A109754 -id:A109754 - cp's OEIS frontend

A118654 Square array T(n,k) read by antidiagonals: T(n,k) = 2^n*Fibonacci(k) - Fibonacci(k-2).

1, 1, 0, 1, 1, 1, 1, 3, 2, 1, 1, 7, 4, 3, 2, 1, 15, 8, 7, 5, 3, 1, 31, 16, 15, 11, 8, 5, 1, 63, 32, 31, 23, 18, 13, 8, 1, 127, 64, 63, 47, 38, 29, 21, 13, 1, 255, 128, 127, 95, 78, 61, 47, 34, 21, 1, 511, 256, 255, 191, 158, 125, 99, 76, 55, 34

Offset: 0

Ross La Haye, May 17 2006

Inverse binomial transform (by columns) of A090888.

			T(2,3) = 7 because 2^2(Fibonacci(3)) - Fibonacci(3-2) = 4*2 - 1 = 7.
{1};
{1,  0};
{1,  1,  1};
{1,  3,  2,  1};
{1,  7,  4,  3,  2};
{1, 15,  8,  7,  5,  3};
{1, 31, 16, 15, 11,  8,  5};
{1, 63, 32, 31, 23, 18, 13,  8};

Rows: T(0,k) = A000045(k-1), for k > 0; T(1,k) = A000045(k+1); T(2,k) = A000032(k+1); T(3,k) = A022097(k); T(4,k) = A022105(k); T(5,k) = A022401(k).

Columns: T(n,1) = A000225(n); T(n,2) = A000079(n); T(n,3) = A000225(n+1); T(n,4) = A055010(n+1); T(n,5) = A051633(n); a(T,6) = A036563(n+3).

T(n,k) = 2^n*Fibonacci(k) - Fibonacci(k-2).
T(n,k) = (2^n-2)*Fibonacci(k) + Fibonacci(k+1).
T(n,0) = 1; T(n,1) = 2^n - 1; T(n,k) = T(n,k-1) + T(n,k-2), for k > 1.
T(0,k) = Fibonacci(k-1); T(1,k) = Fibonacci(k+1); T(n,k) = 3T(n-1,k) - 2T(n-2,k), for n > 1.
T(n,k) = 2T(n-1,k) + Fibonacci(k-2), for n > 0.
T(n,k) = A109754(2^n-2, k+1) = A101220(2^n-2, 0, k+1), for n > 0.
O.g.f. (by rows) = (1+(-2+2^n)x)/(1-x-x^2).
Sum_{k=0..n} T(n-k,k) = A119587(n+1). - Ross La Haye, May 31 2006

A007502 Les Marvin sequence: a(n) = F(n) + (n-1)*F(n-1), F() = Fibonacci numbers.

Original entry on oeis.org

1, 2, 4, 9, 17, 33, 61, 112, 202, 361, 639, 1123, 1961, 3406, 5888, 10137, 17389, 29733, 50693, 86204, 146246, 247577, 418299, 705479, 1187857, 1997018, 3352636, 5621097, 9412937, 15744681, 26307469, 43912648

Offset: 1

N. J. A. Sloane, Robert G. Wilson v

Denominators of convergents of the continued fraction with the n partial quotients: [1;1,1,...(n-1 1's)...,1,n], starting with [1], [1;2], [1;1,3], [1;1,1,4], ... Numerators are A088209(n-1). - Paul D. Hanna, Sep 23 2003

			a(7) = F(7) + 6*F(6) = 13 + 6*8 = 61.

Les Marvin, Problem, J. Rec. Math., Vol. 10 (No. 3, 1976-1977), p. 213.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..500
Ignas Gasparavičius, Andrius Grigutis, and Juozas Petkelis, Picturesque convolution-like recurrences and partial sums' generation, arXiv:2507.23619 [math.NT], 2025. See p. 27.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Cf. A000045, A045925, A088209 (numerators), A101220, A109754.

a007502 n = a007502_list !! (n-1)
a007502_list = zipWith (+) a045925_list $ tail a000045_list
-- Reinhard Zumkeller, Oct 01 2012, Mar 04 2012

# The function 'fibrec' is defined in A354044.
function A007502(n)
    n == 0 && return BigInt(1)
    a, b = fibrec(n-1)
    (n-1)*a + b
end
println([A007502(n) for n in 1:32]) # Peter Luschny, May 18 2022

