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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A227973 Minimum composite squarefree numbers k such that p(i)-n divides k+n, for n=1, 2, 3, 4,..., where p(i) are the prime factors of k.

Original entry on oeis.org

15, 273, 77, 6, 21, 6, 33, 10, 15, 14, 21, 33, 35, 22, 33, 26, 39, 57, 65, 34, 51, 38, 57, 551, 95, 46, 69, 203, 115, 145, 161, 58, 87, 62, 93, 629, 155, 697, 217, 74, 111, 518, 185, 82, 123, 86, 129, 2537, 215, 94, 141, 689, 235, 4366, 329, 106, 159, 1247, 265
Offset: 1

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Author

Paolo P. Lava, Aug 02 2013

Keywords

Examples

			For n=185 the minimum k is 543. Prime factors of 543 are 3 and 181. We have: 543 + 185 = 728, 3 - 185 = -182 and 728 / (-182) = -4, 181 - 185 = -4 and 728 / (-4) = 182.
		

Crossrefs

Programs

  • Maple
    with(numtheory); P:=proc(i) local c, d, k, n, ok, p; for k from 1 to i do
    for n from 2 to i do if not isprime(n) then p:=ifactors(n)[2]; ok:=1;
    for d from 1 to nops(p) do if p[d][2]>1 or p[d][1]=k then ok:=0; break; fi;
    if  not type((n+k)/(p[d][1]-k), integer) then ok:=0; break; fi; od;
    if ok=1 then print(n); break; fi; fi; od; od; end: P(10^6);

A227974 Minimum composite squarefree numbers k such that p(i)+n divides k-n, for n=1, 2, 3, 4,..., where p(i) are the prime factors of k .

Original entry on oeis.org

385, 182, 195, 1054, 165, 6, 1015, 4958, 2193, 10, 5159, 113937, 5593, 14, 15, 196009, 3657, 6318638, 2755, 1227818, 21, 22, 2795, 152358, 12121, 26, 21827, 17578, 36569, 30, 38335, 457907, 33, 34, 35
Offset: 1

Views

Author

Paolo P. Lava, Aug 02 2013

Keywords

Comments

Fixed points are the squarefree semiprimes.

Examples

			For n=3 the minimum k is 195. Prime factors of 195 are 3, 5 and 13. We have: 195 - 3 = 192, 3 + 3 = 6 and 192 / 6 = 32, 5 + 3 = 8 and 192 / 8 = 24, 13 + 3 = 16 and 192 / 16 = 12.
		

Crossrefs

Programs

  • Maple
    with(numtheory); P:=proc(i) local c, d, k, n, ok, p; for k from 1 to i do
    for n from 2 to i do if not isprime(n) then p:=ifactors(n)[2]; ok:=1;
    for d from 1 to nops(p) do if p[d][2]>1 then ok:=0; break; fi;
    if  not type((n-k)/(p[d][1]+k), integer) then ok:=0; break; fi; od;
    if ok=1 then print(n); break; fi; fi; od; od; end: P(10^6);
Showing 1-2 of 2 results.