A099930 -id:A099930 - cp's OEIS frontend

A084175 Jacobsthal oblong numbers.

0, 1, 3, 15, 55, 231, 903, 3655, 14535, 58311, 232903, 932295, 3727815, 14913991, 59650503, 238612935, 954429895, 3817763271, 15270965703, 61084037575, 244335800775, 977343902151, 3909374210503, 15637499638215, 62549992960455

Offset: 0

Paul Barry, May 18 2003

Inverse binomial transform is A001019 doubled up.
Binomial transform is A084177.
Partial sums of A003683.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, arXiv:math/0509316 [math.NT], 2005-2006.
N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.
Ronald Orozco López, Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function, arXiv:2408.08943 [math.CO], 2024. See p. 11.
Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 6.
Index entries for linear recurrences with constant coefficients, signature (3,6,-8).

Except for initial terms, same as A015249 and A084152.

Cf. A001654, A001656, A084158, A084159, A099930.

List([0..30], n-> (2^(2*n+1) -(-2)^n -1)/9); # G. C. Greubel, Sep 21 2019

[(2*4^n-(-2)^n-1)/9: n in [0..30]]; // Vincenzo Librandi, Jun 04 2011

for n from 1 to 25 do print(round(2^n/3)*round(2^(n+1)/3)) od; # Gary Detlefs, Feb 10 2010

Table[(2*4^n -(-2)^n -1)/9, {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011, modified by G. C. Greubel, Sep 21 2019 *)
LinearRecurrence[{3,6,-8}, {0,1,3}, 25] (* Jean-François Alcover, Sep 21 2017 *)

a(n)=(2*4^n-(-2)^n-1)/9 \\ Charles R Greathouse IV, Sep 24 2015

def A084175(n): return ((m:=1<Chai Wah Wu, Apr 25 2025

[gaussian_binomial(n, 2, -2) for n in range(1, 26)] # Zerinvary Lajos, May 28 2009

a(n) = A001045(n)*A001045(n+1).
a(n) = (2*4^n - (-2)^n - 1)/9;
a(n) = 3*a(n-1) + 6*a(n-2) - 8*a(n-3), a(0)=0, a(1)=1, a(2)=3.
G.f.: x/((1+2*x)*(1-x)*(1-4*x)).
E.g.f.: (2*exp(4*x) - exp(x) - exp(-2*x))/9.
a(n+1) - 4*a(n) = 1, -1, 3, -5, 11, ... = A001045(n+1) signed. - Paul Curtz, May 19 2008
a(n) = round(2^n/3) * round(2^(n+1)/3). - Gary Detlefs, Feb 10 2010
From Peter Bala, Mar 30 2015: (Start)
The shifted o.g.f. A(x) := 1/( (1 + 2*x)*(1 - x)*(1 - 4*x) ) = 1/(1 - 3*x - 6*x^2 + 8*x^3). Hence A(x) == 1/(1 - 3*x + 3*x^2 - x^3) (mod 9) == 1/(1 - x)^3 (mod 9). It follows by Theorem 1 of Heninger et al. that (A(x))^(1/3) = 1 + x + 4*x^2 + 10*x^3 + ... has integral coefficients.
Sum_{n >= 0} a(n+1)*x^n = exp( Sum_{n >= 1} J(3*n)/J(n)*x^n/n ), where J(n) = A001045(n) are the Jacobsthal numbers. Cf. A001656, A099930. (End)

A099927 Pellonomial triangle P(k,n) read by rows.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 5, 5, 1, 1, 12, 30, 12, 1, 1, 29, 174, 174, 29, 1, 1, 70, 1015, 2436, 1015, 70, 1, 1, 169, 5915, 34307, 34307, 5915, 169, 1, 1, 408, 34476, 482664, 1166438, 482664, 34476, 408, 1, 1, 985, 200940, 6791772, 39618670, 39618670, 6791772, 200940, 985, 1

Offset: 0

Ralf Stephan, Nov 03 2004

Also (signed) coefficients of solutions to 0 = Sum[i=0..k+1, x(i)*Pell(m+i)^k ].

