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A319357
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Filter sequence combining A003415(d) from all proper divisors d of n, where A003415(d) = arithmetic derivative of d.
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3
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1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 8, 2, 9, 4, 4, 2, 10, 3, 4, 11, 12, 2, 13, 2, 14, 4, 4, 4, 15, 2, 4, 4, 16, 2, 17, 2, 18, 19, 4, 2, 20, 3, 21, 4, 22, 2, 23, 4, 24, 4, 4, 2, 25, 2, 4, 26, 27, 4, 28, 2, 29, 4, 30, 2, 31, 2, 4, 32, 33, 4, 34, 2, 35, 36, 4, 2, 37, 4, 4, 4, 38, 2, 39, 4, 40, 4, 4, 4, 41, 2, 42, 43, 44, 2, 45, 2, 46, 47
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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 A319356.
The only duplicates in range 1..65537 with a(n) > 4 are the following six pairs: a(1445) = a(2783), a(4205) = a(11849), a(5819) = a(8381), a(6727) = a(15523), a(8405) = a(31211) and a(28577) = a(44573). All these have prime signature p^2 * q^1. If all the other duplicates respect the prime signature as well, then also the last implication given below is valid.
For all i, j:
a(i) = a(j) => A101296(i) = A101296(j). [Conjectural, see notes above]
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LINKS
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EXAMPLE
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Proper divisors of 1445 are [1, 5, 17, 85, 289], while the proper divisors of 2783 are [1, 11, 23, 121, 253]. 1 contributes 0 and primes contribute 1, so only the last two matter in each set. We have A003415(85) = 22 = A003415(121) and A003415(289) = 34 = A003415(253), thus the value of arithmetic derivative coincides for all proper divisors, thus a(1445) = a(2783).
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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; };
A003415(n) = {my(fac); if(n<1, 0, fac=factor(n); sum(i=1, matsize(fac)[1], n*fac[i, 2]/fac[i, 1]))}; \\ From A003415
A319356(n) = { my(m=1); fordiv(n, d, if(d<n, m *= prime(1+A003415(d)))); (m); };
v319357 = rgs_transform(vector(up_to, n, A319356(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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