

A295887


Filter sequence combining A003557(n) and A173557(n); the restricted growth sequence transform of A291756.


10



1, 1, 2, 3, 4, 2, 5, 6, 7, 4, 8, 9, 10, 5, 11, 12, 13, 7, 14, 15, 10, 8, 16, 17, 18, 10, 19, 20, 21, 11, 22, 23, 24, 13, 25, 26, 27, 14, 25, 28, 29, 10, 30, 31, 32, 16, 33, 34, 35, 18, 36, 37, 38, 19, 29, 39, 27, 21, 40, 41, 42, 22, 43, 44, 45, 24, 46, 47, 48, 25, 49, 50, 51, 27, 52, 53, 42, 25, 54, 55, 56, 29, 57, 37, 58, 30, 59, 60, 61, 32, 51, 62, 42, 33, 51
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OFFSET

1,3


COMMENTS

First define function f(n) = (1/2)*(2 + ((A003557(n) + A173557(n))^2)  A003557(n)  3*A173557(n)), or in short, f(n) = P(A003557(n), A173557(n)), where P(n,k) is triangular table sequence A000027 used as an injective N x N > N pairing function. Then apply the restricted growth sequence transform to the sequence f(1), f(2), f(3), ... See the examplesection.
This is also the restricted growth sequence transform of sequence A291756, as A291756(n) = P(A003557(n), A000010(n)), where again P(n,k) is sequence A000027 used as a pairing function. Given either an ordered pair (A003557(n),A000010(n)) or (A003557(n),A173557(n)), the other one can be computed because A000010(n) = A003557(n)*A173557(n).
Note that the exact pairing function P used is not important, as long as it provides an injective mapping N x N > N. So instead of Cantor's mapping we could as well used bitinterleaving A054238 (Morton code) to pack together A003557(n) and A173557(n), or equally, A000010(n) and A003557(n).


LINKS

Antti Karttunen, Table of n, a(n) for n = 1..65537
Eric Weisstein's World of Mathematics, Pairing Function
Wikipedia, Pairing Function


EXAMPLE

The first ten terms of the sequence f(n) = (1/2)*(2 + ((A003557(n) + A173557(n))^2)  A003557(n)  3*A173557(n)) are 1, 1, 2, 3, 7, 2, 16, 10, 9, 7. When we assign to each newly occurring term the least unused number k so far (starting by giving k=1 for the initial term, this k increases by one for each new distinct term produced by f(n) when n grows), and for each repeated term the same number it was given the previous time (equal to the number it was given for the first time), we obtain 1, 1, 2, 3, 4, 2, 5, 6, 7, 4, the first 10 terms of this sequence. Note how f(10) = 7 gets 4 because when seven occurred for the first time at f(5), it was the 4th distinct new number in that sequence.
This is also true for the sequence A291756 although there the terms are different: 1, 1, 2, 5, 7, 2, 16, 25, 31, 7.


PROG

(PARI)
allocatemem(2^30);
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; };
write_to_bfile(start_offset, vec, bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)1, " ", vec[n])); }
A003557(n) = { my(f=factor(n)); for (i=1, #f~, f[i, 2] = max(0, f[i, 2]1)); factorback(f); };
A173557(n) = my(f=factor(n)[, 1]); prod(k=1, #f, f[k]1); \\ This function from Michel Marcus, Oct 31 2017
Anotsubmitted7(n) = (1/2)*(2 + ((A003557(n)+A173557(n))^2)  A003557(n)  3*A173557(n));
write_to_bfile(1, rgs_transform(vector(up_to, n, Anotsubmitted7(n))), "b295887.txt");


CROSSREFS

Cf. A000010, A003557, A054238, A173557, A291750, A291756, A295300, A295886.
Sequence in context: A286547 A157000 A026346 * A293450 A322990 A120636
Adjacent sequences: A295884 A295885 A295886 * A295888 A295889 A295890


KEYWORD

nonn


AUTHOR

Antti Karttunen, Dec 03 2017


STATUS

approved



