A001911 -id:A001911 - cp's OEIS frontend

A202874 Symmetric matrix based on (1,2,3,5,8,13,...), by antidiagonals.

1, 2, 2, 3, 5, 3, 5, 8, 8, 5, 8, 13, 14, 13, 8, 13, 21, 23, 23, 21, 13, 21, 34, 37, 39, 37, 34, 21, 34, 55, 60, 63, 63, 60, 55, 34, 55, 89, 97, 102, 103, 102, 97, 89, 55, 89, 144, 157, 165, 167, 167, 165, 157, 144, 89, 144, 233, 254, 267, 270, 272, 270, 267, 254

Offset: 1

Clark Kimberling, Dec 26 2011

Let s=(1,2,3,5,8,13,...)=(F(k+1)), where F=A000045, and let T be the infinite square matrix whose n-th row is formed by putting n-1 zeros before the terms of s. Let T' be the transpose of T. Then A202874 represents the matrix product M=T'*T. M is the self-fusion matrix of s, as defined at A193722. See A202875 for characteristic polynomials of principal submatrices of M, with interlacing zeros.

			Northwest corner:
1....2....3....5....8....13
2....5....8....13...21...34
3....8....14...23...37...60
5....13...23...39...63...102
8....21...37...63...102..167

Cf. A202875.

s[k_] := Fibonacci[k + 1];
U = NestList[Most[Prepend[#, 0]] &, #, Length[#] - 1] &[Table[s[k], {k, 1, 15}]];
L = Transpose[U]; M = L.U; TableForm[M]
m[i_, j_] := M[[i]][[j]];
Flatten[Table[m[i, n + 1 - i], {n, 1, 12}, {i, 1, n}]]
f[n_] := Sum[m[i, n], {i, 1, n}] + Sum[m[n, j], {j, 1, n - 1}]
Table[f[n], {n, 1, 12}]
Table[Sqrt[f[n]], {n, 1, 12}]  (* A001911 *)
Table[m[1, j], {j, 1, 12}]     (* A000045 *)
Table[m[j, j], {j, 1, 12}]     (* A119996 *)
Table[m[j, j + 1], {j, 1, 12}] (* A180664 *)
Table[Sum[m[i, n + 1 - i], {i, 1, n}], {n, 1, 12}]  (* A002940 *)

A226538 a(2t) = a(2t-1) + 1, a(2t+1) = a(2t) + a(2t-2) for t >= 1, with a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 2, 3, 4, 6, 7, 11, 12, 19, 20, 32, 33, 53, 54, 87, 88, 142, 143, 231, 232, 375, 376, 608, 609, 985, 986, 1595, 1596, 2582, 2583, 4179, 4180, 6763, 6764, 10944, 10945, 17709, 17710, 28655, 28656, 46366, 46367, 75023, 75024, 121391, 121392, 196416, 196417

Offset: 0

V. T. Jayabalaji, Jun 10 2013

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,0,0,-1).

a226538 n = a226538_list !! n
a226538_list = concat $ transpose [drop 2 a000071_list, tail a001911_list]
-- Reinhard Zumkeller, Jun 18 2013

f:= proc(n) option remember;
      if n <= 1 then 1
    elif n mod 2 = 0 then f(n-1)+1
    else f(n-1)+f(n-3)
      fi
    end:
t21:=[seq(f(n),n=0..60)];

LinearRecurrence[{0, 2, 0, 0, 0, -1}, {1, 1, 2, 3, 4, 6}, 50] (* Jean-François Alcover, Feb 13 2018 *)

a(2n) = A000071(n+3), a(2n+1) = A001911(n+1). - Philippe Deléham, Jun 18 2013
G.f.: (1+x+x^3)/((1-x)*(1+x)*(1-x^2-x^4)). - Philippe Deléham, Jun 18 2013
a(n) = a(n-1) + a(n-3)*(1-(-1)^n)/2 + (1+(-1)^n)/2. - Paolo P. Lava, Jun 27 2013

Edited by N. J. A. Sloane, Jun 18 2013

A246104 Least m > 0 for which (s(m), ..., s(n+m-1)) = (s(0), ..., s(n)), the first n+1 terms of the infinite Fibonacci word A003849.