Sagan and Savage give two combinatorial interpretations for entry T(n,k) in terms of statistics on integer partitions fitting inside a k x (n-k) rectangle. They also relate the values T(n,k) to q-binomial coefficients evaluated at q = -(3 + 2*sqrt(2)). - Peter Bala, Mar 15 2013

			Triangle starts:
  1;
  1,   1;
  1,   2,    1;
  1,   5,    5,     1;
  1,  12,   30,    12,     1;
  1,  29,  174,   174,    29,    1;
  1,  70, 1015,  2436,  1015,   70,   1;
  1, 169, 5915, 34307, 34307, 5915, 169, 1;
  ...

Alois P. Heinz, Rows n = 0..56, flattened
Tom Edgar and Michael Z. Spivey, Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.
S. Falcon, On The Generating Functions of the Powers of the K-Fibonacci Numbers, Scholars Journal of Engineering and Technology (SJET), 2014; 2 (4C):669-675.
B. Sagan and C. Savage, Combinatorial Interpretations of Binomial Coefficient Analogues Related to Lucas Sequences, arXiv:0911.3159 [math.CO], 2009.
B. Sagan and C. Savage, Combinatorial Interpretations of Binomial Coefficient Analogues Related to Lucas Sequences, Integers 10 (2010), 697-703, A52.

Columns include A000129, A084158, A099930, A099931, A383719.
Row sums are in A099928. Central column is in A099929.
Cf. A010048, A256832, A383715.

p:= proc(n) p(n):= `if`(n<2, n, 2*p(n-1)+p(n-2)) end:
f:= proc(n) f(n):= `if`(n=0, 1, p(n)*f(n-1)) end:
T:= (n, k)-> f(n)/(f(k)*f(n-k)):
seq(seq(T(n, k), k=0..n), n=0..10); # Alois P. Heinz, Aug 15 2013

p[n_] := p[n] = If[n<2, n, 2*p[n-1] + p[n-2]]; f[n_] := f[n] = If[n == 0, 1, p[n] * f[n-1]]; T[n_, k_] := f[n]/(f[k]*f[n-k]); Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Feb 19 2015, after Alois P. Heinz *)

P(k, n) = Prod[i=k-n+1..k, Pell(i)] / Prod[i=1..n, Pell(i)], with Pell(n) = A000129(n).
From Peter Bala, Mar 15 2013: (Start)
In terms of the Pell numbers, Pell(n) = A000129(n), the triangle entry T(n,k) = [n]!/([k]!*[n-k]!), where [n]! := Pell(1)*...*Pell(n) for n >= 1, with the convention [0]! = 1.
Define E(x) = 1 + sum {n>=0} x^n/[n]!. Then a generating function for this triangle is E(z)*E(x*z) = 1 + (1 + x)*z + (1 + 2*x + x^2)*z^2/[2]! + (1 + 5*x + 5*x^2 + x^3)*z^3/[3]! + ... (End)
G.f. of column k: x^k * exp( Sum_{j>=1} Pell((k+1)*j)/Pell(j) * x^j/j ). - Seiichi Manyama, May 07 2025

A001656 Fibonomial coefficients.

Original entry on oeis.org

1, 5, 40, 260, 1820, 12376, 85085, 582505, 3994320, 27372840, 187628376, 1285992240, 8814405145, 60414613805, 414088493560, 2838203264876, 19453338487220, 133335155341960, 913892777190965, 6263914210945105

Offset: 0

N. J. A. Sloane

			G.f. = 1 + 5*x + 40*x^2 + 260*x^3 + 1820*x^4 + 12376*x^5 + 85085*x^6 + ... .

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Alfred Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 17.
Nadia Heninger, E. M. Rains, and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.
Thomas Koshy, Infinite Sums Involving Jacobsthal Polynomial Products Revisited, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 3-14.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Cf. A001622, A001654, A001655, A001657, A001658, A084175, A099930.

with (combinat): a:=n->1/6*fibonacci(n)*fibonacci(n+1)*fibonacci(n+2)*fibonacci(n+3): seq(a(n), n=1..18); # Zerinvary Lajos, Oct 07 2007
A001656:=-1/(z-1)/(z**2-7*z+1)/(z**2+3*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation

