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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A112677 Sum of digits of the sum of the previous 4 terms.

Original entry on oeis.org

1, 1, 1, 1, 4, 7, 4, 7, 4, 4, 10, 7, 7, 10, 7, 4, 10, 4, 7, 7, 10, 10, 7, 7, 7, 4, 7, 7, 7, 7, 10, 4, 10, 4, 10, 10, 7, 4, 4, 7, 4, 10, 7, 10, 4, 4, 7, 7, 4, 4, 4, 10, 4, 4, 4, 4, 7, 10, 7, 10, 7, 7, 4, 10, 10, 4, 10, 7, 4, 7, 10, 10, 4, 4, 10, 10, 10, 7, 10, 10, 10, 10
Offset: 0

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Author

Jonathan Vos Post, Dec 30 2005

Keywords

Comments

This is to the tetranacci sequence as A112661 is to the tribonacci and as A030132 is to Fibonacci. A000288 is the tetranacci sequence (A000078) but starting with values (1,1,1,1). Andrew Carmichael Post (andrewpost(AT)gmail.com) wrote the program that generated this sequence and showed that for any 4 initial integers a(0),a(1),a(2),a(3) the length of the cycle eventually entered is a factor of 312. For instance, starting with (6,6,6,6) continues in a cycle of length 1 since SOD(6+6+6+6) = SOD(24) = 6; and 1 divides 312. For the SOD(tribonacci) which is A112661, the length of any cycle eventually entered is a factor of 78.
All terms for n >= 4 are 4, 7, or 10. The sequence has period 78; the 78 terms after the initial 1,1,1,1 repeat forever. - Nathaniel Johnston, May 04 2011

Examples

			a(0)=a(1)=a(2)=a(3)=1.
a(4) = SOD(1+1+1+1) = SOD(4) = 4.
a(5) = SOD(1+1+1+4) = SOD(7) = 7.
a(10) = SOD(4+7+4+4) = SOD(19) = 10, note that we do not iterate SOD to reduce 10 to 1.

Crossrefs

Cf. A000078, A000288, A004090, A007953, A010888, A030132, A112661.

Programs

  • Maple
    A112677 := proc(n) option remember: if(n<=3)then return 1:fi: return add(d,d=convert(procname(n-1) + procname(n-2) + procname(n-3) + procname(n-4),base,10)): end: seq(A112677(n),n=0..100); # Nathaniel Johnston, May 04 2011
  • Mathematica
    nxt[{a_,b_,c_,d_}]:={b,c,d,Total[IntegerDigits[a+b+c+d]]}; Transpose[ NestList[ nxt,{1,1,1,1},90]][[1]] (* or *) PadRight[{1,1,1,1},120,{10,10,10,10,4,7,4,7,4,4,10,7,7,10,7,4,10,4,7,7,10,10,7,7,7,4,7,7,7,7,10,4,10,4,10,10,7,4,4,7,4,10,7,10,4,4,7,7,4,4,4,10,4,4,4,4,7,10,7,10,7,7,4,10,10,4,10,7,4,7,10,10,4,4,10,10,10,7}](* Harvey P. Dale, Mar 05 2016 *)

Formula

a(0)=a(1)=a(2)=a(3)=1. a(n) = SumDigits(a(n-1) + a(n-2) + a(n-3) + a(n-4)).
a(n) = SumDigits(A000288(n)).
a(n) = A007953(a(n-1) + a(n-2) + a(n-3) + a(n-4)). - Nathaniel Johnston, May 04 2011

Extensions

Name corrected by Nathaniel Johnston, May 04 2011

A112678 Sum of digits of previous 5 terms.

Original entry on oeis.org

1, 1, 1, 1, 1, 5, 9, 8, 6, 11, 12, 10, 11, 5, 13, 6, 9, 8, 5, 5, 6, 6, 3, 7, 9, 4, 11, 7, 11, 6, 12, 11, 11, 6, 10, 5, 7, 12, 4, 11, 12, 10, 13, 5, 6, 10, 8, 6, 8, 11, 7, 4, 9, 12, 7, 12, 8, 12, 6, 9, 11, 10, 12, 12, 9, 9, 7, 13, 5, 7, 5, 10, 4, 4, 3, 8, 11, 3, 11, 9, 6, 4, 6, 9, 7, 5, 4, 4, 11, 4
Offset: 0

Views

Author

Jonathan Vos Post, Dec 30 2005

Keywords

Comments

This is to the pentanacci sequence A001591 as A112661 is to the tribonacci and as A030132 is to Fibonacci. A000322 is the pentanacci sequence (A001591) but starting with values (1,1,1,1,1). Andrew Carmichael Post (andrewpost(AT)gmail.com) wrote the program that generated this sequence and showed that for any 5 initial integers a(0),a(1),a(2),a(3),a(4) the length of the cycle eventually entered is a factor of 2184. For the SOD(teranacci) the limit cycle length is always a factor of 312. For the SOD(tribonacci) which is A112661, the length of any cycle eventually entered is a factor of 78.

Examples

			a(0)=a(1)=a(2)=a(3)=a(4)=1.
a(5) = SOD(1+1+1+1+1) = SOD(5) = 5.
a(6) = SOD(1+1+1+1+5) = SOD(9) = 9.
a(7) = SOD(1+1+1+5+9) = SOD(17) = 8.
a(8) = SOD(1+1+5+9+8) = SOD(24) = 6.
a(9) = SOD(1+5+9+8+6) = SOD(29) = 11, note that we do not iterate SOD to reduce 11 to 2.

Crossrefs

Cf. A000322, A001591, A004090, A007953, A010888, A030132, A112661.

Formula

a(0)=a(1)=a(2)=a(3)=a(4)=1. a(n) = SumDigits(a(n-1)+a(n-2)+a(n-3)+a(n-4)+a(n-5)). a(n) = SumDigits(A000322(n)).

A378384 Digital root of the sum of the previous 3 terms; a(0) = a(1) = a(2) = 1.

Original entry on oeis.org

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

Views

Author

James C. McMahon, Nov 24 2024

Keywords

Comments

This differs from A112661 which is sum of digits of sum of previous 3 terms.
Digital root of A000213 (tribonacci numbers beginning {1,1,1}).
This has a period of 39 beginning with the first term.
Decimal expansion of 12373315960504936995263080863765792902/111111111111111111111111111111111111111 = 0.[111359843644544432957367727773892136118] (periodic).

Links

Crossrefs

Cf. A000213, A010888, A112661.

Programs

  • Mathematica
    Nest[Append[#,ResourceFunction["AdditiveDigitalRoot"][Total[Take[#,-3]]]]&,{1,1,1},85]

Formula

a(n) = A010888(A000213(n)).
Showing 1-3 of 3 results.

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