A096940 -id:A096940 - cp's OEIS frontend

A022096 Fibonacci sequence beginning 1, 6.

1, 6, 7, 13, 20, 33, 53, 86, 139, 225, 364, 589, 953, 1542, 2495, 4037, 6532, 10569, 17101, 27670, 44771, 72441, 117212, 189653, 306865, 496518, 803383, 1299901, 2103284, 3403185, 5506469, 8909654, 14416123, 23325777, 37741900, 61067677, 98809577, 159877254

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(6;n-1-k,k), n>=1, with a(-1)=5. These are the sums of the SW-NE diagonals in P(6;n,k), the (6,1) Pascal triangle A093563. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also sums of SW-NE diagonals in (1,5)-Pascal triangle A096940.
Subsequence of primes: 7, 13, 53, 139, 953, 44771, 189653, 1494692464747, ... - R. J. Mathar, Aug 09 2012
a(n) is the sum of seven consecutive Fibonacci numbers. a(n) = F(n-4) + F(n-3) + F(n-2) + F(n-1) + F(n) + F(n+1) + F(n+2), where F(n)=A000045(n), extended so that F(-1)=1, F(-2)=-1, F(-3)=2, and F(-4)=-3. - Graeme McRae, Apr 24 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Jia Huang, Hecke algebras of simply-laced type with independent parameters, arXiv:1902.11139 [math.RT], 2019.
Tanya Khovanova, Recursive Sequences
José L. Ramírez, Gustavo N. Rubiano, and Rodrigo de Castro, A Generalization of the Fibonacci Word Fractal and the Fibonacci Snowflake, arXiv preprint arXiv:1212.1368 [cs.DM], 2012-2014.
J. L. Ramírez and G. N. Rubiano, Properties and Generalizations of the Fibonacci Word Fractal, The Mathematica Journal, Vol. 16 (2014).
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000045, A093563, A096940, A101220, A109754.

a0:=1; a1:=6; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

CoefficientList[Series[(1+5 x)/(1-x-x^2), {x,0,40}], x] (* Vincenzo Librandi, Apr 25 2014 *)
LinearRecurrence[{1,1},{1,6},40] (* Harvey P. Dale, Aug 07 2023 *)

a(n)=([0,1; 1,1]^n*[1;6])[1,1] \\ Charles R Greathouse IV, Jan 29 2016

A022096=BinaryRecurrenceSequence(1,1,1,6)
print([A022096(n) for n in range(41)]) # G. C. Greubel, Jun 02 2025

a(n) = a(n-1) + a(n-2), n>=2, a(0)=1, a(1)=6.
G.f.: (1+5*x)/(1-x-x^2).
a(n) = A109754(5, n+1).
a(n) = 5*Fibonacci(n+2) - 4*Fibonacci(n+1). - Gary Detlefs, Dec 21 2010
a(n) = (2^(-1-n)*((1 - sqrt(5))^n*(-11 + sqrt(5)) + (1 + sqrt(5))^n*(11 + sqrt(5))))/sqrt(5). - Herbert Kociemba, Dec 18 2011
a(n) = Fibonacci(n+3) - Fibonacci(n-4). - Greg Dresden and Sam Neale, Mar 08 2022

Spelling correction by Jason G. Wurtzel, Aug 22 2010

A095666 Pascal (1,4) triangle.

Original entry on oeis.org

4, 1, 4, 1, 5, 4, 1, 6, 9, 4, 1, 7, 15, 13, 4, 1, 8, 22, 28, 17, 4, 1, 9, 30, 50, 45, 21, 4, 1, 10, 39, 80, 95, 66, 25, 4, 1, 11, 49, 119, 175, 161, 91, 29, 4, 1, 12, 60, 168, 294, 336, 252, 120, 33, 4, 1, 13, 72, 228, 462, 630, 588, 372, 153, 37, 4, 1, 14, 85, 300, 690, 1092

