cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A028387 a(n) = n + (n+1)^2.

Original entry on oeis.org

1, 5, 11, 19, 29, 41, 55, 71, 89, 109, 131, 155, 181, 209, 239, 271, 305, 341, 379, 419, 461, 505, 551, 599, 649, 701, 755, 811, 869, 929, 991, 1055, 1121, 1189, 1259, 1331, 1405, 1481, 1559, 1639, 1721, 1805, 1891, 1979, 2069, 2161, 2255, 2351, 2449, 2549, 2651
Offset: 0

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Author

Keywords

Comments

a(n+1) is the least k > a(n) + 1 such that A000217(a(n)) + A000217(k) is a square. - David Wasserman, Jun 30 2005
Values of Fibonacci polynomial n^2 - n - 1 for n = 2, 3, 4, 5, ... - Artur Jasinski, Nov 19 2006
A127701 * [1, 2, 3, ...]. - Gary W. Adamson, Jan 24 2007
Row sums of triangle A135223. - Gary W. Adamson, Nov 23 2007
Equals row sums of triangle A143596. - Gary W. Adamson, Aug 26 2008
a(n-1) gives the number of n X k rectangles on an n X n chessboard (for k = 1, 2, 3, ..., n). - Aaron Dunigan AtLee, Feb 13 2009
sqrt(a(0) + sqrt(a(1) + sqrt(a(2) + sqrt(a(3) + ...)))) = sqrt(1 + sqrt(5 + sqrt(11 + sqrt(19 + ...)))) = 2. - Miklos Kristof, Dec 24 2009
When n + 1 is prime, a(n) gives the number of irreducible representations of any nonabelian group of order (n+1)^3. - Andrew Rupinski, Mar 17 2010
a(n) = A176271(n+1, n+1). - Reinhard Zumkeller, Apr 13 2010
The product of any 4 consecutive integers plus 1 is a square (see A062938); the terms of this sequence are the square roots. - Harvey P. Dale, Oct 19 2011
Or numbers not expressed in the form m + floor(sqrt(m)) with integer m. - Vladimir Shevelev, Apr 09 2012
Left edge of the triangle in A214604: a(n) = A214604(n+1,1). - Reinhard Zumkeller, Jul 25 2012
Another expression involving phi = (1 + sqrt(5))/2 is a(n) = (n + phi)(n + 1 - phi). Therefore the numbers in this sequence, even if they are prime in Z, are not prime in Z[phi]. - Alonso del Arte, Aug 03 2013
a(n-1) = n*(n+1) - 1, n>=0, with a(-1) = -1, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 5 for b = 2*n+1. In general D = b^2 - 4ac > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
a(n) has prime factors given by A038872. - Richard R. Forberg, Dec 10 2014
A253607(a(n)) = -1. - Reinhard Zumkeller, Jan 05 2015
An example of a quadratic sequence for which the continued square root map (see A257574) produces the number 2. There are infinitely many sequences with this property - another example is A028387. See Popular Computing link. - N. J. A. Sloane, May 03 2015
Left edge of the triangle in A260910: a(n) = A260910(n+2,1). - Reinhard Zumkeller, Aug 04 2015
Numbers m such that 4m+5 is a square. - Bruce J. Nicholson, Jul 19 2017
The numbers represented as 131 in base n: 131_4 = 29, 131_5 = 41, ... . If 'digits' larger than the base are allowed then 131_2 = 11 and 131_1 = 5 also. - Ron Knott, Nov 14 2017
From Klaus Purath, Mar 18 2019: (Start)
Let m be a(n) or a prime factor of a(n). Then, except for 1 and 5, there are, if m is a prime, exactly two squares y^2 such that the difference y^2 - m contains exactly one pair of factors {x,z} such that the following applies: x*z = y^2 - m, x + y = z with
x < y, where {x,y,z} are relatively prime numbers. {x,y,z} are the initial values of a sequence of the Fibonacci type. Thus each a(n) > 5, if it is a prime, and each prime factor p > 5 of an a(n) can be assigned to exactly two sequences of the Fibonacci type. a(0) = 1 belongs to the original Fibonacci sequence and a(1) = 5 to the Lucas sequence.
But also the reverse assignment applies. From any sequence (f(i)) of the Fibonacci type we get from its 3 initial values by f(i)^2 - f(i-1)*f(i+1) with f(i-1) < f(i) a term a(n) or a prime factor p of a(n). This relation is also valid for any i. In this case we get the absolute value |a(n)| or |p|. (End)
a(n-1) = 2*T(n) - 1, for n>=1, with T = A000217, is a proper subsequence of A089270, and the terms are 0,-1,+1 (mod 5). - Wolfdieter Lang, Jul 05 2019
a(n+1) is the number of wedged n-dimensional spheres in the homotopy of the neighborhood complex of Kneser graph KG_{2,n}. Here, KG_{2,n} is a graph whose vertex set is the collection of subsets of cardinality 2 of set {1,2,...,n+3,n+4} and two vertices are adjacent if and only if they are disjoint. - Anurag Singh, Mar 22 2021
Also the number of squares between (n+2)^2 and (n+2)^4. - Karl-Heinz Hofmann, Dec 07 2021
(x, y, z) = (A001105(n+1), -a(n-1), -a(n)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 8. - XU Pingya, Apr 25 2022
The least significant digit of terms of this sequence cycles through 1, 5, 1, 9, 9. - Torlach Rush, Jun 05 2024

Examples

			From _Ilya Gutkovskiy_, Apr 13 2016: (Start)
Illustration of initial terms:
                                        o               o
                        o           o   o o           o o
            o       o   o o       o o   o o o       o o o
    o   o   o o   o o   o o o   o o o   o o o o   o o o o
o   o o o   o o o o o   o o o o o o o   o o o o o o o o o
n=0  n=1       n=2           n=3               n=4
(End)
From _Klaus Purath_, Mar 18 2019: (Start)
Examples:
a(0) = 1: 1^1-0*1 = 1, 0+1 = 1 (Fibonacci A000045).
a(1) = 5: 3^2-1*4 = 5, 1+3 = 4 (Lucas A000032).
a(2) = 11: 4^2-1*5 = 11, 1+4 = 5 (A000285); 5^2-2*7 = 11, 2+5 = 7 (A001060).
a(3) = 19: 5^2-1*6 = 19, 1+5 = 6 (A022095); 7^2-3*10 = 19, 3+7 = 10 (A022120).
a(4) = 29: 6^2-1*7 = 29, 1+6 = 7 (A022096); 9^2-4*13 = 29, 4+9 = 13 (A022130).
a(11)/5 = 31: 7^2-2*9 = 31, 2+7 = 9 (A022113); 8^2-3*11 = 31, 3+8 = 11 (A022121).
a(24)/11 = 59: 9^2-2*11 = 59, 2+9 = 11 (A022114); 12^2-5*17 = 59, 5+12 = 17 (A022137).
(End)
		

Crossrefs

Complement of A028392. Third column of array A094954.
Cf. A000217, A002522, A062392, A062786, A127701, A135223, A143596, A052905, A162997, A062938 (squares of this sequence).
A110331 and A165900 are signed versions.
Cf. A002327 (primes), A094210.
Frobenius number for k successive numbers: this sequence (k=2), A079326 (k=3), A138984 (k=4), A138985 (k=5), A138986 (k=6), A138987 (k=7), A138988 (k=8).