A007502:= func< n | Fibonacci(n) +(n-1)*Fibonacci(n-1) >;
[A007502(n): n in [1..40]]; // G. C. Greubel, Aug 26 2025

Table[Fibonacci[n]+(n-1)*Fibonacci[n-1], {n,40}] (* or *) LinearRecurrence[ {2,1,-2,-1}, {1,2,4,9}, 40](* Harvey P. Dale, Jul 13 2011 *)
f[n_] := Denominator@  FromContinuedFraction@ Join[ Table[1, {n}], {n + 1}]; Array[f, 30, 0] (* Robert G. Wilson v, Mar 04 2012 *)

Vec((1-x^2+x^3)/(1-x-x^2)^2+O(x^99)) \\ Charles R Greathouse IV, Mar 04 2012

def A007502(n): return fibonacci(n) +(n-1)*fibonacci(n-1)
print([A007502(n) for n in range(1,41)]) # G. C. Greubel, Aug 26 2025

G.f.: (1-x^2+x^3)/(1-x-x^2)^2. - Paul D. Hanna, Sep 23 2003
a(n+1) = A109754(n, n+1) = A101220(n, 0, n+1). - N. J. A. Sloane, May 19 2006
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) for n>3, a(0)=1, a(1)=2, a(2)=4, a(3)=9. - Harvey P. Dale, Jul 13 2011
E.g.f.: exp(x/2)*( ((3 + 2*x)/sqrt(5))*sinh(sqrt(5)*x/2) - cosh(sqrt(5)*x/2) ) + 1. - G. C. Greubel, Aug 26 2025

A022099 Fibonacci sequence beginning 1, 9.

Original entry on oeis.org

1, 9, 10, 19, 29, 48, 77, 125, 202, 327, 529, 856, 1385, 2241, 3626, 5867, 9493, 15360, 24853, 40213, 65066, 105279, 170345, 275624, 445969, 721593, 1167562, 1889155, 3056717, 4945872, 8002589, 12948461, 20951050, 33899511, 54850561, 88750072, 143600633, 232350705

Offset: 0

N. J. A. Sloane, Jun 14 1998

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(9;n-1-k,k) with n>=1, a(-1)=8. These are the SW-NE diagonals in P(9;n,k), the (9,1) Pascal triangle A093644. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have: b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
Pisano period lengths:  1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (perhaps the same as A001175). - R. J. Mathar, Aug 10 2012
No, it is not the same as in A001175. The Pisano periods are different for moduli 71 and 142, where they are 35 and 105 instead of 70 and 210. Otherwise they coincide with those of the Fibonacci sequence. - Klaus Purath, Jun 26 2022

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000045, A001175, A093644, A101220, A109754.

a0:=1; a1:=9; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1, 1}, {1, 9}, 36] (* Robert G. Wilson v, Apr 11 2014 *)

a(n) = a(n-1) + a(n-2), n>=2, a(0)=1, a(1)=9. a(-1):=8.
G.f.: (1+8*x)/(1-x-x^2).
a(n) = A109754(8, n+1) = A101220(8, 0, n+1).
a(n+1) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5))+ 4*((1 + sqrt(5))^(n-1) - (1 - sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) =  8*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012
a(n) = 10*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
a(n) = Lucas(n+3) + Fibonacci(n-4). - Greg Dresden and Mary Beth Pittman, Mar 12 2022

A022100 Fibonacci sequence beginning 1, 10.

Original entry on oeis.org

1, 10, 11, 21, 32, 53, 85, 138, 223, 361, 584, 945, 1529, 2474, 4003, 6477, 10480, 16957, 27437, 44394, 71831, 116225, 188056, 304281, 492337, 796618, 1288955, 2085573, 3374528, 5460101, 8834629, 14294730, 23129359, 37424089, 60553448, 97977537, 158530985

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(10; n-1-k, k), n >= 1, with a(-1)=9. These are the SW-NE diagonals in P(10; n, k), the (10,1) Pascal triangle A093645. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have:
b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1, 1).

