A015441 -id:A015441 - cp's OEIS frontend

A172351 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=6.

1, 1, 1, 1, 1, 1, 1, 7, 7, 1, 1, 13, 91, 13, 1, 1, 55, 715, 715, 55, 1, 1, 133, 7315, 13585, 7315, 133, 1, 1, 463, 61579, 483835, 483835, 61579, 463, 1, 1, 1261, 583843, 11093017, 46931995, 11093017, 583843, 1261, 1, 1, 4039, 5093179, 336877411

Offset: 0

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

Start from the generalized Fibonacci sequence A015441 and its partial products c(n) = 1, 1, 1, 7, 91, 5005, 665665, 308202895, 388643850595, 1569732512553205... Then t(n,k) = c(n)/(c(k)*c(n-k)).

Row sums are 1, 2, 3, 16, 119, 1542, 28483, 1091756, 70288239, 7576979362, 1345651717403,..

			1;
1, 1;
1, 1, 1;
1, 7, 7, 1;
1, 13, 91, 13, 1;
1, 55, 715, 715, 55, 1;
1, 133, 7315, 13585, 7315, 133, 1;
1, 463, 61579, 483835, 483835, 61579, 463, 1;
1, 1261, 583843, 11093017, 46931995, 11093017, 583843, 1261, 1;
1, 4039, 5093179, 336877411, 3446515051, 3446515051, 336877411, 5093179, 4039, 1;

Cf. A010048 (m=1), A015109 (m=2), A172350 (m=5), A172352 (m=7).

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}]

A232015 Expansion of (1-2x)/((1+2x)(1-3x)).

Original entry on oeis.org

1, -1, 5, -1, 29, 23, 197, 335, 1517, 3527, 12629, 33791, 109565, 312311, 969701, 2843567, 8661773, 25723175, 77693813, 232032863, 698195741, 2090392919, 6279567365, 18821924879, 56499329069, 169430878343, 508426852757, 1525012122815, 4575573239357

Offset: 0

Philippe Deléham, Nov 17 2013

Index entries for linear recurrences with constant coefficients, signature (1,6).

Cf. A078008, A108561.

Table[(3^n + 4 (-2)^n)/5, {n, 0, 30}] (* Bruno Berselli, Nov 18 2013 *)
CoefficientList[Series[(1-2x)/((1+2x)(1-3x)),{x,0,40}],x] (* or *) LinearRecurrence[ {1,6},{1,-1},30] (* Harvey P. Dale, Apr 20 2017 *)

Vec((1-2*x)/((1+2*x)*(1-3*x))+O(x^20)) \\ Edward Jiang, Sep 06 2014

G.f.: (1 - 2*x) / (1 - x - 6*x^2).
a(n) = a(n-1) + 6*a(n-2) for n>1, a(0)=1, a(1)=-1.
a(n) = sum_{k=0..n} A108561(n,k)*2^k.
a(n) = A102901(n) - A015441(n).
From Bruno Berselli, Nov 18 2013: (Start)
a(n) = (3^n + 4*(-2)^n)/5.
a(n+1) + a(n) = 4*A015441(n).
a(n+1) - a(n) = -2*(-1)^n*A165405(n).
Sum(a(i), i=0..n) = A091001(n+1). (End)

A369490 a(n) = 3^(n+1) + 2*(-2)^(n+1).

Original entry on oeis.org

-1, 17, 11, 113, 179, 857, 1931, 7073, 18659, 61097, 173051, 539633, 1577939, 4815737, 14283371, 43177793, 128878019, 387944777, 1161212891, 3488881553, 10456158899, 31389448217, 94126401611, 282463090913, 847221500579, 2542000046057

Offset: 0

Philippe Deléham, Jan 24 2024

Index entries for linear recurrences with constant coefficients, signature (1, 6).

Cf. A000079, A000244, A003063, A015441, A085279.

LinearRecurrence[{1,6},{-1,17},26] (* James C. McMahon, Jan 30 2024 *)

def A369490(n): return 3**(n+1)+(1<Chai Wah Wu, Feb 25 2024

a(n) = a(n-1) + 6*a(n-2); a(0) = -1, a(1) = 17.
G.f.: (18*x-1)/((1+2*x)*(1-3*x)).
a(2*n) = A003063(2*n+2).
a(2*n+1) = A085279(2*n+3).
a(n) = 18*A015441(n) - A015441(n+1).

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.

A304252 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 6*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, 6, 1, 12, 1, 18, 36, 1, 24, 108, 1, 30, 216, 216, 1, 36, 360, 864, 1, 42, 540, 2160, 1296, 1, 48, 756, 4320, 6480, 1, 54, 1008, 7560, 19440, 7776, 1, 60, 1296, 12096, 45360, 46656, 1, 66, 1620, 18144, 90720, 163296, 46656, 1, 72, 1980, 25920, 163296, 435456, 326592, 1, 78, 2376, 35640

Offset: 0

Zagros Lalo, May 09 2018

The numbers in rows of the triangle are along skew diagonals pointing top-right in center-justified triangle given in A013613 ((1+6*x)^n).

