A292568 a(n) = a(n-1) + sum of base-1000 digits of a(n-1), a(0)=1.

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 1049, 1099, 1199, 1399, 1799, 2599, 3200, 3403, 3809, 4621, 5246, 5497, 5999, 7003, 7013, 7033, 7073, 7153, 7313, 7633, 8273, 8554, 9116, 9241, 9491, 9991, 10991, 11992, 12995, 14002, 14018, 14050, 14114, 14242, 14498

Offset: 0

Author

Peter Weiss, Sep 19 2017

Keywords

• nonn
• base

Comments

In Germany you just write Q3 for the base-1000 digit sum (see book: "Taschenbuch der Mathematik" by Bronstein, Semendjajew, Musiol, Mühlig, p. 332) and you need it for the so-called "Teilbarkeitskriterium" for the number 37. If you add Q3 to a number you can also find this rule for the number 37.

Sum of base-1000 digits of m can also be described as "break the digit-string of m into triples starting at the right, and add these 3-digit numbers". For example, 1234567 -> 567 + 234 + (00)1 = 802.

None of the numbers of this sequence is divisible by 3 or 37.

The general form of this sequence is n + sum of base-(10^m) digits of n.

m=1: 1, 2, 4, 8, 16, 23, 28, 38, 49, 62, 70, 77, 91, 101, 103, ... (Cf. A004207.)

m=2: 1, 2, 4, 8, 16, 32, 64, 128, 157, 215, 232, 266, 334, 371, ... (Cf. A286660.)

m=3: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 1049, 1099, ... (this sequence)

Examples

			a(16) = 2599 = 2 * 1000^1 + 599 * 1000^0. The sum of digits of a(17 - 1) = 2599 in base 1000 is therefore 2 + 599 = 601. a(17) = a(16) + the sum of digits of a(60) in base 1000 is therefore 2599 + 601 = 3200.
		

Crossrefs

Cf. A004207, A286660.

Programs

Mathematica

NestList[Total[IntegerDigits[#,1000]]+#&,1,50] (* Harvey P. Dale, Dec 12 2018 *)

PARI

a(n) = if (n==0, 1, prev = a(n-1); prev + sumdigits(prev, 1000)); \\ Michel Marcus, Sep 20 2017