A147722 -id:A147722 - cp's OEIS frontend

A015448 a(0) = 1, a(1) = 1, and a(n) = 4*a(n-1) + a(n-2) for n >= 2.

1, 1, 5, 21, 89, 377, 1597, 6765, 28657, 121393, 514229, 2178309, 9227465, 39088169, 165580141, 701408733, 2971215073, 12586269025, 53316291173, 225851433717, 956722026041, 4052739537881, 17167680177565, 72723460248141, 308061521170129, 1304969544928657, 5527939700884757

Offset: 0

Olivier Gérard

If one deletes the leading 0 in A084326, takes the inverse binomial transform, and adds a(0)=1 in front, one obtains this sequence here. - Al Hakanson (hawkuu(AT)gmail.com), May 02 2009
For n >= 1, row sums of triangle
  m |k=0     1     2     3     4     5     6     7
====+=============================================
  0 |  1
  1 |  1     4
  2 |  1     4    16
  3 |  1     8    16    64
  4 |  1     8    48    64   256
  5 |  1    12    48   256   256  1024
  6 |  1    12    96   256  1280  1024  4096
  7 |  1    16    96   640  1280  6144  4096 16384
which is triangle for numbers 4^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012
a(n) = a(n;-2) = 3^n*Sum_{k=0..n} binomial(n,k)*F(k+1)*(-2/3)^k, where a(n;d), n=0,1,...,d, denotes the delta-Fibonacci numbers defined in comments to A000045 (see also the papers of Witula et al.). We note that (see A033887) F(3n+1) = 3^n*a(n,2/3) = Sum_{k=0..n} binomial(n,k)*F(k-1)*(-2/3)^k, which implies F(3n+1) + 3^(-n)*a(n) = Sum_{k=0..n} binomial(n,k)*L(k)*(-2/3)^k, where L(k) denotes the k-th Lucas number. - Roman Witula, Jul 12 2012
a(n+1) is (for n >= 0) the number of length-n strings of 5 letters {0,1,2,3,4} with no two adjacent nonzero letters identical. The general case (strings of L letters) is the sequence with g.f. (1+x)/(1-(L-1)*x-x^2). - Joerg Arndt, Oct 11 2012
Starting with offset 1 the sequence is the INVERT transform of (1, 4, 4*3, 4*3^2, 4*3^3, ...); i.e., of A003946: (1, 4, 12, 36, 108, ...). - Gary W. Adamson, Aug 06 2016
a(n+1) equals the number of quinary sequences of length n such that no two consecutive terms differ by 3. - David Nacin, May 31 2017

T. D. Noe, Table of n, a(n) for n = 0..200
Joerg Arndt, Matters Computational (The Fxtbook), pp. 313-315.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876 (see Corollary 1 (vi)).
Gary Detlefs and Wolfdieter Lang, Improved Formula for the Multi-Section of the Linear Three-Term Recurrence Sequence, arXiv:2304.12937 [math.CO], 2023.
Amelia Gilson, Hadley Killen, Steven J. Miller, Nadia Razek, Joshua M. Siktar, and Liza Sulkin, Zeckendorf's Theorem Using Indices in an Arithmetic Progression, arXiv:2005.10396 [math.NT], 2020.
Edyta Hetmaniok, Bozena Piatek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Math. 15 (2017), 477-485.
I. M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013) 13.4.5.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Bahar Kuloğlu, Engin Özkan, and Marin Marin, Fibonacci and Lucas Polynomials in n-gon, An. Şt. Univ. Ovidius Constanţa (Romania 2023) Vol. 31, No 2, 127-140.
Roman Witula, Binomials transformation formulae of scaled Lucas numbers, Demonstratio Math. 46 (2013), 15-27.
R. Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009) 310-329, MR2555042
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A001076, A147722 (INVERT transform), A109499 (INVERTi transform), A154626 (Binomial transform), A086344 (inverse binomial transform), A003946, A049310.

Cf. A000032, A000045, A014445, A014448, A033887, A046854, A055830, A055870, A084326, A167808.