Original entry on oeis.org

2, 3, 5, 5, 8, 8, 8, 13, 13, 13, 13, 13, 21, 21, 21, 21, 21, 21, 21, 21, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 89, 89, 89, 89, 89, 89, 89, 89, 89, 89, 89, 89

Offset: 0

Clark Kimberling, Aug 14 2014

If n is a term of A001911, then a(n) = n+2, otherwise a(n) > n+2. - Ivan Neretin, Sep 30 2017

			In A003849, the initial segment (s(0), ..., s(6)) = (0,1,0,0,1,0,1) first repeats at (s(8), ..., s(14)), so that a(6) = 8.

Ivan Neretin, Table of n, a(n) for n = 0..10944

Cf. A000045, A003849, A241422, A246105.

seq(combinat:-fibonacci(n)$combinat:-fibonacci(n-2),n=2..12); # Robert Israel, Oct 01 2017

s = Flatten[Nest[{#, #[[1]]} &, {0, 1}, 12]]; b[m_, n_] := b[m, n] = Take[s, {m, n}]; q = -1 + Flatten[Table[Select[n + Range[2, 1600], b[#, n + # - 1] == b[1, n] &, 1], {n, 1, 120}]]
Flatten@Table[ConstantArray[Fibonacci[n + 1], Fibonacci[n - 1]], {n, 10}] (* Ivan Neretin, Sep 30 2017 *)

Concatenation of F(n - 2) copies of F(n), for n >= 1, where F = A000045 (Fibonacci numbers).

A345123 Number T(n,k) of ordered subsequences of {1,...,n} containing at least k elements and such that the first differences contain only odd numbers; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 2, 1, 4, 3, 1, 7, 6, 3, 1, 12, 11, 7, 3, 1, 20, 19, 14, 8, 3, 1, 33, 32, 26, 17, 9, 3, 1, 54, 53, 46, 34, 20, 10, 3, 1, 88, 87, 79, 63, 43, 23, 11, 3, 1, 143, 142, 133, 113, 83, 53, 26, 12, 3, 1, 232, 231, 221, 196, 156, 106, 64, 29, 13, 3, 1, 376, 375, 364, 334, 279, 209, 132, 76, 32, 14, 3, 1

Offset: 0

Alois P. Heinz, Jun 08 2021

The sequence of column k satisfies a linear recurrence with constant coefficients of order 2k if k >= 2 and of order 3 for k in {0, 1}.

			T(0,0) = 1: [].
T(1,1) = 1: [1].
T(2,2) = 1: [1,2].
T(3,1) = 6: [1], [2], [3], [1,2], [2,3], [1,2,3].
T(4,0) = 12: [], [1], [2], [3], [4], [1,2], [1,4], [2,3], [3,4], [1,2,3], [2,3,4], [1,2,3,4].
T(6,3) = 17: [1,2,3], [1,2,5], [1,4,5], [2,3,4], [2,3,6], [2,5,6], [3,4,5], [4,5,6], [1,2,3,4], [1,2,3,6], [1,2,5,6], [1,4,5,6], [2,3,4,5], [3,4,5,6], [1,2,3,4,5], [2,3,4,5,6], [1,2,3,4,5,6].
Triangle T(n,k) begins:
    1;
    2,   1;
    4,   3,   1;
    7,   6,   3,   1;
   12,  11,   7,   3,   1;
   20,  19,  14,   8,   3,   1;
   33,  32,  26,  17,   9,   3,  1;
   54,  53,  46,  34,  20,  10,  3,  1;
   88,  87,  79,  63,  43,  23, 11,  3,  1;
  143, 142, 133, 113,  83,  53, 26, 12,  3, 1;
  232, 231, 221, 196, 156, 106, 64, 29, 13, 3, 1;
  ...

Chu, Hung Viet, Various Sequences from Counting Subsets, Fib. Quart., 59:2 (May 2021), 150-157.

Alois P. Heinz, Rows n = 0..140, flattened
Chu, Hung Viet, Various Sequences from Counting Subsets, arXiv:2005.10081 [math.CO], 2021.

