A292324 -id:A292324 - cp's OEIS frontend

A094686 A Fibonacci convolution.

1, 0, 1, 2, 2, 4, 7, 10, 17, 28, 44, 72, 117, 188, 305, 494, 798, 1292, 2091, 3382, 5473, 8856, 14328, 23184, 37513, 60696, 98209, 158906, 257114, 416020, 673135, 1089154, 1762289, 2851444, 4613732, 7465176, 12078909, 19544084, 31622993, 51167078, 82790070

Offset: 0

Paul Barry, May 19 2004

Convolution of A000045 and A049347.
Diagonal sums of number triangle A116088. - Paul Barry, Feb 04 2006
Let (b(n)) be the p-INVERT of (1,1,0,0,0,0,0,0,...) using p(S) = 1 - S^2; then b(n) = a(n+1) for n >=0. See A292324. - Clark Kimberling, Sep 15 2017

G. C. Greubel, Table of n, a(n) for n = 0..1000
Elena Barcucci, Antonio Bernini, Stefano Bilotta, and Renzo Pinzani, Non-overlapping matrices, arXiv:1601.07723 [cs.DM], 2016 (see 1st column of Table 1 p. 8).
Stefano Bilotta, Variable-length Non-overlapping Codes, arXiv preprint arXiv:1605.03785 [cs.IT], 2016 [See Table 2].
Joshua P. Bowman, Compositions with an Odd Number of Parts, and Other Congruences, J. Int. Seq (2024) Vol. 27, Art. 24.3.6. See p. 19.
David Broadhurst, Multiple Deligne values: a data mine with empirically tamed denominators, arXiv:1409.7204 [hep-th], 2014 (see p. 10).
Leonard Rozendaal, Pisano word, tesselation, plane-filling fractal, Preprint, 2017.
Index entries for linear recurrences with constant coefficients, signature (0,1,2,1).

Cf. A000045, A005252, A049347, A093040, A116088, A292324.

[(Fibonacci(n+1) +((n+2) mod 3) -1)/2: n in [0..40]]; // G. C. Greubel, Feb 09 2023

LinearRecurrence[{0,1,2,1}, {1,0,1,2}, 40] (* Jean-François Alcover, Sep 21 2017 *)

Vec(1/((1-x-x^2)*(1+x+x^2)) + O(x^50)) \\ Michel Marcus, Sep 27 2014

[(fibonacci(n+1) + (n+2)%3 - 1)/2 for n in range(41)] # G. C. Greubel, Feb 09 2023

G.f.: 1/((1-x-x^2)*(1+x+x^2)).
a(n) = 2*sqrt(3)*Sum_{k=0..n} Fibonacci(k+1)*cos((4*(n-k)+1)*Pi/6)/3.
a(n) = a(n-2) + 2*a(n-3) + a(n-4).
From Paul Barry, Jan 13 2005: (Start)
a(n) = A005252(n) - (-cos((2*n+1)*Pi/3)/2 - sqrt(3)*sin((2*n+1)*Pi/3)/6 + sqrt(3)*cos(Pi*n/3+Pi/6)/6 + sin((2*n+1)*Pi/6)/2).
a(n) = Sum_{k=0..floor(n/2)} if(mod(n-k, 2)=0, binomial(n-k, k), 0).
a(n) = A093040(n-1) - Fibonacci(n). (End)
a(n) = Sum_{k=0..floor(n/2)} C(n-k, k)*(1+(-1)^(n-k))/2. - Paul Barry, Sep 09 2005
From Paul Barry, Feb 04 2006: (Start)
a(n) = Sum_{k=0..floor(n/2)} C(2*k, n-2*k).
a(n) = Sum_{k=0..floor(n/2)} C(n-k,k)*C(3*k,n-k)/C(3*k,k). (End)
2*a(n) = A000045(n+1) + A049347(n). - R. J. Mathar, Feb 13 2020
a(n) = (1/2)*(A000045(n+1) + A049347(n)). - G. C. Greubel, Feb 09 2023

A073778 a(n) = Sum_{k=0..n} T(k)*T(n-k), where T is A000073; convolution of A000073 with itself.

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 5, 12, 26, 56, 118, 244, 499, 1010, 2027, 4040, 8004, 15776, 30956, 60504, 117845, 228818, 443057, 855732, 1649022, 3171128, 6086626, 11662252, 22309543, 42614178, 81286743, 154856528, 294660040, 560052736, 1063367384, 2017030256

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 10 2002

Number of binary sequences of length n+1 that have exactly one subsequence 000. Example: a(4)=5 because we have 00010,00011,01000,10001 and 11000. Column 1 of A118390. - Emeric Deutsch, Apr 27 2006

Let (b(n)) be the p-INVERT of (1,1,1,0,0,0,0,0,0,...) using p(S) = 1 - S^2; then b(n) = a(n+3) for n >= 0. See A292324. - Clark Kimberling, Sep 15 2017