Cf. A000045.

a(n) = A109754(9, n+1) = A101220(9, 0, n+1).

a0:=1; a1:=10; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1,1},{1,10},40] (* Harvey P. Dale, May 17 2017 *)

a(n) = a(n-1) + a(n-2) for n >= 2, a(0)=1, a(1)=10, a(-1):=9.
G.f.: (1 + 9*x)/(1 - x - x^2).
a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*(9*binomial(1, k) - 8*binomial(0, k)). - Paul Barry, May 05 2005
a(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)) + (9/2)*((1+sqrt(5))^(n-1) - (1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
From Bruno Berselli, Feb 20 2017: (Start)
a(n) =  9*A000045(n) + A000045(n+1).
a(n) = 11*A000045(n) - A000045(n-2). (End)

A088209 Numerators of convergents of the continued fraction with the n+1 partial quotients: [1;1,1,...(n 1's)...,1,n+1], starting with [1], [1;2], [1;1,3], [1;1,1,4], ...

Original entry on oeis.org

1, 3, 7, 14, 28, 53, 99, 181, 327, 584, 1034, 1817, 3173, 5511, 9527, 16402, 28136, 48109, 82023, 139481, 236631, 400588, 676822, 1141489, 1921993, 3231243, 5424679, 9095126, 15230452, 25475429, 42566379, 71052157, 118489383

Offset: 0

Paul D. Hanna, Sep 23 2003

Denominators form the Les Marvin sequence: A007502(n+1).

			a(3)/A007502(4) = [1;1,1,4] = 14/9.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

a(n) = A109754(n, n+2) = A101220(n, 0, n+2).

Cf. A007502 (the denominators), A000045, A045925.

a088209 n = a088209_list !! n
a088209_list = zipWith (+) a000045_list $ tail a045925_list
-- Reinhard Zumkeller, Oct 01 2012, Mar 04 2012

# The function 'fibrec' is defined in A354044.
function A088209(n)
    n == 0 && return BigInt(1)
    a, b = fibrec(n)
    a + (n + 1)*b
end
println([A088209(n) for n in 0:32]) # Peter Luschny, May 18 2022

f[n_] := Numerator@  FromContinuedFraction@ Join[ Table[1, {n}], {n + 1}]; Array[f, 30, 0] (* Robert G. Wilson v, Mar 04 2012 *)
CoefficientList[Series[(1+x-x^3)/(-1+x+x^2)^2,{x,0,40}],x] (* or *) LinearRecurrence[{2,1,-2,-1},{1,3,7,14},40] (* Harvey P. Dale, Jul 13 2021 *)

G.f.: (1+x-x^3)/(1-x-x^2)^2. [Corrected by Georg Fischer, Aug 16 2021]
a(n) = Fibonacci(n) + (n+1)*Fibonacci(n+1). - Paul Barry, Apr 20 2004
a(n) = a(n-1) + a(n-2) + Lucas(n). - Yuchun Ji, Apr 23 2023

A022103 Fibonacci sequence beginning 1, 13.

Original entry on oeis.org

1, 13, 14, 27, 41, 68, 109, 177, 286, 463, 749, 1212, 1961, 3173, 5134, 8307, 13441, 21748, 35189, 56937, 92126, 149063, 241189, 390252, 631441, 1021693, 1653134, 2674827, 4327961, 7002788, 11330749, 18333537, 29664286, 47997823, 77662109, 125659932, 203322041, 328981973

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(13;n-1-k,k) for n>=1, a(-1)=12. These are the SW-NE diagonals in P(13;n,k), the (13,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have:
b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
Pisano period lengths: 1, 3, 8, 6, 4, 24, 16, 12, 24, 12, 10, 24, 28, 48, 8, 24, 36, 24, 18, 12, ... (is this A106291?). - R. J. Mathar, Aug 10 2012

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1).

a(n) = A109754(12, n+1) = A101220(12, 0, n+1).

Cf. A000032, A000045.

a0:=1; a1:=13; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1, 1}, {1, 13}, 40] (* or *) Table[LucasL[n + 5] - 5 LucasL[n], {n, 0, 40}] (* Bruno Berselli, Dec 30 2016 *)

a(n) = a(n-1) + a(n-2) for n>=2, a(0)=1, a(1)=13, and a(-1):=12.
G.f.: (1 + 12*x)/(1 - x - x^2).
a(n) = ((1 + sqrt(5))^n-(1 - sqrt(5))^n)/(2^n*sqrt(5))+ 6*((1 + sqrt(5))^(n-1)-(1 - sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)) for n>0. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) = 12*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012
a(n) = 14*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
a(n) = Lucas(n+5) - 5*Lucas(n). - Bruno Berselli, Dec 30 2016

A094588 a(n) = n*F(n-1) + F(n), where F = A000045.