Offset: 0

Wolfdieter Lang, Jun 11 2004

This is the fourth member, q=4, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with a(0,0)=2, not 1), A095660 (q=3), A096940 (q=5), A096956 (q=6).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) := Sum_{m=0..n} a(n,m)*x^m is G(z,x) = g(z)/(1 - x*z*f(z)). Here: g(x) = (4-3*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (4-3*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022095(n-2), n >= 2, with n=1 value 4. [Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.]
T(2*n,n) = A029609(n) for n > 0, A029609 are the central terms of the Pascal (2,3) triangle A029600. - Reinhard Zumkeller, Apr 08 2012

			Triangle begins:
  [4];
  [1,4];
  [1,5,4];
  [1,6,9,4];
  [1,7,15,13,4];
  ...

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Wolfdieter Lang, First 10 rows.

Row sums: A020714(n-1), n >= 1, 4 if n=0.
Alternating row sums are [4, -3, followed by 0's].
Column sequences (without leading zeros) give for m=1..9, with n >= 0: A000027(n+4), A055999(n+1), A060488(n+3), A095667-71, A095819.

a095666 n k = a095666_tabl !! n !! k
a095666_row n = a095666_tabl !! n
a095666_tabl = [4] : iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,4]
-- Reinhard Zumkeller, Apr 08 2012

a(n,k):=(1+3*k/n)*binomial(n,k) # Mircea Merca, Apr 08 2012

A095666[n_, k_] := If[n == k,  4, (3*k/n + 1)*Binomial[n, k]];
Table[A095666[n, k], {n, 0, 12}, {k, 0, n}] (* Paolo Xausa, Apr 14 2025 *)

Recursion: a(n, m) = 0 if m > n, a(0, 0) = 4; a(n, 0) = 1 if n>=1; a(n, m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (4-3*x)/(1-x)^(m+1), m >= 0.
a(n,k) = (1 + 3*k/n)*binomial(n,k). - Mircea Merca, Apr 08 2012

A096956 Pascal (1,6) triangle.

Original entry on oeis.org

6, 1, 6, 1, 7, 6, 1, 8, 13, 6, 1, 9, 21, 19, 6, 1, 10, 30, 40, 25, 6, 1, 11, 40, 70, 65, 31, 6, 1, 12, 51, 110, 135, 96, 37, 6, 1, 13, 63, 161, 245, 231, 133, 43, 6, 1, 14, 76, 224, 406, 476, 364, 176, 49, 6, 1, 15, 90, 300, 630, 882, 840, 540, 225, 55, 6, 1, 16, 105, 390, 930

Offset: 0

Wolfdieter Lang, Aug 13 2004

Except for the first row this is the row reversed (6,1)-Pascal triangle A093563.
This is the sixth member, q=6, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with a(0,0)=2, not 1), A095660 (q=3), A095666 (q=4), A096940 (q=5).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=g(z)/(1-x*z*f(z)). Here: g(x)=(6-5*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(6-5*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k) = A022097(n-2), n >= 2, with n=1 value 6. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins:
  [0]  6;
  [1]  1,  6;
  [2]  1,  7,  6;
  [3]  1,  8, 13,  6;
  [4]  1,  9, 21, 19,  6;
  [5]  1, 10, 30, 40, 25,  6;
  ...

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).
Wolfdieter Lang, First 10 rows.

Row sums: A005009(n-1), n>=1, 6 if n=0; g.f.: (6-5*x)/(1-2*x). Alternating row sums are [6, -5, followed by 0's].

Column sequences (without leading zeros) give for m=1..9, with n >= 0: A000027(n+6), A056115, A096957-9, A097297-A097300.

a(n,k):=piecewise(n=0,6,0Mircea Merca, Apr 08 2012

A096956[n_, k_] := If[n == k, 6, (5*k/n + 1)*Binomial[n, k]];
Table[A096956[n, k], {n, 0, 12}, {k, 0, n}] (* Paolo Xausa, Apr 14 2025 *)

Recursion: a(n,m)=0 if m > n, a(0,0) = 6; a(n,0) = 1 if n >= 1; a(n,m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (6-5*x)/(1-x)^(m+1), m >= 0.
a(n,k) = (1+5*k/n)*binomial(n,k), for n > 0. - Mircea Merca, Apr 08 2012