Programs

Formula

a(n) = sqrt(A062938(n)). - Floor van Lamoen, Oct 08 2001
a(0) = 1, a(1) = 5, a(n) = (n+1)*a(n-1) - (n+2)*a(n-2) for n > 1. - Gerald McGarvey, Sep 24 2004
a(n) = A105728(n+2, n+1). - Reinhard Zumkeller, Apr 18 2005
a(n) = A109128(n+2, 2). - Reinhard Zumkeller, Jun 20 2005
a(n) = 2*T(n+1) - 1, where T(n) = A000217(n). - Gary W. Adamson, Aug 15 2007
a(n) = A005408(n) + A002378(n); A084990(n+1) = Sum_{k=0..n} a(k). - Reinhard Zumkeller, Aug 20 2007
Binomial transform of [1, 4, 2, 0, 0, 0, ...] = (1, 5, 11, 19, ...). - Gary W. Adamson, Sep 20 2007
G.f.: (1+2*x-x^2)/(1-x)^3. a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - R. J. Mathar, Jul 11 2009
a(n) = (n + 2 + 1/phi) * (n + 2 - phi); where phi = 1.618033989... Example: a(3) = 19 = (5 + .6180339...) * (3.381966...). Cf. next to leftmost column in A162997 array. - Gary W. Adamson, Jul 23 2009
a(n) = a(n-1) + 2*(n+1), with n > 0, a(0) = 1. - Vincenzo Librandi, Nov 18 2010
For k < n, a(n) = (k+1)*a(n-k) - k*a(n-k-1) + k*(k+1); e.g., a(5) = 41 = 4*11 - 3*5 + 3*4. - Charlie Marion, Jan 13 2011
a(n) = lower right term in M^2, M = the 2 X 2 matrix [1, n; 1, (n+1)]. - Gary W. Adamson, Jun 29 2011
G.f.: (x^2-2*x-1)/(x-1)^3 = G(0) where G(k) = 1 + x*(k+1)*(k+4)/(1 - 1/(1 + (k+1)*(k+4)/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 16 2012
Sum_{n>0} 1/a(n) = 1 + Pi*tan(sqrt(5)*Pi/2)/sqrt(5). - Enrique Pérez Herrero, Oct 11 2013
E.g.f.: exp(x) (1+4*x+x^2). - Tom Copeland, Dec 02 2013
a(n) = A005408(A000217(n)). - Tony Foster III, May 31 2016
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2).
Product_{n>=1} (1 - 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2)/6. (End)
a(5*n+1)/5 = A062786(n+1). - Torlach Rush, Jun 05 2024

Extensions

Minor edits by N. J. A. Sloane, Jul 04 2010, following suggestions from the Sequence Fans Mailing List

A035513 Wythoff array read by falling antidiagonals.

Original entry on oeis.org

1, 2, 4, 3, 7, 6, 5, 11, 10, 9, 8, 18, 16, 15, 12, 13, 29, 26, 24, 20, 14, 21, 47, 42, 39, 32, 23, 17, 34, 76, 68, 63, 52, 37, 28, 19, 55, 123, 110, 102, 84, 60, 45, 31, 22, 89, 199, 178, 165, 136, 97, 73, 50, 36, 25, 144, 322, 288, 267, 220, 157, 118, 81, 58, 41, 27, 233, 521
Offset: 1