[Fibonacci(3*n-1): n in [0..40]]; // Vincenzo Librandi, Jul 04 2015

with(combinat): a:=n->fibonacci(n,4)-3*fibonacci(n-1,4): seq(a(n), n=1..23); # Zerinvary Lajos, Apr 04 2008

Fibonacci/@(3*Range[0,30]-1) (* Vladimir Joseph Stephan Orlovsky, Mar 01 2010 *)
LinearRecurrence[{4,1},{1,1},30] (* Harvey P. Dale, May 15 2019 *)

a[0]:1$
a[1]:1$
a[n]:=4*a[n-1]+a[n-2]$
A015448(n):=a[n]$
makelist(A015448(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

a(n) = fibonacci(3*n-1); \\ Altug Alkan, Dec 10 2015

a(n) = Fibonacci(3n-1) = ( (1+sqrt(5))*(2-sqrt(5))^n - (1-sqrt(5))*(2+sqrt(5))^n )/ (2*sqrt(5)).
O.g.f.: (1-3*x)/(1-4*x-x^2). - Len Smiley, Dec 09 2001
a(n) = Sum_{k=0..n} 3^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
a(n) = upper left term in the 2 X 2 matrix [1,2; 2,3]^n. - Gary W. Adamson, Mar 02 2008
[a(n), A001076(n)] = [1,4; 1,3]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = A167808(3*n-1) for n > 0. - Reinhard Zumkeller, Nov 12 2009
a(n) = Fibonacci(3n+1) mod Fibonacci(3n), n > 0.
a(n) = (A000032(3*n)-Fibonacci(3*n))/2 = (A014448(n)-A014445(n))/2.
For n >= 2, a(n) = F_n(4) + F_(n+1)(4), where F_n(x) is a Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(n) = A001076(n+1) - 3*A001076(n). - R. J. Mathar, Jul 12 2012
From Gary Detlefs and Wolfdieter Lang, Aug 20 2012: (Start)
a(n) = (5*F(n)^3 + 5*F(n-1)^3 + 3*(-1)^n*F(n-2))/2,
a(n) = (F(n+1)^3 + 2*F(n)^3 - F(n-2)^3)/2, n >= 0, with F(-1) = 1 and F(-2) = -1. Second line from first one with 3*(-1)^n* F(n-2) = F(n-1)^3 - 4*F(n-2)^3 - F(n-3)^3 (in Koshy's book, p. 89, 32. (with a - sign) and 33. For the Koshy reference see A000045) and the F^3 recurrence (see row n=4 of A055870, or Koshy p. 87, 1.). First line from the preceding R. J. Mathar formula with F(3*n) = 5*F(n)^3 + 3*(-1)^n*F(n) (Koshy p. 89, 46.) and the above mentioned formula, Koshy's 32. and 33., with n -> n+2 in order to eliminate F(n+1)^3. (End)
For n > 0, a(n) = L(n-1)*L(n)*F(n) + F(n+1)*(-1)^n with L(n)=A000032(n). - J. M. Bergot, Dec 10 2015
For n > 1, a(n)^2 is the denominator of continued fraction [4,4,...,4, 6, 4,4,...4], which has n-1 4's before, and n-1 4's after, the middle 6. - Greg Dresden, Sep 18 2019
From Gary Detlefs and Wolfdieter Lang, Mar 06 2023: (Start)
a(n) = A001076(n) + A001076(n-1), with A001076(-1) = 1. See the R. J. Mathar formula above.
a(n+1) = i^n*(S(n-1,-4*i) - i*S(n-2,-4*i)), for n >= 0, with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(n, -1) = 0. From the simplified Fibonacci trisection formula for {F(3*n+2)}_{n>=0}. (End)
a(n) = Sum_{k=0..n} A046854(n-1,k)*4^k. - R. J. Mathar, Feb 10 2024
E.g.f.: exp(2*x)*(5*cosh(sqrt(5)*x) - sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, Jun 03 2024

A203378 T(n,k)=Number of (n+1)X(k+1) 0..1 arrays with every 2X2 subblock having equal diagonal elements or equal antidiagonal elements.