The coefficients in the expansion of 1/(1-x-6*x^2) are given by the sequence generated by the row sums.

			Triangle begins:
  1;
  1;
  1,  6;
  1, 12;
  1, 18,   36;
  1, 24,  108;
  1, 30,  216,   216;
  1, 36,  360,   864;
  1, 42,  540,  2160,   1296;
  1, 48,  756,  4320,   6480;
  1, 54, 1008,  7560,  19440,    7776;
  1, 60, 1296, 12096,  45360,   46656;
  1, 66, 1620, 18144,  90720,  163296,   46656;
  1, 72, 1980, 25920, 163296,  435456,  326592;
  1, 78, 2376, 35640, 272160,  979776, 1306368,  279936;
  1, 84, 2808, 47520, 427680, 1959552, 3919104, 2239488;

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

Zagros Lalo, Left-justified triangle
Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (1+6x)^n

Row sums give A015441.

Cf. A013613.

t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, t[n - 1, k] + 6 t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 13}, {k, 0, Floor[n/2]}] // Flatten (* Robert G. Wilson v, May 19 2018 *)
Table[6^k Binomial[n - k, k], {n, 0, 13}, {k, 0, Floor[n/2]}] // Flatten

T(n,k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, T(n-1,k) + 6*T(n-2,k-1)));
tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n,k), ", ")); print); \\ Michel Marcus, May 10 2018

T(n,k) = 6^k*binomial(n-k,k), n >= 0, 0 <= k <= floor(n/2).

A102765 Array read by antidiagonals: T(n, k) = ((n+4)^k-(n-1)^k)/5.

Original entry on oeis.org

0, 0, 1, 0, 1, 3, 0, 1, 5, 13, 0, 1, 7, 25, 51, 0, 1, 9, 43, 125, 205, 0, 1, 11, 67, 259, 625, 819, 0, 1, 13, 97, 477, 1555, 3125, 3277, 0, 1, 15, 133, 803, 3355, 9331, 15625, 13107, 0, 1, 17, 175, 1261, 6505, 23517, 55987, 78125, 52429, 0, 1, 19, 223, 1875, 11605

Offset: 0

Lambert Klasen (lambert.klasen(AT)gmx.net) and Gary W. Adamson, Feb 10 2005

Consider a 5x5 matrix M =
[n, 1, 1, 1, 1]
[1, n, 1, 1, 1]
[1, 1, n, 1, 1]
[1, 1, 1, n, 1]
[1, 1, 1, 1, n].
The n-th row of the array contains the values of the non diagonal elements of M^k, k=0,1,.... (Corresponding diagonal entry = non diagonal entry + (n-1)^k.)
For row r we have polynomial ((r+4)^n-(r-1)^n)/5. Corresponding g.f.s: x/((1-(r-1)x)(1-(r+4)x))
If r(n) denotes a row sequence, r(n+1)/r(n) converges to n+4.
Triangle T(n, k) = (4^(n-k-1)-(-1)^(n-k-1))/5*(binomial(k+(n-k-1),n-k-1)) gives coefficients for polynomials for the columns of the array. First four polynomial are:
  1
  3 + 2*k
  13 + 9*k + 3*k^2
  51 + 52*k + 18*k^2 + 4*k^3
  ...

			Array begins:
  0, 1,  3, 13,  51,  205, ...
  0, 1,  5, 25, 125,  625, ...
  0, 1,  7, 43, 259, 1555, ...
  0, 1,  9, 67, 477, 3355, ...
  0, 1, 11, 97, 803, 6505, ...
  ...

Cf. A015521 (for n=0), A000351 (for n=1), A003464 (for n=2), A016130 (for n=3), A016140 (for n=4), A016153 (for n=5), A016164 (for n=6), A016174 (for n=7), A016184 (for n=8), A015441 (for n=-1), A091005 (for n=-2).