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Author

Keywords

Comments

T(0,0)=1, T(0,1)=2,...; y^2-x^2-xy
Inverse of sequence A064274 considered as a permutation of the nonnegative integers. - Howard A. Landman, Sep 25 2001
The Wythoff array W consists of all the Wythoff pairs (x(n),y(n)), where x=A000201 and y=A001950, so that W contains every positive integer exactly once. The differences T(i,2j+1)-T(i,2j) form the Wythoff difference array, A080164, which also contains every positive integer exactly once. - Clark Kimberling, Feb 08 2003
For n>2 the determinant of any n X n contiguous subarray of A035513 (as a square array) is 0. - Gerald McGarvey, Sep 18 2004
From Clark Kimberling, Nov 14 2007: (Start)
Except for initial terms in some cases:
(Row 1) = A000045
(Row 2) = A000032
(Row 3) = A006355
(Row 4) = A022086
(Row 5) = A022087
(Row 6) = A000285
(Row 7) = A022095
(Row 8) = A013655 (sum of Fibonacci and Lucas numbers)
(Row 9) = A022112
(Column 1) = A003622 = AA Wythoff sequence
(Column 2) = A035336 = BA Wythoff sequence
(Column 3) = A035337 = ABA Wythoff sequence
(Column 4) = A035338 = BBA Wythoff sequence
(Column 5) = A035339 = ABBA Wythoff sequence
(Column 6) = A035340 = BBBA Wythoff sequence
Main diagonal = A020941. (End)
The Wythoff array is the dispersion of the sequence given by floor(n*x+x-1), where x=(golden ratio). See A191426 for a discussion of dispersions. - Clark Kimberling, Jun 03 2011
If u and v are finite sets of numbers in a row of the Wythoff array such that (product of all the numbers in u) = (product of all the numbers in v), then u = v. See A160009 (row 1 products), A274286 (row 2), A274287 (row 3), A274288 (row 4). - Clark Kimberling, Jun 17 2016
All columns of the Wythoff array are compound Wythoff sequences. This follows from the main theorem in the 1972 paper by Carlitz, Scoville and Hoggatt. For an explicit expression see Theorem 10 in Kimberling's paper from 2008 in JIS. - Michel Dekking, Aug 31 2017
The Wythoff array can be viewed as an infinite graph over the set of nonnegative integers, built as follows: start with an empty graph; for all n = 0, 1, ..., create an edge between n and the sum of the degrees of all i < n. Finally, remove vertex 0. In the resulting graph, the connected components are chains and correspond to the rows of the Wythoff array. - Luc Rousseau, Sep 28 2017
Suppose that h < k are consecutive terms in a row of the Wythoff array. If k is in an even numbered column, then h = floor(k/tau); otherwise, h = -1 + floor(k/tau). - Clark Kimberling, Mar 05 2020
From Clark Kimberling, May 26 2020: (Start)
For k > = 0, column k shows the numbers m having F(k+1) as least term in the Zeckendorf representation of m. For n >= 1, let r(n,k) be the number of terms in column k that are <= n. Then n/r(n,k) = n/(F(k+1)*tau + F(k)*(n-1)), by Bottomley's formula, so that the limiting ratio is 1/(F(k+1)*tau + F(k)). Summing over all k gives Sum_{k>=0} 1/(F(k+1)*tau + F(k)) = 1. Thus, in the limiting sense:
38.19...% of the numbers m have least term 1;
23.60...% have least term 2;
14.58...% have least term 3;
9.01...% have least term 5, etc. (End)
Named after the Dutch mathematician Willem Abraham Wythoff (1865-1939). - Amiram Eldar, Jun 11 2021
From Clark Kimberling, Jun 04 2025: (Start)
Let u(n) = (T(n,1),T(n,2)) mod 2. The positive integers (A000027) are partitioned into 4 sets (sequences):
{n : u(n) = (0,0)} = (3, 5, 9, 15, 19, 25, 29,...) = 1 + 2*A190429
{n: u(n) = (0,1)} = (2, 6, 12, 16, 18, 22, 28,...) = A191331
{n : u(n) = (1,0)} = (1, 7, 11, 13, 17, 21, 23,...) = A086843
{n: u(n) = (1,1)} = (4, 8, 10, 14, 20, 24, 26,...) = A191330.
Let v(n) = (T(n,1),T(n,2)) mod 3. The positive integers are partitioned into 9 sets (sequences):
{n : v(n) = (0,0)} = (4, 13, 19, 28, 43, 52,...) = 1 + 3*A190434
{n: v(n) = (0,1)} = (3, 12, 27, 36, 42, 51,...) = 3*A140399
{n : v(n) = (0,2)} = (5, 11, 20, 35, 44, 50,...) = 2 + 3*A190439
{n: v(n) = (1,0)} = (9, 18, 24, 33, 48, 57,...) = 3*A140400
{n: v(n) = (1,1)} = (2, 8, 17, 26, 32, 41,...) = A384601
{n : v(n) = (1,2)} = (1, 10, 16, 25, 34, 40,...) = A384602
{n: v(n) = (2,0)} = (14, 23, 29, 38, 47, 53,...) = 2 + 3*A190438
{n: v(n) = (2,1)} = (7, 22, 31, 37, 46, 61,...) = 1 + 3*A190433
{n : v(n) = (2,2)} = (6, 15, 21, 30, 39, 45,...) = 3*A140398.
Conjecture: If m >= 2, then {(T(n,1), T(n,2)) mod m} has cardinality m^2. (End)

Examples

			The Wythoff array begins:
   1    2    3    5    8   13   21   34   55   89  144 ...
   4    7   11   18   29   47   76  123  199  322  521 ...
   6   10   16   26   42   68  110  178  288  466  754 ...
   9   15   24   39   63  102  165  267  432  699 1131 ...
  12   20   32   52   84  136  220  356  576  932 1508 ...
  14   23   37   60   97  157  254  411  665 1076 1741 ...
  17   28   45   73  118  191  309  500  809 1309 2118 ...
  19   31   50   81  131  212  343  555  898 1453 2351 ...
  22   36   58   94  152  246  398  644 1042 1686 2728 ...
  25   41   66  107  173  280  453  733 1186 1919 3105 ...
  27   44   71  115  186  301  487  788 1275 2063 3338 ...
  ...
The extended Wythoff array has two extra columns, giving the row number n and A000201(n), separated from the main array by a vertical bar:
0     1  |   1    2    3    5    8   13   21   34   55   89  144   ...
1     3  |   4    7   11   18   29   47   76  123  199  322  521   ...
2     4  |   6   10   16   26   42   68  110  178  288  466  754   ...
3     6  |   9   15   24   39   63  102  165  267  432  699 1131   ...
4     8  |  12   20   32   52   84  136  220  356  576  932 1508   ...
5     9  |  14   23   37   60   97  157  254  411  665 1076 1741   ...
6    11  |  17   28   45   73  118  191  309  500  809 1309 2118   ...
7    12  |  19   31   50   81  131  212  343  555  898 1453 2351   ...
8    14  |  22   36   58   94  152  246  398  644 1042 1686 2728   ...
9    16  |  25   41   66  107  173  280  453  733 1186 1919 3105   ...
10   17  |  27   44   71  115  186  301  487  788 1275 2063 3338   ...
11   19  |  30   49   79   ...
12   21  |  33   54   87   ...
13   22  |  35   57   92   ...
14   24  |  38   62   ...
15   25  |  40   65   ...
16   27  |  43   70   ...
17   29  |  46   75   ...
18   30  |  48   78   ...
19   32  |  51   83   ...
20   33  |  53   86   ...
21   35  |  56   91   ...
22   37  |  59   96   ...
23   38  |  61   99   ...
24   40  |  64   ...
25   42  |  67   ...
26   43  |  69   ...
27   45  |  72   ...
28   46  |  74   ...
29   48  |  77   ...
30   50  |  80   ...
31   51  |  82   ...
32   53  |  85   ...
33   55  |  88   ...
34   56  |  90   ...
35   58  |  93   ...
36   59  |  95   ...
37   61  |  98   ...
38   63  |     ...
   ...
Each row of the extended Wythoff array also satisfies the Fibonacci recurrence, and may be extended to the left using this recurrence backwards.
From _Peter Munn_, Jun 11 2021: (Start)
The Wythoff array appears to have the following relationship to the traditional Fibonacci rabbit breeding story, modified for simplicity to be a story of asexual reproduction.
Give each rabbit a number, 0 for the initial rabbit.
When a new round of rabbits is born, allocate consecutive numbers according to 2 rules (the opposite of many cultural rules for inheritance precedence): (1) newly born child of Rabbit 0 gets the next available number; (2) the descendants of a younger child of any given rabbit precede the descendants of an older child of the same rabbit.
Row n of the Wythoff array lists the children of Rabbit n (so Rabbit 0's children have the Fibonacci numbers: 1, 2, 3, 5, ...). The generation tree below shows rabbits 0 to 20. It is modified so that each round of births appears on a row.
                                                                 0
                                                                 :
                                       ,-------------------------:
                                       :                         :
                       ,---------------:                         1
                       :               :                         :
              ,--------:               2               ,---------:
              :        :               :               :         :
        ,-----:        3         ,-----:         ,-----:         4
        :     :        :         :     :         :     :         :
     ,--:     5     ,--:     ,---:     6     ,---:     7     ,---:
     :  :     :     :  :     :   :     :     :   :     :     :   :
  ,--:  8  ,--:  ,--:  9  ,--:  10  ,--:  ,--:  11  ,--:  ,--:  12
  :  :  :  :  :  :  :  :  :  :   :  :  :  :  :   :  :  :  :  :   :
  : 13  :  : 14  : 15  :  : 16   :  : 17  : 18   :  : 19  : 20   :
The extended array's nontrivial extra column (A000201) gives the number that would have been allocated to the first child of Rabbit n, if Rabbit n (and only Rabbit n) had started breeding one round early.
(End)
		

