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A319693
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Filter sequence combining sopfr(d) from all proper divisors d of n, where sopfr(d) is A001414(d) = sum of primes dividing d with repetition.
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4
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1, 2, 2, 3, 2, 4, 2, 5, 6, 7, 2, 8, 2, 9, 10, 11, 2, 12, 2, 13, 14, 15, 2, 16, 17, 18, 19, 20, 2, 21, 2, 22, 23, 24, 25, 26, 2, 27, 28, 29, 2, 30, 2, 31, 32, 33, 2, 34, 35, 36, 37, 38, 2, 39, 40, 41, 42, 43, 2, 44, 2, 45, 46, 47, 48, 49, 2, 50, 51, 52, 2, 53, 2, 54, 55, 56, 57, 58, 2, 59, 60, 61, 2, 62, 63, 64, 65, 66, 2, 67, 68, 69, 70, 71, 72, 73, 2, 74, 75, 76, 2, 77, 2, 78, 79, 80, 2, 73, 2, 81, 82, 83, 2, 84, 85
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OFFSET
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1,2
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COMMENTS
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Restricted growth sequence transform of A319692.
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LINKS
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EXAMPLE
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The proper divisors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, while
the proper divisors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54.
It happens that sopfr(8) = sopfr(9), sopfr(16) = sopfr(18), sopfr(24) = sopfr(27), sopfr(32) = sopfr(36) and sopfr(48) = sopfr(54), and the rest of proper divisors (1, 2, 3, 4, 6, 12) are shared by both numbers, from which follows that by taking product of sopfr over proper divisors gives an identical result for both, thus a(96) = a(108). Here sopfr = A001414.
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PROG
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(PARI)
up_to = 65537;
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; };
A319692(n) = { my(m=1); fordiv(n, d, if(d<n, m *= prime(1+A001414(d)))); (m); };
v319693 = rgs_transform(vector(up_to, n, A319692(n)));
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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