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.

Showing 1-10 of 13 results. Next

A010077 a(n) = sum of digits of a(n-1) + sum of digits of a(n-2); a(0) = 0, a(1) = 1.

Original entry on oeis.org

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
Offset: 0

Views

Author

N. J. A. Sloane, Leonid Broukhis

Keywords

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
a(n) == A004090(n) (mod 9) (A004090(n) = digital sum of Fibonacci(n)). - 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

Links

Crossrefs

Cf. A007953, A007612, A065076.
Cf. A000045, A010073, A010074, A010075, A010076.
Cf. A131294, A131295, A131296, A131297, A131318, A131319, A131320.

Programs

  • 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 *)
  • 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

Formula

Periodic from n=3 with period 24. - Franklin T. Adams-Watters, Mar 13 2006
a(n) = A030132(n-4) + A030132(n-3) for n>3. - Reinhard Zumkeller, Jul 04 2007
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) = A059995(a(n-1)) + A059995(a(n-2)) + A010879(a(n-1)) + A010879(a(n-2)). - 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

A131294 a(n)=ds_3(a(n-1))+ds_3(a(n-2)), a(0)=0, a(1)=1; where ds_3=digital sum base 3.

Original entry on oeis.org

0, 1, 1, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3
Offset: 0

Views

Author

Hieronymus Fischer, Jun 27 2007

Keywords

Comments

The digital sum analog (in base 3) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(2)=3.
a(n) and Fib(n)=A000045(n) are congruent modulo 2 which implies that (a(n) mod 2) is equal to (Fib(n) mod 2)=A011655(n). Thus (a(n) mod 2) is periodic with the Pisano period A001175(2)=3 too.
For general bases p>2, we have the inequality 2<=a(n)<=2p-3 (for n>2). Actually, a(n)<=3=A131319(3) for the base p=3.

Examples

			a(5)=3, since a(3)=2, ds_3(2)=2, a(4)=3=10(base 3),
ds_3(3)=1 and so a(5)=2+1.

Links

Crossrefs

Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131295, A131296, A131297, A131318, A131319, A131320.

Programs

  • Mathematica
    nxt[{a_,b_}]:={b,Total[IntegerDigits[a,3]]+Total[IntegerDigits[b,3]]}; Transpose[NestList[nxt,{0,1},100]][[1]] (* Harvey P. Dale, Aug 02 2016 *)

Formula

a(n) = a(n-1)+a(n-2)-2*(floor(a(n-1)/3)+floor(a(n-2)/3)).
a(n) = floor(a(n-1)/3)+floor(a(n-2)/3)+(a(n-1)mod 3)+(a(n-2)mod 3).
a(n) = A002264(a(n-1))+A002264(a(n-2))+A010872(a(n-1))+A010872(a(n-2)).
a(n) = Fib(n)-2*sum{1A000045(n).

Extensions

Incorrect comment removed by Michel Marcus, Apr 29 2018

A010074 a(n) = sum of base-7 digits of a(n-1) + sum of base-7 digits of a(n-2).

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 7, 3, 4, 7, 5, 6, 11, 11, 10, 9, 7, 4, 5, 9, 8, 5, 7, 6, 7, 7, 2, 3, 5, 8, 7, 3, 4, 7, 5, 6, 11, 11, 10, 9, 7, 4, 5, 9, 8, 5, 7, 6, 7, 7, 2, 3, 5, 8, 7, 3, 4, 7, 5, 6, 11, 11, 10, 9, 7, 4, 5, 9, 8, 5, 7, 6, 7, 7
Offset: 0

Views

Author

N. J. A. Sloane, Leonid Broukhis

Keywords

Comments

The digital sum analog (in base 7) of the Fibonacci recurrence. - Hieronymus Fischer, Jun 27 2007
a(n) and Fib(n)=A000045(n) are congruent modulo 6 which implies that (a(n) mod 6) is equal to (Fib(n) mod 6) = A082117(n-1) (for n>0). Thus (a(n) mod 6) is periodic with the Pisano period A001175(6)=24. - Hieronymus Fischer, Jun 27 2007
For general bases p>2, the inequality 2<=a(n)<=2p-3 holds (for n>2). Actually, a(n)<=11=A131319(7) for the base p=7. - Hieronymus Fischer, Jun 27 2007

Links

Crossrefs

Cf. A000045, A010073, A010075, A010076, A010077, A131294, A131295, A131296, A131297, A131318, A131319, A131320.

