cp's OEIS Frontend

A361350 A variant of A359143 which includes the intermediate terms before digits are deleted (see Comments for precise definition).

%I A361350 #15 Mar 18 2023 08:07:58
%S A361350 11,112,1124,11248,1124816,2486,248620,4860,486018,48601827,
%T A361350 4860182736,486018273645,8601827365,860182736546,86018273654656,
%U A361350 8601827365465667,860182736546566780,601273654656670,60127365465667064,-1273545704,-127354570438,-12735457043849,-1273545704384962,-127354570438496270,1273545743849627,127354574384962777,12735457438496277791,273545743849627779
%N A361350 A variant of A359143 which includes the intermediate terms before digits are deleted (see Comments for precise definition).
%C A361350 This is essentially the same sequence as A359143 (so this too is a finite sequence), the difference being that it includes the terms before any digits are cancelled. Let S be the digit string of a(n), replacing a minus sign if present by 0.
%C A361350 Let T = S concatenated with the digit-sum of S.
%C A361350 If the leading digit of T is not present in the digit-sum of S, then a(n+1) = A359142(T), as in A359143.
%C A361350 If the leading digit of T is present in the digit-sum of S, then we add two new terms instead of one: a(n+1) = a(n) concatenated with the digit-sum of S, and a(n+2) = A359142(T), as in A359143.
%H A361350 Michael De Vlieger, <a href="/A361350/b361350.txt">Table of n, a(n) for n = 0..10000</a>
%H A361350 Michael De Vlieger, <a href="/A361350/a361350_2.png">Scatterplot of log_10(abs(a(n)))</a>, n = 1..10^3, showing negative terms in red.
%H A361350 Michael De Vlieger, <a href="/A361350/a361350_3.png">Scatterplot of log_10(abs(a(n)))</a>, n = 1..10^4, showing negative terms in red.
%H A361350 Michael De Vlieger, <a href="/A361350/a361350_4.png">Scatterplot of log_10(abs(a(n)))</a>, showing all terms, with negative terms in red.
%e A361350 The digit strings for the initial terms are:
%e A361350   11,
%e A361350   112,
%e A361350   1124,
%e A361350   11248,
%e A361350   1124816,
%e A361350   2486,
%e A361350   248620,
%e A361350   4860,
%e A361350   486018,
%e A361350   48601827,
%e A361350   4860182736,
%e A361350   486018273645,
%e A361350   8601827365,
%e A361350   860182736546,
%e A361350   86018273654656,
%e A361350   8601827365465667,
%e A361350   860182736546566780,
%e A361350   601273654656670,
%e A361350   60127365465667064,
%e A361350   01273545704,
%e A361350   0127354570438,
%e A361350   012735457043849,
%e A361350   01273545704384962,
%e A361350   0127354570438496270,
%e A361350   1273545743849627,
%e A361350   127354574384962777,
%e A361350   12735457438496277791,
%e A361350   273545743849627779, ...
%e A361350 The sequence itself is obtained by replacing the leading zeros by minus signs.
%e A361350 For example, after the term 601273654656670, we first append its digit-sum 64, getting 60127365465667064. Since the leading digit 6 is present in 64, we cancel all the 6's, getting 01273545704. The corresponding term in the sequence is -1273545704.
%t A361350 a[1] = {1, 1}; nn = 28;
%t A361350 Do[Which[ListQ[m], k = m; Clear[m],
%t A361350       FreeQ[#3, #2], Set[k, #1~Join~#3],
%t A361350       True, Set[k, #1~Join~#3];
%t A361350       Set[m, DeleteCases[#1~Join~#3, #2]]] & @@
%t A361350        {#, First[#], IntegerDigits@ Total[#]} &[a[n - 1]];
%t A361350  Set[a[n], k], {n, 2, nn}];
%t A361350 Array[(1 - 2 Boole[First[#] == 0])*FromDigits@ # &@ a[#] &, nn] (* _Michael De Vlieger_, Mar 16 2023 *)
%Y A361350 Cf. A359142, A359143.
%K A361350 sign,base,fini
%O A361350 0,1
%A A361350 _N. J. A. Sloane_, Mar 16 2023
%E A361350 More than the usual number of terms are shown in order to clarify the differences from A359143.