Original entry on oeis.org

0, 1, 3, 5, 11, 20, 38, 69, 125, 223, 395, 694, 1212, 2105, 3639, 6265, 10747, 18376, 31330, 53277, 90385, 153011, 258523, 436010, 734136, 1234225, 2072043, 3474029, 5817515, 9730748, 16258910, 27139509, 45258917, 75408775, 125538539

Offset: 0

Paul Barry, May 13 2004

This is the transform of the Fibonacci numbers under the inverse of the signed permutations matrix (see A094587).

Vincenzo Librandi, Table of n, a(n) for n = 0..250
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Cf. A000045, A007502, A045925, A088209, A354044.

a094588 n = a094588_list !! n
a094588_list = 0 : zipWith (+) (tail a000045_list)
                               (zipWith (*) [1..] a000045_list)
-- Reinhard Zumkeller, Mar 04 2012

# The function 'fibrec' is defined in A354044.
function A094588(n)
    n == 0 && return BigInt(0)
    a, b = fibrec(n - 1)
    a*n + b
end
println([A094588(n) for n in 0:34]) # Peter Luschny, May 16 2022

[n*Fibonacci(n-1)+Fibonacci(n): n in [0..60]]; // Vincenzo Librandi, Apr 23 2011

CoefficientList[Series[x (1+x-2x^2)/(1-x-x^2)^2,{x,0,40}],x]  (* Harvey P. Dale, Apr 16 2011 *)

Vec((1+x-2*x^2)/(1-x-x^2)^2+O(x^99)) \\ Charles R Greathouse IV, Mar 04 2012

G.f. : x*(1 + x - 2*x^2)/(1 - x - x^2)^2.
a(n) = A101220(n, 0, n) - Ross La Haye, Jan 28 2005
a(n) = A109754(n, n). - Ross La Haye, Aug 20 2005
a(n) = (sin(c*n)*i - n*sin(c*(n - 1)))/(i^n*sqrt(5/4)) where c = arccos(i/2). - Peter Luschny, May 16 2022

A022101 Fibonacci sequence beginning 1, 11.

Original entry on oeis.org

1, 11, 12, 23, 35, 58, 93, 151, 244, 395, 639, 1034, 1673, 2707, 4380, 7087, 11467, 18554, 30021, 48575, 78596, 127171, 205767, 332938, 538705, 871643, 1410348, 2281991, 3692339, 5974330, 9666669, 15640999, 25307668, 40948667, 66256335, 107205002, 173461337, 280666339

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(11;n-1-k,k) with n >= 1, a(-1)=10. These are the SW-NE diagonals in P(11;n,k), the (11,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have:
b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (is this A001175?). - R. J. Mathar, Aug 10 2012

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1, 1).

Cf. A000032, A000045.

a(n) = A109754(10, n+1) = A101220(10, 0, n+1).

a0:=1; a1:=11; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1,1},{1,11},40] (* Harvey P. Dale, Aug 16 2015 *)

a(n) = 10*fibonacci(n)+fibonacci(n+1) \\ Charles R Greathouse IV, Jun 11 2015

a(n) = a(n-1) + a(n-2), n >= 2, a(0)=1, a(1)=11. a(-1)=10.
G.f.: (1+10*x)/(1-x-x^2).
a(n-1) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)) + 5*((1+sqrt(5))^(n-1) - (1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) = 10*A000045(n) + A000045(n+1). - R. J. Mathar, Apr 07 2011
a(n) = 12*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
a(n) = A000045(n+4) + A000032(n-4) for n > 0. - Bruno Berselli, Sep 27 2017

A117501 Triangle generated from an array of generalized Fibonacci-like terms.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 3, 3, 1, 4, 4, 5, 5, 1, 5, 5, 7, 8, 8, 1, 6, 6, 9, 11, 13, 13, 1, 7, 7, 11, 14, 18, 21, 21, 1, 8, 8, 13, 17, 23, 29, 34, 34, 1, 9, 9, 15, 20, 28, 37, 47, 55, 55, 1, 10, 10, 17, 23, 33, 45, 60, 76, 89, 89, 1, 11, 11, 19, 26, 38, 53, 73, 97, 123, 144, 144

Offset: 1

Gary W. Adamson, Mar 23 2006

Difference terms of the array columns in triangle format becomes A117502.
Row sums of the triangle are A104161: (1, 2, 5, 10, 19, 34, 59, ...), generated by a(k) = a(k-1) + a(k-2) + n.
This is the lower triangular version of A109754 (without a row and column 0). - Ross La Haye, Apr 12 2006