References

  • John H. Conway, Posting to Math Fun Mailing List, Nov 25 1996.
  • Clark Kimberling, "Stolarsky interspersions," Ars Combinatoria 39 (1995) 129-138.

Crossrefs

See comments above for more cross-references.
Cf. A003622, A064274 (inverse), A083412 (transpose), A000201, A001950, A080164, A003603, A265650, A019586 (row that contains n).
For two versions of the extended Wythoff array, see A287869, A287870.

Programs

  • Maple
    W:= proc(n,k) Digits:= 100; (Matrix([n, floor((1+sqrt(5))/2* (n+1))]). Matrix([[0,1], [1,1]])^(k+1))[1,2] end: seq(seq(W(n, d-n), n=0..d), d=0..10); # Alois P. Heinz, Aug 18 2008
    A035513 := proc(r, c)
        option remember;
        if c = 1 then
            A003622(r) ;
        else
            A022342(1+procname(r, c-1)) ;
        end if;
    end proc:
    seq(seq(A035513(r,d-r),r=1..d-1),d=2..15) ; # R. J. Mathar, Jan 25 2015
  • Mathematica
    W[n_, k_] := Fibonacci[k + 1] Floor[n*GoldenRatio] + (n - 1) Fibonacci[k]; Table[ W[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten
  • PARI
    T(n,k)=(n+sqrtint(5*n^2))\2*fibonacci(k+1) + (n-1)*fibonacci(k)
    for(k=0,9,for(n=1,k, print1(T(n,k+1-n)", "))) \\ Charles R Greathouse IV, Mar 09 2016
    
  • Python
    from sympy import fibonacci as F, sqrt
    import math
    tau = (sqrt(5) + 1)/2
    def T(n, k): return F(k + 1)*int(math.floor(n*tau)) + F(k)*(n - 1)
    for n in range(1, 11): print([T(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Apr 23 2017
    
  • Python
    from math import isqrt, comb
    from gmpy2 import fib2
    def A035513(n):
        a = (m:=isqrt(k:=n<<1))+(k>m*(m+1))
        x = n-comb(a,2)
        b, c = fib2(a-x+2)
        return b*(x+isqrt(5*x*x)>>1)+c*(x-1) # Chai Wah Wu, Jun 26 2025

Formula

T(n, k) = Fib(k+1)*floor[n*tau]+Fib(k)*(n-1) where tau = (sqrt(5)+1)/2 = A001622 and Fib(n) = A000045(n). - Henry Bottomley, Dec 10 2001
T(n,-1) = n-1. T(n,0) = floor(n*tau). T(n,k) = T(n,k-1) + T(n,k-2) for k>=1. - R. J. Mathar, Sep 03 2016

Extensions

Comments about the extended Wythoff array added by N. J. A. Sloane, Mar 07 2016

A101220 a(n) = Sum_{k=0..n} Fibonacci(n-k)*n^k.

Original entry on oeis.org

0, 1, 3, 14, 91, 820, 9650, 140601, 2440317, 49109632, 1123595495, 28792920872, 816742025772, 25402428294801, 859492240650847, 31427791175659690, 1234928473553777403, 51893300561135516404, 2322083099525697299278
Offset: 0

Author

Ross La Haye, Dec 14 2004

Keywords

Comments

In what follows a(i,j,k) denotes a three-dimensional array, the terms a(n) are defined as a(n,n,n) in that array. - Joerg Arndt, Jan 03 2021
Previous name was: Three-dimensional array: a(i,j,k) = expansion of x*(1 + (i-j)*x)/((1-j*x)*(1-x-x^2)), read by a(n,n,n).
a(i,j,k) = the k-th value of the convolution of the Fibonacci numbers (A000045) with the powers of i = Sum_{m=0..k} a(i-1,j,m), both for i = j and i > 0; a(i,j,k) = a(i-1,j,k) + a(j,j,k-1), for i,k > 0; a(i,1,k) = Sum_{m=0..k} a(i-1,0,m), for i > 0. With F = Fibonacci and L = Lucas, then a(1,1,k) = F(k+2) - 1; a(2,1,k) = F(k+3) - 2; a(3,1,k) = L(k+2) - 3; a(4,1,k) = 4*F(k+1) + F(k) - 4; a(1,2,k) = 2^k - F(k+1); a(2,2,k) = 2^(k+1) - F(k+3); a(3,2,k) = 3(2^k - F(k+2)) + F(k); a(4,2,k) = 2^(k+2) - F(k+4) - F(k+2); a(1,3,k) = (3^k + L(k-1))/5, for k > 0; a(2,3,k) = (2 * 3^k - L(k)) /5, for k > 0; a(3,3,k) = (3^(k+1) - L(k+2))/5; a(4,3,k) = (4 * 3^k - L(k+2) - L(k+1))/5, etc..

Examples

			a(1,3,3) = 6 because a(1,3,0) = 0, a(1,3,1) = 1, a(1,3,2) = 2 and 4*2 - 2*1 - 3*0 = 6.
		