Columns k=0-3 give: A000071(n+3), A001911, A001924(n-1), A344004.
T(2n,n) give A340766.
Cf. A073044, A105438.

b:= proc(n, l, t) option remember; `if`(n=0, `if`(t=0, 1, 0), `if`(0
      in [l, irem(1+l-n, 2)], b(n-1, n, max(0, t-1)), 0)+b(n-1, l, t))
    end:
T:= (n, k)-> b(n, 0, k):
seq(seq(T(n, k), k=0..n), n=0..10);
# second Maple program:
g:= proc(n, k) option remember; `if`(k>n, 0,
     `if`(k in [0, 1], n^k, g(n-1, k-1)+g(n-2, k)))
    end:
T:= proc(n, k) option remember;
     `if`(k>n, 0, g(n, k)+T(n, k+1))
    end:
seq(seq(T(n, k), k=0..n), n=0..10);
# third Maple program:
T:= proc(n, k) option remember; `if`(k>n, 0, binomial(iquo(n+k, 2), k)+
      `if`(k>0, binomial(iquo(n+k-1, 2), k), 0)+T(n, k+1))
    end:
seq(seq(T(n, k), k=0..n), n=0..10);

T[n_, k_] := T[n, k] = If[k > n, 0, Binomial[Quotient[n+k, 2], k] +
     If[k > 0, Binomial[Quotient[n+k-1, 2], k], 0] + T[n, k+1]];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 11}] // Flatten (* Jean-François Alcover, Nov 06 2021, after 3rd Maple program *)

A360259 a(0) = 0, and for any n > 0, let k > 0 be as small as possible and such that F(2) + ... + F(1+k) >= n (where F(m) denotes A000045(m), the m-th Fibonacci number); a(n) = k + a(F(2) + ... + F(1+k) - n).

Original entry on oeis.org

0, 1, 3, 2, 6, 4, 3, 10, 6, 7, 5, 4, 15, 8, 9, 11, 7, 8, 6, 5, 21, 10, 11, 13, 12, 16, 9, 10, 12, 8, 9, 7, 6, 28, 12, 13, 15, 14, 18, 16, 15, 22, 11, 12, 14, 13, 17, 10, 11, 13, 9, 10, 8, 7, 36, 14, 15, 17, 16, 20, 18, 17, 24, 20, 21, 19, 18, 29, 13, 14, 16

Offset: 0

Rémy Sigrist, Jan 31 2023

See A095791 for the corresponding k's.

This sequence has similarities with A227192; here we use Fibonacci numbers, there powers of 2.

			The first terms, alongside the corresponding k's, are:
  n      a(n)  k
  -----  ----  ---
      0     0  N/A
      1     1    1
      2     3    2
      3     2    2
      4     6    3
      5     4    3
      6     3    3
      7    10    4
      8     6    4
      9     7    4
     10     5    4
     11     4    4
     12    15    5

Rémy Sigrist, Table of n, a(n) for n = 0..10946

See A095791, A360260 and A360265 for similar sequences.

Cf. A000045, A001911, A227192.

{ t = k = 0; print1 (0); for (n = 1, #a = vector(70), if (n > t, t += fibonacci(1+k++);); print1 (", "a[n] = k+if (t==n, 0, a[t-n]));); }

a(A001911(n)) = n.

A371592 Irregular table T(n, k), n >= 0, k = 1..2^n, read by rows; the n-th row lists the nonnegative integers whose dual Zeckendorf-binary representation has n ones.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 7, 6, 8, 9, 10, 12, 13, 15, 20, 11, 14, 16, 17, 18, 21, 22, 23, 25, 26, 28, 33, 34, 36, 41, 54, 19, 24, 27, 29, 30, 31, 35, 37, 38, 39, 42, 43, 44, 46, 47, 49, 55, 56, 57, 59, 60, 62, 67, 68, 70, 75, 88, 89, 91, 96, 109, 143

Offset: 0

Rémy Sigrist, Mar 28 2024

As a flat sequence, this is a permutation of the nonnegative integers with inverse A371593.

			Table T(n, k) begins:
    0
    1, 2
    3, 4, 5, 7
    6, 8, 9, 10, 12, 13, 15, 20
    11, 14, 16, 17, 18, 21, 22, 23, 25, 26, 28, 33, 34, 36, 41, 54
    ...

Rémy Sigrist, PARI program
Index entries for sequences that are permutations of the natural numbers

See A371590 for a similar sequence.

Cf. A001911, A035508, A112310, A371593 (inverse).

PARI
```
\\ See Links section.
```

A112310(T(n, k)) = n.
T(n, 1) = A001911(n).
T(n, 2^n) = A035508(n).

A053809 Second partial sums of A001891.

Original entry on oeis.org

1, 6, 21, 57, 133, 281, 554, 1039, 1878, 3302, 5686, 9638, 16143, 26796, 44179, 72471, 118435, 193015, 313920, 509805, 827036, 1340636, 2171996, 3517532, 5695053, 9218786, 14920769, 24147269, 39076593, 63233317, 102320326

Offset: 0

Barry E. Williams, Mar 27 2000

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-9,6,1,-3,1).

