A081667 -id:A081667 - cp's OEIS frontend

A033192 a(n) = binomial(Fibonacci(n) + 1, 2).

0, 1, 1, 3, 6, 15, 36, 91, 231, 595, 1540, 4005, 10440, 27261, 71253, 186355, 487578, 1276003, 3339820, 8742471, 22885995, 59912931, 156848616, 410626153, 1075018896, 2814412825, 7368190921, 19290113571, 50502074766, 132215989335, 346145696820, 906220783315

Offset: 0

Simon P. Norton

a(n) is the sum of n-th row in Wythoff array A003603. [Reinhard Zumkeller, Jan 26 2012]

A subsequence of the triangular numbers A000217. In fact, binomial(F(n)+1,2) = A000217(F(n)). - M. F. Hasler, Jan 27 2012

Alois P. Heinz, Table of n, a(n) for n = 0..2394
James P. Jones and Péter Kiss, Representation of integers as terms of a linear recurrence with maximal index, Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae, 25. (1998) pp. 21-37. See Theorem 3.7 p. 33.
Kálmán Liptai and László Szalay, Random inhomogeneous binary recurrences, Annales Univ. Sci. Budapest, Sect. Comp. 54 (2023) 253-263. See p. 262.
Index entries for linear recurrences with constant coefficients, signature (3,1,-5,-1,1).

Cf. A000045, A000217, A033191, A081667.

a:= n-> (f-> f*(f+1)/2)((<<0|1>, <1|1>>^n)[1, 2]):
seq(a(n), n=0..35);  # Alois P. Heinz, Sep 06 2008

Table[Binomial[Fibonacci[n] + 1, 2], {n, 0, 50}] (* Alonso del Arte, Jan 26 2012 *)
LinearRecurrence[{3,1,-5,-1,1},{0,1,1,3,6},40] (* Harvey P. Dale, Apr 04 2020 *)

a(n)=binomial(fibonacci(n)+1,2) \\ Charles R Greathouse IV, Jan 26 2012

G.f.: x(x^3-x^2-2x+1)/[(1+x)(1-3x+x^2)(1-x-x^2)].
a(n) = ((Fibonacci(n)+Fibonacci(n)^2)/2). - Gary Detlefs, Dec 24 2010
Equals A000217 o A000045. - M. F. Hasler, Jan 27 2012
a(n) = A032441(n) - 1. - Filip Zaludek, Oct 30 2016

A054783 (n^2)-th Fibonacci number.

Original entry on oeis.org

0, 1, 3, 34, 987, 75025, 14930352, 7778742049, 10610209857723, 37889062373143906, 354224848179261915075, 8670007398507948658051921, 555565404224292694404015791808, 93202207781383214849429075266681969, 40934782466626840596168752972961528246147

Offset: 0

Jeff Burch, May 22 2000

nonn

The sequence (5*a(n+1)){n>=1} = (5, 15, 170, 4935, ...) is realizable in the sense that there is a self-map on a set T:X->X with the property that a(n) = #{x in X:T^nx=x} for all n >= 1. This is the simplest illustrative example of two different phenomena. The Fibonacci sequence sampled along an odd power cannot be made realizable after multiplication by a constant; the Fibonacci sequence sampled along an even power becomes realizable after multiplication by 5 (the discriminant of the sequence). This is now known to be an instance of a more general phenomenon in the following sense. If (a(n)) is a linear recurrence sequence whose characteristic polynomial F has simple zeros then the sequence (Ma(n^s)) satisfies the Dold congruence, where M=|discriminant(F)| and s is an integer multiple of the exponent of the Galois group of the splitting field of F over the rationals. Under an additional hypothesis on the signs of the coefficients of F, the sequence (Ma(n^s)) is realizable. - _Thomas Ward, May 06 2022

Seiichi Manyama, Table of n, a(n) for n = 0..69
Jakub Byszewski, Grzegorz Graff and Thomas Ward, Dold sequences, periodic points, and dynamics, arXiv:2007.04031 [math.DS], 2020-2021; Bull. Lond. Math. Soc. 53 (2021), no. 5, 1263-1298.
T. Kotek and J. A. Makowsky, Recurrence Relations for Graph Polynomials on Bi-iterative Families of Graphs, arXiv preprint arXiv:1309.4020 [math.CO], 2013.
Florian Luca and Tom Ward, On (almost) realizable subsequences of linearly recurrent sequences, arXiv:2204.02711 [math.NT], 2022.
Piotr Miska and Tom Ward, Stirling numbers and periodic points, arXiv:2102.07561 [math.NT], 2021; Acta Arith. 201 (2021), no. 4, 421-435.
Patrick Moss and Tom Ward, Fibonacci along even powers is (almost) realizable, arXiv:2011.13068 [math.NT], 2020; Fibonacci Quart. 60 (2022), no. 1, 40-47.

