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%I A169920 #16 Oct 21 2022 14:24:52
%S A169920 0,1,4,9,16,25,36,49,64,81,100,111,144,199,306,455,646,179,424,731,
%T A169920 400,441,564,769,1126,505,606,829,1124,1481,900,991,12640,17190,1066,
%U A169920 1355,1086,1259,1304,1651,1000,3440,6120,1749,2126,2605,2886,3569,3864,4841
%N A169920 a(n) = n*n in the arithmetic where digits are multiplied in base 10 (as usual) but when digits are to be added they are also multiplied in base 10.
%C A169920 How should the carry digits be handled? In this version they have been handled by simply adding them in the old way, which is a bit worrisome. For example, in the calculation below, when the column containing 5 and 4 is "added", i.e. multiplied, there is a carry of 2, which here has been added to the 1 to get 3.
%H A169920 David Consiglio, Jr., <a href="/A169920/b169920.txt">Table of n, a(n) for n = 0..10000</a>
%e A169920 a(14) = 14*14 = 306:
%e A169920 ....14
%e A169920 ....14
%e A169920 ------
%e A169920 ....56
%e A169920 ...14.
%e A169920 ------
%e A169920 ...306
%e A169920 ------
%o A169920 (Python)
%o A169920 from math import prod
%o A169920 def A169920(m):
%o A169920     n = len(str(m*m))+1
%o A169920     hold = list(zip(*[list(str(int(b)*m).ljust(n-1-a,"X").rjust(n-1,"X")) for a,b in enumerate(str(m))]))#List of products of long multiplication
%o A169920     store = []
%o A169920     for a,c in enumerate(hold):
%o A169920         if c.count('X') == len(c):
%o A169920             store.append(0)
%o A169920         else:
%o A169920             store.append(prod([int(b) for b in c if b.isdigit()])*10**(len(hold)-a-1))
%o A169920     return(sum(store))
%o A169920 # _David Consiglio, Jr._, Oct 21 2022
%Y A169920 The four versions are A000290, A169919, A169920, A169921.
%K A169920 nonn,base
%O A169920 0,3
%A A169920 _David Applegate_, _Marc LeBrun_ and _N. J. A. Sloane_, Jul 20 2010
%E A169920 More terms from _David Consiglio, Jr._, Oct 21 2022

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