Cf. A001911, A001891, A053808.

Right-hand column 9 of triangle A011794. Pairwise sums of A014166.

List([0..40], n-> Fibonacci(n+10) - (2*n^3 + 27*n^2 + 145*n + 324)/6) # G. C. Greubel, Jul 06 2019

[Fibonacci(n+10) - (2*n^3 + 27*n^2 + 145*n + 324)/6: n in [0..40]]; // G. C. Greubel, Jul 06 2019

Table[Fibonacci[n+10] - (2*n^3+27*n^2+145*n+324)/6, {n,0,40}] (* G. C. Greubel, Jul 06 2019 *)

vector(40, n, n--; fibonacci(n+10) - (2*n^3 + 27*n^2 + 145*n + 324)/6) \\ G. C. Greubel, Jul 06 2019

[fibonacci(n+10) - (2*n^3 + 27*n^2 + 145*n + 324)/6 for n in (0..40)] # G. C. Greubel, Jul 06 2019

a(n) = a(n-1) + a(n-2) + (2*n+3)*C(n+2, 2)/3; a(-x)=0.
a(n) = Fibonacci(n+10) - (2*n^3 + 27*n^2 + 145*n + 324)/6.
G.f.: (1+x)/((1-x)^4*(1-x-x^2)).
a(n) = 5*a(n-1) - 9*a(n-2) + 6*a(n-3) + a(n-4) - 3*a(n-5) + a(n-6). - Wesley Ivan Hurt, Apr 21 2021

A054470 Partial sums of A054469.

Original entry on oeis.org

1, 8, 36, 121, 339, 838, 1891, 3983, 7953, 15225, 28183, 50779, 89518, 155053, 264767, 446952, 747572, 1241207, 2048762, 3366122, 5510518, 8995550, 14652578, 23827138, 38696751, 62785150, 101794318, 164950755, 267183785, 432650132

Offset: 0

Barry E. Williams, Mar 31 2000

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. F. Horadam, Special Properties of the Sequence W(n){a,b; p,q}, Fib. Quart., Vol. 5, No. 5 (1967), pp. 424-434.
A. K. Whitford, Binet's Formula Generalized, Fibonacci Quarterly, Vol. 15, No. 1, 1979, pp. 21, 24, 29.
Index entries for linear recurrences with constant coefficients, signature (7,-20,29,-20,1,8,-5,1).

Cf. A000045, A001891, A001911, A054469.

Right-hand column 13 of triangle A011794.

A054470:= func< n | Fibonacci(n+14) - (45120 +21458*n +4925*n^2 +680*n^3 +55*n^4 +2*n^5)/120 >;
[A054470(n): n in [0..40]]; // G. C. Greubel, Oct 21 2024

Accumulate[RecurrenceTable[{a[0]==1,a[1]==7,a[n]==a[n-1]+a[n-2]+(n+2) Binomial[n+3,3]/2},a,{n,40}]] (* Harvey P. Dale, Sep 22 2013 *)
CoefficientList[Series[(1+x)/((1-x)^6*(1-x-x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 23 2013 *)

def A054470(n): return fibonacci(n+14) -(45120 +21458*n +4925*n^2 +680*n^3 +55*n^4 +2*n^5)//120
[A054470(n) for n in range(41)] # G. C. Greubel, Oct 21 2024

a(n) = a(n-1) + a(n-2) + (2*n+5)*C(n+4, 4)/5, with a(-n) = 0.
a(n) = Sum_{j=1..[(n+2)/2]} binomial(n+6-j, n+2-2*j) + 2*Sum_{j=1..[(n+1)/2]} binomial(n+6-j, n+1-2*j), where [x]=greatest integer in x.
G.f.: (1+x) / ((1-x)^6*(1-x-x^2)). - Colin Barker, Jun 11 2013
From G. C. Greubel, Oct 21 2024: (Start)
a(n) = Fibonacci(n+14) - Sum_{j=0..5} Fibonacci(13-2*j)*binomial(n+j,j).
a(n) = Fibonacci(n+14) - (1/120)*(45120 + 21458*n + 4925*n^2 + 680*n^3 + 55*n^4 + 2*n^5). (End)

A094102 Triangle read by rows in which row n contains Fib(1), ..., Fib(n-1), Fib(n), Fib(n-1), ..., Fib(1).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 2, 1, 1, 1, 1, 2, 3, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 8, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 8, 13, 8, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 8, 13, 21, 13, 8, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 8, 13, 21, 34, 21, 13, 8, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 34, 21, 13, 8, 5, 3, 2, 1, 1

Offset: 1

Alysia Veenhof (ladyluck1899(AT)hotmail.com), May 05 2004

			Triangle begins:
1
1 1 1
1 1 2 1 1
1 1 2 3 2 1 1
1 1 2 3 5 3 2 1 1
1 1 2 3 5 8 5 3 2 1 1

Row sums are in A001911. Cf. A094103.

