A113225 -id:A113225 - cp's OEIS frontend

A113224 a(2n) = A002315(n), a(2n+1) = A082639(n+1).

1, 2, 7, 16, 41, 98, 239, 576, 1393, 3362, 8119, 19600, 47321, 114242, 275807, 665856, 1607521, 3880898, 9369319, 22619536, 54608393, 131836322, 318281039, 768398400, 1855077841, 4478554082, 10812186007, 26102926096, 63018038201

Offset: 0

Creighton Dement, Oct 18 2005

From Paul D. Hanna, Oct 22 2005: (Start)
The logarithmic derivative of this sequence is twice the g.f. of A113282, where a(2*n) = A113282(2*n), a(4*n+1) = A113282(4*n+1) - 3, a(4*n+3) = A113282(4*n+3) - 1.
Equals the self-convolution of integer sequence A113281. (End)
With an offset of 1, this sequence is the case P1 = 2, P2 = 0, Q = -1 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 19 2015
Floretion Algebra Multiplication Program, FAMP Code: -2ibaseiseq[B*C], B = - .5'i + .5'j - .5i' + .5j' - 'kk' - .5'ik' - .5'jk' - .5'ki' - .5'kj'; C = + .5'i + .5i' + .5'ii' + .5e

Creighton Dement, Floretion Online Multiplier. [broken link]
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (2,2,-2,-1).

Cf. A113225, A002315, A082639, A100828, A113281, A113282, A113283, A113284, A129744.

[Floor((1+Sqrt(2))^(n+1)/2): n in [0..30]]; // Vincenzo Librandi, Mar 20 2015

a[n_] := n*Sum[ Sum[ Binomial[i, n-k-i]*Binomial[k+i-1, k-1], {i, Ceiling[(n-k)/2], n-k}]*(1-(-1)^k)/(2*k), {k, 1, n}]; Table[a[n], {n, 1, 29}] (* Jean-François Alcover, Feb 26 2013, after Vladimir Kruchinin *)
CoefficientList[Series[(1 + x^2) / ((x^2 - 1) (x^2 + 2 x - 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Mar 20 2015 *)
LinearRecurrence[{2,2,-2,-1},{1,2,7,16},30] (* Harvey P. Dale, Oct 10 2017 *)

a(n):=n*sum(sum(binomial(i,n-k-i)*binomial(k+i-1,k-1),i,ceiling((n-k)/2),n-k)*(1-(-1)^k)/(2*k),k,1,n); /* Vladimir Kruchinin, Apr 11 2011 */

{a(n)=local(x=X+X*O(X^n));polcoeff((1+x^2)/(1-x^2)/(1-2*x-x^2),n,X)} \\ Paul D. Hanna

G.f.: (1+x^2)/((x-1)*(x+1)*(x^2+2*x-1)).
a(n+2) - a(n+1) - a(n) = A100828(n+1).
a(n) = -(u^(n+1)-1)*(v^(n+1)-1)/2 with u = 1+sqrt(2), v = 1-sqrt(2). - Vladeta Jovovic, May 30 2007
a(n) = n * Sum_{k=1..n} Sum_{i=ceiling((n-k)/2)..n-k} binomial(i,n-k-i)*binomial(k+i-1,k-1)*(1-(-1)^k)/(2*k). - Vladimir Kruchinin, Apr 11 2011
a(n) = A001333(n+1) - A000035(n). - R. J. Mathar, Apr 12 2011
a(n) = floor((1+sqrt(2))^(n+1)/2). - Bruno Berselli, Feb 06 2013
From Peter Bala, Mar 19 2015: (Start)
a(n) = (1/2) * A129744(n+1).
exp( Sum_{n >= 1} 2*a(n-1)*x^n/n ) = 1 + 2*Sum_{n >= 1} Pell(n) *x^n. (End)
a(n) = A105635(n-1) + A105635(n+1). - R. J. Mathar, Mar 23 2023

A181864 a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the squares of the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

Original entry on oeis.org

1, 2, 7, 207, 700207, 207000000700207, 70020700000000000000207000000700207, 2070000007002070000000000000000000000000000000000070020700000000000000207000000700207

Offset: 1

Peter Bala, Nov 28 2010

The calculations for the first few values of the sequence are
... 2^2 = 4 so a(3) = 14/2 = 7
... 7^2 = 49 so a(4) = 1449/7 = 207
... 207^2 = 42849 so a(5) = 144942849/207 = 700207.
For similarly defined sequences see A181754 through A181756 and A181865 through
 A181870.

A000129, A113225, A181754, A181755, A181756, A181865, A181866,A181867, A181868, A181869, A181870

#A181864
M:=8: a:=array(1..M):s:=array(1..M):
a[1]:=1:a[2]:=2:
s[1]:=convert(a[1]^2,string):
s[2]:=cat(s[1],convert(a[2]^2,string)):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(s[n-1],convert(a[n]^2,string));
end do:
seq(a[n],n = 1..M);

DEFINITION
a(1) = 1, a(2) = 2, and for n >= 3
(1)... a(n) = concatenate(a(1)^2,a(2)^2,...,a(n-1)^2)/a(n-1).
RECURRENCE RELATION
For n >= 2
(2)...a(n+2) = a(n+1) + 10^F(n,2)*a(n) = a(n+1) + 10^Pell(n)*a(n),
where F(n,2) is the Fibonacci polynomial F(n,x) evaluated at x = 2
and where Pell(n) = A000129(n).
RELATION WITH OTHER SEQUENCES
a(n) has A113225(n-2) digits.
a(n)^2 has Pell(n-1) digits.