Crossrefs

a(0, j, k) = A000045(k).
a(1, 2, k+1) - a(1, 2, k) = A099036(k).
a(3, 2, k+1) - a(3, 2, k) = A104004(k).
a(4, 2, k+1) - a(4, 2, k) = A027973(k).
a(1, 3, k+1) - a(1, 3, k) = A099159(k).
a(i, 0, k) = A109754(i, k).
a(i, i+1, 3) = A002522(i+1).
a(i, i+1, 4) = A071568(i+1).
a(2^i-2, 0, k+1) = A118654(i, k), for i > 0.
Sequences of the form a(n, 0, k): A000045(k+1) (n=1), A000032(k) (n=2), A000285(k-1) (n=3), A022095(k-1) (n=4), A022096(k-1) (n=5), A022097(k-1) (n=6), A022098(k-1) (n=7), A022099(k-1) (n=8), A022100(k-1) (n=9), A022101(k-1) (n=10), A022102(k-1) (n=11), A022103(k-1) (n=12), A022104(k-1) (n=13), A022105(k-1) (n=14), A022106(k-1) (n=15), A022107(k-1) (n=16), A022108(k-1) (n=17), A022109(k-1) (n=18), A022110(k-1) (n=19), A088209(k-2) (n=k-2), A007502(k) (n=k-1), A094588(k) (n=k).
Sequences of the form a(1, n, k): A000071(k+2) (n=1), A027934(k-1) (n=2), A098703(k) (n=3).
Sequences of the form a(2, n, k): A001911(k) (n=1), A008466(k+1) (n=2), A106517(k-1) (n=3).
Sequences of the form a(3, n, k): A027961(k) (n=1), A094688(k) (n=3).
Sequences of the form a(4, n, k): A053311(k-1) (n=1), A027974(k-1) (n=2).

Programs

  • Magma
    A101220:= func< n | (&+[n^k*Fibonacci(n-k): k in [0..n]]) >;
    [A101220(n): n in [0..30]]; // G. C. Greubel, Jun 01 2025
    
  • Mathematica
    Join[{0}, Table[Sum[Fibonacci[n-k]*n^k, {k, 0, n}], {n, 1, 20}]] (* Vaclav Kotesovec, Jan 03 2021 *)
  • PARI
    a(n)=sum(k=0,n,fibonacci(n-k)*n^k) \\ Joerg Arndt, Jan 03 2021
    
  • SageMath
    def A101220(n): return sum(n^k*fibonacci(n-k) for k in range(n+1))
    print([A101220(n) for n in range(31)]) # G. C. Greubel, Jun 01 2025

Formula

a(i, j, 0) = 0, a(i, j, 1) = 1, a(i, j, 2) = i+1; a(i, j, k) = ((j+1)*a(i, j, k-1)) - ((j-1)*a(i, j, k-2)) - (j*a(i, j, k-3)), for k > 2.
a(i, j, k) = Fibonacci(k) + i*a(j, j, k-1), for i, k > 0.
a(i, j, k) = (Phi^k - (-Phi)^-k + i(((j^k - Phi^k) / (j - Phi)) - ((j^k - (-Phi)^-k) / (j - (-Phi)^-1)))) / sqrt(5), where Phi denotes the golden mean/ratio (A001622).
i^k = a(i-1, i, k) + a(i-2, i, k+1).
A104161(k) = Sum_{m=0..k} a(k-m, 0, m).
a(i, j, 0) = 0, a(i, j, 1) = 1, a(i, j, 2) = i+1, a(i, j, 3) = i*(j+1) + 2; a(i, j, k) = (j+2)*a(i, j, k-1) - 2*j*a(i, j, k-2) - a(i, j, k-3) + j*a(i, j, k-4), for k > 3. a(i, j, 0) = 0, a(i, j, 1) = 1; a(i, j, k) = a(i, j, k-1) + a(i, j, k-2) + i * j^(k-2), for k > 1.
G.f.: x*(1 + (i-j)*x)/((1-j*x)*(1-x-x^2)).
a(n, n, n) = Sum_{k=0..n} Fibonacci(n-k) * n^k. - Ross La Haye, Jan 14 2006
Sum_{m=0..k} binomial(k,m)*(i-1)^m = a(i-1,i,k) + a(i-2,i,k+1), for i > 1. - Ross La Haye, May 29 2006
From Ross La Haye, Jun 03 2006: (Start)
a(3, 3, k+1) - a(3, 3, k) = A106517(k).
a(1, 1, k) = A001924(k) - A001924(k-1), for k > 0.
a(2, 1, k) = A001891(k) - A001891(k-1), for k > 0.
a(3, 1, k) = A023537(k) - A023537(k-1), for k > 0.
Sum_{j=0..i+1} a(i-j+1, 0, j) - Sum_{j=0..i} a(i-j, 0, j) = A001595(i). (End)
a(i,j,k) = a(j,j,k) + (i-j)*a(j,j,k-1), for k > 0.
a(n) ~ n^(n-1). - Vaclav Kotesovec, Jan 03 2021

Extensions

New name from Joerg Arndt, Jan 03 2021

A109754 Matrix defined by: a(i,0) = 0, a(i,j) = i*Fibonacci(j-1) + Fibonacci(j), for j > 0; read by ascending antidiagonals.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 3, 3, 0, 1, 4, 4, 5, 5, 0, 1, 5, 5, 7, 8, 8, 0, 1, 6, 6, 9, 11, 13, 13, 0, 1, 7, 7, 11, 14, 18, 21, 21, 0, 1, 8, 8, 13, 17, 23, 29, 34, 34, 0, 1, 9, 9, 15, 20, 28, 37, 47, 55, 55, 0, 1, 10, 10, 17, 23, 33, 45, 60, 76, 89, 89
Offset: 0

Author

Ross La Haye, Aug 11 2005; corrected Apr 14 2006

Keywords

Comments

Lower triangular version is at A117501. - Ross La Haye, Apr 12 2006

Examples

			Table starts:
[0] 0, 1,  1,  2,  3,  5,  8, 13,  21,  34, ...
[1] 0, 1,  2,  3,  5,  8, 13, 21,  34,  55, ...
[2] 0, 1,  3,  4,  7, 11, 18, 29,  47,  76, ...
[3] 0, 1,  4,  5,  9, 14, 23, 37,  60,  97, ...
[4] 0, 1,  5,  6, 11, 17, 28, 45,  73, 118, ...
[5] 0, 1,  6,  7, 13, 20, 33, 53,  86, 139, ...
[6] 0, 1,  7,  8, 15, 23, 38, 61,  99, 160, ...
[7] 0, 1,  8,  9, 17, 26, 43, 69, 112, 181, ...
[8] 0, 1,  9, 10, 19, 29, 48, 77, 125, 202, ...
[9] 0, 1, 10, 11, 21, 32, 53, 85, 138, 223, ...
		

Crossrefs

Rows: A000045(j); A000045(j+1), for j > 0; A000032(j), for j > 0; A000285(j-1), for j > 0; A022095(j-1), for j > 0; A022096(j-1), for j > 0; A022097(j-1), for j > 0. Diagonals: a(i, i) = A094588(i); a(i, i+1) = A007502(i+1); a(i, i+2) = A088209(i); Sum[a(i-j, j), {j=0...i}] = A104161(i). a(i, j) = A101220(i, 0, j).
Rows 7 - 19: A022098(j-1), for j > 0; A022099(j-1), for j > 0; A022100(j-1), for j > 0; A022101(j-1), for j > 0; A022102(j-1), for j > 0; A022103(j-1), for j > 0; A022104(j-1), for j > 0; A022106(j-1), for j > 0; A022107(j-1), for j > 0; A022108(j-1), for j > 0; A022109(j-1), for j > 0; A022110(j-1), for j > 0.
a(2^i-2, j+1) = A118654(i, j), for i > 0.
Cf. A117501.

