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Lexicographically earliest infinite sequence such that a(i) = a(j) => A007814(f(i)) = A007814(f(j)) and A278222(f(i)) = A278222(f(j)), for all i, j >= 1, where f(k) = A109812(k).
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%I #18 Apr 07 2022 12:14:19

%S 1,2,3,4,5,6,7,8,9,6,7,10,6,11,12,13,14,6,15,16,15,16,17,18,7,16,19,

%T 20,21,22,18,7,16,22,6,22,18,17,23,24,25,15,16,26,16,27,28,23,29,30,

%U 23,22,6,31,16,7,32,33,15,33,18,22,18,27,33,16,22,34,23,17,25,27,16,35,36,37,38,32,28,32,39,18,40,16

%N Lexicographically earliest infinite sequence such that a(i) = a(j) => A007814(f(i)) = A007814(f(j)) and A278222(f(i)) = A278222(f(j)), for all i, j >= 1, where f(k) = A109812(k).

%C Restricted growth sequence transform of the ordered pair [A007814(A109812(n)), A046523(A005940(1+A109812(n)))].

%C The sequence allots a new distinct number for each newly encountered combination of the 2-adic valuation of A109812 (A351964), and the multiset of the lengths of 1-runs in the odd part of A109812 (A351965). See the examples.

%C For all i, j: a(i) = a(j) => A352889(i) = A352889(j).

%H Antti Karttunen, <a href="/A351578/b351578.txt">Table of n, a(n) for n = 1..100000</a>

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

%e n A109812(n) [base-2] A351964(n) Lengths of a(n)

%e (# of trailing 0's) 1-runs (allotted #)

%e -----+----------------------------------------------------------------------

%e 1 : 1 [1], 0 [1] 1

%e 2 : 2 [10], 1 [1] 2

%e 3 : 4 [100], 2 [1] 3

%e 4 : 3 [11], 0 [2] 4

%e 5 : 8 [1000], 3 [1] 5

%e 6 : 5 [101], 0 [1,1] 6

%e 7 : 10 [1010], 1 [1,1] 7

%e 8 : 16 [10000], 4 [1] 8

%e 9 : 6 [110], 1 [2] 9

%e 10 : 9 [1001], 0 [1,1] 6

%e 11 : 18 [10010], 1 [1,1] 7

%e Because the combinations of the multiset of 1-runs in the binary expansion of A109812(n) and the number of trailing zeros in it (A351964) are unique for n = 1 .. 9, a unique increasing number (starting from 1) is allotted for each, and a(n) = n for n <= 9. On the other hand, at n=10, the binary expansion is [1001], for which these two measures are equal to that of binary expansion [101] found first time at n=6, therefore the rgs-transform allots for 10 the same number as for 6, and a(10) = a(6) = 6. At n=11, the binary expansion is [10010], where these two measures coincide with that of [1010] found first time at n=7, therefore a(10) = a(7) = 7.

%o (PARI)

%o rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };

%o v109812 = readvec("b109812_to10e5.txt"); \\ Prepared from b-file data with gawk ' { print $2 } '

%o up_to = #v109812;

%o A109812(n) = v109812[n];

%o A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); (t); };

%o A007814(n) = valuation(n,2);

%o A046523(n) = { my(f=vecsort(factor(n)[, 2], , 4), p); prod(i=1, #f, (p=nextprime(p+1))^f[i]); }; \\ From A046523

%o v351578 = rgs_transform(vector(up_to, n, [A007814(A109812(n)), A046523(A005940(1+A109812(n)))]));

%o A351578(n) = v351578[n];

%Y Cf. A007814, A109812, A278222 (A286622), A351963, A351964, A351965, A352888, A352889.

%K nonn,base

%O 1,2

%A _Antti Karttunen_, Apr 07 2022