Cf. (n^k)-th Fibonacci number: A000045 (k=1), this sequence (k=2), A182149 (k=3), A250490 (k=4), A250491 (k=5), A250492 (k=6), A250493 (k=7), A250494 (k=8).
Cf. A081667.
Cf. A341617 shows a similar property for the Stirling numbers of the second kind.

[Fibonacci(n^2): n in [0..50]]; // Vincenzo Librandi, Apr 09 2011

a:= n-> (<<0|1>, <1|1>>^(n^2))[1, 2]:
seq(a(n), n=0..15);  # Alois P. Heinz, Jun 10 2018

Table[Fibonacci[n^2], {n, 15}] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)

a(n)=fibonacci(n^2) \\ Charles R Greathouse IV, Oct 07 2016

a(n) = Sum_{k=1..T(n-1)+1} binomial(T(n-1), k-1)*F(n-1+k), where F(n) is A000045 and T(n) is A000217. - Tony Foster III, Sep 03 2018

A045995 Rows of Fibonacci-Pascal triangle.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 8, 3, 1, 1, 5, 55, 55, 5, 1, 1, 8, 610, 6765, 610, 8, 1, 1, 13, 10946, 9227465, 9227465, 10946, 13, 1, 1, 21, 317811, 225851433717, 190392490709135, 225851433717, 317811, 21, 1, 1, 34, 14930352

Offset: 0

N. J. A. Sloane

			1,
1,  1,
1,  1,      1,
1,  2,      2,            1,
1,  3,      8,            3,               1,
1,  5,     55,           55,               5,            1,
1,  8,    610,         6765,             610,            8,      1,
1, 13,  10946,      9227465,         9227465,        10946,     13,  1,
1, 21, 317811, 225851433717, 190392490709135, 225851433717, 317811, 21, 1,
...

Reinhard Zumkeller, Rows n=0..14 of triangle, flattened
R. Whitney, Problem H-254, Fib. Quart., 13 (1975), p. 281.

Cf. A000045, A007318, A006449 (row sums), A081667.

Main diagonal gives A281450.

a045995 n k = a045995_tabl !! n !! k
a045995_row n = a045995_tabl !! n
a045995_tabl = map (map (a000045 . fromInteger)) a007318_tabl
-- Reinhard Zumkeller, Dec 29 2011

A045995 := proc(n,k)
    combinat[fibonacci](binomial(n,k)) ;
end proc: # R. J. Mathar, Dec 03 2014

Flatten[Table[Fibonacci[Binomial[n,k]],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Dec 31 2013 *)

Take Pascal triangle (A007318) and replace each i by Fibonacci(i): a(n,k)=Fibonacci(binomial(n,k)).

More terms from David W. Wilson

A125752 Moessner triangle using the Fibonacci terms.

Original entry on oeis.org

1, 1, 2, 4, 9, 8, 26, 69, 77, 55, 261, 806, 1088, 920, 610, 4062, 14362, 22887, 22856, 17034, 10946, 98912, 395253, 728605, 847832, 721756, 502606, 317811, 3809193, 17008391, 35644614, 47557978, 46166656, 35655012, 23828383, 14930352

Offset: 1

Gary W. Adamson, Dec 06 2006

A Moessner triangle is generated with the recurrence described in A125714, starting from a first row M(1,c) filled with the Fibonacci numbers M(1,c) = A000045(c), c >= 1.
Subsequent rows n are generated from the numbers in their previous rows with the rule:
Mark/circle all elements M(n-1, A000217(t)) of the previous row n-1, t >= 1.
Define the elements M(n,.) as the partial sums of the M(n-1,.) that have not been marked:
M(n,c) = Sum_{j=1..c} M(n-1,A014132(j)), c >= 1. The T(n,m) are then defined by reading the marked/circled terms "along antidiagonals": T(n,m) = M(n+m-1, A000217(m)), n >= 1, 1 <= m <= n.