A119468 Triangle read by rows: T(n,k) = Sum_{j=0..n-k} binomial(n,2j)*binomial(n-2j,k).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 4, 6, 3, 1, 8, 16, 12, 4, 1, 16, 40, 40, 20, 5, 1, 32, 96, 120, 80, 30, 6, 1, 64, 224, 336, 280, 140, 42, 7, 1, 128, 512, 896, 896, 560, 224, 56, 8, 1, 256, 1152, 2304, 2688, 2016, 1008, 336, 72, 9, 1, 512, 2560, 5760, 7680, 6720, 4032, 1680, 480, 90, 10, 1

Offset: 0

Paul Barry, May 21 2006

Product of Pascal's triangle A007318 and A119467. Row sums are A007051. Diagonal sums are A113225.
Variant of A080928, A115068 and A082137. - R. J. Mathar, Feb 09 2010
Matrix inverse of the Euler tangent triangle A081733. - Peter Luschny, Jul 18 2012
Central column: T(2*n,n) = A069723(n). - Peter Luschny, Jul 22 2012
Subtriangle of the triangle in A198792. - Philippe Deléham, Nov 10 2013

			Triangle begins
    1;
    1,    1;
    2,    2,    1;
    4,    6,    3,    1;
    8,   16,   12,    4,    1;
   16,   40,   40,   20,    5,    1;
   32,   96,  120,   80,   30,    6,    1;
   64,  224,  336,  280,  140,   42,    7,   1;
  128,  512,  896,  896,  560,  224,   56,   8,  1;
  256, 1152, 2304, 2688, 2016, 1008,  336,  72,  9,  1;
  512, 2560, 5760, 7680, 6720, 4032, 1680, 480, 90, 10, 1;

A082137 read as triangle with rows reversed.

Cf. A000182, A009006, A038207, A081733, A155585.

A119468_row := proc(n) local s,t,k;
  s := series(exp(z*x)/(1-tanh(x)),x,n+2);
  t := factorial(n)*coeff(s,x,n); seq(coeff(t,z,k), k=(0..n)) end:
for n from 0 to 7 do A119468_row(n) od; # Peter Luschny, Aug 01 2012
# Alternatively:
T := (n, k) -> 2^(n-k-1+0^(n-k))*binomial(n,k):
for n from 0 to 9 do seq(T(n,k), k=0..n) od; # Peter Luschny, Nov 10 2017

A[k_] := Table[If[m < n, 1, -1], {m, k}, {n, k}]; a = Join[{{1}}, Table[(-1)^n*CoefficientList[CharacteristicPolynomial[A[n], x], x], {n, 1, 10}]]; Flatten[a] (* Roger L. Bagula and Gary W. Adamson, Jan 25 2009 *)
Table[Sum[Binomial[n,2j]Binomial[n-2j,k],{j,0,n-k}],{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, Dec 14 2022 *)

R = PolynomialRing(QQ, 'x')
def p(n,x) :
  return 1 if n==0 else add((-1)^n*binomial(n,k)*(x^(n-k)-1) for k in range(n))
def A119468_row(n):
    x = R.gen()
    return [abs(cf) for cf in list((p(n,x-1)-p(n,x+1))/2+x^n)]
for n in (0..8) : print(A119468_row(n)) # Peter Luschny, Jul 22 2012

G.f.: (1 - x - xy)/(1 - 2x - 2x*y + 2x^2*y + x^2*y^2).
Number triangle T(n,k) = Sum_{j=0..n} binomial(n,j)*binomial(j,k)*(1+(-1)^(j-k))/2.
Define matrix: A(n,m,k) = If[m < n, 1, -1];
  p(x,k) = CharacteristicPolynomial[A[n,m,k],x]; then t(n,m) = coefficients(p(x,n)). - Roger L. Bagula and Gary W. Adamson, Jan 25 2009
E.g.f.: exp(x*z)/(1-tanh(x)). - Peter Luschny, Aug 01 2012
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - 2*T(n-2,k-1) - T(n-2,k-2) for n >= 2, T(0,0) = T(1,0) = T(1,1) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 10 2013
E.g.f.: [(e^(2t)+1)/2] e^(tx) = e^(P.(x)t), so this is an Appell sequence with lowering operator D = d/dx and raising operator R = x + 2/(e^(-2D)+1), i.e., D P_n(x) = n P_{n-1}(x) and R p_n(x) = P_{n+1}(x) where P_n(x) = [(x+2)^n + x^n]/2. Also, (P.(x)+y)^n = P_n(x+y), umbrally. R = x + 1 + D - 2 D^3/3! + ... contains the e.g.f.(D) mod signs of A009006 and A155585 and signed, aerated A000182, the zag numbers, so the unsigned differential component 2/[e^(2D)+1] = 2 Sum_{n >= 0} Eta(-n) (-2D)^n/n!, where Eta(s) is the Dirichlet eta function, and 2 *(-2)^n Eta(-n) = (-1)^n (2^(n+1)-4^(n+1)) Zeta(-n) = (2^(n+1)-4^(n+1)) B(n+1)/(n+1) with Zeta(s), the Riemann zeta function, and B(n), the Bernoulli numbers. The polynomials PI_n(x) of A081733 are the umbral compositional inverses of this sequence, i.e., P_n(PI.(x)) = x^n = PI_n(P.(x)) under umbral composition. Aside from the signs and the main diagonals, multiplying this triangle by 2 gives the face-vectors of the hypercubes A038207. - Tom Copeland, Sep 27 2015
T(n,k) = 2^(n-k-1+0^(n-k))*binomial(n, k). - Peter Luschny, Nov 10 2017

cp's OEIS Frontend

A113224 a(2n) = A002315(n), a(2n+1) = A082639(n+1).

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A181864 a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the squares of the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

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A119468 Triangle read by rows: T(n,k) = Sum_{j=0..n-k} binomial(n,2j)*binomial(n-2j,k).

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