A103143 -id:A103143 - cp's OEIS frontend

A373445 Triple convolution of the three tribonacci-like sequences A000073(n), A077947(n-2), and A103143(n).

0, 0, 0, 0, 0, 0, 1, 3, 9, 28, 75, 195, 498, 1229, 2978, 7115, 16756, 39031, 90089, 206228, 468795, 1059197, 2380257, 5323610, 11856514, 26306896, 58172254, 128246136, 281957282, 618367332, 1353112803

Offset: 0

Greg Dresden and Xiaoyuan Wang, Jun 05 2024

nonn

If we set b(n)=A000073(n), c(n)=A077947(n-2) with c(0)=c(1)=0, and d(n)=A103143(n), then all three sequences b(n), c(n), and d(n) start with the terms 0,0,1,1,2 and have signatures {1,1,1}, {1,1,2}, and {1,1,3} respectively. The triple convolution is defined as a(n) = Sum_{i+j+k=n} b(i)*c(j)*d(k).

			For n=7 the triple convolution of the three sequences b(n)=A000073(n), c(n)=A077947(n-2) with c(0)=c(1)=0, and d(n)=A103143(n) has only three nonzero terms in the sum: b(2)*c(2)*d(3), b(2)*c(3)*d(2), and b(3)*c(2)*c(2). All three terms are 1, so the triple convolution adds up to 3. Hence, a(7) = 3.

Cf. A000073, A077947, A103143.

CoefficientList[Series[x^6/((1-x-x^2-x^3)(1-x-x^2-2x^3)(1-x-x^2-3x^3)), {x, 0, 30}], x]

a(n) = (A000073(n+2) + A103143(n+2))/2 - A077947(n).
a(n) = 3*a(n-1) + a(n-3) - 12*a(n-4) - 3*a(n-5) + 2*a(n-6) + 17*a(n-7) + 11*a(n-8) + 6*a(n-9).
G.f.: x^6/((1 - 2*x)*(1 + x + x^2)*(1 - x - x^2 - x^3)*(1 - x - x^2 - 3*x^3)).

A081578 Pascal-(1,3,1) array.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 9, 9, 1, 1, 13, 33, 13, 1, 1, 17, 73, 73, 17, 1, 1, 21, 129, 245, 129, 21, 1, 1, 25, 201, 593, 593, 201, 25, 1, 1, 29, 289, 1181, 1921, 1181, 289, 29, 1, 1, 33, 393, 2073, 4881, 4881, 2073, 393, 33, 1, 1, 37, 513, 3333, 10497, 15525, 10497, 3333, 513, 37, 1

Offset: 0

Paul Barry, Mar 23 2003

One of a family of Pascal-like arrays. A007318 is equivalent to the (1,0,1)-array. A008288 is equivalent to the (1,1,1)-array. Rows include A016813, A081585, A081586. Coefficients of the row polynomials in the Newton basis are given by A013611.

As a number triangle, this is the Riordan array (1/(1-x), x*(1+3*x)/(1-x)). It has row sums A015518(n+1) and diagonal sums A103143. - Paul Barry, Jan 24 2005

			Square array begins as:
  1,  1,   1,   1,    1, ... A000012;
  1,  5,   9,  13,   17, ... A016813;
  1,  9,  33,  73,  129, ... A081585;
  1, 13,  73, 245,  593, ... A081586;
  1, 17, 129, 593, 1921, ...
As a triangle this begins:
  1;
  1,  1;
  1,  5,   1;
  1,  9,   9,    1;
  1, 13,  33,   13,     1;
  1, 17,  73,   73,    17,     1;
  1, 21, 129,  245,   129,    21,     1;
  1, 25, 201,  593,   593,   201,    25,    1;
  1, 29, 289, 1181,  1921,  1181,   289,   29,   1;
  1, 33, 393, 2073,  4881,  4881,  2073,  393,  33,  1;
  1, 37, 513, 3333, 10497, 15525, 10497, 3333, 513, 37, 1; - _Philippe Deléham_, Mar 15 2014

