Displaying 1-10 of 26 results found.
Indices of records of A004090(n) - n, where A004090 is the sum of digits of the Fibonacci numbers A000045.
+20
2
0, 6, 11, 16, 58, 178, 195, 273, 695, 862, 989, 2477, 2619
COMMENTS
For indices n = 1, 5, 83, 156 and 512, the value of the difference A004090(n) - n is the same as for the preceding record (see A278834). So the sequence of indices of records in the weak sense of >= would be (0, 1, 5, 6, 11, 16, 58, 83, 156, 178, 195, 273, 512, 695, 862, 989, 2477, 2619). Conjectured to be finite and complete. Indeed it appears that 0.9*n < A004090(n) < n for all sufficiently large n.
MATHEMATICA
Function[t, Flatten@ Map[First@ Position[t, #] - 1 &, Union@ Rest@ FoldList[Max, 0, t]]]@ Table[Plus @@ IntegerDigits@ Fibonacci@n - n, {n, 0, 10^4}] (* Michael De Vlieger, Dec 28 2016 *)
PROG
(PARI) m=-1; for(k=0, 1e4, sumdigits(fibonacci(k))-k>m&&print1(k", ")+m=sumdigits(fibonacci(k))-k)
CROSSREFS
Cf. A000045 (Fibonacci numbers), A004090 (their digital sums), A278834 (the record values corresponding to the indices listed here). Except for the initial 0, this appears to be a subsequence of A264935.
0, 2, 6, 8, 15, 30, 32, 40, 44, 46, 51, 57, 92
COMMENTS
For indices n = 1, 5, 83, 156 and 512, the value of the difference A004090(n) - n is the same as for the preceding record, namely (0, 0, 15, 15, 40), respectively. So the sequence of records in the weak sense of >= would be (0, 0, 0, 2, 6, 8, 15, 15, 15, 30, 32, 40, 40, 44, 46, 51, 57, 92). Conjectured to be finite and complete. Indeed it appears that 0.9*n < A004090(n) < n for all sufficiently large n.
MATHEMATICA
Union@ Rest@ FoldList[Max, 0, #] &@ Table[Plus @@ IntegerDigits@ Fibonacci@ n - n, {n, 0, 10^4}] (* Michael De Vlieger, Dec 28 2016 *)
PROG
(PARI) m=-1; for(k=0, 1e4, sumdigits(fibonacci(k))-k>m && print1(m=sumdigits(fibonacci(k))-k, ", "))
CROSSREFS
Cf. A000045 (Fibonacci numbers), A004090 (their digital sums), A278833 (the indices corresponding to the record values listed here), A264935.
1, 2, 3, 4, 5, 6, 7, 10, 12, 16, 20, 21, 35, 56, 78, 84, 97, 125, 138, 184, 189, 300, 418, 437, 550
COMMENTS
No further terms between 550 and 12000. [ R. J. Mathar, Sep 27 2009]
PROG
(PARI) isok(n) = my(f=fibonacci(n), d=digits(f)); !(vecsum(d) % #d); \\ Michel Marcus, Apr 28 2018
0, 0, 1, 1, 1, 0, -2, 3, 5, 2, 0, -6, 3, 5, -3, 8, -8, -5, -1, 5, -4, 1, 5, -5, -3, 6, 7, -2, 7, 6, 13, 0, 2, -1, -3, 0, 9, 2, -6, -4, 16, 10, -4, 2, 11, 16, 11, 10, -6, -6, 4, 22, 4, 12, 1, -3, 8, 5, -15, 15, 6, 8, 0, 2, 13, -2, -7, 8, 17, 4, 8, 25, 0, 9, -8, 10, 10, -9, -2, 21, -4, 2, 18, -15, 12, -4, 6, -10, 19, -5, 17, 23, 14, 28, 5, 4, 6, -3, 16, -2
COMMENTS
Conjectured to increase to infinity. It appears that the slope of A004090(n) is roughly 0.93, at least in the range 0..10^5. I conjecture that this sequence takes its minimum at a(2619) = -92. - M. F. Hasler, Dec 30 2016
MATHEMATICA
Table[n - Total@ IntegerDigits@ Fibonacci@ n, {n, 0, 99}] (* Michael De Vlieger, Dec 28 2016 *)
PROG
(PARI) A280185(n)=n-sumdigits(fibonacci(n)) /* To produce the b-file; can be used for searches or similar purpose, this is faster than to compute fib(n) anew for each term. */
b=-a=1; for(n=0, 1e5, write("/tmp/ A280185.txt", n" ", n-sumdigits(a=b+b=a)))
Digital root of Fibonacci(n).
+10
19
0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8
COMMENTS
Any initial pair (a(0), a(1)) of nonzero single-digit numbers enters a cycle of length 24, except for the 8 cases where 3 divides both a(0), a(1) and (a(0), a(1)) != (9, 9), which enter a cycle of length 8 and (9, 9), which is immediately periodic of period length 1. - Jonathan Vos Post, Dec 29 2005 [Corrected by Jianing Song, Apr 17 2021] First term that differs from A004090 is a(10). In general, all terms of A004090 having one digit are the same in this sequence. - Alonso del Arte, Sep 16 2012 Decimal expansion of 12484270798876404618091 / 1111111111111111111111110 = 0.0[112358437189887641562819] (periodic). - Daniel Forgues, Feb 27 2017
LINKS
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1).