Programs

  • Maple
    A := (n, k) -> ifelse(k = 0, 0,
          n*combinat:-fibonacci(k-1) + combinat:-fibonacci(k)):
    seq(seq(A(n - k, k), k = 0..n), n = 0..6); # Peter Luschny, May 28 2022
  • Mathematica
    T[n_, 0]:= 0; T[n_, 1]:= 1; T[n_, 2]:= n - 1; T[n_, 3]:= n - 1; T[n_, n_]:= Fibonacci[n]; T[n_, k_]:= T[n, k] = T[n - 1, k - 1] + T[n - 2, k - 2]; Table[T[n, k], {n, 0, 15}, {k, 0, n}] (* G. C. Greubel, Jan 07 2017 *)

Formula

a(i, 0) = 0, a(i, j) = i*Fibonacci(j-1) + Fibonacci(j), for j > 0.
a(i, 0) = 0, a(i, 1) = 1, a(i, 2) = i+1, a(i, j) = a(i, j-1) + a(i, j-2), for j > 2.
G.f.: (x*(1 + ix))/(1 - x - x^2).
Sum_{j=0..i+1} a(i-j+1, j) - Sum_{j=0..i} a(i-j, j) = A001595(i). - Ross La Haye, Jun 03 2006

Extensions

More terms from G. C. Greubel, Jan 07 2017

A093563 (6,1)-Pascal triangle.

Original entry on oeis.org

1, 6, 1, 6, 7, 1, 6, 13, 8, 1, 6, 19, 21, 9, 1, 6, 25, 40, 30, 10, 1, 6, 31, 65, 70, 40, 11, 1, 6, 37, 96, 135, 110, 51, 12, 1, 6, 43, 133, 231, 245, 161, 63, 13, 1, 6, 49, 176, 364, 476, 406, 224, 76, 14, 1, 6, 55, 225, 540, 840, 882, 630, 300, 90, 15, 1, 6, 61, 280, 765, 1380
Offset: 0

Author

Wolfdieter Lang, Apr 22 2004

Keywords

Comments

The array F(6;n,m) gives in the columns m >= 1 the figurate numbers based on A016921, including the octagonal numbers A000567, (see the W. Lang link).
This is the sixth member, d=6, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-2, for d=1..5.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). 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)=(1+5*z)/(1-(1+x)*z).
The SW-NE diagonals give A022096(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 5. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

Examples

			Triangle begins
  1;
  6,  1;
  6,  7,  1;
  6, 13,  8,  1;
  6, 19, 21,  9,  1;
  6, 25, 40, 30, 10,  1;
  ...
		

References

  • Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
  • Ivo Schneider: Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Crossrefs

Row sums: A005009(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 5 for n=2 and 0 else.
The column sequences give for m=1..9: A016921, A000567 (octagonal), A002414, A002419, A051843, A027810, A034265, A054487, A055848.

Programs

  • Haskell
    a093563 n k = a093563_tabl !! n !! k
    a093563_row n = a093563_tabl !! n
    a093563_tabl = [1] : iterate
                   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [6, 1]
    -- Reinhard Zumkeller, Aug 31 2014
    
  • Mathematica
    lim = 11; s = Series[(1 + 5*x)/(1 - x)^(m + 1), {x, 0, lim}]; t = Table[ CoefficientList[s, x], {m, 0, lim}]; Flatten[ Table[t[[j - k + 1, k]], {j, lim + 1}, {k, j, 1, -1}]] (* Jean-François Alcover, Sep 16 2011, after g.f. *)
  • Python
    from math import comb, isqrt
    def A093563(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+5*(r-a))//r if n else 1 # Chai Wah Wu, Nov 12 2024

Formula

a(n, m)=F(6;n-m, m) for 0<= m <= n, otherwise 0, with F(6;0, 0)=1, F(6;n, 0)=6 if n>=1 and F(6;n, m):= (6*n+m)*binomial(n+m-1, m-1)/m if m>=1.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=6 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+5*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 5*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(6 + 13*x + 8*x^2/2! + x^3/3!) = 6 + 19*x + 40*x^2/2! + 70*x^3/3! + 110*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A096940 Pascal (1,5) triangle.

Original entry on oeis.org

5, 1, 5, 1, 6, 5, 1, 7, 11, 5, 1, 8, 18, 16, 5, 1, 9, 26, 34, 21, 5, 1, 10, 35, 60, 55, 26, 5, 1, 11, 45, 95, 115, 81, 31, 5, 1, 12, 56, 140, 210, 196, 112, 36, 5, 1, 13, 68, 196, 350, 406, 308, 148, 41, 5, 1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5, 1, 15, 95, 345, 810, 1302
Offset: 0

Author

Wolfdieter Lang, Jul 16 2004

Keywords

Comments

This is the fifth member, q=5, 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), 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)=(5-4*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(5-4*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022096(n-2), n>=2, with n=1 value 5. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Examples

			Triangle begins:
  5;
  1,  5;
  1,  6,  5;
  1,  7, 11,   5;
  1,  8, 18,  16,   5;
  1,  9, 26,  34,  21,   5;
  1, 10, 35,  60,  55,  26,   5;
  1, 11, 45,  95, 115,  81,  31,   5;
  1, 12, 56, 140, 210, 196, 112,  36,   5;
  1, 13, 68, 196, 350, 406, 308, 148,  41,  5;
  1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5; etc.
		

Crossrefs

Row sums: A007283(n-1), n>=1, 5 if n=0; g.f.: (5-4*x)/(1-2*x). Alternating row sums are [5, -4, followed by 0's].
Column sequences (without leading zeros) give for m=1..9, with n>=0: A000027(n+5), A056000(n-1), A096941-7.

Programs

  • Maple
    a(n,k):=piecewise(n=0,5,0Mircea Merca, Apr 08 2012
  • PARI
    a(n) = {if(n <= 1, return(5 - 4*(n==1))); my(m = (sqrtint(8*n + 1) - 1)\2, t = n - binomial(m + 1, 2)); (1+4*t/m)*binomial(m,t)} \\ David A. Corneth, Aug 28 2019

Formula

Recursion: a(n, m)=0 if m>n, a(0, 0)= 5; 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): (5-4*x)/(1-x)^(m+1), m>=0.
a(n,k) = (1+4*k/n)*binomial(n,k), for n>0. - Mircea Merca, Apr 08 2012

A352744 Array read by ascending antidiagonals. Generalized Fibonacci numbers F(n, k) = (psi^k*(phi - n) - phi^k*(psi - n)) / (phi - psi) where phi = (1 + sqrt(5))/2 and psi = (1 - sqrt(5))/2. F(n, k) for n >= 0 and k >= 0.