Vincenzo Librandi, Rows n = 0..100, flattened
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
P. Barry, A note on Krawtchouk Polynomials and Riordan Arrays, JIS 11 (2008) 08.2.2

Cf. Pascal (1,m,1) array: A123562 (m = -3), A098593 (m = -2), A000012 (m = -1), A007318 (m = 0), A008288 (m = 1), A081577 (m = 2), A081579 (m = 4), A081580 (m = 5), A081581 (m = 6), A081582 (m = 7), A143683 (m = 8).

a081578 n k = a081578_tabl !! n !! k
a081578_row n = a081578_tabl !! n
a081578_tabl = map fst $ iterate
   (\(us, vs) -> (vs, zipWith (+) (map (* 3) ([0] ++ us ++ [0])) $
                      zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 1])
-- Reinhard Zumkeller, Mar 16 2014

A081578:= func< n,k,q | (&+[Binomial(k, j)*Binomial(n-j, k)*q^j: j in [0..n-k]]) >;
[A081578(n,k,3): k in [0..n], n in [0..12]]; // G. C. Greubel, May 26 2021

Table[Hypergeometric2F1[-k, k-n, 1, 4], {n,0,10}, {k,0,n}]//Flatten (* Jean-François Alcover, May 24 2013 *)

flatten([[hypergeometric([-k, k-n], [1], 4).simplify() for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 26 2021

Square array T(n, k) defined by T(n, 0) = T(0, k) = 1, T(n, k) = T(n, k-1) + 3*T(n-1, k-1) + T(n-1, k).
Rows are the expansions of (1+3*x)^k/(1-x)^(k+1).
T(n,k) = Sum_{j=0..n} binomial(k,j-k)*binomial(n+k-j,k)*3^(j-k). - Paul Barry, Oct 23 2006
E.g.f. for the n-th subdiagonal of the triangle, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(4*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 8*x + 16*x^2/2) = 1 + 9*x + 33*x^2/2! + 73*x^3/3! + 129*x^4/4! + 201*x^5/5! + .... - Peter Bala, Mar 05 2017
From G. C. Greubel, May 26 2021: (Start)
T(n, k, m) = Hypergeometric2F1([-k, k-n], [1], m+1), for m = 3.
T(n, k, m) = Sum_{j=0..n-k} binomial(k,j)*binomial(n-j,k)*m^j, for m = 3.
Sum_{k=0..n} T(n, k, 3) = A015518(n+1). (End)

A247594 a(n) = a(n-1) + a(n-2) + 3*a(n-3) with a(0) = 1, a(1) = 2, a(2) = 5.

Original entry on oeis.org

1, 2, 5, 10, 21, 46, 97, 206, 441, 938, 1997, 4258, 9069, 19318, 41161, 87686, 186801, 397970, 847829, 1806202, 3847941, 8197630, 17464177, 37205630, 79262697, 168860858, 359740445, 766389394, 1632712413, 3478323142, 7410203737, 15786664118, 33631837281

Offset: 0

Michael Somos, Sep 20 2014

a(n) is the number of words of length n in {A,B,C} such that no two consecutive letters are B and every letter C is adjacent to exactly one letter B.

			G.f. = 1 + 2*x + 5*x^2 + 10*x^3 + 21*x^4 + 46*x^5 + 97*x^6 + 206*x^7 + ...
a(3) = 10 with words [AAA, AAB, ABA, ABC, ACB, BAA, BAB, BCA, CBA, CBC].

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
David Beckwith, Problem 11776, The American Mathematical Monthly, 121 (2014), 455. See solution, 123 (May, 2016), 508-510.
Roberto Tauraso, Solution of Problem 11776.
Index entries for linear recurrences with constant coefficients, signature (1,1,3).

