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

%I

%S 1,2,4,8,16,32,64,128,256,512,1024,1049,1099,1199,1399,1799,2599,3200,

%T 3403,3809,4621,5246,5497,5999,7003,7013,7033,7073,7153,7313,7633,

%U 8273,8554,9116,9241,9491,9991,10991,11992,12995,14002,14018,14050,14114,14242,14498

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

%C In Germany you just write Q3 for the base-1000 digit sum (see book: "Taschenbuch der Mathematik" by Bronstein, Semendjajew, Musiol, MÃ¼hlig, 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.

%C 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.

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

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

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

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

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

%e 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.

%t NestList[Total[IntegerDigits[#,1000]]+#&,1,50] (* _Harvey P. Dale_, Dec 12 2018 *)

%o (PARI) a(n) = if (n==0, 1, prev = a(n-1); prev + sumdigits(prev, 1000)); \\ _Michel Marcus_, Sep 20 2017

%Y Cf. A004207, A286660.

%K nonn,base

%O 0,2

%A _Peter Weiss_, Sep 19 2017

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Last modified September 24 08:31 EDT 2021. Contains 347623 sequences. (Running on oeis4.)