A087936 -id:A087936 - cp's OEIS frontend

A306646 Square array A(n,k), n >= 0, k >= 1, read by antidiagonals, where column k is the expansion of g.f. (k+1-x^k)/(1-x^k-x^(k+1)).

2, 3, 1, 4, 0, 3, 5, 0, 2, 4, 6, 0, 0, 3, 7, 7, 0, 0, 3, 2, 11, 8, 0, 0, 0, 4, 5, 18, 9, 0, 0, 0, 4, 0, 5, 29, 10, 0, 0, 0, 0, 5, 3, 7, 47, 11, 0, 0, 0, 0, 5, 0, 7, 10, 76, 12, 0, 0, 0, 0, 0, 6, 0, 4, 12, 123, 13, 0, 0, 0, 0, 0, 6, 0, 4, 3, 17, 199

Offset: 0

Seiichi Manyama, Mar 03 2019

			A(6,1) = 6*Sum_{j=1..6} binomial(j,6-j)/j = 6*(1/3+3/2+1+1/6) = 18.
A(6,2) = 6*Sum_{j=1..3} binomial(j,6-2*j)/j = 6*(1/2+1/3) = 5.
Square array begins:
    2,  3, 4, 5, 6, 7, 8, 9, 10, 11, ...
    1,  0, 0, 0, 0, 0, 0, 0,  0,  0, ...
    3,  2, 0, 0, 0, 0, 0, 0,  0,  0, ...
    4,  3, 3, 0, 0, 0, 0, 0,  0,  0, ...
    7,  2, 4, 4, 0, 0, 0, 0,  0,  0, ...
   11,  5, 0, 5, 5, 0, 0, 0,  0,  0, ...
   18,  5, 3, 0, 6, 6, 0, 0,  0,  0, ...
   29,  7, 7, 0, 0, 7, 7, 0,  0,  0, ...
   47, 10, 4, 4, 0, 0, 8, 8,  0,  0, ...
   76, 12, 3, 9, 0, 0, 0, 9,  9,  0, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns 1-9 give A000032, A001608, A050443, A087935, A087936, A306755, A306756, A306757, A306758.

Cf. A306713, A306735.

T[0, k_] := k + 1; T[n_, k_] := n *Sum[Binomial[j, n - k*j]/j, {j, 1, Floor[n/k]}]; Table[T[k, n - k + 1], {n, 0, 11}, {k, 0, n}] // Flatten (* Amiram Eldar, Jun 21 2021 *)

A(0,k) = k+1 and A(n,k) = n*Sum_{j=1..floor(n/k)} binomial(j,n-k*j)/j for n > 0.

A(n,k) = (k+1)*A306713(n,k) - A306713(n-k,k) for n >= k.

A050443 a(0)=4, a(1)=0, a(2)=0, a(3)=3; thereafter a(n) = a(n-3) + a(n-4).

Original entry on oeis.org

4, 0, 0, 3, 4, 0, 3, 7, 4, 3, 10, 11, 7, 13, 21, 18, 20, 34, 39, 38, 54, 73, 77, 92, 127, 150, 169, 219, 277, 319, 388, 496, 596, 707, 884, 1092, 1303, 1591, 1976, 2395, 2894, 3567, 4371, 5289, 6461, 7938, 9660, 11750, 14399, 17598, 21410, 26149, 31997

Offset: 0

Tony Davie (ad(AT)dcs.st-and.ac.uk), Dec 23 1999

Related to Perrin sequence. a(p) is divisible by p for primes p.
Wells states that Mihaly Bencze [Beneze] (1998) proved the divisibility property for this sequence: that a(n) is always divisible by n when n is prime. - Gary W. Adamson, Nov 14 2006
As a(n) = trace(M^n) where M = [0,1,0,0; 0,0,1,0; 0,0,0,1; 1,1,0,0], the previous property comes from the fact that trace(M^n) = trace(M) (= 0) mod n for n prime. - Robert FERREOL, Apr 09 2024

			a(11) = 11 because a(7) = 7 and a(8) = 4.

David Wells, "Prime Numbers, the Most Mysterious Figures in Math", John Wiley & Sons, Inc.; 2005, p. 103.

Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from T. D. Noe)
Sadjia Abbad and Hacène Belbachir, The r-Fibonacci polynomial and its companion sequences linked with some classical sequences, Integers (2025), Vol. 25, Art. No. A38. See p. 17.
Mihaly Bencze, Dan Saracino, and Allen Stenger, Solution of Problem 10655: A Recurrence Generating Multiples of Primes, American Mathematical Monthly 107 (2000) 281-282.
Johann Cigler, Recurrences for certain sequences of binomial sums in terms of (generalized) Fibonacci and Lucas polynomials, arXiv:2212.02118 [math.NT], 2022.
Gregory T. Minton, Linear recurrence sequences satisfying congruence conditions, Proc. Amer. Math. Soc. (2014), Vol. 142, No. 7, 2337-2352. MR3195758.
Index entries for linear recurrences with constant coefficients, signature (0,0,1,1).