FORMULA
a(n + 1) = sum of digits of (a(n) + a(n - 1)).
Periodic with period 24 = A001175(9) given by {1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9}. G.f.: x*( -1 -x -2*x^2 -3*x^3 -5*x^4 -8*x^5 -4*x^6 -3*x^7 -7*x^8 -x^9 -8*x^10 -9*x^11 -8*x^12 -8*x^13 -7*x^14 -6*x^15 -4*x^16 -x^17 -5*x^18 -6*x^19 -2*x^20 -8*x^21 -x^22 -9*x^23 ) / ( (x-1) *(1+x+x^2) *(1+x) *(1-x+x^2) *(1+x^2) *(x^4-x^2+1) *(1+x^4) *(x^8-x^4+1) ). - R. J. Mathar, Feb 08 2013
EXAMPLE
a(10) = 1 because F(10) = 55, and since 5 + 5 = 10 and 1 + 0 = 1 is the digital root of 55.
MATHEMATICA
digitalRoot[n_Integer?Positive] := FixedPoint[Plus@@IntegerDigits[#]&, n]; Table[If[n == 0, 0, digitalRoot[Fibonacci[n]]], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, May 02 2011 *) Table[NestWhile[Total[IntegerDigits[#]]&, Fibonacci[n], # > 9 &], {n, 0, 90}] (* Harvey P. Dale, May 07 2012 *) PadRight[{0}, 120, {9, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1}] (* Harvey P. Dale, Jul 20 2024 *)
PROG
(Haskell)
a030132 n = a030132_list !! n
a030132_list =
0 : 1 : map a007953 (zipWith (+) a030132_list (tail a030132_list))
AUTHOR
youngelder(AT)webtv.net (Ana)
a(n) = sum of digits of a(n-1) + sum of digits of a(n-2); a(0) = 0, a(1) = 1.
+10
16
0, 1, 1, 2, 3, 5, 8, 13, 12, 7, 10, 8, 9, 17, 17, 16, 15, 13, 10, 5, 6, 11, 8, 10, 9, 10, 10, 2, 3, 5, 8, 13, 12, 7, 10, 8, 9, 17, 17, 16, 15, 13, 10, 5, 6, 11, 8, 10, 9, 10, 10, 2, 3, 5, 8, 13, 12, 7, 10, 8, 9, 17, 17, 16, 15, 13, 10, 5, 6, 11, 8, 10, 9, 10, 10
COMMENTS
The digital sum analog (in base 10) of the Fibonacci recurrence. - Hieronymus Fischer, Jun 27 2007 a(n) and Fibonacci(n) = A000045(n) are congruent modulo 9 which implies that (a(n) mod 9) is equal to (Fibonacci(n) mod 9) = A007887(n). Thus (a(n) mod 9) is periodic with the Pisano period A001175(9)=24. - Hieronymus Fischer, Jun 27 2007 For general bases p > 2, we have the inequality 2 <= a(n) <= 2p-3 (for n > 2). Actually, a(n) <= 17 = A131319(10) for the base p=10. - Hieronymus Fischer, Jun 27 2007
FORMULA
a(n) = a(n-1) + a(n-2) - 9*(floor(a(n-1)/10) + floor(a(n-2)/10)). - Hieronymus Fischer, Jun 27 2007 a(n) = floor(a(n-1)/10) + floor(a(n-2)/10) + (a(n-1) mod 10) + (a(n-2) mod 10). - Hieronymus Fischer, Jun 27 2007 a(n) = Fibonacci(n) - 9*Sum_{k=2..n-1} Fibonacci(n-k+1)*floor(a(k)/10) where Fibonacci(n) = A000045(n). - Hieronymus Fischer, Jun 27 2007
MATHEMATICA
a[0] = 0; a[1] = 1; a[n_] := a[n] = Apply[ Plus, IntegerDigits[ a[n - 1] ]] + Apply[ Plus, IntegerDigits[ a[n - 2] ]]; Table[ a[n], {n, 0, 100} ]
nxt[{a_, b_}]:={b, Total[IntegerDigits[a]]+Total[IntegerDigits[b]]}; NestList[ nxt, {0, 1}, 80][[All, 1]] (* Harvey P. Dale, Apr 15 2018 *)
PROG
(PARI) first(n) = {n = max(n, 2); my(res = vector(n)); res[2] = 1; for(i = 3, n, res[i] = sumdigits(res[i-1]) + sumdigits(res[i-2]) ); res } \\ David A. Corneth, May 26 2021
a(n)=ds_4(a(n-1))+ds_4(a(n-2)), a(0)=0, a(1)=1; where ds_4=digital sum base 4.