Cf. A103143, A123102.

a247594 n = a247594_list !! n
a247594_list = 1 : 2 : 5 : zipWith (+)
   (tail $ zipWith (+) a247594_list $ tail a247594_list)
   (map (* 3) a247594_list)
-- Reinhard Zumkeller, Sep 21 2014

I:=[1,2,5]; [n le 3 select I[n] else Self(n-1)+Self(n-2)+3*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Aug 13 2015

LinearRecurrence[{1, 1, 3}, {1, 2, 5}, 40] (* Vincenzo Librandi, Aug 13 2015 *)

{a(n) = if( n<0, polcoeff( (2*x + x^2 + x^3) / (3 + x + x^2 - x^3) + x * O(x^-n), -n), polcoeff( (1 + x + 2*x^2) / (1 - x - x^2 - 3*x^3) + x * O(x^n), n))};

first(m)={my(v=vector(m));v[1]=1;v[2]=2;v[3]=5;for(i=4,m,v[i]=v[i-1]+v[i-2]+3*v[i-3]);v;} /* Anders Hellström, Aug 12 2015 */

G.f.: (1 + x + 2*x^2) / (1 - x - x^2 - 3*x^3).
0 = a(n) - a(n-1) - a(n-2) - 3*a(n-3) for all n in Z.
From Greg Dresden, Aug 05 2022: (Start)
a(n) = b(n+3) - b(n) for b(n) = A103143(n).
a(n) = c(n+2) - 2*c(n-1) for c(n) = A123102(n). (End)

A237358 The number of tilings of the 3 X 4 X n room with 1 X 2 X 3 boxes.

Original entry on oeis.org

1, 1, 11, 64, 296, 1716, 9123, 48761, 264457, 1420548, 7652666, 41237256, 222050029, 1196138637, 6442843111, 34702528552, 186921714672, 1006820870616, 5423072856651, 29210535955209, 157337764568209, 847474515870020, 4564784961695166, 24587476389796440

Offset: 0

R. J. Mathar, Feb 07 2014

The count compiles all arrangements without respect to symmetry: Stacks that are equivalent after rotations or flips through any of the 3 axes or 3 planes are counted with multiplicity.

The rational generating function is the main body of the Maple program.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
R. J. Mathar, Tilings of rectangular regions by rectangular tiles: counts derived from transfer matrices, arXiv:1406.7788 [math.CO], eq. (58).
Index entries for linear recurrences with constant coefficients, signature (2, 14, 42, -42, -237, -504, -103, 487, 1012, 448, -306, -74, -915, 450, -873, -54, 162).

Cf. A000079 (2 X 2 X n rooms), A103143 (2 X 3 X n rooms).

A237358 := proc(n)
    (1-x)*(1+x)*(1-3*x)*(3*x^2+2*x+1)*(1-x^2-7*x^3+9*x^6)/
    (504*x^6 +306*x^11 +1 -1012*x^9 +103*x^7 -2*x +54*x^16 -162*x^17
    -450*x^14 +74*x^12 -14*x^2 -487*x^8 -42*x^3 -448*x^10 +915*x^13
    +237*x^5 +873*x^15 +42*x^4) ;
    coeftayl(%,x=0,n) ;
end proc:
seq(A237358(n),n=0..20) ;

CoefficientList[Series[(1 - x) (1 + x) (1 - 3 x) (3 x^2 + 2 x + 1) (1 - x^2 - 7 x^3 + 9 x^6)/(504 x^6 + 306 x^11 + 1 - 1012 x^9 + 103 x^7 - 2 x + 54 x^16 - 162 x^17 - 450 x^14 + 74 x^12 - 14 x^2 - 487 x^8 - 42 x^3 - 448 x^10 + 915 x^13 + 237 x^5 + 873 x^15 + 42 x^4), {x, 0, 30}], x] (* Vincenzo Librandi, Feb 08 2014 *)

A237357 The number of tilings of the 3 X 3 X (2n) room with 1 X 2 X 3 boxes.

