A112455 -id:A112455 - cp's OEIS frontend

A112458 Let b(n) = A112455(n). Then b(n)/n is an integer iff n is prime (at least for the first few values, as for the Perrin sequence). This sequence is the values of b(p)/p, where p is the n th prime.

1, 1, -1, 1, 0, 1, 3, -3, 2, -16, 13, 58, 93, -48, 257, -508, -2439, 3751, 3120, 21824, -28485, -49886, -184653, -158325, -1238859, 2621639, -6281879, -10666638, 5587636, -32300736, -541428896, -309499443, -3404250996, 4487895627, 30910925690, -33820854920, -104685213761, 219810779538

Offset: 1

Anthony C Robin, Dec 13 2005

sign

See Perrin sequences

A112455 := proc(n) option remember; if n <= 3 then op(n+1,[ -3,0,2,3]) ; else -A112455(n-2)-A112455(n-3) ; fi ; end: A112458 := proc(n) local p ; p := ithprime(n) ; A112455(p)/p ; end: seq(A112458(n),n=1..80) ; # R. J. Mathar, Aug 24 2007

More terms from R. J. Mathar, Aug 24 2007

A218439 a(n) = A001609(n)^2, where g.f. of A001609 is x(1+3x^2)/(1-x-x^3).

Original entry on oeis.org

1, 1, 16, 25, 36, 100, 225, 441, 961, 2116, 4489, 9604, 20736, 44521, 95481, 205209, 440896, 946729, 2033476, 4368100, 9381969, 20151121, 43283241, 92968164, 199685161, 428904100, 921243904, 1978737289, 4250127249, 9128847025, 19607840784, 42115658841

Offset: 1

Paul D. Hanna, Oct 28 2012

nonn

A001609 equals the logarithmic derivative of Narayana's cows sequence A000930.

			O.g.f.: A(x) = x + x^2 + 16*x^3 + 25*x^4 + 36*x^5 + 100*x^6 + 225*x^7 +...
L.g.f.: L(x) = x + x^2/2 + 16*x^3/3 + 25*x^4/4 + 36*x^5/5 + 100*x^6/6 + 225*x^7/7 +...
where exponentiation yields the g.f. of A218438:
exp(L(x)) = 1 + x + x^2 + 6*x^3 + 12*x^4 + 19*x^5 + 48*x^6 + 110*x^7 +...

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,3,1,-1,-1).

Cf. A000930, A001609, A002478, A112455, A218438.

Rest[CoefficientList[Series[x*(1 + 14*x^2 + 5*x^3 - 9*x^4 - 9*x^5)/((1 + x^2 - x^3)*(1 - x - 2*x^2 - x^3)), {x, 0, 50}], x]] (* G. C. Greubel, Apr 28 2017 *)

{a(n)=polcoeff(x*(1+14*x^2+5*x^3-9*x^4-9*x^5)/((1+x^2-x^3)*(1-x-2*x^2-x^3+x*O(x^n))),n)}
for(n=1,40,print1(a(n),", "))

O.g.f.: x*(1 + 14*x^2 + 5*x^3 - 9*x^4 - 9*x^5)/((1 + x^2 - x^3)*(1 - x - 2*x^2 - x^3)).
Logarithmic derivative of A218438.
a(n) = -2*(-1)^n*A112455(n) +3*A002478(n) -2*A002478(n-1)-2*A002478(n-2), n>1. - R. J. Mathar, Oct 28 2012

A321196 Riordan triangle T = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)).

Original entry on oeis.org

1, 0, 1, -1, 0, 1, 1, -2, 0, 1, 1, 2, -3, 0, 1, -2, 3, 3, -4, 0, 1, 0, -6, 6, 4, -5, 0, 1, 3, -1, -12, 10, 5, -6, 0, 1, -2, 12, -4, -20, 15, 6, -7, 0, 1, -3, -7, 30, -10, -30, 21, 7, -8, 0, 1, 5, -16, -15, 60, -20, -42, 28, 8, -9, 0, 1

