cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A353485 Numbers k such that no x with an odd arithmetic derivative is encountered when repeatedly prime shifting from k down to 1 with the map x -> A064989(x).

Original entry on oeis.org

1, 4, 8, 9, 16, 25, 32, 36, 49, 64, 72, 81, 100, 108, 121, 128, 144, 169, 196, 200, 216, 225, 256, 288, 289, 324, 361, 392, 400, 432, 441, 484, 512, 529, 576, 625, 648, 676, 729, 784, 800, 841, 864, 900, 961, 968, 972, 1024, 1089, 1152, 1156, 1225, 1296, 1352, 1369, 1444, 1521, 1568, 1600, 1681, 1728, 1764, 1800, 1849
Offset: 1

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Author

Antti Karttunen, Apr 23 2022

Keywords

Comments

Note that k itself must also be in A235992 to be included here. All terms must be powerful (in A001694) because otherwise at some point in the shifting process a number of the form 4u+2 would be encountered, and they are all in A235991.

Crossrefs

Positions of zeros in A353484.
Cf. A000290 (subsequence), A003415, A064989, A165560, A235991, A235992.
Subsequence of A001694.

Programs

  • PARI
    A000265(n) = (n>>valuation(n,2));
    A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
    A064989(n) = { my(f=factor(A000265(n))); for(i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f); };
    isA353485(n) = { while(n>1, if(A003415(n)%2, return(0)); n = A064989(n)); (1); };

A353515 The length of the shortest path from n to 1 when using the transitions x -> A003415(x) and x -> A003961(x), or -1 if no 1 can ever be reached from n.

Original entry on oeis.org

0, 1, 1, 4, 1, 2, 1, 7, 3, 2, 1, 6, 1, 4, 7, 8, 1, 4, 1, 6, 3, 2, 1, 7, 3, 6, 7, 8, 1, 2, 1, 10, 5, 2, 6, 5, 1, 4, 4, 6, 1, 2, 1, 6, 5, 4, 1, 9, 4, 5, 5, 8, 1, 6, 8, 8, 3, 2, 1, 4, 1, 6, 5, 10, 5, 2, 1, 5, 4, 2, 1, 8, 1, 5, 6, 6, 5, 2, 1, 9, 7, 2, 1, 4, 3, 5, 7, 8, 1, 5, 7, 7, 3, 5, 4, 9, 1, 6, 7, 6, 1, 3, 1, 7, 2
Offset: 1

Views

Author

Antti Karttunen, Apr 23 2022

Keywords

Comments

This is a variant of A327969 that seems to be less in need of an escape clause. Note that enough prime shifts with A003961 will eventually transform every term of A100716 (which is a subsequence of A099309) to a term of A048103, and that A051903(A003961(n)) = A051903(n). See also the array A344027.
Records 0, 1, 4, 7, 8, 10, 12, 13, 14, 15, 16, 19, ... occur at 1, 2, 4, 8, 16, 32, 128, 256, 768, 1024, 2048, 4096, ..., etc.

Examples

			From n = 4, we can reach 1 with just four steps as A003961(4) = 9, A003415(9) = 6, A003415(6) = 5 and A003415(5) = 1, and because there are no shorter paths we have a(4) = 4.
From n = 8, we can reach 1 with seven steps, as A003415(8) = 12, A003961(12) = 45, A003415(45) = 39, A003961(39) = 85, A003415(85) = 22, A003415(22) = 13, A003415(13) = 1, and because there are no shorter paths we have a(8) = 7.
For n = 15, as A003415(15) = 8, we know that a(15) is at most a(8)+1, i.e., 8. But we can do better, as A003961(15) = 35, A003961(35) = 77, A003415(77) = 18, A003415(18) = 21, A003415(21) = 10, A003415(10) = 7, A003415(7) = 1, and because there are no shorter paths we have a(15) = 7.
From n = 49, we can reach 1 in four steps, as A003961(49) = 121, A003415(121) = 22, A003415(22) = 13, A003415(13) = 1. Note that this is less than A099307(49)-1, as it would take 5 steps to reach 1 if using the arithmetic derivative only, 49 -> 14 -> 9 -> 6 -> 5 -> 1.
		

Crossrefs

Programs

  • PARI
    A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
    A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); }; \\ From A003961
    A353515(n) = if(1==n,0,my(xs=Set([n]),newxs,a,b,u); for(k=1,oo, newxs=Set([]); for(i=1,#xs,u = xs[i]; a = A003415(u); if(1==a, return(k)); if(isprime(a), return(k+1)); b = A003961(u); newxs = setunion([a],newxs); newxs = setunion([b],newxs)); xs = newxs));

Formula

a(1) = 0, a(p^p) = 1 + a(A003961(p^p)) for primes p, and for other numbers, a(n) = 1 + min(a(A003415(n)), a(A003961(n))).
a(p) = 1 for all primes p.
a(n) < A099307(n), unless A099307(n) = 0. [I.e., for all n in A099308]
Showing 1-2 of 2 results.