A101907 -id:A101907 - cp's OEIS frontend

A111035 Numbers n that divide the sum of the first n nonzero Fibonacci numbers.

1, 2, 24, 48, 72, 77, 96, 120, 144, 192, 216, 240, 288, 319, 323, 336, 360, 384, 432, 480, 576, 600, 648, 672, 720, 768, 864, 960, 1008, 1080, 1104, 1152, 1200, 1224, 1296, 1320, 1344, 1368, 1440, 1517, 1536, 1680, 1728, 1800, 1920, 1944, 2016, 2064, 2160

Offset: 1

Joseph L. Pe, Oct 05 2005

nonn

The sum of the first n nonzero Fibonacci numbers is F(n+2)-1, sequence A000071. Knott discusses the factorization of these numbers. Most of the terms are divisible by 24. - T. D. Noe, Oct 10 2005, edited by M. F. Hasler, Mar 01 2020

All terms are either multiples of 24 (cf. A124455) or odd (cf. A331976) or congruent to 2 (mod 12), cf. A331870 where this statement is proved. - M. F. Hasler, Mar 01 2020

			2 | 4, 24 | 121392, 48 | 12586269024, ... [Corrected by _M. F. Hasler_, Feb 06 2020]

Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Ron Knott, The Mathematical Magic of the Fibonacci Numbers
Daniel Yaqubi and Amirali Fatehizadeh, Some results on average of Fibonacci and Lucas sequences, arXiv:2001.11839 [math.CO], 2020.

See A101907 for another version.
Cf. A111058 (the analog for Lucas numbers).
Cf. A124455 (k for a(n) = 24k), A124456 (other a(n)), A331976 (odd a(n)), A331870 (even a(n) != 24k).

Filtered([1..3000], n-> ((Fibonacci(n+2)-1) mod n)=0 ); # G. C. Greubel, Feb 03 2020

[1] cat [n: n in [1..3000] | Fibonacci(n+2) mod n eq 1 ]; // G. C. Greubel, Feb 03 2020

select(n-> irem(combinat[fibonacci](n+2)-1, n)=0, [$1..3000])[]; # G. C. Greubel, Feb 03 2020

Select[Range[3000], Mod[Fibonacci[ #+2]-1, # ]==0&] (*  T. D. Noe, Oct 06 2005 *)

is(n)=((Mod([1,1;1,0],n))^(n+2))[1,2]==1 \\ Charles R Greathouse IV, Feb 04 2013

[n for n in (1..3000) if mod(fibonacci(n+2), n)==1 ] # G. C. Greubel, Feb 03 2020

{n: n| A000071(n+2)}. - R. J. Mathar, Feb 05 2020

More terms from Rick L. Shepherd and T. D. Noe, Oct 06 2005

A219612 Numbers k that divide the sum of the first k Fibonacci numbers (beginning with F(0)).

Original entry on oeis.org

1, 4, 6, 9, 11, 19, 24, 29, 31, 34, 41, 46, 48, 59, 61, 71, 72, 79, 89, 94, 96, 100, 101, 106, 109, 120, 129, 131, 139, 144, 149, 151, 166, 179, 181, 191, 192, 199, 201, 211, 214, 216, 220, 226, 229, 239, 240, 241, 249, 251, 269, 271, 274, 281, 288, 311

Offset: 1

Alex Ratushnyak, May 03 2013

Numbers k such that A000045(k+1) == 1 (mod k). - Robert Israel, Oct 13 2015

			Sum of first 6 Fibonacci numbers is 0+1+1+2+3+5 = 12. Because 6 divides 12, 6 is in the sequence.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000045, A000071, A023172, A045345, A101907.

fmod:= proc(a, b) local A, n, f1, f2, f;
  uses LinearAlgebra[Modular];
  A:= Mod(b, <<1, 1>|<1, 0>>, integer[8]);
  MatrixPower(b, M, a)[1, 2];
end proc:
1, op(select(t -> fmod(t+1,t) = 1, [$2..10^4])); # Robert Israel, Oct 13 2015

okQ[n_] := n == 1 || Mod[Fibonacci[n+1], n] == 1;
Select[Range[1000], okQ] (* Jean-François Alcover, Feb 04 2023 *)

lista(nn) = {sf = 0; for (n=0, nn, sf += fibonacci(n); if (sf % (n+1) == 0, print1(n+1, ", ")););} \\ Michel Marcus, Jun 05 2013

sum_, prpr, prev = 0, 0, 1
for i in range(1, 1000):
  sum_ += prpr
  if sum_ % i == 0:  print(i, end=', ')
  prpr, prev = prev, prpr+prev

a(n) = A101907(n) + 1. - Altug Alkan, Dec 29 2015

A140971 Numbers k such that arithmetic mean of squares of the first k nonzero Fibonacci numbers is an integer.

Original entry on oeis.org

1, 2, 3, 5, 7, 12, 13, 15, 17, 23, 24, 25, 36, 37, 43, 47, 48, 50, 53, 60, 67, 72, 73, 75, 83, 96, 97, 103, 107, 108, 110, 113, 120, 125, 127, 137, 144, 157, 163, 167, 168, 170, 173, 175, 180, 192, 193, 195, 197, 216, 223, 227, 233, 240, 257, 263, 277, 283, 288, 293

Offset: 1

Ctibor O. Zizka, Jul 27 2008

Is the root mean square RMS(F(0),...,F(k-1)) an integer for some k?

			k = 2: (A000045(1)^2+A000045(2)^2)/2 = 1.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)

Cf. A000045, A001654, A101907, A140972.

nn=300;With[{fib2=Fibonacci[Range[nn]]^2},Flatten[Position[Table[ Mean[ Take[ fib2,n]],{n,nn}],?IntegerQ]]] (* _Harvey P. Dale, May 02 2013 *)

Numbers k such that A001654(k)/k is an integer.

