A348408 -id:A348408 - cp's OEIS frontend

A348433 a(1) = 1; a(n+1) = 2*a(n) if the digit sum of a(n) is already in the sequence, otherwise a(n+1) = digitsum(a(n)).

1, 2, 4, 8, 16, 7, 14, 5, 10, 20, 40, 80, 160, 320, 640, 1280, 11, 22, 44, 88, 176, 352, 704, 1408, 13, 26, 52, 104, 208, 416, 832, 1664, 17, 34, 68, 136, 272, 544, 1088, 2176, 4352, 8704, 19, 38, 76, 152, 304, 608, 1216, 2432, 4864, 9728, 19456, 25, 50, 100, 200

Offset: 1

Rodolfo Kurchan, Oct 18 2021

There are no multiples of 3 in this sequence.

Will all other positive integers appear in this sequence?

			The sum of the digits of a(4) = 8 is 8, which is already in the sequence, so a(5) = 2*8 = 16.
The sum of the digits of a(5) = 16 is 7, which is not yet in the sequence, so a(6) = 7.
From _Omar E. Pol_, Oct 19 2021: (Start)
Written as an irregular triangle the sequence begins (see A348408):
   1,  2,  4,   8,  16;
   7, 14;
   5, 10, 20,  40,  80, 160,  320,  640, 1280;
  11, 22, 44,  88, 176, 352,  704, 1408;
  13, 26, 52, 104, 208, 416,  832, 1664;
  17, 34, 68, 136, 272, 544, 1088, 2176, 4352, 8704;
  19, 38, 76, 152, 304, 608, 1216, 2432, 4864, 9728, 19456;
... (End)

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program for A348433

Cf. A001651, A007953, A008585, A331440 (similar), A348400, A348408, A348483.

seq[len_] := Module[{s = {1}, k, d}, While[Length[s] < len, k = s[[-1]]; If[MemberQ[s, (d = Plus @@ IntegerDigits[k])], AppendTo[s, 2*k], AppendTo[s, d]]]; s]; seq[50] (* Amiram Eldar, Oct 19 2021 *)

lista(nn) = my(s, v=List([1])); for(n=1, nn, if(setsearch(vecsort(v), s=sumdigits(v[n])), listput(v, 2*v[n]), listput(v, s))); v \\ Jinyuan Wang, Oct 21 2021

PARI
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Definition and examples clarified by N. J. A. Sloane, Oct 24 2021

A348624 a(n) = sum of row n of A348433 expressed as an irregular triangle.

Original entry on oeis.org

31, 21, 2555, 2805, 3315, 17391, 38893, 104857575, 59363, 2097120, 31713, 376809, 117440484, 18790481885, 197132241, 2885681109, 42991575, 4966055899, 13153337295, 3959422917, 120946279055305, 4191888080835, 3729543441416139, 321057395310519, 84662395338675, 294669116243901

Offset: 1

Michael De Vlieger, Oct 25 2021

The binary expansion w of a(n) has an interesting appearance shown by the bitmap in links. We may divide w with length m into 3 parts: the most significant part includes all bits including the last 0 before the middle of the word, m/2, a central run of k 1's that includes all but the last 1 before a 0, and a least significant part that includes the last 1 in the central run of 1s and an assortment of 0's. For example, a(3) = 2555 -> 100.11111.1011, which we may partition as shown by "." so as to preserve the otherwise-leading 0 in the last part. The central run of 1s generally increases in length as n increases.

			Table showing the first 5 rows of A348433 each having A348408(n) terms, and their sum a(n):
n\k 1  2  3   4   5   6   7    8    9     a(n)   binary(a(n))
--------------------------------------------------------------
1:  1  2  4   8  16                   ->   31 ->        11111
2:  7 14                              ->   21 ->        10101
3:  5 10 20  40  80 160 320  640 1280 -> 2555 -> 100111111011
4: 11 22 44  88 176 352 704 1408      -> 2805 -> 101011110101
5: 13 26 52 104 208 416 832 1664      -> 3315 -> 110011110011

Michael De Vlieger, Table of n, a(n) for n = 1..1540
Michael De Vlieger, Log log scatterplot of a(n) for n=1..120.
Michael De Vlieger, Log log scatterplot of a(n) for n=1..3961.
Michael De Vlieger, Big-endian bitmap of the binary expansion of a(n) for n=1..3961 showing 1s in black and 0s in white.

Cf. A348408, A348433.

c[1] = m = q = 1; Most@ Reap[Do[If[IntegerQ[c[#]], Set[n, 2 m], Set[n, #]] &@ Total@ IntegerDigits[m]; If[m > n, Sow[q]; Set[q, n], q += n]; Set[c[n], 1]; m = n, 650]][[-1, -1]]
(* Extract up to 3961 terms from bitmap: *)
Block[{s = ImageData[ColorNegate@ Import["https://oeis.org/A348624/a348624_2.png"], "Bit"]}, Array[FromDigits[s[[#]], 2] &, 26]] (* Michael De Vlieger, Oct 26 2021 *)

cp's OEIS Frontend

A348433 a(1) = 1; a(n+1) = 2*a(n) if the digit sum of a(n) is already in the sequence, otherwise a(n+1) = digitsum(a(n)).

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