A131294 -id:A131294 - cp's OEIS frontend

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

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

N. J. A. Sloane, Leonid Broukhis

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

David A. Corneth, Table of n, a(n) for n = 0..9999
Index entries for Colombian or self numbers and related sequences

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

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 *)

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

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

A131295 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.

Original entry on oeis.org

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

Offset: 0

Hieronymus Fischer, Jun 27 2007

The digital sum analog (in base 4) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(3)=8.
Also a(n)==A004090(n) modulo 3 (A004090(n)=digital sum of Fib(n)).
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

			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.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for Colombian or self numbers and related sequences

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

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 *)

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)=A002265(a(n-1))+A002265(a(n-2))+A010873(a(n-1))+A010873(a(n-2)).
a(n)=Fib(n)-3*sum{1A000045(n).

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

N. J. A. Sloane, Leonid Broukhis

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

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for Colombian or self numbers and related sequences

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

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 *)

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)

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

N. J. A. Sloane, Leonid Broukhis

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

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for Colombian or self numbers and related sequences

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

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 *)

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)

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

Hieronymus Fischer, Jun 27 2007

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.

			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.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for Colombian or self numbers and related sequences

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

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

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).

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

Hieronymus Fischer, Jun 27 2007

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.

			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.

Index entries for Colombian or self numbers and related sequences

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

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 *)

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).

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

Hieronymus Fischer, Jul 08 2007

The inequality a(n)>=3 holds for n>2.

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

			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.

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).

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

N. J. A. Sloane, Leonid Broukhis

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

Index entries for Colombian or self numbers and related sequences
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,1).

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

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

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)

Incorrect comment removed by Michel Marcus, Apr 29 2018

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

Hieronymus Fischer, Jul 03

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

			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.

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).

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

Hieronymus Fischer, Jul 08 2007

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.

			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.

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).

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

a(n)=2n-A131320(n).

cp's OEIS Frontend

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

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A131295 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.

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

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

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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.

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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.

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A131320 2*n - maximal value arising in the sequence S(n) representing the digital sum analog base n of the Fibonacci recurrence.

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

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