A112661 Sum of digits of sum of previous 3 terms.

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, 10, 3, 6, 10, 10, 8, 10, 10, 10, 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, 10, 3, 6, 10, 10, 8, 10, 10, 10, 3, 5, 9, 8, 4, 3, 6, 4, 4, 5, 4, 4, 4

Offset: 0

Author

Jonathan Vos Post and Andrew Carmichael Post (andrewpost(AT)gmail.com), Dec 29 2005

Keywords

• base
• easy
• nonn

Comments

Sum of digits, not iterated (i.e., not digital sum, reducing to a single digit) as we twice get a term of 10 which we do not reduce to 1. This is to tribonacci (A000073) as A030132 is to Fibonacci (A000045). This sequence has a preamble of 3 terms (1, 1, 1), then enters a cycle of length 39 (ending with 10, 10, 10).

Links

• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1).

Crossrefs

Cf. A000073, A004090, A007953, A010888, A030132.

Programs

Mathematica

a[0] = a[1] = a[2] = 1; a[n_] := a[n] = Total@ IntegerDigits[a[n-1] + a[n-2] + a[n-3]]; a /@ Range[0, 93] (* Giovanni Resta, Jun 17 2016 *)

Formula

a(n+2) = sum of digits of (a(n) + a(n-1) + a(n-2)). a(n+2) = A007953(a(n) + a(n-1) + a(n-2)).

Extensions

Data and name corrected by Giovanni Resta, Jun 17 2016