A103651 is a WRONG version of A068318
zak seidov
zakseidov at yahoo.com
Sat Dec 15 16:32:10 CET 2007
Neil,
A103651 is a WRONG version of A068318
(terms for square semiprimes are WRONG in A103651),
Sorry, my bug...
Zak
%I A103651
%S A103651
4,5,9,7,9,8,10,13,25,15,14,19,12,21,16,25,49,20,16,22,31,33,18,26,39,
%N A103651 Sum of factors of n-th semiprime.
%I A068318
%S A068318
4,5,6,7,9,8,10,13,10,15,14,19,12,21,16,25,14,20,16,22,31,33,18,26,39,
%N A068318 Sum of prime factors of n-th semiprime.
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Dear Seqfans,
article about the sequences of equations mentioned therein:
A045944 <http://www.research.att.com/%7Enjas/sequences/A045944>
Rhombic matchstick numbers: n*(3*n+2).
+20
7
0, 5, 16, 33, 56, 85, 120, 161, 208, 261, 320, 385, 456, 533, 616, 705, 800,
901, 1008,
The equations 1 + 2 = 3 and 3^2 + 4^2 =5^2 set the stage for considering
whether or not it is also true that 5^3 + 6^3 = 7^3 and 7^4 + 8^4 = 9^4.
Reflecting on Fermat's Last Theorem or resorting to a calculator dispels any
hope that either of the two equations could be correct. However, 5^3 + 6^3
+ 2 does equal 7^3 and 7^4 + 8^4 + 64 equals 9^4. More significantly, each
of these equations is the first of an infinite sequence of equations
featuring consecutive integers that conform to the spirit of the equations
mentioned in A000384. For n>0,
a(n)^3+(a(n)+1)^3 +...+(a(n)+n)^3 +2*A000217(n)^2=
(a(n)+n+1)^3+...+(a(n)+2n)^3;
e.g., 5^3+6^3+2*1^2=7^3; 16^3+17^3+18^3+2*3^2=19^3+20^3; see also A033954
A033954 <http://www.research.att.com/%7Enjas/sequences/A033954>
n*(4*n+3). Also, second 10-gonal (or decagonal) numbers.
+20
16
0, 7, 22, 45, 76, 115, 162, 217, 280, 351, 430, 517, 612, 715, 826, 945,
1072, 1207,
The equations 1 + 2 = 3 and 3^2 + 4^2 =5^2 set the stage for considering
whether or not it is also true that 5^3 + 6^3 = 7^3 and 7^4 + 8^4 = 9^4.
Reflecting on Fermat's Last Theorem or resorting to a calculator dispels any
hope that either of the two equations could be correct. However, 5^3 + 6^3
+ 2 does equal 7^3 and 7^4 + 8^4 + 64 equals 9^4. More significantly, each
of these equations is the first of an infinite sequence of equations
featuring consecutive integers that conform to the spirit of the equations
mentioned in A000384. For n>0,
a(n)^4+(a(n)+1)^4 +...+(a(n)+n)^4 +(4*A000217(n))^3 =
(a(n)+n+1)^4+...+(a(n)+2n)^4;
e.g., 7^4+8^4+(4*1)^3=9^4; 22^4+23^4+24^4+(4*3)^3=25^4+26^4; see also 045944
would greatly appreciate it.
Charlie Marion
Yorktown Heights NY 10598
--Boundary_(ID_Cawq0M8R67EttM/F18XbtQ)
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<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'>Dear Seqfans,<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'> I have submitted the following two comments to
OEIS and am planning an article about the sequences of equations mentioned therein:<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
<table class=MsoNormalTable border=0 cellspacing=0 cellpadding=0 width="100%"
</table>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt;display:none'><o:p> </o:p></span></font></p>
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</table>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face="Courier New"><span style='font-size:10.0pt;
font-family:"Courier New"'>The equations 1 + 2 = 3 and 3^2 + 4^2 =5^2 set the
7^4 + 8^4 = 9^4. Reflecting on Fermat’s Last Theorem or resorting
to a calculator dispels any hope that either of the two equations could be
correct. However, 5^3 + 6^3 + 2 does equal 7^3 and 7^4 + 8^4 + 64 equals
9^4. More significantly, each of these equations is the first of an
infinite sequence of equations featuring consecutive integers that conform to
the spirit of the equations mentioned in A000384. For n>0,<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face="Courier New"><span style='font-size:10.0pt;
font-family:"Courier New"'>a(n)^3+(a(n)+1)^3 +...+(a(n)+n)^3 +2*A000217(n)^2= (a(n)+n+1)^3+...+(a(n)+2n)^3;<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face="Courier New"><span style='font-size:10.0pt;
font-family:"Courier New"'>e.g., 5^3+6^3+2*1^2=7^3;
16^3+17^3+18^3+2*3^2=19^3+20^3; see also A033954<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face="Courier New"><span style='font-size:10.0pt;
font-family:"Courier New"'>see also A033954 <tt><font face="Courier New">-
Charlie Marion (charliemath at optonline.net), Dec 8 2007</font></tt><o:p></o:p></span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
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12.0pt;display:none'><o:p> </o:p></span></font></p>
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</table>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 face="Courier New"><span style='font-size:10.0pt;
font-family:"Courier New"'>The equations 1 + 2 = 3 and 3^2 + 4^2 =5^2 set the
7^4 + 8^4 = 9^4. Reflecting on Fermat’s Last Theorem or resorting
to a calculator dispels any hope that either of the two equations could be
correct. However, 5^3 + 6^3 + 2 does equal 7^3 and 7^4 + 8^4 + 64 equals
9^4. More significantly, each of these equations is the first of an
infinite sequence of equations featuring consecutive integers that conform to
the spirit of the equations mentioned in A000384. For n>0,<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face="Courier New"><span style='font-size:10.0pt;
font-family:"Courier New"'>a(n)^4+(a(n)+1)^4 +...+(a(n)+n)^4 +(4*A000217(n))^3 =
(a(n)+n+1)^4+...+(a(n)+2n)^4;<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 face="Courier New"><span style='font-size:10.0pt;
font-family:"Courier New"'>e.g., 7^4+8^4+(4*1)^3=9^4; 22^4+23^4+24^4+(4*3)^3=25^4+26^4;
<p class=MsoNormal><font size=2 face="Courier New"><span style='font-size:10.0pt;
font-family:"Courier New"'> <tt><font face="Courier New">- Charlie Marion
(charliemath at optonline.net), Dec 8 2007</font></tt><o:p></o:p></span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'> If anyone could supply me with a past reference to
these sequences, I would greatly appreciate it.<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'> Thanks very much.<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'>Charlie Marion<o:p></o:p></span></font></p>
<p class=MsoNormal><st1:place w:st="on"><st1:City w:st="on"><font size=3
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
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