A374452 Iterated rascal triangle R3: T(n,k) = Sum_{m=0..3} binomial(n-k,m)*binomial(k,m).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 8, 28, 56, 69, 56, 28, 8, 1, 1, 9, 36, 84, 121, 121, 84, 36, 9, 1, 1, 10, 45, 120, 195, 226, 195, 120, 45, 10, 1

Offset: 0

Kolosov Petro, Jul 08 2024

Triangle T(n,k) is the third triangle R3 among the rascal-family triangles; A077028 is triangle R1, A374378 is triangle R2.
Triangle T(n,k) equals Pascal's triangle A007318 through row 2i+1, i=2 (i.e., row 7).
Triangle T(n,k) equals Pascal's triangle A007318 through column i, i=2 (i.e., column 3).

			Triangle begins:
--------------------------------------------------
k=     0   1   2   3    4    5    6   7   8   9 10
--------------------------------------------------
n=0:   1
n=1:   1   1
n=2:   1   2   1
n=3:   1   3   3   1
n=4:   1   4   6   4    1
n=5:   1   5  10  10    5    1
n=6:   1   6  15  20   15    6    1
n=7:   1   7  21  35   35   21    7   1
n=8:   1   8  28  56   69   56   28   8   1
n=9:   1   9  36  84  121  121   84  36   9   1
n=10:  1  10  45 120  195  226  195  120  45  10  1

Jena Gregory, Brandt Kronholm and Jacob White, Iterated rascal triangles, Aequationes mathematicae, 2023.
Jena Gregory, Iterated rascal triangles, Theses and Dissertations. 1050., The University of Texas Rio Grande Valley, 2022.
Amelia Gibbs and Brian K. Miceli, Two Combinatorial Interpretations of Rascal Numbers, arXiv:2405.11045 [math.CO], 2024.
Philip K. Hotchkiss, Student Inquiry and the Rascal Triangle, arXiv:1907.07749 [math.HO], 2019.
Philip K. Hotchkiss, Generalized Rascal Triangles, Journal of Integer Sequences, Vol. 23, 2020.
Petro Kolosov, Identities in Iterated Rascal Triangles, 2024.

Cf. A077028, A000027, A000217, A000292, A005894, A247608, A008860, A374378, A007318, A096943, A096940

t[n_, k_] := Sum[Binomial[n - k, m]*Binomial[k, m], {m, 0, 3}]; Column[Table[t[n, k], {n, 0, 12}, {k, 0, n}], Left]

T(n,k) = 1 + k*(n-k) + 1/4*(k-1)*k*(n-k-1)*(n-k) + 1/36*(k-2)*(k-1)*k*(n-k-2)*(n-k-1)*(n-k).
Row sums give A008860(n).
Diagonal T(n+1, n) gives A000027(n).
Diagonal T(n+2, n) gives A000217(n).
Diagonal T(n+3, n) gives A000292(n).
Diagonal T(n+4, n) gives A005894(n).
Diagonal T(n+6, n) gives A247608(n).
Column k=4 difference binomial(n+8, 4) - T(n+8, 4) gives C(n+4,4)=A007318(n+4,4).
Column k=5 difference binomial(n+9, 5) - T(n+9, 5) gives sixth column of (1,5)-Pascal triangle A096943.
G.f.: (1 + 4*x^6*y^3 - 3*x*(1 + y) - 6*x^5*y^2*(1 + y) + 2*x^4*y*(2 + 7*y+ 2*y^2) + x^2*(3 + 10*y + 3*y^2) - x^3*(1 + 11*y + 11*y^2 + y^3))/((1 - x)^4*(1 - x*y)^4). - Stefano Spezia, Jul 09 2024

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A022096 Fibonacci sequence beginning 1, 6.

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A095666 Pascal (1,4) triangle.

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A096956 Pascal (1,6) triangle.

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A374452 Iterated rascal triangle R3: T(n,k) = Sum_{m=0..3} binomial(n-k,m)*binomial(k,m).

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