Table[(Fibonacci[n+3]*Fibonacci[n+2]*Fibonacci[n+1]*Fibonacci[n])/6,{n,0,50}] (* Vladimir Joseph Stephan Orlovsky, Nov 23 2009 *)
LinearRecurrence[{5,15,-15,-5,1},{1,5,40,260,1820},20] (* Vincenzo Librandi, Aug 02 2012 *)
Times@@@Partition[Fibonacci[Range[30]],4,1]/6 (* Harvey P. Dale, Oct 13 2016 *)

b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
vector(20, n, b(n-1, 4))  \\ Joerg Arndt, May 08 2016

a(n) = ((4+n, 4)) (see A010048), or fibonomial(4+n, 4).
G.f.: 1/(1-5*x-15*x^2+15*x^3+5*x^4-x^5) = 1/((1-x)*(1+3*x+x^2)*(1-7*x+x^2)) (see Comments to A055870). a(n)= 7*a(n-1)-a(n-2)+((-1)^n)*fibonomial(n+2, 2), n >= 2; a(0)=1, a(1)=5; fibonomial(n+2, 2)= A001654(n+1).
a(n) = Product_{k=1..n} Fibonacci(k+4)/Fibonacci(k). - Gary Detlefs, Feb 06 2011
a(n) = (F(n+3)^2-F(n+2)^2)*F(n+3)*F(n+2)/6, where F(n) is the n-th Fibonacci number. - Gary Detlefs, Oct 12 2011
a(n) = a(-5-n) for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(+a(n+1) - 2*a(n+2)) + a(n+1)*(-5*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
From Peter Bala, Mar 30 2015: (Start)
The o.g.f. A(x) = 1/(1 - 5*x - 15*x^2 + 15*x^3 + 5*x^4 - x^5). Hence A(x) (mod 25) = 1/(1 - 5*x + 10*x^2 - 10^x^3 + 5*x^4 - x^5) (mod 25) = 1/(1 - x)^5 (mod 25). It follows by Theorem 1 of Heninger et al. that A(x)^(1/5) = 1 + x + 6*x^2 + 26*x^3 + ... has integral coefficients.
Sum_{n >= 0} a(n)*x^n = exp( Sum_{n >= 1} Fibonacci(5*n)/Fibonacci(n)*x^n/n ). Cf. A084175, A099930. (End)
Sum_{n>=0} 1/a(n) = 51/2 - 15*phi, where phi is the golden ratio (A001622) (Koshy, 2022, section 3.3, p. 9). - Amiram Eldar, Jan 23 2025

Corrected and extended by Wolfdieter Lang, Jun 27 2000

More terms from Vladimir Joseph Stephan Orlovsky, Nov 23 2009

A383740 a(0) = 4; a(n) = Pell(4*n)/Pell(n) for n > 0.

Original entry on oeis.org

4, 12, 204, 2772, 39236, 551532, 7761996, 109216308, 1536797956, 21624369228, 304278011724, 4281516425748, 60245508232004, 847718631046572, 11928306344398284, 167844007448966772, 2361744410638758916, 33232265756370284172, 467613464999874177996, 6579820775754484587348

Offset: 0

Seiichi Manyama, May 07 2025

Index entries for linear recurrences with constant coefficients, signature (12,30,-12,-1).

Row n=4 of A383742.

Cf. A000129, A099930.

a[n_] := Fibonacci[4*n, 2]/Fibonacci[n, 2]; a[0] = 4; Array[a, 20, 0] (* Amiram Eldar, May 08 2025 *)

my(N=30, x='x+O('x^N)); Vec(4*(1-9*x-15*x^2+3*x^3)/((1+2*x-x^2)*(1-14*x-x^2)))

a(n) = 12*a(n-1) + 30*a(n-2) - 12*a(n-3) - a(n-4).

G.f.: 4 * (1-9*x-15*x^2+3*x^3)/((1+2*x-x^2) * (1-14*x-x^2)).

cp's OEIS Frontend

A084175 Jacobsthal oblong numbers.

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A099927 Pellonomial triangle P(k,n) read by rows.

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A001656 Fibonomial coefficients.

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A383740 a(0) = 4; a(n) = Pell(4*n)/Pell(n) for n > 0.

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