Programs

  • Mathematica
    nxt[{a_,b_}]:={b,Total[IntegerDigits[a,7]]+Total[IntegerDigits[b,7]]}; Transpose[NestList[nxt,{0,1},80]][[1]] (* Harvey P. Dale, Oct 12 2013 *)

Formula

Periodic from n=3 with period 24. - Franklin T. Adams-Watters, Mar 13 2006
From Hieronymus Fischer, Jun 27 2007: (Start)
a(n) = a(n-1)+a(n-2)-6*(floor(a(n-1)/7)+floor(a(n-2)/7)).
a(n) = floor(a(n-1)/7)+floor(a(n-2)/7)+(a(n-1)mod 7)+(a(n-2)mod 7).
a(n) = (a(n-1)+a(n-2)+6*(A010876(a(n-1))+A010876(a(n-2))))/7.
a(n) = Fib(n)-6*sum{1A000045(n). (End)

Extensions

Incorrect comment removed by Michel Marcus, Apr 29 2018

A010075 a(n) = sum of base-8 digits of a(n-1) + sum of base-8 digits of a(n-2).

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8
Offset: 0

Views

Author

N. J. A. Sloane, Leonid Broukhis

Keywords

Comments

The digital sum analog (in base 8) of the Fibonacci recurrence. - Hieronymus Fischer, Jun 27 2007
a(n) and Fib(n)=A000045(n) are congruent modulo 7 which implies that (a(n) mod 7) is equal to (Fib(n) mod 7). Thus (a(n) mod 7) is periodic with the Pisano period A001175(7)=16. - Hieronymus Fischer, Jun 27 2007
For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=11=A131319(8) for the base p=8. - Hieronymus Fischer, Jun 27 2007

Links

Crossrefs

Cf. A000045, A010073, A010074, A010076, A010077, A131294, A131295, A131296, A131297, A131318, A131319, A131320.

Programs

  • Mathematica
    nxt[{a_,b_}]:={b,Total[IntegerDigits[a,8]]+Total[IntegerDigits[b,8]]}; NestList[ nxt,{0,1},100][[All,1]] (* or *) PadRight[{0,1,1},100,{7,8,8,2,3,5,8,6,7,13,13,12,11,9,6,8}] (* Harvey P. Dale, Apr 19 2020 *)

Formula

Periodic from n=3 with period 16. - Franklin T. Adams-Watters, Mar 13 2006
From Hieronymus Fischer, Jun 27 2007: (Start)
a(n) = a(n-1)+a(n-2)-7*(floor(a(n-1)/8)+floor(a(n-2)/8)).
a(n) = floor(a(n-1)/8)+floor(a(n-2)/8)+(a(n-1)mod 8)+(a(n-2)mod 8).
a(n) = (a(n-1)+a(n-2)+7*(A010877(a(n-1))+A010877(a(n-2))))/8.
a(n) = Fib(n)-7*sum{1A000045(n). (End)

Extensions

Incorrect comment removed by Michel Marcus, Apr 29 2018

A131296 a(n) = ds_5(a(n-1))+ds_5(a(n-2)), a(0)=0, a(1)=1; where ds_5=digital sum base 5.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3
Offset: 0

Views

Author

Hieronymus Fischer, Jun 27 2007

Keywords

Comments

The digital sum analog (in base 5) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(4)=6.
a(n) and Fib(n)=A000045(n) are congruent modulo 4 which implies that (a(n) mod 4) is equal to (Fib(n) mod 4)=A079343(n). Thus (a(n) mod 4) is periodic with the Pisano period A001175(4)=6 too.
For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=5=A131319(5) for the base p=5.

Examples

			a(10)=3, since a(8)=5=10(base 5), ds_5(5)=1,
a(9)=2, ds_5(2)=2 and so a(10)=1+2.

Links

Crossrefs

Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131294, A131295, A131297, A131318, A131319, A131320.

Programs

  • Mathematica
    nxt[{a_,b_}]:={b,Total[IntegerDigits[a,5]]+Total[IntegerDigits[b,5]]}; NestList[nxt,{0,1},100][[;;,1]] (* Harvey P. Dale, Sep 01 2024 *)

Formula

a(n) = a(n-1)+a(n-2)-4*(floor(a(n-1)/5)+floor(a(n-2)/5)).
a(n) = floor(a(n-1)/5)+floor(a(n-2)/5)+(a(n-1)mod 5)+(a(n-2)mod 5).
a(n) = A002266(a(n-1))+A002266(a(n-2))+A010874(a(n-1))+A010874(a(n-2)).
a(n) = Fib(n)-4*sum{1A000045(n).

