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%I #14 Nov 07 2019 14:39:27
%S 1,1,1,1,2,1,1,1,1,1,1,2,1,2,1,1,1,1,1,1,1,1,2,3,2,3,2,1,1,1,3,3,3,3,
%T 1,1,1,2,1,6,3,6,1,2,1,1,1,1,1,3,3,1,1,1,1,1,2,1,2,1,6,1,2,1,2,1,1,1,
%U 1,1,1,1,1,1,1,1,1,1,1,2,3,2,3,2,1,2,3,2,3,2,1,1,1,3,3,3,3,1,1,3,3,3,3,1,1
%N Square array read by antidiagonals: A(n, k) = A327936(A276086(n) * A276086(k)).
%C Array A(n, k) is symmetric, and is read as (n,k) = (0, 0), (1, 0), (0, 1), (2, 0), (1, 1), (0, 2), (3, 0), (2, 1), (1, 2), (0, 3), ...
%C If A(n, k) is 1, it tells that adding of n and k do not generate any carries, when done in primorial base (A049345). If A(n, k) is larger than one, then its prime factors indicate in which specific moduli (digit positions) the sum was larger than allowed for that position.
%H Antti Karttunen, <a href="/A329041/b329041.txt">Table of n, a(n) for n = 0..10439; the first 144 antidiagonals of the array</a>
%H <a href="/index/Pri#primorialbase">Index entries for sequences related to primorial base</a>
%F A(n, k) = A327936(A276086(n) * A276086(k)).
%F For all n, A(A328841(n), A328842(n)) = 1 and A(A328770(n), A328770(n)) = 1.
%e The top left corner of the array:
%e 0 1 2 3 4 5 6 7 8 9 10 11 12
%e +--------------------------------------
%e 0: | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%e 1: | 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, ...
%e 2: | 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 3, 3, 1, ...
%e 3: | 1, 2, 1, 2, 3, 6, 1, 2, 1, 2, 3, 6, 1, ...
%e 4: | 1, 1, 3, 3, 3, 3, 1, 1, 3, 3, 3, 3, 1, ...
%e 5: | 1, 2, 3, 6, 3, 6, 1, 2, 3, 6, 3, 6, 1, ...
%e 6: | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%e 7: | 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, ...
%e 8: | 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 3, 3, 1, ...
%e 9: | 1, 2, 1, 2, 3, 6, 1, 2, 1, 2, 3, 6, 1, ...
%e 10: | 1, 1, 3, 3, 3, 3, 1, 1, 3, 3, 3, 3, 1, ...
%e 11: | 1, 2, 3, 6, 3, 6, 1, 2, 3, 6, 3, 6, 1, ...
%e 12: | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%e ...
%e A(11,25) = A(25,11) = 10 because 11 is written in primorial base representation (A049345) as "121" and 25 as "401", and when these are added together digit by digit, we see that the maximal allowed digits "421" for the rightmost three positions are exceeded in positions 1 and 3, with the 1st and 3rd primes 2 and 5 as their moduli, thus A(11,25) = 2*5 = 10.
%o (PARI)
%o up_to = 105;
%o A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
%o A327936(n) = { my(f = factor(n)); for(k=1, #f~, f[k,2] = (f[k,2]>=f[k,1])); factorback(f); };
%o A329041sq(row,col) = A327936(A276086(row)*A276086(col));
%o A329041list(up_to) = { my(v = vector(up_to), i=0); for(a=0,oo, for(col=0,a, if(i++ > up_to, return(v)); v[i] = A329041sq(a-col,col))); (v); };
%o v329041 = A329041list(up_to);
%o A329041(n) = v329041[1+n];
%Y Cf. A049345, A276086, A327936.
%Y Cf. also A317836, A324351, A328770, A328841, A328842.
%K nonn,tabl
%O 0,5
%A _Antti Karttunen_, Nov 03 2019
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