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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A131319 Maximal value arising in the sequence S(n) representing the digital sum analog base n of the Fibonacci recurrence.

Original entry on oeis.org

1, 2, 3, 5, 5, 9, 11, 13, 13, 17, 19, 13, 19, 25, 27, 26, 25, 33, 35, 32, 33, 34, 35, 45, 41, 49, 51, 53, 43, 34, 54, 51, 56, 56, 67, 61, 55, 73, 55, 67, 69, 81, 65, 85, 67, 82, 91, 93, 89, 97, 99, 88, 89, 105, 107, 89, 97, 97, 89, 98, 111, 121, 109, 118, 105, 129, 112
Offset: 1

Views

Author

Hieronymus Fischer, Jul 08 2007

Keywords

Comments

The respective period lengths of S(n) are given by A001175(n-1) (which is the Pisano period to n-1) for n>=2.
The inequality a(n)<=2n-3 holds for n>2.
a(n)=2n-3 infinitely often; lim sup a(n)/n=2 for n-->oo.

Examples

			a(3)=3, since the digital sum analog base 3 of the Fibonacci sequence is S(3)=0,1,1,2,3,3,2,3,3,... where the pattern {2,3,3} is the periodic part (see A131294) and so has a maximal value of 3.
a(9)=13 because the pattern base 9 is {2,3,5,8,13,13,10,7,9,8,9,9} (see A010076) where the maximal value is 13.

Crossrefs

Cf. A000032, A000045, A131318, A131320.
See A010074, A010075, A010076, A010077, A131294, A131295, A131296, A131297 for the definition of the digital sum analog of the Fibonacci recurrence(in different bases).

Formula

For n=Lucas(2m)=A000032(2m) with m>0, we have a(n)=2n-3.
a(n)=2n-A131320(n).

A010073 a(n) = sum of base-6 digits of a(n-1) + sum of base-6 digits of a(n-2); a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 8, 6, 4, 5, 9, 9, 8, 7, 5, 7, 7, 4, 6, 5, 6, 6, 2, 3, 5, 8, 8, 6, 4, 5, 9, 9, 8, 7, 5, 7, 7, 4, 6, 5, 6, 6, 2, 3, 5, 8, 8, 6, 4, 5, 9, 9, 8, 7, 5, 7, 7, 4, 6, 5, 6, 6, 2, 3, 5, 8, 8, 6, 4, 5, 9, 9, 8, 7, 5, 7, 7, 4, 6, 5, 6, 6
Offset: 0

Views

Author

N. J. A. Sloane, Leonid Broukhis

Keywords

Comments

The digital sum analog (in base 6) of the Fibonacci recurrence. - 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) <= 9 = A131319(6) for the base p=6. - Hieronymus Fischer, Jun 27 2007
a(n) and Fibonacci(n)=A000045(n) are congruent modulo 5 which implies that (a(n) mod 5) is equal to (Fibonacci(n) mod 5) = A082116(n) (for n > 0). Thus (a(n) mod 6) is periodic with the Pisano period A001175(5)=20. - Hieronymus Fischer, Jun 27 2007

Links

Crossrefs

Cf. A000045, A010074, A010075, A010076, A010077, A131294, A131295, A131296, A131297, A131318, A131319, A131320.

Programs

  • Magma
    [0] cat [n le 2 select 1 else Self(n-1)+Self(n-2)-5*((Self(n-1) div 6)+(Self(n-2) div 6)): n in [1..100]]; // Vincenzo Librandi, Jul 11 2015
  • Mathematica
    nxt[{a_,b_,c_}]:={b,c,Total[IntegerDigits[c,6]]+Total[ IntegerDigits[ b,6]]}; Transpose[NestList[nxt,{0,1,1},90]][[1]] (* Harvey P. Dale, Oct 09 2014 *)
  • PARI
    lista(nn) = {va = vector(nn); va[2] = 1; for (n=3, nn, va[n] = sumdigits(va[n-1], 6) + sumdigits(va[n-2], 6);); va;} \\ Michel Marcus, Apr 24 2018