Extensions

Incorrect comment removed by Michel Marcus, Apr 29 2018

A131297 a(n) = ds_11(a(n-1))+ds_11(a(n-2)), a(0)=0, a(1)=1; where ds_11=digital sum base 11.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 13, 11, 4, 5, 9, 14, 13, 7, 10, 17, 17, 14, 11, 5, 6, 11, 7, 8, 15, 13, 8, 11, 9, 10, 19, 19, 18, 17, 15, 12, 7, 9, 16, 15, 11, 6, 7, 13, 10, 13, 13, 6, 9, 15, 14, 9, 13, 12, 5, 7, 12, 9, 11, 10, 11, 11, 2, 3, 5, 8, 13, 11, 4, 5, 9, 14, 13, 7, 10, 17, 17, 14, 11
Offset: 0

Views

Author

Hieronymus Fischer, Jun 27 2007

Keywords

Comments

The digital sum analog (in base 11) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(10)=60.
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 too.
a(n)==A074867(n) modulo 10 (A074867(n)=digital product analog base 10 of the Fibonacci recurrence).
For general bases p>2, we have the inequality 2<=a(n)<=2p-3 (for n>2). Actually, a(n)<=19=A131319(11) for the base p=11.

Examples

			a(10)=5, since a(8)=11=10(base 11), ds_11(11)=1,
a(9)=4, ds_11(4)=4 and so a(10)=1+4.

Links

Crossrefs

Cf. A131297 A000045, A010073, A010074, A010075, A010076, A010077, A131294, A131295, A131296, A131318, A131319, A131320.

Programs

  • Mathematica
    nxt[{a_,b_}]:={b,Total[IntegerDigits[a,11]]+Total[IntegerDigits[b,11]]}; NestList[nxt,{0,1},80][[All,1]] (* or *) PadRight[{0,1,1},80,{10,11,11,2,3,5,8,13,11,4,5,9,14,13,7,10,17,17,14,11,5,6,11,7,8,15,13,8,11,9,10,19,19,18,17,15,12,7,9,16,15,11,6,7,13,10,13,13,6,9,15,14,9,13,12,5,7,12,9,11}] (* Harvey P. Dale, Jul 24 2017 *)

Formula

a(n) = a(n-1)+a(n-2)-10*(floor(a(n-1)/11)+floor(a(n-2)/11)).
a(n) = floor(a(n-1)/11)+floor(a(n-2)/11)+(a(n-1)mod 11)+(a(n-2)mod 11).
a(n) = Fib(n)-10*sum{1A000045(n).

Extensions

Incorrect comment removed by Michel Marcus, Apr 29 2018

A131320 2*n - 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, 3, 5, 3, 3, 3, 5, 3, 3, 11, 7, 3, 3, 6, 9, 3, 3, 8, 9, 10, 11, 3, 9, 3, 3, 3, 15, 26, 8, 13, 10, 12, 3, 11, 19, 3, 23, 13, 13, 3, 21, 3, 23, 10, 3, 3, 9, 3, 3, 16, 17, 3, 3, 23, 17, 19, 29, 22, 11, 3, 17, 10, 25, 3, 22, 3, 35, 30, 11, 29, 57, 3, 3, 17, 65, 16, 13, 20, 21, 3, 3
Offset: 1

Views

Author

Hieronymus Fischer, Jul 08 2007

Keywords

Comments

The inequality a(n)>=3 holds for n>2.
a(n)=3 arises infinitely often; lim inf a(n)=3 for n-->oo.

Examples

			a(3)=3, since the digital sum analog base 3 of the Fibonacci sequence is 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 which implies 2*3-3=3. a(9)=5, because the pattern here is {2,3,5,8,13,13,10,7,9,8,9,9} (see A010076) where the maximal value is 13 and so 2*9-13=5.

Crossrefs

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

Formula

a(n)=2n-A131319(n).
a(Lucas(2n))=3 where Lucas(n)=A000032(n).

A010076 a(n) = sum of base-9 digits of a(n-1) + sum of base-9 digits of a(n-2).