Edited and extended by R. J. Mathar, Aug 05 2008

A282772 Starting from F(n), minimum number, greater than 1, of consecutive Fibonacci numbers whose average is an integer.

Original entry on oeis.org

4, 2, 3, 12, 2, 13, 3, 2, 6, 5, 2, 12, 4, 2, 3, 12, 2, 24, 3, 2, 6, 24, 2, 12, 4, 2, 3, 12, 2, 5, 3, 2, 6, 13, 2, 12, 4, 2, 3, 5, 2, 24, 3, 2, 5, 24, 2, 12, 4, 2, 3, 12, 2, 24, 3, 2, 6, 24, 2, 5, 4, 2, 3, 12, 2, 24, 3, 2, 6, 5, 2, 12, 4, 2, 3, 12, 2, 24, 3, 2, 6

Offset: 0

Paolo P. Lava, Mar 03 2017

Entries are 2, 3, 4, 5, 6, 12, 13 and 24.

Periodic with period equal to 420.

			a(0) = 4 because F(0) + F(1) + F(2) + F(3) = 0 + 1 + 1 + 2 = 4 and 4/4 = 1;
a(1) = 2 because F(1) + F(2) = 1 + 1 = 2 and 2/2 = 1;
a(2) = 3 because F(2) + F(3) + F(4) = 1 + 2 + 3 = 6 and 6/3 = 2;
a(3) = 12 because F(3) + F(4) + ... + F(13) + F(14) =  2 + 3 + ... + 233 + 377 = 984 and 984/12 = 82.

Paolo P. Lava, Table of n, a(n) for n = 0..1000

Cf. A000045, A101907, A111035, A254141.

with(combinat): P:=proc(q) local a,k,n; for k from 0 to q do a:=fibonacci(k); for n from 1 to q do a:=a+fibonacci(k+n);
if type(a/(n+1),integer) then print(n+1); break; fi; od; od; end: P(10^3);

Table[k = 1; While[! IntegerQ@ Mean@ Take[#, n ;; n + k], k++]; k + 1, {n, Length@ # - 24}] &@ Fibonacci@ Range[0, 419] (* Michael De Vlieger, Mar 06 2017 *)

a(3*k + 1) = 2;
a(12*k + 2) = a(12*k + 6) = 3;
a(12*k) = 4;
a(30*k + 9) = a(30*k + 29) = a(60*k + 44) = 5;
a(60*k + 8) = a(60*k + 20) = a(60*k + 32) = a(60*k + 56) = 6;
a(60*k + 3) = a(60*k + 11) = a(60*k + 15) = a(60*k + 23) = a(60*k + 27) = a(60*k + 35) = a(60*k + 47) = a(60*k + 51) = 12;
a(420*k + 5) = a(420*k + 33) = a(420*k + 117) = a(420*k + 173) = a(420*k + 201) = a(420*k + 257) = a(420*k + 285) = a(420*k + 341) = 13;
a(420*k + x) = 24, with x = 17, 21, 41, 45, 53, 57, 65, 77, 81, 93, 101, 105, 113, 125, 137, 141, 153, 161, 165, 177, 185, 197, 213, 221, 225, 233, 237, 245, 261, 273, 281, 293, 297, 305, 317, 321, 333, 345, 353, 365, 377, 381, 393, 401, 405, 413, 417.

A266960 Integer averages of first n Fibonacci numbers (beginning with F(0)).

Original entry on oeis.org

0, 1, 2, 6, 13, 356, 3126, 28691, 70268, 271396, 6534495, 64591632, 162057126, 26237436541, 66438353080, 7020479040553, 11201604625686, 296414282891996, 32360305554728271, 339791857819043616, 871053578019254406, 5731478440138170841, 9181907843495831675

Offset: 1

Altug Alkan, Jan 07 2016

It seems only 0, 1, 2, 13 are Fibonacci numbers.
Are there other Fibonacci numbers of the form (Fibonacci(k) - 1) / (k - 1)?
2 and 13 are the prime numbers. Are there other prime numbers in this sequence?

			1 is a term because (Fibonacci(0) + Fibonacci(1) + Fibonacci(2) + Fibonacci(3)) / 4 = 4 / 4 = 1.
2 is a term because (Fibonacci(0) + Fibonacci(1) + Fibonacci(2) + Fibonacci(3) + Fibonacci(4) + Fibonacci(5)) / 6 = 12 / 6 = 2.

Cf. A101907, A219612.

Table[Mean@ Fibonacci@ Range[0, n], {n, 0, 100}] /. Rational -> Nothing (* _Michael De Vlieger, Jan 07 2016 *)
Module[{nn=100,fibs},fibs=Accumulate[Fibonacci[Range[0,nn]]];Select[ #[[1]] / #[[2]]&/@Thread[{fibs,Range[nn+1]}],IntegerQ]] (* Harvey P. Dale, Nov 15 2020 *)

m(n) = sum(k=0, n, fibonacci(k)) % (n+1);
b(n) = sum(k=0, n, fibonacci(k)) / (n+1);
for(n=0, 1e2, if(m(n)==0, print1(b(n), ", ")));

a(n) = A000071(A219612(n) + 1) / A219612(n).

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A111035 Numbers n that divide the sum of the first n nonzero Fibonacci numbers.

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A219612 Numbers k that divide the sum of the first k Fibonacci numbers (beginning with F(0)).

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A140971 Numbers k such that arithmetic mean of squares of the first k nonzero Fibonacci numbers is an integer.

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A282772 Starting from F(n), minimum number, greater than 1, of consecutive Fibonacci numbers whose average is an integer.

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A266960 Integer averages of first n Fibonacci numbers (beginning with F(0)).

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