MM(n,N)=local(M);M=matrix(n,n);for(i=1,n, for(j=1,n,if(i==j,M[i,j]=N,M[i,j]=1)));M
for(k=0,10, for(i=0,10,print1((MM(5,k)^i)[1,2],","));print())

p(n,k)=((n+4)^k-(n-1)^k)/5
for(k=0,10, for(i=0,10,print1(p(k,i),","));print())

for(k=0,10, for(i=0,10,print1(polcoeff(x/((1-(k-1)*x)*(1-(k+4)*x)),i),","));print())

A277604 Array of coefficients T(k,n) of the formal power series A(k,x) read by upwards antidiagonals, where A(k,x) = sqrt(1 + 2xA(k,x) + (4k+1)x^2*(A(k,x))^2), k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 5, 1, 1, 1, 7, 9, 13, 1, 1, 1, 9, 13, 37, 25, 1, 1, 1, 11, 17, 73, 81, 61, 1, 1, 1, 13, 21, 121, 169, 301, 125, 1, 1, 1, 15, 25, 181, 289, 841, 729, 295, 1, 1, 1, 17, 29, 253, 441, 1801, 2197, 2549, 625, 1, 1, 1, 19, 33, 337, 625, 3301, 4913, 10123, 6561, 1447, 1

Offset: 0

Werner Schulte, Oct 29 2016

For k = 0 see A000012, for k = 1 see A098615, and for k = 2 see A200376.
It will be interesting using the formulae for k < 0 (attention: signed terms!). Especially for k = -1 see A157674.
If G is the g.f. of central binomial coefficients (see A000984) and B(k,x) = G(k*x^2), then B(k,x) = A(k,x)/(1+x*A(k,x)) and A(k,x) = B(k,x) / (1-x*B(k,x)) for k >= 0. - Werner Schulte, Aug 07 2017

			The terms define the array T(k,n) for k >= 0 and n >= 0, i.e.,
k\n  0  1   2   3    4     5      6      7       8        9  . . .
0:   1  1   1   1    1     1      1      1       1        1  . . .
1:   1  1   3   5   13    25     61    125     295      625  . . .
2:   1  1   5   9   37    81    301    729    2549     6561  . . .
3:   1  1   7  13   73   169    841   2197   10123    28561  . . .
4:   1  1   9  17  121   289   1801   4913   28057    83521  . . .
5:   1  1  11  21  181   441   3301   9261   63071   194481  . . .
6:   1  1  13  25  253   625   5461  15625  123565   390625  . . .
7:   1  1  15  29  337   841   8401  24389  219619   707281  . . .
8:   1  1  17  33  433  1089  12241  35937  362993  1185921  . . .
9:   1  1  19  37  541  1369  17101  50653  567127  1874161  . . .
etc.

Cf. A000012, A000045, A000108, A000984, A015441, A046521, A098615, A111959, A200376.

A(k,x) = (x + sqrt(1 - 4*k*x^2))/(1 - (4*k+1)*x^2) for k >= 0.
T(k,0) = 1 and T(k,2*n+2) = (4*k+1)^(n+1)-2*(Sum_{i=0..n} A000108(i)*k^(i+1)* (4*k+1)^(n-i)), and T(k,2*n+1) = (4*k+1)^n for k >= 0 and n >= 0.
A(k,x) = 1/(1 - x - 2*k*x^2*C(k*x^2)), k >= 0, where C is the g.f. of A000108.
Conjecture: If B(k,n) satisfy B(k,0) = B(k,1) = 1 and B(k,n+2) = B(k,n+1) + k*B(k,n) for k >= 0 and n >= 0 (generalized Fibonacci numbers, see A015441) and G(k,x) = Sum_{n>=0} A000108(n)*B(k,n)*x^n for k >= 0, then you will have (1): A(k,x*G(k,x)) = G(k,x) and (2): G(k,x/A(k,x)) = A(k,x) for k >= 0. Especially for k = 1 see A098615 and for k = 2 see A200376.
Conjecture: T(k,2*n) = Sum_{i=0..n} A046521(n,i)*k^(n-i) for k, n >= 0. - Werner Schulte, Aug 02 2017
Recurrence: T(k,2*n+2) = (4*k+1)*T(k,2*n)-2*k^(n+1)*A000108(n) with initial value T(k,0) = 1 for k >= 0 and n >= 0. - Werner Schulte, Aug 09 2017
T(k,n) = Sum_{i=0..n} A111959(n,i)*k^((n-i)/2) for k >= 0 and n >= 0. - Werner Schulte, Aug 09 2017

cp's OEIS Frontend

A172351 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=6.

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A232015 Expansion of (1-2*x)/((1+2*x)*(1-3*x)).

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A369490 a(n) = 3^(n+1) + 2*(-2)^(n+1).

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Python

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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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A304252 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 6*T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

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A102765 Array read by antidiagonals: T(n, k) = ((n+4)^k-(n-1)^k)/5.

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A277604 Array of coefficients T(k,n) of the formal power series A(k,x) read by upwards antidiagonals, where A(k,x) = sqrt(1 + 2*x*A(k,x) + (4*k+1)*x^2*(A(k,x))^2), k >= 0.

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A232015 Expansion of (1-2x)/((1+2x)(1-3x)).

A277604 Array of coefficients T(k,n) of the formal power series A(k,x) read by upwards antidiagonals, where A(k,x) = sqrt(1 + 2xA(k,x) + (4k+1)x^2*(A(k,x))^2), k >= 0.