Michael De Vlieger, Table of n, a(n) for n = 0..3770, terms n = 2..1002 from Vincenzo Librandi.
Daniel Birmajer, Juan B. Gil, and Michael D. Weiner, Linear recurrence sequences and their convolutions via Bell polynomials, arXiv:1405.7727 [math.CO], 2014; see also, J. Int. Seq. 18 (2015) # 15.1.2.
Michael Dairyko, Lara Pudwell, Samantha Tyner and Casey Wynn. Non-contiguous pattern avoidance in binary trees. Electron. J. Combin. 19 (2012), no. 3, Paper 22, 21 pp. MR2967227. - From _N. J. A. Sloane_, Feb 01 2013
Omar Khadir, László Németh, and László Szalay, Tiling of dominoes with ranked colors, Results in Math. (2024) Vol. 79, Art. No. 253. See p. 2.
Emrah Kilic and Helmut Prodinger, A Note on the Conjecture of Ramirez and Sirvent, J. Int. Seq. 17 (2014) # 14.5.8.
László Németh and László Szalay, Explicit solution of system of two higher-order recurrences, arXiv:2408.12196 [math.NT], 2024. See p. 10.
José L. Ramirez and Víctor F. Sirvent, Incomplete Tribonacci Numbers and Polynomials, Journal of Integer Sequences, Vol. 17, 2014, #14.4.2.
Index entries for linear recurrences with constant coefficients, signature (2,1,0,-3,-2,-1).

Cf. A000073, A118390.

A073778:=proc(n) coeftayl(x^4/(1-x-x^2-x^3)^2, x=0, n); end proc: seq(A073778(n), n=0..40); # Wesley Ivan Hurt, Nov 17 2014

CoefficientList[Series[x^4/(1-x-x^2-x^3)^2, {x, 0, 40}], x]

a(n):= sum((k+1)*sum(binomial(j,n-3*k+2*j-4)*binomial(k,j), j,0,k), k,0,n-4);
makelist(a(n), n, 0, 30); /* Vladimir Kruchinin, Dec 14 2011 */

T(n) = ([0, 1, 0; 0, 0, 1; 1, 1, 1]^n)[1, 3]; \\ A000073
a(n) = sum(k=0, n, T(k)*T(n-k)); \\ Michel Marcus, Oct 20 2021

[( x^4/(1-x-x^2-x^3)^2 ).series(x,n+1).list()[n] for n in (0..40)] # Zerinvary Lajos, Jun 02 2009; modified by G. C. Greubel, Dec 15 2021

G.f.: x^4/(1 - x - x^2 - x^3)^2.
a(n) = Sum_{k=0..n-4} (k+1)*Sum_{j=0..k} binomial(j,n-3*k+2*j-4)*binomial(k,j). - Vladimir Kruchinin, Dec 14 2011
(n-2)*a(n) - (n-1)*a(n-1) - n*a(n-2) - (n+1)*a(n-3) = 0, n > 2. - Michael D. Weiner, Nov 18 2014

Two initial zeros inserted by Hans J. H. Tuenter, Oct 20 2021

A292402 p-INVERT of (1,0,0,1,0,0,0,0,0,0,...), where p(S) = 1 - S^2.

Original entry on oeis.org

0, 1, 0, 1, 2, 1, 4, 2, 6, 7, 8, 16, 14, 29, 32, 47, 70, 82, 136, 162, 244, 331, 440, 650, 834, 1220, 1632, 2262, 3176, 4261, 6056, 8175, 11414, 15747, 21568, 30121, 41094, 57210, 78644, 108521, 150300, 206456, 286288, 393865, 544424, 751675, 1035980

Offset: 0

Clark Kimberling, Sep 30 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0, 1, 0, 0, 2, 0, 0, 1)

Cf. A292324, A292403, A292404.

z = 60; s = x + x^4; p = 1 - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292402 *)

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

a(n) = a(n-2) + 2*a(n-5) + a(n-8) for n >= 9.

A099444 A Chebyshev transform of Fib(2n+2).

Original entry on oeis.org

1, 3, 7, 15, 32, 69, 149, 321, 691, 1488, 3205, 6903, 14867, 32019, 68960, 148521, 319873, 688917, 1483735, 3195552, 6882329, 14822619, 31923791, 68754951, 148079008, 318920925, 686866813, 1479319737, 3186042539, 6861847920

Offset: 0

Paul Barry, Oct 16 2004

The denominator is a parameterization of the Alexander polynomial for the knot 6_2 (Miller Institute knot). The g.f. is the image of the g.f. of Fib(2n+2) under the Chebyshev transform A(x)->(1/(1+x^2))A(x/(1+x^2)).

This sequence is the p-INVERT of A010892 using p(S) = 1 - S - S^2; see A292324. - Clark Kimberling, Sep 26 2017

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Dror Bar-Natan, The Rolfsen Knot Table
Index entries for linear recurrences with constant coefficients, signature (3,-3,3,-1).

Cf. A001906.

