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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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

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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

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).

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

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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

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

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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

Crossrefs

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

A169732 a(1) = 1000; for n>1, a(n) = a(n-1) - digitsum(a(n-1)).

Original entry on oeis.org

1000, 999, 972, 954, 936, 918, 900, 891, 873, 855, 837, 819, 801, 792, 774, 756, 738, 720, 711, 702, 693, 675, 657, 639, 621, 612, 603, 594, 576, 558, 540, 531, 522, 513, 504, 495, 477, 459, 441, 432, 423, 414, 405, 396, 378, 360, 351, 342, 333, 324, 315, 306, 297, 279, 261, 252, 243, 234, 225, 216, 207, 198, 180, 171, 162, 153, 144, 135, 126, 117, 108, 99, 81, 72, 63, 54, 45, 36, 27, 18, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 1

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Author

N. J. A. Sloane, May 01 2010, based on a suggestion from Chris Cole

Keywords

Crossrefs

Programs

  • Maple
    f:=proc(n) global S; option remember; if n=1 then RETURN(S) else RETURN(f(n-1)-digsum(f(n-1))); fi; end; S:=1000; [seq(f(n),n=1..120)];
  • Mathematica
    NestList[#-Total[IntegerDigits[#]]&,1000,100] (* Harvey P. Dale, Mar 28 2020 *)

A134268 a(n+2) = sum_digits[a(n+1)] + sum_digits[a(n)] + sum_digits[a(n+1)]*sum_digits[a(n)], with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 3, 19, 54, 109, 120, 55, 153, 109, 120, 55, 153, 109, 120, 55, 153, 109, 120, 55, 153, 109, 120, 55, 153, 109, 120, 55, 153, 109, 120, 55, 153, 109, 120, 55, 153, 109, 120, 55, 153, 109, 120, 55, 153, 109, 120, 55, 153, 109, 120, 55, 153, 109, 120, 55, 153
Offset: 0

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Author

Keywords

Comments

Period [153,109,120,55] after 6 terms. - Georg Fischer, Aug 22 2020

Crossrefs

Cf. A010077.

Programs

  • Maple
    P:=proc(n)local a0,a1,i,k,w,x; a0:=0; a1:=1; print(a0);print(a1); for i from 1 by 1 to n do w:=0; k:=a0+a1; while k>0 do w:=w+k-(trunc(k/10)*10); k:=trunc(k/10); od; x:=0; k:=a1; while k>0 do x:=x+k-(trunc(k/10)*10); k:=trunc(k/10); od; a0:=a1; a1:=w+x+w*x; print(a1); od; end: P(100);
  • Mathematica
    PadRight[{0,1,3,19,54},100,{153,109,120,55}] (* Harvey P. Dale, Aug 03 2023 *)
Previous Showing 11-15 of 15 results.