			First few rows of the array T(n,k) are:
       k=1 k=2 k=3 k=4 k=5 k=6
  n=1:  1,  1,  2,  3,  5,  8, ...
  n=2:  1,  2,  3,  5,  8, 13, ...
  n=3:  1,  3,  4,  7, 11, 18, ...
  n=4:  1,  4,  5,  9, 14, 23, ...
  n=5:  1,  5,  6, 11, 17, 28, ...
First few rows of the triangle are:
  1;
  1, 1;
  1, 2, 2;
  1, 3, 3,  3;
  1, 4, 4,  5,  5;
  1, 5, 5,  7,  8,  8;
  1, 6, 6,  9, 11, 13, 13;
  1, 7, 7, 11, 14, 18, 21, 21; ...

G. C. Greubel, Rows n = 1..100 of triangle, flattened

Cf. A000045, A001595, A104161 (diagonal sums), A109754 (with column of 0's), A117502.

F:=Fibonacci;; Flat(List([1..15], n-> List([1..n], k-> (n-k+1)*F(k-1) + F(k-2) ))); # G. C. Greubel, Jul 13 2019

F:=Fibonacci; [(n-k+1)*F(k-1) + F(k-2): k in [1..n], n in [1..15]]; // G. C. Greubel, Jul 13 2019

a[n_, k_] := a[n, k] = If[k==1, 1, If[k==2, n, a[n, k-1] + a[n, k-2]]]; Table[a[n-k+1, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Aug 15 2017 *)
T[n_, k_]:= n*Fibonacci[k-1] + Fibonacci[k-2]; Table[T[n-k+1, k], {n, 15}, {k, n}]//Flatten (* G. C. Greubel, Jul 13 2019 *)

T(n,k) = n*fibonacci(k-1) + fibonacci(k-2);
for(n=1,15, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jul 13 2019

from sympy.core.cache import cacheit
@cacheit
def a(n, k):
    return 1 if k==1 else n if k==2 else a(n, k - 1) + a(n, k - 2)
for n in range(1, 21): print([a(n - k + 1, k) for k in range(1, n + 1)]) # Indranil Ghosh, Aug 19 2017

f=fibonacci; [[(n-k+1)*f(k-1) + f(k-2) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Jul 13 2019

The triangle by rows = antidiagonals of an array in which n-th row is generated by a Fibonacci-like operation: (1, n...then a(k+1) = a(k) + a(k-1)).

T(n,k) = n*Fibonacci(k-1) + Fibonacci(k-2). - G. C. Greubel, Jul 13 2019

Row sums comment corrected by Philippe Deléham, Nov 18 2013

A022102 Fibonacci sequence beginning 1, 12.

Original entry on oeis.org

1, 12, 13, 25, 38, 63, 101, 164, 265, 429, 694, 1123, 1817, 2940, 4757, 7697, 12454, 20151, 32605, 52756, 85361, 138117, 223478, 361595, 585073, 946668, 1531741, 2478409, 4010150, 6488559, 10498709

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(12;n-1-k,k) with n>=1, a(-1)=11. These are the SW-NE diagonals in P(12;n,k), the (12,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have:
b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (is this A001175?). - R. J. Mathar, Aug 10 2012

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1, 1).

Cf. A000045, A001175, A093645, A101220, A109754.

a0:=1; a1:=12; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1,1},{1,12},40] (* Harvey P. Dale, Jan 23 2012 *)

a(n) = if(n==0, 1, if(n==1, 12, a(n-1)+a(n-2))) \\ Felix Fröhlich, Jun 09 2022

Vec((1+11*x)/(1-x-x^2) + O(x^20)) \\ Felix Fröhlich, Jun 09 2022

a(n) = a(n-1) + a(n-2), n >= 2, a(0)=1, a(1)=12. a(-1):=11.
G.f.: (1+11*x)/(1-x-x^2).
a(n) = A109754(11, n+1) = A101220(11, 0, n+1).
a(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)) + (11/2)*((1+sqrt(5))^(n-1)-(1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). Offset 1. a(3)=13. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) = 11*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012
a(n) = 13*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
a(n) = F(n+5) + F(n-1) - F(n-5) for F(n) the Fibonacci number A000045(n). - Greg Dresden and Aamen Muharram, Jun 09 2022

cp's OEIS Frontend

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