Column 3 of A306646.
Cf. A001608, A052338.
Cf. A087935, A087936.

a:=[4,0,0,3];; for n in [5..60] do a[n]:=a[n-3]+a[n-4]; od; Print(a); # Muniru A Asiru, Mar 09 2019

I:=[4,0,0,3]; [n le 4 select I[n] else Self(n-3) +Self(n-4): n in [1..60]]; // G. C. Greubel, Mar 04 2019

LinearRecurrence[{0,0,1,1}, {4,0,0,3}, 60] (* G. C. Greubel, Mar 04 2019 *)

polsym(x^4-x-1,55) \\ Joerg Arndt, Mar 04 2019

((4-x^3)/(1-x^3-x^4)).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Mar 04 2019

G.f.: (4-x^3)/(1-x^3-x^4). - Christian G. Bower, Dec 23 1999
a(n) = (x_1)^n + (x_2)^n + (x_3)^n + (x_4)^n where (x_i) 1 <= i <= 4 are the roots of x^4 = x + 1. - Benoit Cloitre, Oct 27 2003
Let M = the 4 X 4 matrix [0,1,0,0; 0,0,1,0; 0,0,0,1; 1,1,0,0]; then a(n) = the leftmost term of M^n * [4,0,0,3]. Example: a(13) = 13 since M^13 * [4,0,0,3] = [13,21,18,20]. - Gary W. Adamson, Nov 14 2006
a(0) = 4 and a(n) = n*Sum_{k=1..floor(n/3)} binomial(k,n-3*k)/k for n > 0. - Seiichi Manyama, Mar 04 2019
From Aleksander Bosek, Mar 10 2019: (Start)
a(n+10) = a(n+5) + 2*a(n+3) + a(n).
a(n+11) = a(n+6) + 3*a(n+1) + 2*a(n).
a(n+12) = a(n+10) + 5*a(n+5) + a(n).
a(n+12) = 3*a(n+5) + a(n+3) + a(n).
a(n+13) = 3*a(n+6) + 2*a(n+1) + a(n).
a(n+14) = 2*a(n+8) + 3*a(n+3) + a(n).
a(n+15) = 2*a(n+7) + 4*a(n+5) + a(n).
a(n+15) = 2*a(n+9) + 4*a(n+1) + 3*a(n).
a(n+19) = a(n+17) + 5*a(n+5) + a(n).
a(n+20) = 5*a(n+10) + 6*a(n+5) + a(n).
a(n+22) = a*(n+21) + 5*a(n+5) + a(n).
a(n+25) = 2*a(n+21) + 5*a(n+5) + a(n).
a((s+4)*n+m) = Sum_{l=0..n} binomial(n-l,l)*a(s*n+l+m) for every m,s > 0.
a(m) = Sum_{l=0..n}(-1)^{n-l}*binomial(n-l,l)*a(m+n+3*l) for every m > 0. (End)
a(n) = 4*A017817(n) - A017817(n-3). - R. J. Mathar, Aug 10 2021

More terms from Christian G. Bower, Dec 23 1999

More terms from Benoit Cloitre, Oct 27 2003

A087935 Perrin sequence of order 5.

Original entry on oeis.org

5, 0, 0, 0, 4, 5, 0, 0, 4, 9, 5, 0, 4, 13, 14, 5, 4, 17, 27, 19, 9, 21, 44, 46, 28, 30, 65, 90, 74, 58, 95, 155, 164, 132, 153, 250, 319, 296, 285, 403, 569, 615, 581, 688, 972, 1184, 1196, 1269, 1660, 2156, 2380, 2465, 2929, 3816, 4536, 4845, 5394, 6745, 8352, 9381

Offset: 0

Benoit Cloitre, Oct 27 2003

If p is prime, p divides a(p).

Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from T. D. Noe)
Sadjia Abbad and Hacène Belbachir, The r-Fibonacci polynomial and its companion sequences linked with some classical sequences, Integers (2025), Vol. 25, Art. No. A38. See p. 17.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,1,1).

Column 4 of A306646.
Cf. A001608, A050443.
Cf. A087936.

a:=[5,0,0,0,4];; for n in [6..60] do a[n]:=a[n-4]+a[n-5]; od; Print(a); # Muniru A Asiru, Mar 06 2019

I:=[5,0,0,0,4]; [n le 5 select I[n] else Self(n-4) +Self(n-5): n in [1..60]]; // G. C. Greubel, Mar 06 2019

seq(coeff(series((x^4-5)/(x^5+x^4-1),x,n+1), x, n), n = 0 .. 60); # Muniru A Asiru, Mar 06 2019

LinearRecurrence[{0,0,0,1,1},{5,0,0,0,4},60] (* Harvey P. Dale, Oct 03 2016 *)

my(x='x+O('x^60)); Vec((5-x^4)/(1-x^4-x^5)) \\ G. C. Greubel, Mar 06 2019

polsym(x^5-x-1,66) \\ Joerg Arndt, Mar 10 2019

((5-x^4)/(1-x^4-x^5)).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Mar 06 2019

a(n) = a(n-4) + a(n-5), with a(0)=5, a(1)=a(2)=a(3)=0.
a(n) = (x_1)^n + (x_2)^n + (x_3)^n + (x_4)^n + (x_5)^n where (x_i) 1 <= i <= 5 are the roots of x^5=x+1.
G.f.: (5 - x^4)/(1 -x^4 -x^5). - Colin Barker, Jun 16 2013
a(0) = 5 and a(n) = n*Sum_{k=1..floor(n/4)} binomial(k,n-4*k)/k for n > 0. - Seiichi Manyama, Mar 04 2019
From Aleksander Bosek, Mar 06 2019: (Start)
a((s+5)*n + m) = Sum_{j=0..n} binomial(n-j,j)*a(s*n+j+m) for all s > 0, m > 0.
a(m) = Sum_{j=0..n} (-1)^(n-j)*binomial(n-j,j)*a(m+n+4*j) for all m > 0. (End)

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A306646 Square array A(n,k), n >= 0, k >= 1, read by antidiagonals, where column k is the expansion of g.f. (k+1-x^k)/(1-x^k-x^(k+1)).

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A050443 a(0)=4, a(1)=0, a(2)=0, a(3)=3; thereafter a(n) = a(n-3) + a(n-4).

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A087935 Perrin sequence of order 5.

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