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 1, 2, 2, 1, 1, 3, 3, 3, 2, 1, 4, 4, 5, 5, 3, 1, 5, 5, 7, 8, 8, 5, 1, 6, 6, 9, 11, 13, 13, 8, 1, 7, 7, 11, 14, 18, 21, 21, 13, 1, 8, 8, 13, 17, 23, 29, 34, 34, 21, 1, 9, 9, 15, 20, 28, 37, 47, 55, 55, 34, 1, 10, 10, 17, 23, 33, 45, 60, 76, 89, 89, 55
Offset: 0

Author

Peter Luschny, Apr 01 2022

Keywords

Comments

The definition declares the Fibonacci numbers for all integers n and k. An alternative version is A353595.
The identity F(n, k) = (-1)^k*F(1 - n, -k) holds for all integers n, k. Proof:
F(n, k)*(2+phi) = (phi^k*(n*phi + 1) - (-phi)^(-k)*((n-1)*phi - 1))
= (-1)^k*(phi^(-k)*((1-n)*phi+1) - (-phi)^k*(-n*phi-1))
= (-1)^k*F(1-n, -k)*(2+phi).
This identity can be seen as an extension of Cassini's theorem of 1680 and of an identity given by Graham, Knuth and Patashnik in 'Concrete Mathematics' (6.106 and 6.107). The beginning of the full array with arguments in Z x Z can be found in the linked note.
The enumeration is the result of the simple form of the chosen definition. The classical positive Fibonacci numbers starting with 1, 1, 2, 3,... are in row n = 1 with offset 0. The nonnegative Fibonacci numbers starting 0, 1, 1, 2, 3,... are in row 0 with offset 1. They prolong towards -infinity with an index shifted by 1 compared to the enumeration used by Knuth. A characteristic of our enumeration is F(n, 0) = 1 for all integer n.
Fibonacci numbers vanish only for (n,k) in {(-1,2), (0,1), (1,-1), (2,-2)}. The zeros correspond to the identities (phi + 1)*psi^2 = (psi + 1)*phi^2, psi*phi = phi*psi, (phi - 1)*phi = (psi - 1)*psi and (phi - 2)*phi^2 = (psi - 2)*psi^2.
For divisibility properties see A352747.
For any fixed k, the sequence F(n, k) is a linear function of n. In other words, an arithmetic progression. This implies that F(n+1, k) = 2*F(n, k) - F(n-1, k) for all n in Z. Special case of this is Fibonacci(n+1) = 2 *Fibonacci(n) - Fibonacci(n-2). - Michael Somos, May 08 2022

Examples

			Array starts:
n\k 0, 1,  2,  3,  4,  5,  6,   7,   8,   9, ...
---------------------------------------------------------
[0] 1, 0,  1,  1,  2,  3,  5,   8,  13,  21, ... A212804
[1] 1, 1,  2,  3,  5,  8, 13,  21,  34,  55, ... A000045 (shifted once)
[2] 1, 2,  3,  5,  8, 13, 21,  34,  55,  89, ... A000045 (shifted twice)
[3] 1, 3,  4,  7, 11, 18, 29,  47,  76, 123, ... A000032 (shifted once)
[4] 1, 4,  5,  9, 14, 23, 37,  60,  97, 157, ... A000285
[5] 1, 5,  6, 11, 17, 28, 45,  73, 118, 191, ... A022095
[6] 1, 6,  7, 13, 20, 33, 53,  86, 139, 225, ... A022096
[7] 1, 7,  8, 15, 23, 38, 61,  99, 160, 259, ... A022097
[8] 1, 8,  9, 17, 26, 43, 69, 112, 181, 293, ... A022098
[9] 1, 9, 10, 19, 29, 48, 77, 125, 202, 327, ... A022099
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, sec. 6.6.
  • Donald Ervin Knuth, The Art of Computer Programming, Third Edition, Vol. 1, Fundamental Algorithms. Chapter 1.2.8 Fibonacci Numbers. Addison-Wesley, Reading, MA, 1997.

Crossrefs

Diagonals: A088209 (main), A007502, A066982 (antidiagonal sums).
Cf. A352747, A353595 (alternative version), A354265 (generalized Lucas numbers).
Similar arrays based on the Catalan and the Bell numbers are A352680 and A352682.

Programs

  • Julia
    # Time complexity is O(lg n).
    function fibrec(n::Int)
        n == 0 && return (BigInt(0), BigInt(1))
        a, b = fibrec(div(n, 2))
        c = a * (b * 2 - a)
        d = a * a + b * b
        iseven(n) ? (c, d) : (d, c + d)
    end
    function Fibonacci(n::Int, k::Int)
        k == 0 && return BigInt(1)
        k  < 0 && return (-1)^k*Fibonacci(1 - n, -k)
        a, b = fibrec(k - 1)
        a + b*n
    end
    for n in -6:6
        println([Fibonacci(n, k) for k in -6:6])
    end
    
  • Maple
    f := n -> combinat:-fibonacci(n + 1): F := (n, k) -> (n-1)*f(k-1) + f(k):
    seq(seq(F(n-k, k), k = 0..n), n = 0..9);
    # The next implementation is for illustration only but is not recommended
    # as it relies on floating point arithmetic.
    phi := (1 + sqrt(5))/2: psi := (1 - sqrt(5))/2:
    F := (n, k) -> (psi^k*(phi - n) - phi^k*(psi - n)) / (phi - psi):
    for n from -6 to 6 do lprint(seq(simplify(F(n, k)), k = -6..6)) od;
  • Mathematica
    Table[LinearRecurrence[{1, 1}, {1, n}, 10], {n, 0, 9}] // TableForm
    F[ n_, k_] := (MatrixPower[{{0, 1}, {1, 1}}, k].{{1}, {n}})[[1, 1]]; (* Michael Somos, May 08 2022 *)
    c := Pi/2 - I*ArcSinh[1/2]; (* Based on a remark from Bill Gosper. *)
    F[n_, k_] := 2 (I (n-1) Sin[k c] + Sin[(k+1) c]) / (I^k Sqrt[5]);
    Table[Simplify[F[n, k]], {n, -6, 6}, {k, -6, 6}] // TableForm (* Peter Luschny, May 10 2022 *)
  • PARI
    F(n, k) = ([0, 1; 1, 1]^k*[1; n])[1, 1]
    