Original entry on oeis.org

12, 36, 36, 108, 164, 108, 324, 748, 748, 324, 972, 3412, 5208, 3412, 972, 2916, 15564, 36300, 36300, 15564, 2916, 8748, 70996, 253068, 387324, 253068, 70996, 8748, 26244, 323852, 1764360, 4136292, 4136292, 1764360, 323852, 26244, 78732, 1477268

Offset: 1

R. H. Hardin Dec 31 2011

Table starts
....12......36......108........324..........972...........2916.............8748
....36.....164......748.......3412........15564..........70996...........323852
...108.....748.....5208......36300.......253068........1764360.........12301020
...324....3412....36300.....387324......4136292.......44183028........471988172
...972...15564...253068....4136292.....67731264.....1109832180......18189909480
..2916...70996..1764360...44183028...1109832180....27913330472.....702420750924
..8748..323852.12301020..471988172..18189909480...702420750924...27149753088792
.26244.1477268.85762188.5042151644.298154846436.17679917387404.1049837466171436

			Some solutions for n=4 k=3
..0..0..1..0....1..0..1..1....1..1..0..0....1..1..1..1....1..0..0..0
..0..1..1..1....0..1..1..0....1..1..1..0....1..1..0..1....0..1..0..0
..1..0..1..1....0..0..1..1....1..0..1..1....0..1..1..1....1..0..1..0
..1..1..0..1....1..0..0..1....0..1..1..1....1..1..0..1....1..1..1..1
..1..1..1..1....0..1..0..0....0..0..1..1....1..0..0..0....1..1..0..1

R. H. Hardin, Table of n, a(n) for n = 1..420

Column 1 is A003946(n+1)

Column 2 is A147722(n+2)

A147720 Riordan array (1, x(1-x)/(1-3x)).

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 6, 4, 1, 0, 18, 16, 6, 1, 0, 54, 60, 30, 8, 1, 0, 162, 216, 134, 48, 10, 1, 0, 486, 756, 558, 248, 70, 12, 1, 0, 1458, 2592, 2214, 1168, 410, 96, 14, 1, 0, 4374, 8748, 8478, 5160, 2150, 628, 126, 16

Offset: 0

Paul Barry, Nov 11 2008

Array [0,2,1,0,0,0,....] DELTA [1,0,0,0,......] for Deléham DELTA as in A084938.
Row sums are A001835. Diagonal sums are related to A030186.
Row sums of inverse are essentially A091593. A147720*A007318 is A147721.

			Triangle begins
1;
0,   1;
0,   2,   1;
0,   6,   4,   1;
0,  18,  16,   6,   1;
0,  54,  60,  30,   8,   1;
0, 162, 216, 134,  48,  10,   1;

Indranil Ghosh, Rows 0..100, flattened

nmax=9; Flatten[CoefficientList[Series[CoefficientList[Series[(1-3*x)/(1-(3+y)*x+y*x^2), {x, 0, nmax}],x],{y,0,nmax}],y]] (* Indranil Ghosh, Mar 10 2017, after Philippe Deléham *)

Sum_{k=0..n} T(n,k)*x^k = A000007(n), A001835(n), A147722(n), A084120(n) for x = 0, 1, 2, 3 respectively. - Philippe Deléham, Nov 15 2008

G.f.: (1-3*x)/(1-(3+y)*x+y*x^2). - Philippe Deléham, Feb 15 2012

A147721 a(n) = C(2,n) DELTA C(0,n).

Original entry on oeis.org

1, 1, 1, 3, 4, 1, 11, 17, 7, 1, 41, 72, 40, 10, 1, 153, 301, 208, 72, 13, 1, 571, 1244, 1021, 446, 113, 16, 1, 2131, 5093, 4819, 2525, 813, 163, 19, 1, 7953, 20688, 22104, 13452, 5218, 1336, 222, 22, 1, 29681, 83481, 99192, 68568, 30986, 9586, 2042, 290, 25, 1

Offset: 0

Paul Barry, Nov 11 2008

Triangle T equal to [1,2,1,0,0,0,...] DELTA [1,0,0,0,...] for Deléham DELTA as in A084938.

T = A147720*A007318. Row sums are A147722.