Formula

Periodic from n=3 with period 20. - Franklin T. Adams-Watters, Mar 13 2006
a(n) = a(n-1) + a(n-2) - 5*(floor(a(n-1)/6) + floor(a(n-2)/6)). - Hieronymus Fischer, Jun 27 2007
a(n) = floor(a(n-1)/6) + floor(a(n-2)/6) + (a(n-1) mod 6) + (a(n-2) mod 6). - Hieronymus Fischer, Jun 27 2007
a(n) = (a(n-1) + a(n-2) + 5*(A010875(a(n-1)) + A010875(a(n-2))))/6. - Hieronymus Fischer, Jun 27 2007
a(n) = Fibonacci(n) - 5*Sum_{k=2..n-1} Fibonacci(n-k+1)*floor(a(k)/6). - Hieronymus Fischer, Jun 27 2007

Extensions

Incorrect comment removed by Michel Marcus, Apr 28 2018

A074867 a(n) = M(a(n-1)) + M(a(n-2)) where a(1)=a(2)=1 and M(k) is the product of the digits of k in base 10.

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 13, 11, 4, 5, 9, 14, 13, 7, 10, 7, 7, 14, 11, 5, 6, 11, 7, 8, 15, 13, 8, 11, 9, 10, 9, 9, 18, 17, 15, 12, 7, 9, 16, 15, 11, 6, 7, 13, 10, 3, 3, 6, 9, 15, 14, 9, 13, 12, 5, 7, 12, 9, 11, 10, 1, 1, 2, 3, 5, 8, 13, 11, 4, 5, 9, 14, 13, 7, 10, 7, 7, 14, 11, 5, 6, 11, 7, 8, 15, 13
Offset: 1

Views

Author

Felice Russo, Sep 11 2002

Keywords

Comments

Periodic with least period 60. - Christopher N. Swanson (cswanson(AT)ashland.edu), Jul 22 2003
From Hieronymus Fischer, Jul 01 2007: (Start)
The digital product analog (in base 10) of the Fibonacci recurrence.
a(n) and Fib(n)=A000045(n) are congruent modulo 10 which implies that (a(n) mod 10) is equal to (Fib(n) mod 10) = A003893(n). Thus (a(n) mod 10) is periodic with the Pisano period A001175(10)=60.
a(n)==A131297(n) modulo 10 (A131297(n)=digital sum analog base 11 of the Fibonacci recurrence).
For general bases p>1, we have the inequality 1<=a(n)<=2p-2 (for n>0). Actually, a(n)<=18.
(End)

Links

Crossrefs

Cf. A000045, A010073, A010074, A010075, A010076, A131294, A131295, A131296, A131297, A131318, A131319, A131320.

Programs

  • Mathematica
    nxt[{a_,b_}]:={b,Times@@IntegerDigits[a]+Times@@IntegerDigits[b]}; Transpose[ NestList[nxt,{1,1},90]][[1]] (* Harvey P. Dale, Feb 01 2015 *)

Formula

From Hieronymus Fischer, Jul 01 2007: (Start)
a(n) = a(n-1)+a(n-2)-10*(floor(a(n-1)/10)+floor(a(n-2)/10)). This is valid, since a(n)<100.
a(n) = ds_10(a(n-1))+ds_10(a(n-2))-(floor(a(n-1)/10)+floor(a(n-2)/10)) where ds_10(x) is the digital sum of x in base 10.
a(n) = (a(n-1)mod 10)+(a(n-2)mod 10) = A010879(a(n-1))+A010879(a(n-2)).
a(n) = A131297(n) if A131297(n)<=10.
a(n) = Fib(n)-10*sum{1A000045(n).
a(n) = A000045(n)-10*sum{1A000045(n-k+1)*A059995(a(k))}. (End)

Extensions

More terms from Christopher N. Swanson (cswanson(AT)ashland.edu), Jul 22 2003
Definition adapted to offset by Georg Fischer, Jun 18 2021
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Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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