  • PARI
    {F(n, k) = n*fibonacci(k) + fibonacci(k-1)}; /* Michael Somos, May 08 2022 */

Formula

F(n, k) = F(n, k-1) + F(n, k-2) for k >= 2, otherwise 1, n for k = 0, 1.
F(n, k) = (n-1)*f(k-1) + f(k) where f(n) = A000045(n+1), the Fibonacci numbers starting with f(0) = 1.
F(n, k) = ((phi^k*(n*phi + 1) - (-phi)^(-k)*((n - 1)*phi - 1)))/(2 + phi).
F(n, k) = [x^k] (1 + (n - 1)*x)/(1 - x - x^2) for k >= 0.
F(k, n) = [x^k] (F(0, n) + F(0, n-1)*x)/(1 - x)^2 for k >= 0.
F(n, k) = (k!/sqrt(5))*[x^k] ((n-psi)*exp(phi*x) - (n-phi)*exp(psi*x)) for k >= 0.
F(n, k) - F(n-1, k) = sign(k)^(n-1)*f(k) for all n, k in Z, where A000045 is extended to negative integers by f(-n) = (-1)^(n-1)*f(n) (CMath 6.107). - Peter Luschny, May 09 2022
F(n, k) = 2*((n-1)*i*sin(k*c) + sin((k+1)*c))/(i^k*sqrt(5)) where c = Pi/2 - i*arcsinh(1/2), for all n, k in Z. Based on a remark from Bill Gosper. - Peter Luschny, May 10 2022

A131777 5*A065941 - 4*A049310.

Original entry on oeis.org

1, 5, 1, 1, 5, 1, 5, -3, 10, 1, 1, 5, 3, 10, 1, 5, -7, 20, -1, 15, 1, 1, 5, 1, 20, 10, 15, 1, 5, -11, 30, -15, 50, 6, 20, 1, 1, 5, -5, 30, 15, 50, 22, 20, 1, 5, -15, 40, -45, 105, -9, 100, 18, 25, 1
Offset: 1

Author

Gary W. Adamson, Jul 14 2007

Keywords

Comments

Row sums = A022096(n-1).

Examples

			First few rows of the triangle are:
1;
5, 1
1, 5, 1;
5, -3, 10, 1;
1, 5, 3, 10, 1;
5, -7, 20, -1, 15, 1;
1, 5, 1, 20, 10, 15, 1;
...
		

Formula

5*A065941 - 4*A049310 as infinite lower triangular matrices.

Extensions

Corrected A-number in row sums reference R. J. Mathar, Jun 16 2009

A022368 Fibonacci sequence beginning 2, 12.

Original entry on oeis.org

2, 12, 14, 26, 40, 66, 106, 172, 278, 450, 728, 1178, 1906, 3084, 4990, 8074, 13064, 21138, 34202, 55340, 89542, 144882, 234424, 379306, 613730, 993036, 1606766, 2599802, 4206568, 6806370, 11012938
Offset: 0

Keywords

Programs

  • Mathematica
    a={};b=2;c=12;AppendTo[a, b];AppendTo[a, c];Do[b=b+c;AppendTo[a, b];c=b+c;AppendTo[a, c], {n, 4!}];a (* Vladimir Joseph Stephan Orlovsky, Sep 18 2008 *)
    Table[2*(Fibonacci[n + 2] + 4*Fibonacci[n]), {n,0,50}] (* or *) LinearRecurrence[{1,1}, {2,12}, 50] (* G. C. Greubel, Aug 27 2017 *)
  • PARI
    for(n=0,50, print1(2*(fibonacci(n+2) + 4*fibonacci(n)), ", ")) \\ G. C. Greubel, Aug 27 2017

Formula

G.f.: 2*(1+5*x)/(1-x-x^2). - Philippe Deléham, Nov 19 2008
a(n) = 2*A022096(n). - R. J. Mathar, Sep 27 2014
a(n) = 2*(Fibonacci(n+2) + 4*Fibonacci(n)). - G. C. Greubel, Aug 27 2017

A127830 a(n) = Sum_{k=0..n} (binomial(floor(k/2),n-k) mod 2).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 2, 3, 3, 3, 2, 2, 3, 2, 3, 5, 5, 4, 4, 5, 4, 3, 3, 3, 4, 4, 3, 4, 5, 3, 5, 8, 8, 7, 6, 7, 7, 5, 6, 8, 7, 6, 5, 5, 5, 4, 4, 5, 6, 5, 5, 7, 6, 4, 5, 6, 7, 7, 5, 6, 8, 5, 8, 13, 13, 11, 10, 12, 11, 8, 9, 11, 11, 10, 8, 9, 10, 7, 9, 13, 12
Offset: 0

Author

Paul Barry, Feb 01 2007

Keywords

Comments

Row sums of number triangle A127829.
From Johannes W. Meijer, Jun 05 2011: (Start)
The Ze3 and Ze4 triangle sums, see A180662 for their definitions, of Sierpinski's triangle A047999 equal this sequence.
The sequences A127830(2^n-p), p>=0, are apparently all Fibonacci like sequences, i.e., the next term is the sum of the two nonzero terms that precede it; see the crossrefs. (End)

Crossrefs

Cf.: A000045 (p=0), A000204 (p=7), A001060 (p=13), A000285 (p=14), A022095 (p=16), A022120 (p=24), A022121 (p=25), A022113 (p=28), A022096 (p=30), A022097 (p=31), A022098 (p=32), A022130 (p=44), A022137 (p=48), A022138 (p=49), A022122 (p=52), A022114 (p=53), A022123 (p=56), A022115 (p=60), A022100 (p=62), A022101 (p=63), A022103 (p=64), A022136 (p=79), A022388 (p=80), A022389 (p=88). - Johannes W. Meijer, Jun 05 2011

Programs

  • Maple
    A127830 := proc(n) local k: option remember: add(binomial(floor(k/2), n-k) mod 2, k=0..n) end: seq(A127830(n), n=0..80); # Johannes W. Meijer, Jun 05 2011
  • Mathematica
    Table[Sum[Mod[Binomial[Floor[k/2],n-k],2],{k,0,n}],{n,0,80}] (* James C. McMahon, Jan 04 2025 *)
  • Python
    def A127830(n): return sum(not ~(k>>1)&n-k for k in range(n+1)) # Chai Wah Wu, Jul 29 2025

Formula

a(2^n) = F(n); a(2^(n+1)+1) = L(n).
a(n) mod 2 = A000931(n+5) mod 2 = A011656(n+4).
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