+10
14
0, 1, 1, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3
COMMENTS
The digital sum analog (in base 4) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(3)=8. For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=5= A131319(4) for the base p=4. a(n) and Fib(n)= A000045(n) are congruent modulo 3 which implies that (a(n) mod 3) is equal to (Fib(n) mod 3)= A082115(n-1) (for n>0). Thus (a(n) mod 3) is periodic with the Pisano period = A001175(3)=8 too. - Hieronymus Fischer
FORMULA
a(n)=a(n-1)+a(n-2)-3*(floor(a(n-1)/4)+floor(a(n-2)/4)).
a(n)=floor(a(n-1)/4)+floor(a(n-2)/4)+(a(n-1)mod 4)+(a(n-2)mod 4).
a(n)=Fib(n)-3*sum{1<k<n, Fib(n-k+1)*floor(a(k)/4)}, where Fib(n)= A000045(n).
EXAMPLE
a(8)=3, since a(6)=5=11(base 4), ds_4(5)=2,
a(7)=4=10(base 4), ds_4(4)=1 and so a(8)=2+1.
MATHEMATICA
nxt[{a_, b_}]:={b, Total[IntegerDigits[a, 4]]+Total[IntegerDigits[b, 4]]}; NestList[ nxt, {0, 1}, 110][[All, 1]] (* Harvey P. Dale, Jul 30 2018 *)
CROSSREFS
Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131294, A131296, A131297, A131318, A131319, A131320.
Product of the digits of the n-th Fibonacci number.
+10
7
0, 1, 1, 2, 3, 5, 8, 3, 2, 12, 25, 72, 16, 18, 147, 0, 504, 315, 320, 32, 1260, 0, 49, 3360, 3456, 0, 162, 1728, 168, 720, 0, 7776, 0, 33600, 0, 30240, 0, 15680, 0, 311040, 0, 0, 326592, 435456, 0, 0, 0, 0, 0, 0, 0, 0, 0, 102060, 3951360, 24883200, 1411200
COMMENTS
Probably, the last nonzero term is a(184). - Giovanni Resta, Jul 14 2015
EXAMPLE
Fibonacci(7) = 13, thus a(7) = 1*3 = 3.
MATHEMATICA
Array[Times@@IntegerDigits@Fibonacci[#]&, 100, 0] (* Vincenzo Librandi, Jan 04 2020 *)
Numbers k such that the sum of the digits of Fibonacci(k) is k.
+10
6
0, 1, 5, 10, 31, 35, 62, 72, 175, 180, 216, 251, 252, 360, 494, 504, 540, 946, 1188, 2222
COMMENTS
Since the number of digits in the k-th Fibonacci number ~ k*log_10(Golden Ratio), theoretically this sequence is infinite, but then the average density of those digits = ~ 0.208987. - Robert G. Wilson vRobert Dawson of Saint Mary's University says it is likely that 2222 is the last term, as (assuming that the digits are equally distributed) the expected digit sum is ~ 0.9*k. - Stefan Steinerberger, Mar 12 2006 [Assuming that the average digit is (0+1+2+...+9)/10 = 9/2, the expected digit sum is ~ (9/2)*log_10((1+sqrt(5))/2)*k = 0.94044438...*k. - Jon E. Schoenfield, Aug 28 2022] Bankoff's short paper lists the first seven terms. - T. D. Noe, Mar 19 2012
REFERENCES
Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Prometheus Books, NY, 2007, page 209.
EXAMPLE
Fibonacci(10) = 55 and 5+5 = 10.
MATHEMATICA
Do[ If[ Apply[ Plus, IntegerDigits[ Fibonacci[n]]] == n, Print[n]], {n, 1, 10^5} ] (* Sven Simon *) Do[ If[ Mod[ Fibonacci[n], 9] == Mod[n, 9], If[ Plus @@ IntegerDigits[ Fibonacci[n]] == n, Print[n]]], {n, 0, 10^6}] (* Robert G. Wilson v *) Select[Range[0, 10^5], Plus @@ IntegerDigits[Fibonacci[ # ]] == # &] (* Ron Knott, Oct 30 2010 *)
PROG
(PARI) isok(n) = sumdigits(fibonacci(n)) == n; \\ Michel Marcus, Feb 18 2015
Digit sum of Lucas numbers.
+10
4
2, 1, 3, 4, 7, 2, 9, 11, 11, 13, 6, 19, 7, 8, 15, 14, 11, 16, 27, 25, 16, 23, 21, 26, 20, 28, 21, 22, 25, 29, 36, 20, 38, 40, 24, 28, 34, 26, 33, 23, 38, 34, 54, 43, 52, 41, 30, 62, 47, 46, 39, 49, 43, 47, 45, 47, 47, 58, 33, 73, 43, 53, 33, 68, 56, 70, 45, 43
EXAMPLE
15127 is the 20th Lucas number ( A000032(20)) with digit sum 16, so a(20)=16.
MATHEMATICA
Table[Total[IntegerDigits[LucasL[n]]], {n, 0, 100}] (* T. D. Noe, Oct 28 2013 *)
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