A036066 The summarize Lucas sequence: summarize the previous two terms, start with 1, 3.

1, 3, 1311, 2331, 331241, 14432231, 34433241, 54533231, 2544632221, 163534435221, 263544436231, 363554634231, 463554733221, 17364544733221, 37263554634231, 37363554734231, 37364544933221, 1937263554933221, 3927263544835231, 391827264534836231, 293827363544836231

Offset: 0

Author

Floor van Lamoen

Keywords

• base
• easy
• nice
• nonn

Comments

After the 26th term the sequence goes into a cycle of 46 terms.

"Summarize" uses here method C = A244112: in order of decreasing digit value.

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000
• Index to sequences related to say what you see

Crossrefs

Cf. A036059.

Cf. A244112 (summarizing as used here: by decreasing digit value), A047842 (alternative summarizing method: by increasing digit value), A047843 (another method: don't omit missing digits between smallest and largest one).

Programs

Maple

a:= proc(n) option remember; `if`(n<2, 2*n+1, (p-> parse(cat(seq((c->
     `if`(c=0, [][], [c, 9-i][]))(coeff(p, x, 9-i)), i=0..9))))(
      add(x^i, i=map(x-> convert(x, base, 10)[], [a(n-1),a(n-2)]))))
    end:
seq(a(n), n=0..20);  # Alois P. Heinz, Jun 18 2022

Mathematica

a[0] = 1; a[1] = 3; a[n_] := a[n] = FromDigits @ Flatten @ Reverse @ Select[ Transpose @ { DigitCount[a[n-1]] + DigitCount[a[n-2]], Append[ Range[9], 0]}, #[[1]] > 0 &];
Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Dec 30 2017 *)

PARI

{a=[1,3]; for(n=1,50,a=concat(a,A244112(eval(Str(a[n],a[n+1]))))); a} \\ M. F. Hasler, Feb 25 2018

Formula

a(n+1) = A244112(concat(a(n),a(n-1))). - M. F. Hasler, Feb 25 2018