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 13, 13, 10, 7, 9, 8, 9, 9, 2, 3, 5, 8, 13, 13, 10, 7, 9, 8, 9, 9, 2, 3, 5, 8, 13, 13, 10, 7, 9, 8, 9, 9, 2, 3, 5, 8, 13, 13, 10, 7, 9, 8, 9, 9, 2, 3, 5, 8, 13, 13, 10, 7, 9, 8, 9, 9, 2, 3, 5, 8, 13, 13, 10, 7, 9, 8, 9, 9
Offset: 0

Views

Author

N. J. A. Sloane, Leonid Broukhis

Keywords

Comments

The digital sum analog (in base 9) of the Fibonacci recurrence. - Hieronymus Fischer, Jun 27 2007
a(n) and Fib(n)=A000045(n) are congruent modulo 8 which implies that (a(n) mod 8) is equal to (Fib(n) mod 8) = A079344(n). Thus (a(n) mod 8) is periodic with the Pisano period A001175(8)=12. - 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)<=13=A131319(9) for the base p=9. - Hieronymus Fischer, Jun 27 2007

Links

Crossrefs

Cf. A000045, A001175, A010073, A010074, A010075, A010077, A010878, A079344, A131294, A131295, A131296, A131297, A131318, A131319, A131320.

Programs

  • Mathematica
    PadRight[{0, 1, 1}, 100, {8, 9, 9, 2, 3, 5, 8, 13, 13, 10, 7, 9}] (* Paolo Xausa, Aug 25 2024 *)

Formula

Periodic from n=3 with period 12. - Franklin T. Adams-Watters, Mar 13 2006
From Hieronymus Fischer, Jun 27 2007: (Start)
a(n) = a(n-1)+a(n-2)-8*(floor(a(n-1)/9)+floor(a(n-2)/9)).
a(n) = floor(a(n-1)/9)+floor(a(n-2)/9)+(a(n-1)mod 9)+(a(n-2)mod 9).
a(n) = (a(n-1)+a(n-2)+8*(A010878(a(n-1))+A010878(a(n-2))))/9.
a(n) = Fib(n)-8*sum{1A000045(n). (End)

Extensions

Incorrect comment removed by Michel Marcus, Apr 29 2018

A082115 Fibonacci sequence (mod 3).

Original entry on oeis.org

0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2
Offset: 0

Views

Author

Eric W. Weisstein, Apr 03 2003

Keywords

Links

Crossrefs

Cf. A011655, A079343, A082116, A082117, A105870, A079344.

Programs

Formula

Sequence is periodic with Pisano period 8.
a(n) = 1-floor(n/8)+floor((n-1)/8)+floor((n-3)/8)-2*floor((n-4)/8) +2*floor((n-5)/8)-floor((n-7)/8). - Hieronymus Fischer, Jul 01 2007
a(n) = 1+((n mod 8)+((n+1)mod 8)-2*((n+3)mod 8)+2*((n+4)mod 8) -((n+5)mod 8) -((n+7)mod 8))/8. - Hieronymus Fischer, Jul 01 2007
G.f.: (x+x^2+2x^3+2x^5+2x^6+x^7)/(1-x^8). - Hieronymus Fischer, Jul 01 2007
a(n) = A131295(n) mod 3 (for n>0). - Hieronymus Fischer, Jul 01 2007

Extensions

Added a(0)=0. - Jon Perry, Sep 15 2013

A131318 Sum of terms within one periodic pattern of that sequence representing the digital sum analog base n of the Fibonacci recurrence.

Original entry on oeis.org

1, 2, 8, 30, 24, 120, 156, 126, 96, 234, 640, 88, 264, 416, 700, 630, 352, 680, 468, 304, 1200, 294, 572, 1150, 528, 2600, 2288, 1998, 1176, 290, 3660, 806, 1344, 1122, 1360, 2870, 792, 2960, 532, 2262, 2400, 1722, 1764, 3870, 1056, 5490, 2300, 1598
Offset: 1

Views

Author

Hieronymus Fischer, Jul 03

Keywords

Comments

The respective period lengths are given by A001175(n-1) (which is the Pisano period to n-1) for n>=2.

Examples

			a(3)=8 since the digital sum analog base 3 of the Fibonacci sequence is 0,1,1,2,3,3,2,3,3,... where the pattern {2,3,3} is the periodic part (see A131294) and sums up to 2+3+3=8. a(4)=30 because the pattern base 4 is {2,3,5,5,4,3,4,4} (see A131295) which sums to 30.

Crossrefs

Cf. A000045, A131319, A131320.
See A010073, A010074, A010075, A010076, A010077, A131294, A131295, A131296, A131297 for the definition of the digital sum analog of the Fibonacci sequence (in different bases).
Showing 1-10 of 13 results. Next

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