			The upper left corner of the array M(n,c) is
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, ...
1, 4, 9, 22, 43, 77, 166, 310, 543, 920, 1907, 3504, 6088, 10269, 17034, ...
4, 26, 69, 235, 545, 1088, 2995, 6499, 12587, 22856, 57601, 121003, 230773, ...
26, 261, 806, 3801, 10300, 22887, 80488, 201491, 432264, 847832, 2586423, ...
261, 4062, 14362, 94850, 296341, 728605, 3315028, 9488917, 22445416, ...
4062, 98912, 395253, 3710281, 13199198, 35644614, 213010460, 690899755, ...
and dropping the columns with column numbers in A014132, reading the remaining array by antidiagonals leads to the final triangle T(n,m):
    1;
    1,   2;
    4,   9,    8;
   26,  69,   77,  55;
  261, 806, 1088, 920, 610;
  ...

J. H. Conway and R. K. Guy, "The Book of Numbers", Springer-Verlag, 1996, p. 64.

Joshua Zucker, Table of n, a(n) for n = 1..55
G. S. Kazandzidis, On a conjecture of Moessner and a general problem, Bull. Soc. Math. Grèce (N.S.) 2 (1961), 23-30.
Dexter Kozen and Alexandra Silva, On Moessner's theorem, Amer. Math. Monthly 120(2) (2013), 131-139.
R. Krebbers, L. Parlant, and A. Silva, Moessner's theorem: an exercise in coinductive reasoning in Coq, Theory and practice of formal methods, 309-324, Lecture Notes in Comput. Sci., 9660, Springer, 2016.
Calvin T. Long, Strike it out--add it up, Math. Gaz. 66 (438) (1982), 273-277.
Alfred Moessner, Eine Bemerkung über die Potenzen der natürlichen Zahlen, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss., 29, 1951.
Ivan Paasche, Ein neuer Beweis des Moessnerschen Satzes S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss. 1952 (1952), 1-5 (1953). [Two years are listed at the beginning of the journal issue.]
Ivan Paasche, Beweis des Moessnerschen Satzes mittels linearer Transformationen, Arch. Math. (Basel) 6 (1955), 194-199.
Ivan Paasche, Eine Verallgemeinerung des Moessnerschen Satzes, Compositio Math. 12 (1956), 263-270.
Hans Salié, Bemerkung zu einem Satz von A. Moessner, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss. 1952 (1952), 7-11 (1953). [Two years are listed at the beginning of the journal issue.]
Oskar Perron, Beweis des Moessnerschen Satzes, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss., 31-34, 1951.

Cf. A125714, A125750, A125751, A081667.

T(n,n) = A081667(n-1).

More terms from Joshua Zucker, Jun 17 2007

Description of starting row corrected, comments detailed with formulas by R. J. Mathar, Sep 17 2009

A297965 a(n) = Fibonacci(binomial(n+3, 3)).

Original entry on oeis.org

1, 3, 55, 6765, 9227465, 225851433717, 160500643816367088, 5358359254990966640871840, 13598018856492162040239554477268290, 4244200115309993198876969489421897548446236915, 263621064469290555679241849789653324393054271110084140201023

Offset: 0

Vincenzo Librandi, Jan 10 2018

Fourth diagonal of A045995.

Cf. A000045, A000292, A045995, A081667.

[Fibonacci(Binomial(n+3,3)): n in [0..20]];

Table[Fibonacci[Binomial[n + 3, 3]], {n, 0, 20}]

a(n) = fibonacci(binomial(n+3, 3)) \\ Felix Fröhlich, Jan 12 2018

a(n) = A000045(A000292(n+1)) = A045995(n+3, 3).

cp's OEIS Frontend

A033192 a(n) = binomial(Fibonacci(n) + 1, 2).

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A054783 (n^2)-th Fibonacci number.

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A045995 Rows of Fibonacci-Pascal triangle.

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A125752 Moessner triangle using the Fibonacci terms.

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A297965 a(n) = Fibonacci(binomial(n+3, 3)).

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