14152626, 6191, 1;

Cf. A010048 (m=1), A015109 (m=2), A172349 (m=4), A172351 (m=6).

Clear[f, c, a, t];
f[0, a_] := 0; f[1, a_] := 1;
f[n_, a_] := f[n, a] = f[n - 1, a] + 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}]

A367453 Decimal expansion of (-1 + sqrt(21))/10 = 1/A222134.

Original entry on oeis.org

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

Offset: 0

Wolfdieter Lang, Nov 20 2023

Positive root of the minimal polynomial x^2 + 1/5 - 1/5. The negative root is -(1/5)*A222134 = -0.558257569...
c^n = A(-n) + B(-n)*phi21, and A(n) = S21(n+1) - S21(n) = A365824(n), with phi21 = A222134, and B(n) = S21(n) = A015440(n-1), where S21(n) = sqrt(-5)^(n-1)*S(n-1, 1/sqrt(-5)), with the Chebyshev polynomials {S(n, x)} (see A049310).
The formula for negative indices of S is S(-1, 0) = 0 and S(-n, x) = -S(n-2, x) for n >= 2.

			c = 0.3582575694955840006588047193728008488984456...

Paolo Xausa, Table of n, a(n) for n = 0..10000

Cf. A010477, A049310, A222134, A222135, A365824.

First[RealDigits[(Sqrt[21]-1)/10,10,100]] (* Paolo Xausa, Nov 21 2023 *)

\\ Works in v2.13 and higher; n = 100 decimal places
my(n=100); digits(floor(10^(n-1)*(quadgen(84)-1))) \\ Michal Paulovic, Nov 20 2023

c = 1/phi21 = (1/5)*(1 - phi21), with phi21 = (1 + sqrt(21))/2 = A222134, hence an algebraic number of the real quadratic number field Q(sqrt(21)) but not an algebraic integer like phi24.
Equals (A010477-1)/10. - R. J. Mathar, Nov 21 2023
Equals 2*A222135/10. - Hugo Pfoertner, Mar 21 2024

A085487 a(n) = p^n + q^n, p = (1 + sqrt(21))/2, q = (1 - sqrt(21))/2.

Original entry on oeis.org

1, 11, 16, 71, 151, 506, 1261, 3791, 10096, 29051, 79531, 224786, 622441, 1746371, 4858576, 13590431, 37883311, 105835466, 295252021, 824429351, 2300689456, 6422836211, 17926283491, 50040464546, 139671882001, 389874204731, 1088233614736

Offset: 1

Gary W. Adamson, Jul 02 2003

p + q = 1, p*q = -5, p - q = sqrt(21).

The Lucas sequence V(1,-5) apart from the initial term a(0) = 2. - Peter Bala, Jun 23 2015

			a(5) = 151 = p^5 + q^5, with p = 2.79128...; q = -1.79128...

Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", Wiley, 2001, p. 471.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Wikipedia, Lucas sequence
Index entries for linear recurrences with constant coefficients, signature (1,5).

Cf. A015440.

I:=[ 1,11]; [n le 2 select I[n] else Self(n-1)+5*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Jul 20 2013

CoefficientList[Series[(10 x + 1) / (1 - x - 5 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Jul 20 2013 *)

a(n):=n*sum((binomial(k,n-k)*1^(2*k-n)*(5)^(n-k))/k,k,1,n); /* Dmitry Kruchinin, May 16 2011 */

[lucas_number2(n,1,-5) for n in range(1, 11)] # Zerinvary Lajos, May 14 2009

G.f.: (10*x^2+x)/(1-x-5*x^2).
a(n) = n*sum(k=1..n, (C(k,n-k)*1^(2*k-n)*(5)^(n-k))/k). - Dmitry Kruchinin, May 16 2011
a(n) = a(n-1) + 5a(n-2), n>1.
a(n) = [x^n] ( (1 + x + sqrt(1 + 2*x + 21*x^2))/2 )^n. - Peter Bala, Jun 23 2015

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.

A305838 Triangle read by rows: T(0,0)= 1; T(n,k)= T(n-1,k) + 5*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, 5, 1, 10, 1, 15, 25, 1, 20, 75, 1, 25, 150, 125, 1, 30, 250, 500, 1, 35, 375, 1250, 625, 1, 40, 525, 2500, 3125, 1, 45, 700, 4375, 9375, 3125, 1, 50, 900, 7000, 21875, 18750, 1, 55, 1125, 10500, 43750, 65625, 15625, 1, 60, 1375, 15000, 78750, 175000, 109375

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 A013612 ((1+5*x)^n). The coefficients in the expansion of 1/(1-x-5x^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 2.7913327..., when n approaches infinity. The row sums are A015440 (generalized Fibonacci numbers).

			Triangle begins:
  1;
  1;
  1,  5;
  1, 10;
  1, 15,   25;
  1, 20,   75;
  1, 25,  150,   125;
  1, 30,  250,   500;
  1, 35,  375,  1250,    625;
  1, 40,  525,  2500,   3125;
  1, 45,  700,  4375,   9375,    3125;
  1, 50,  900,  7000,  21875,   18750;
  1, 55, 1125, 10500,  43750,   65625,   15625;
  1, 60, 1375, 15000,  78750,  175000,  109375;
  1, 65, 1650, 20625, 131250,  393750,  437500,   78125;
  1, 70, 1950, 27500, 206250,  787500, 1312500,  625000;
  1, 75, 2275, 35750, 309375, 1443750, 3281250, 2812500,  390625;
  1, 80, 2625, 45500, 446875, 2475000, 7218750, 9375000, 3515625;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 380, 381.

Shara Lalo, Skew diagonals in triangle A013612
Shara Lalo, Right-justified triangle

Row sums give A015440.
Cf. A000012 (column 0), A008587 (column 1), A123296 (column 2), A141480 (column 3).
Cf. A013612.

t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0,  t[n - 1, k] + 5 t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}] // Flatten

G.f.: 1/(1 - t*x - 5*t^2).

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A172350 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=5.

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A367453 Decimal expansion of (-1 + sqrt(21))/10 = 1/A222134.

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A085487 a(n) = p^n + q^n, p = (1 + sqrt(21))/2, q = (1 - sqrt(21))/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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A305838 Triangle read by rows: T(0,0)= 1; T(n,k)= T(n-1,k) + 5*T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

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