			Triangle begins
    1;
    1,   1;
    3,   4,   1;
   11,  17,   7,   1;
   41,  72,  40,  10,   1;
  153, 301, 208,  72,  13,   1;

Indranil Ghosh, Rows 0..100, flattened

Cf. A129267, A007318, A147703.

nmax=9; Flatten[CoefficientList[Series[CoefficientList[Series[(1 - 3*x)/(1 - 4*x + (1 + y)*x^2 - y*x), {x, 0, nmax}], x], {y, 0, nmax}], y]] (* Indranil Ghosh, Mar 10 2017, after Philippe Deléham *)

Riordan array ((1-3x)/(1-4x+x^2), x(1-x)/(1-4x+x^2)).
T(n,k) = 4*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), n > 1. - Philippe Deléham, Feb 13 2012
G.f.: (1-3*x)/(1-4*x+(1+y)*x^2-y*x). - Philippe Deléham, Feb 13 2012
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A001835(n), A147722(n), A084120(n) for x = -1, 0, 1, 2 respectively. - Philippe Deléham, Feb 13 2012

A249581 List of quadruples (r,s,t,u): the matrix M = [[9,24,16][3,10,8][1,4,4]] is raised to successive powers, then (r,s,t,u) are the square roots of M[3,1], M[3,3], M[1,1], M[1,3] respectively.

Original entry on oeis.org

0, 1, 1, 0, 1, 2, 3, 4, 5, 8, 13, 20, 23, 36, 59, 92, 105, 164, 269, 420, 479, 748, 1227, 1916, 2185, 3412, 5597, 8740, 9967, 15564, 25531, 39868, 45465, 70996, 116461, 181860, 207391, 323852, 531243, 829564, 946025, 1477268, 2423293, 3784100

Offset: 0

Russell Walsmith, Nov 03 2014

The general form of these matrices is [[t^2,2tu,u^2][rt,st+ru,su][r^2,2rs,s^2]]. Different symmetries have different properties.

Iff |r * u - s * t| = 1 then terms to the left of a(0) are all integers.

			M^0 = [[1,0,0][0,1,0][0,0,1]]. r = sqrt(M[3,1]) = a(0) = 0; s = sqrt(M[3,3]) = a(1) = 1; t = sqrt(M[1,1]) = a(2) = 1; u = sqrt(M[1,3]) = a(3) = 0.
M^1 = [[9,24,16][3,10,8][1,4,4]]. r = sqrt(M[3,1]) = a(4) = 1; s = sqrt(M[3,3]) = a(5) = 2; t = sqrt(M[1,1]) = a(6) = 3; u = sqrt(M[1,3]) = a(7) = 4.

Colin Barker, Table of n, a(n) for n = 0..1000
Russell Walsmith, DCL-Chemy III: Hyper-Quadratics
Index entries for linear recurrences with constant coefficients, signature (0,0,0,5,0,0,0,-2).

Cf. A249579, A249580, A147722, A052984.

LinearRecurrence[{0,0,0,5,0,0,0,-2},{0,1,1,0,1,2,3,4},50] (* Harvey P. Dale, Aug 01 2016 *)

concat(0, Vec(x*(4*x^6-2*x^5-3*x^4+x^3+x+1)/(2*x^8-5*x^4+1) + O(x^100))) \\ Colin Barker, Nov 04 2014

a(4n) + a(4n + 1) = a(4n + 2).
a(4n + 1) + a(4n + 2) + a(4n + 3) - a(4n) = a(4n + 5)
4a(4n) = a(4n+3).
a(4n+1) = A147722(n), a(4n+2) = A052984(n).
a(n) = 5*a(n-4)-2*a(n-8). - Colin Barker, Nov 04 2014
G.f.: x*(4*x^6-2*x^5-3*x^4+x^3+x+1) / (2*x^8-5*x^4+1). - Colin Barker, Nov 04 2014

cp's OEIS Frontend

A015448 a(0) = 1, a(1) = 1, and a(n) = 4*a(n-1) + a(n-2) for n >= 2.

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A203378 T(n,k)=Number of (n+1)X(k+1) 0..1 arrays with every 2X2 subblock having equal diagonal elements or equal antidiagonal elements.

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A147720 Riordan array (1, x(1-x)/(1-3x)).

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A147721 a(n) = C(2,n) DELTA C(0,n).

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A249581 List of quadruples (r,s,t,u): the matrix M = [[9,24,16][3,10,8][1,4,4]] is raised to successive powers, then (r,s,t,u) are the square roots of M[3,1], M[3,3], M[1,1], M[1,3] respectively.

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