LinearRecurrence[{3,-3,3,-1},{1,3,7,15},30] (* Harvey P. Dale, Sep 30 2018 *)

G.f.: (1+x^2)/(1-3x+3x^2-3x^3+x^4);
a(n) = sum{k=0..floor(n/2), binomial(n-k, k)(-1)^k*Fib(2(n-2k)+2)};
a(n) = sum{k=0..n, binomial((n+k)/2, k)(-1)^((n-k)/2)(1+(-1)^(n+k))Fib(2k+2)/2};
a(n) = sum{k=0..n, A099445(n-k)*binomial(1, k/2)(1+(-1)^k)/2}.

A292326 p-INVERT of (1,1,1,0,0,0,0,0,0,0,0,...), where p(S) = (1 - S)^3.

Original entry on oeis.org

3, 9, 25, 63, 153, 359, 819, 1830, 4018, 8694, 18582, 39298, 82350, 171186, 353338, 724719, 1478061, 2999175, 6057687, 12183945, 24411935, 48740193, 96998325, 192459996, 380812692, 751557756, 1479686972, 2906717460, 5698014924, 11147786740, 21769549380

Offset: 0

Clark Kimberling, Sep 15 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3, 0, -2, -6, 0, 4, 6, 3, 1)

Cf. A292324.

z = 60; s = x + x^2 + x^3; p = (1 - s)^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292326 *)
LinearRecurrence[{3,0,-2,-6,0,4,6,3,1},{3,9,25,63,153,359,819,1830,4018},40] (* Harvey P. Dale, Nov 01 2019 *)

G.f.: -(((1 + x + x^2) (3 - 3 x - 2 x^2 - x^3 + 3 x^4 + 2 x^5 + x^6))/(-1 + x + x^2 + x^3)^3).

a(n) = 3*a(n-1) - 2*a(n-3) - 6*a(n-4) + 4*a(n-6) + 6*a(n-7) + 3*a(n-8) + a(n-9) for n >= 10.

A292325 p-INVERT of (1,0,0,0,1,0,0,0,0,0,...), where p(S) = (1 - S)^2.

Original entry on oeis.org

2, 3, 4, 5, 8, 13, 20, 29, 40, 56, 80, 115, 164, 230, 320, 445, 620, 864, 1200, 1660, 2290, 3155, 4344, 5975, 8206, 11252, 15408, 21078, 28810, 39344, 53680, 73173, 99662, 135640, 184480, 250740, 340578, 462316, 627200, 850420, 1152480, 1561043, 2113420

Offset: 0

Clark Kimberling, Sep 15 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2, -1, 0, 0, 2, -2, 0, 0, 0, -1)

Cf. A079978, A292324.

z = 60; s = x + x^5; p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292325 *)

G.f.: -(((-1 + x) (1 + x^4) (2 + x + x^2 + x^3 + x^4))/((1 - x + x^2)^2 (-1 + x^2 + x^3)^2)).

a(n) = 2*a(n-1) - a(n-2) + 2*a(n-5) - 2*a(n-6) - a(n-10) for n >= 11.

A259074 Triangle T(n,k) = Sum_{j=0..(n-k)/3} C(n-3j-1,k-1)C(n-k-3*j,j).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 2, 4, 6, 4, 1, 3, 7, 10, 10, 5, 1, 4, 12, 18, 20, 15, 6, 1, 5, 19, 33, 39, 35, 21, 7, 1, 7, 28, 58, 76, 75, 56, 28, 8, 1, 10, 42, 96, 144, 156, 132, 84, 36, 9, 1, 14, 64, 156, 260, 315, 294, 217, 120, 45, 10, 1, 19, 97, 253, 455, 610, 630, 518, 338, 165, 55, 11, 1

Offset: 1

Vladimir Kruchinin, Jun 18 2015

			[1]
[1,1]
[1,2,1]
[1,3,3,1]
[2,4,6,4,1]
[3,7,10,10,5,1]

Column k=1 gives A003269, column k=2 A292324.

Row sums give A008999(n-1) for n>0.

Table[Sum[Binomial[n - 3*j - 1, k - 1] Binomial[n - k - 3*j, j], {j, 0, (n - k)/3}], {n, 12}, {k, n}] // Flatten (* Michael De Vlieger, Jun 19 2015 *)

T(n,k):=sum(binomial(n-3*j-1,k-1)*binomial(n-k-3*j,j),j,0,(n-k)/3);

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

cp's OEIS Frontend

A094686 A Fibonacci convolution.

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A073778 a(n) = Sum_{k=0..n} T(k)*T(n-k), where T is A000073; convolution of A000073 with itself.

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A292402 p-INVERT of (1,0,0,1,0,0,0,0,0,0,...), where p(S) = 1 - S^2.

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A099444 A Chebyshev transform of Fib(2n+2).

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A292326 p-INVERT of (1,1,1,0,0,0,0,0,0,0,0,...), where p(S) = (1 - S)^3.

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A292325 p-INVERT of (1,0,0,0,1,0,0,0,0,0,...), where p(S) = (1 - S)^2.

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A259074 Triangle T(n,k) = Sum_{j=0..(n-k)/3} C(n-3*j-1,k-1)*C(n-k-3*j,j).

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A259074 Triangle T(n,k) = Sum_{j=0..(n-k)/3} C(n-3j-1,k-1)C(n-k-3*j,j).