Original entry on oeis.org

1, 6, 64, 616, 5936, 57408, 554624, 5359040, 51781696, 500337216, 4834483264, 46712942656, 451361370176, 4361255727168, 42140406169664, 407179478511680, 3934350491492416, 38015456589811776, 367322368167936064, 3549233239845138496, 34294281215843786816

Offset: 0

R. J. Mathar, Feb 07 2014

The count compiles all arrangements without respect to symmetry: Stacks that are equivalent after rotations or flips through any of the 3 axes or 3 planes are counted with multiplicity.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
R. J. Mathar, Tilings of rectangular regions by rectangular tiles: counts derived from transfer matrices, arXiv:1406.7788 [math.CO], eq. (57).
Index entries for linear recurrences with constant coefficients, signature (7,22,36).

Cf. A000079 (2 X 2 X n rooms), A103143 (2 X 3 X n rooms).

A237357 := proc(n)
    (1-x)/ (-22*x^2-7*x-36*x^3+1) ;
    coeftayl(%,x=0,n) ;
end proc:
seq(A237357(n),n=0..20) ;

CoefficientList[Series[(1 - x)/(-22 x^2 - 7 x - 36 x^3 + 1), {x, 0, 30}], x] (* Vincenzo Librandi, Feb 08 2014 *)
LinearRecurrence[{7,22,36},{1,6,64},30] (* Harvey P. Dale, Mar 20 2024 *)

G.f.: (1-x)/(-22*x^2-7*x-36*x^3+1).

A356411 Sum of powers of roots of x^3 - x^2 - x - 3.

Original entry on oeis.org

3, 1, 3, 13, 19, 41, 99, 197, 419, 913, 1923, 4093, 8755, 18617, 39651, 84533, 180035, 383521, 817155, 1740781, 3708499, 7900745, 16831587, 35857829, 76391651, 162744241, 346709379, 738628573, 1573570675, 3352327385, 7141783779

Offset: 0

Greg Dresden, Aug 05 2022

a(n) is the sum of the n-th powers of the three roots of x^3 - x^2 - x - 3. These roots are c1 = 2.130395..., c2 = -0.5651977... - i*1.0434274..., and c3 = -0.5651977... + i*1.0434274..., and so a(n) = c1^n + c2^n + c3^n. The real parts of c2 and c3 are A273065.

a(n) can also be determined by Vieta's formulas and Newton's identities. For example, a(3) by definition is c1^3 + c2^3 + c3^3, and from Newton's identities this equals e1^3 - 3*e1*e2 + 3*e3 for e1, e2, e3 the elementary symmetric polynomials of x^3 - x^2 - x - 3. From Vieta's formulas we have e1 = 1, e2 = -1, and e3 = 3, giving us e1^3 - 3*e1*e2 + 3*e3 = 1 + 3 + 9 = 13, as expected.

			For n=3, a(3) = (2.130395...)^3 + (-0.5651977... - i*1.0434274...)^3 + (-0.5651977... + i*1.0434274...)^3 = 13.

Index entries for linear recurrences with constant coefficients, signature (1,1,3).

Cf. A103143, A123102, A247594, A356463, A273065 (Re c2,c3).

LinearRecurrence[{1, 1, 3}, {3, 1, 3}, 40]

polsym(x^3 - x^2 - x - 3, 35) \\ Joerg Arndt, Aug 11 2022

a(n) = a(n-1) + a(n-2) + 3*a(n-3) with a(0)=3, a(1)=1, a(2) = 3.

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

cp's OEIS Frontend

A373445 Triple convolution of the three tribonacci-like sequences A000073(n), A077947(n-2), and A103143(n).

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A081578 Pascal-(1,3,1) array.

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A247594 a(n) = a(n-1) + a(n-2) + 3*a(n-3) with a(0) = 1, a(1) = 2, a(2) = 5.

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A237358 The number of tilings of the 3 X 4 X n room with 1 X 2 X 3 boxes.

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A237357 The number of tilings of the 3 X 3 X (2n) room with 1 X 2 X 3 boxes.

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A356411 Sum of powers of roots of x^3 - x^2 - x - 3.

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