Revision History for A057654
(Underlined text is an addition;
strikethrough text is a deletion.)
Showing all changes.
|
|
|
|
#6 by N. J. A. Sloane at Tue Feb 11 19:05:22 EST 2014
|
| AUTHOR
|
_Joseph L. Pe (joseph_l_pe(AT)hotmail.com), _, Feb 12 2002
|
|
|
Discussion
|
Tue Feb 11
| 19:05
| OEIS Server: https://oeis.org/edit/global/2119
|
|
|
|
#5 by Bruno Berselli at Fri Jun 29 04:47:37 EDT 2012
|
|
|
|
#4 by Donovan Johnson at Fri Jun 29 04:29:21 EDT 2012
|
|
|
|
#3 by Donovan Johnson at Fri Jun 29 04:27:31 EDT 2012
|
| NAME
|
Numbers n such that sigma(n)^2- - phi(n)^2 is a perfect square.
|
| DATA
|
1, 95, 99, 146, 323, 377, 1095, 1102, 1351, 1368, 1630, 1919, 1943, 2261, 2639, 2743, 3915, 4607, 4743, 5042, 5045, 5183, 5249, 6347, 6497, 7389, 8265, 8307, 8749, 9407, 9701, 10877, 12535, 13184, 13253, 14344, 14535, 16965, 18405, 19493, 19836
|
| LINKS
|
Donovan Johnson, <a href="/A057654/b057654.txt">Table of n, a(n) for n = 1..1000</a>
|
| EXAMPLE
|
sigma(95)^2- - phi(95)^2 = 120^2 - 72^2 = 96^2, so 95 is a term of the sequence.
|
| STATUS
|
approved
editing
|
|
|
|
#2 by N. J. A. Sloane at Sat Jun 12 03:00:00 EDT 2004
|
| NAME
|
Numbers n such that sigma(n)^2-phi(n)^2 is a perfect square.
|
| KEYWORD
|
easy,nonn,new
|
|
|
|
#1 by N. J. A. Sloane at Fri May 16 03:00:00 EDT 2003
|
| NAME
|
n such that sigma(n)^2-phi(n)^2 is a perfect square.
|
| DATA
|
1, 95, 99, 146, 323, 377, 1095, 1102, 1351, 1368, 1630, 1919, 1943, 2261, 2639, 2743, 3915, 4607, 4743, 5042, 5045, 5183, 5249, 6347, 6497, 7389, 8265, 8307, 8749, 9407, 9701
|
| OFFSET
|
1,2
|
| EXAMPLE
|
sigma(95)^2-phi(95)^2 = 120^2 - 72^2 = 96^2, so 95 is a term of the sequence.
|
| MATHEMATICA
|
Select[Range[10^4], IntegerQ[Sqrt[DivisorSigma[1, # ]^2 - EulerPhi[ # ]^2]] &]
|
| KEYWORD
|
easy,nonn
|
| AUTHOR
|
Joseph L. Pe (joseph_l_pe(AT)hotmail.com), Feb 12 2002
|
| STATUS
|
approved
|
|
|
|
|