Offset: 0

Wolfdieter Lang, Nov 09 2018

This is the (ordinary) convolution triangle based on A077961 (the column k = 0 of T).
The row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k, with R(-1, x) = 0, appear in the Cayley-Hamilton formula for nonnegative powers of a 3 X 3 matrix with Det M = sigma(3; 3) = x1*x2*x3 = +1, sigma(3; 2) := x1*x2 + x1*x*3 + x2*x^3 = +1 and Tr M = sigma(3; 1) = x1 + x2 = x, where x1, x2, and x3 are the eigenvalues of M, and sigma the elementary symmetric functions, as M^n = R(n-2, x)*M^2 + (-R(n-3, x) + R(n-4, x))*M + R(n-3, x)*1_3, for n >= 3, where M^0 = 1_3 is the 3 X 3 unit matrix.
For the Cayley-Hamilton formula for 3 X 3 matrices with Det M = +1, sigma(3,2) = -1 and Tr(M) = x see A104578.
The row sums give A133872 (repeat(1, 1, 0, 0)). The alternating row sums give A057597(n+2), for n >= 0.
The Riordan triangle (1/(1 + x^2 + x^3), x/(1 + x^2 + x^3)) has entries t(n, m) = (-1)^(n-m)*T(n, m) (from the g.f. G(-x, -z), where the g.f. G of T is given below).
The inverse of Riordan T is T^{-1}, given in A321198.

			The triangle T(n, k) begins:
n\k  0   1   2   3   4   5  6  7  8  9 10 ...
---------------------------------------------
0:   1
1:   0   1
2:  -1   0   1
3:   1  -2   0   1
4:   1   2  -3   0   1
5:  -2   3   3  -4   0   1
6:   0  -6   6   4  -5   0  1
7:   3  -1 -12  10   5  -6  0  1
8:  -2  12  -4 -20  15   6 -7  0  1
9:  -3  -7  30 -10 -30  21  7 -8  0  1
10:  5 -16 -15  60 -20 -42 28  8 -9  0  1
...
Cayley-Hamilton formula for the matrix TS(x) =[[x,-1,1], [1,0,0], [0,1,0]] with Det(TS(x)) = +1, sigma(3, 2) = +1, and Tr(TS(x)) = x. For n = 3: TS(x)^3 = R(1, x)*TS(x)^2 + (-R(0, x) + R(-1, x))*TS(x) + R(0, x)*1_3 = x*TS(x)^2 - TS(x) + 1_3. Compare this for x = -1 with r^3 = R(3)*r^2 + (-R(2) + R(1))*r + R(2)*1 = r^2 - r + 1, where r = 1/t = A192918, with the tribonacci constant t = A058265, and R(n) = A057597(n) = R(n-2, -1).
Recurrence: T(5, 2) = T(4, 1) - T(3, 2) + T(2, 2) = 1 -(-1) + 1 = 3.
Boas-Buck type recurrence with B = {0, -2, 3, ...}:
  T(5, 2) = ((2+1)/(5-2))*(3*1 + (-2)*0 + 0*(-3)) = 1*3 = 3.
Z- and A-recurrence with A(n) = {1, 0, -1, 1, -1, ...} and Z(n) = A(n+1):
  T(4, 0) = 0*T(3, 0) - 1*T(3, 1) + 1*T(3, 2) - 1*T(3, 3) = 0 + 2 + 0 - 1 = 1.
  T(5, 2) = 1*T(4, 1) + 0*T(4, 2) - 1*T(4, 3) + 1*T(4, 4) = 2 + 0 + 0 + 1 = 3.

Cf. A057597, A058265, A077961, A001005, A104578, A112455, A133872, A192918, A321197, A321198.

T[n_, k_] /; 0 <= k <= n := T[n, k] = T[n - 1, k - 1] - T[n - 2, k] + T[n - 3, k]; T[0, 0] = 1; T[, ] = 0;
Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* Jean-François Alcover, Jul 06 2019 *)