A166876 a(n) = a(n-1) + Fibonacci(n), a(1)=1983.

Original entry on oeis.org

1983, 1984, 1986, 1989, 1994, 2002, 2015, 2036, 2070, 2125, 2214, 2358, 2591, 2968, 3578, 4565, 6162, 8746, 12927, 19692, 30638, 48349, 77006, 123374, 198399, 319792, 516210, 834021, 1348250, 2180290, 3526559, 5704868, 9229446, 14932333, 24159798

Offset: 1

Geoff Ahiakwo, Oct 22 2009

Starting at some a(1)=s and creating further terms with the recurrence a(n)=a(n-1)+A000045(n) defines a family of sequences with recurrences a(n)= 2*a(n-1) -a(n-3).
The generating functions are x*( s+(1-s)*x+(1-s)*x^2 )/((1-x) * (1-x-x^2)).
The terms are a(n) = A000045(n+2)+s-2 = s + A001911(n-1) = (2*s+1+k)/2 where k=A166863(n-1), n>=1.
Examples: Up to offsets, s=1 yields A000071, s=2 yields A000045 shifted left thrice, s=3 yields A001611 shifted left thrice, s=4 yields A018910.
I appreciate the editing by R. J. Mathar. However I would like further analysis of the following formula. The sequence which I call GAP can have any integer as its first term, not just 1983. Thus a(1) can be 0, 1, 2, 3,... Then a(2) is always a(1)+ 1, while a(3) is a(1) + k(n)/2; where k(n) = k(n-2)+ k(n-1)+4 (This is a separate sequence submitted for consideration). [Geoff Ahiakwo, Nov 19 2009]

			For s=1983, n=3, we have k= A166863(2)= 5, a(3) = (2s+1+k)/2 = (2*1983+1+5)/2 = 1986.
For n=3, a(3)= a(1)+ k(3)/2 = a(1)+ [K(3-2)+ k(3-1)]/2 + 2 = a(1)+ 1 + 2 thus if a(1)is 0, a(3)= 3; if a(1)= 5, a(3)= 8; if a(1)=1983, a(3)= 1986, etc. [_Geoff Ahiakwo_, Nov 19 2009]

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

Cf. A001911, A000071

LinearRecurrence[{2, 0, -1}, {1983, 1984, 1986}, 100] (* G. C. Greubel, May 27 2016 *)

a(n) = 2*a(n-1) - a(n-3).
G.f.: x*(-1983 + 1982*x + 1982*x^2)/((1-x)*(x^2+x-1)).
Let a(n)= a(1)+ k(n)/2, then G.f.: k(n)= k(n-2)+ k(n-1) + 4. - Geoff Ahiakwo, Nov 19 2009

Definition and comments edited by R. J. Mathar, Oct 26 2009

cp's OEIS Frontend

A202874 Symmetric matrix based on (1,2,3,5,8,13,...), by antidiagonals.

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A226538 a(2t) = a(2t-1) + 1, a(2t+1) = a(2t) + a(2t-2) for t >= 1, with a(0) = a(1) = 1.

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A246104 Least m > 0 for which (s(m), ..., s(n+m-1)) = (s(0), ..., s(n)), the first n+1 terms of the infinite Fibonacci word A003849.

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A345123 Number T(n,k) of ordered subsequences of {1,...,n} containing at least k elements and such that the first differences contain only odd numbers; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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A360259 a(0) = 0, and for any n > 0, let k > 0 be as small as possible and such that F(2) + ... + F(1+k) >= n (where F(m) denotes A000045(m), the m-th Fibonacci number); a(n) = k + a(F(2) + ... + F(1+k) - n).

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A371592 Irregular table T(n, k), n >= 0, k = 1..2^n, read by rows; the n-th row lists the nonnegative integers whose dual Zeckendorf-binary representation has n ones.

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A053809 Second partial sums of A001891.

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