login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

If n = Sum_{i} c_i * 10^i then let b(n) = Sum_{i} c_i * (i+1). Order the integers by b(n) and then n.
1

%I #25 Jun 03 2023 10:18:06

%S 0,1,2,10,3,11,100,4,12,20,101,1000,5,13,21,102,110,1001,10000,6,14,

%T 22,30,103,111,200,1002,1010,10001,100000,7,15,23,31,104,112,120,201,

%U 1003,1011,1100,10002,10010,100001,1000000,8,16,24,32,40,105,113,121,202

%N If n = Sum_{i} c_i * 10^i then let b(n) = Sum_{i} c_i * (i+1). Order the integers by b(n) and then n.

%C This is in effect a listing of single-digit (nonnegative) solutions to b + 2c + 3d + 4e + ... = k.

%C The sequence can be considered as an irregular triangle listing partitions in which no part occurs more than 9 times. The row lengths are given by A261776. For example, in row 5 the value 102, corresponds to the partition 1+1+3 (= 2*1 + 0*2 + 1*3). - _Andrew Howroyd_, Apr 25 2023

%H Andrew Howroyd, <a href="/A061196/b061196.txt">Table of n, a(n) for n = 0..96</a>

%H Rasa Smidtaite and Minvydas Ragulskis, <a href="https://doi.org/10.3389/fphy.2022.1094240">Commentary: Multidimensional discrete chaotic maps</a>, Front. Phys. (2022) Vol. 10, 1094240.

%F For n < 10, a(A000070(n)) = n+1 and a(A026905(n)) = 10^(n-1).

%e From _Andrew Howroyd_, Apr 25 2023: (Start)

%e The sequence as a triangle T(n,k) begins:

%e 0 | 0;

%e 1 | 1;

%e 2 | 2, 10;

%e 3 | 3, 11, 100;

%e 4 | 4, 12, 20, 101, 1000;

%e 5 | 5, 13, 21, 102, 110, 1001, 10000;

%e 6 | 6, 14, 22, 30, 103, 111, 200, 1002, 1010, 10001, 100000;

%e ...

%e (End)

%t With[{k = 7}, {{0}}~Join~Values@ PositionIndex[Array[Total@ MapIndexed[#1*First[#2] &, Reverse@ IntegerDigits[#]] &, 10^k]][[1 ;; k]]] // Flatten (* _Michael De Vlieger_, Dec 22 2022, solution only suitable for generating the data field *)

%o (PARI)

%o F(p)={my(v=vector(if(#p, p[#p], 1))); for(i=1, #p, v[p[i]]++); v}

%o row(n)={my(R=[F(p) | p<-partitions(n)]); vecsort([fromdigits(Vecrev(u)) | u<-R, vecmax(u)<=9])}

%o { for(n=0, 7, print(row(n))) } \\ _Andrew Howroyd_, Apr 25 2023

%Y Cf. A000041, A028897, A261776.

%K nonn,base

%O 0,3

%A _Henry Bottomley_, Apr 20 2001