# uses[riordan_array from A256893]
riordan_array(1/(1 + x^2 - x^3), x/(1 + x^2 - x^3), 11) # Peter Luschny, Nov 13 2018

T(n, k) = T(n-1, k-1) - T(n-2, k) + T(n-3, k), T(0, 0) = 1, T(n,k) = 0 if n < k or if k < 0. (Cf. A104578.)
The Riordan property T = (G(x), x*G(x)) with G(x) = 1/(1 + x^2 - x^3) implies the following.
G.f. of row polynomials R(n, x) is G(x, z) = 1/(1 - x*z + z^2 - z^3).
G.f. of column sequence k: x^k/(1 + x^2 - x^3)^(k+1), k >= 0.
Boas-Buck recurrence (see the Aug 10 2017 remark in A046521, also for two references):
T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} B(n-1-j)*T(j, k), for n >= 1, k = 0,1, ..., n-1, and input T(n, n) = 1, for n >= 0. Here B(n) = [x^n]*(d/dx)log(G(x)) = x*(-2 + 3*x)/(1 + x^2 - x^3) = (-1)^n*A112455(n+1), for n >= 0.
Recurrences from the A- and Z- sequences (see the W. Lang link under A006232 with references), which are A(n) = A321197(n) and Z(n) = A(n+1).
  T(0, 0) = 1, T(n, k) = 0 for n < k, and
  T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and
  T(n, k) = Sum_{j=0..n-k} A(j)*T(n-1, k-1+j), for n >= m >= 1.

A362540 Number of chordless cycles of length >= 4 in the n-flower graph.

Original entry on oeis.org

3, 23, 63, 127, 273, 583, 1287, 2975, 6993, 16535, 39525, 95071, 229029, 552199, 1332375, 3215807, 7762611, 18739607, 45240309, 109217983, 263673699, 636563527, 1536798717, 3710157407, 8957109801, 21624374039, 52205854257, 126036078751, 304278008331, 734592089095

Offset: 2

Eric W. Weisstein, Apr 24 2023

The n-flower graph can be defined for n >= 3 without multiple edges. It is a snark for odd n >= 5. The sequence has been extended to n=2 using the recurrence. - Andrew Howroyd, Apr 26 2023

Eric Weisstein's World of Mathematics, Chordless Cycle
Eric Weisstein's World of Mathematics, Flower Graph
Index entries for linear recurrences with constant coefficients, signature (4,-5,4,-2,-4,7,-8,6,4,-6,0,1).

Cf. A362545.

LinearRecurrence[{4, -5, 4, -2, -4, 7, -8, 6, 4, -6, 0, 1}, {3, 23, 63, 127, 273, 583, 1287, 2975, 6993, 16535, 39525, 95071}, 20]
CoefficientList[Series[(-3 - 11 x + 14 x^2 + 22 x^3 + 6 x^4 + 68 x^5 - 9 x^6 - 19 x^7 + 25 x^8 - 13 x^9 - 9 x^10 + x^11)/((-1 + x)^3 (1 - x - x^2 - 3 x^3 - 5 x^4 - 3 x^5 - 4 x^6 + 3 x^8 + x^9)), {x, 0, 20}], x]
Table[(1 + (-1)^n)/2 + 2 (-I)^n ChebyshevT[n, I] + 3 (n - 2) n + RootSum[-1 + # + #^3 &, #^n &] + RootSum[1 + # + #^3 &, #^n &], {n, 2, 20}]

From Andrew Howroyd, Apr 26 2023: (Start)
a(n) = 4*a(n-1) - 5*a(n-2) + 4*a(n-3) - 2*a(n-4) - 4*a(n-5) + 7*a(n-6) - 8*a(n-7) + 6*a(n-8) + 4*a(n-9) - 6*a(n-10) + a(n-12).
G.f.: x^2*(3 + 11*x - 14*x^2 - 22*x^3 - 6*x^4 - 68*x^5 + 9*x^6 + 19*x^7 - 25*x^8 + 13*x^9 + 9*x^10 - x^11)/((1 - x)^3*(1 + x)*(1 - 2*x - x^2)*(1 + x^2 - x^3)*(1 + x^2 + x^3)). (End)
2*a(n) = -2*(-1)^n*A112455(n) +1+6*n^2-12*n+(-1)^n-2*A112455(n)+4*A001333(n). - R. J. Mathar, Feb 18 2024

a(2)-a(4) and a(17) and beyond from Andrew Howroyd, Apr 26 2023

cp's OEIS Frontend

A112458 Let b(n) = A112455(n). Then b(n)/n is an integer iff n is prime (at least for the first few values, as for the Perrin sequence). This sequence is the values of b(p)/p, where p is the n th prime.

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A218439 a(n) = A001609(n)^2, where g.f. of A001609 is x*(1+3*x^2)/(1-x-x^3).

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A321196 Riordan triangle T = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)).

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A362540 Number of chordless cycles of length >= 4 in the n-flower graph.

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Mathematica

Formula

Extensions

A218439 a(n) = A001609(n)^2, where g.f. of A001609 is x(1+3x^2)/(1-x-x^3).