[seqfan] Re: New sequence?

[Richard Guy]

On Sat, 21 Apr 2012, Moses, Peter J. C. wrote:
Well thank you, Peter!  And thank you, Neil --- have you got all
the terms that Peter gives?   R.

> Hi Richard,
> 
> So that you don't feel totally alone ...
> 
> Here are a few more terms of
> A000048
> {1,1,1,1,2,3,5,9,16,28,51,93,170,315,585,1091,2048,3855,7280,13797,26214,49
> 929,95325,182361,349520,671088,1290555,2485504,4793490,9256395,17895679,346
> 36833,67108864,130150493,252645135,490853403,954437120,1857283155,361681456
> 5,7048151355,13743895344,26817356775,52357696365,102280151421,199911205050,
> 390937467284,764877654105,1497207322929,2932031006720,5744387279808,1125899
> 9068416,22076468760335,43303842570870,84973577874915,166799986196480,327534
> 518354199,643371375338640,1264168316450277,2484744621997515,488526061274087
> 7,9607679205048286,18900352534538475,37191016277640225,73201365371846652,14
> 4115188075855872,283796062672454577,558992244657833425,1101298153654301589,
> 2170205185142300190,4277505872164472921,8432797290838652043,166280509960198
> 77513,32794211686594641920,64689951820132126215,127631526564044465235,25185
> 9545753047520816,497091208723120548810,981270957479406797637,19373811211772
> 89913895,3825714619033636628817,7555786372591432341504,14925010118699124785
> 152,29485995600356809139175,58261485282632731311141,11513579234425039599321
> 0,227562507221577256414509,449832863112420158030205,88932474086593410244551
> 5,1758437555803096981390800,3477359660913989536233495,687744466269655707680
> 6224,13603736695443739284605715,26911739984464788584763570,5324473287248947
> 4177664033,105356599088117470204180785,208495164511221941035639215,41264667
> 9761793424944005120,816785180559426160758185055,161690127580131301211314176
> 0,3201137879364215660301826604,6338253001141147007483510784}
> As you can see, your guess of 954437120, 1857283155, 3616814565 is good.
> 
> And as a consequence ... Your sequence ...
> {1,0,0,2,0,2,4,2,0,8,4,2,16,2,4,38,0,2,64,2,16,134,4,2,256,32,4,512,16,2,10
> 84,2,0,2054,4,158,4096,2,4,8198,256,2,16444,2,16,33272,4,2,65536,128,1024,1
> 31078,16,2,262144,2078,256,524294,4,2,1052656,2,4,2097656,0,8222,4194364,2,
> 16,8388614,17404,2,16777216,2,4,33587168,16,2174,67108924,2,65536,134217728
> ,4,2,268439536,131102,4,536870918,256,2,1074003904,8318,16,2147483654,4,524
> 318,4294967296,2,16384,8589935096,1048576}
> 
> I don't have sufficient knowledge of ranks of apparition (i.e. never heard
> of it) to really comment .. but ..
> rank 2 the square prime numbers
> {4,9,25,49,121,169,289,361,529,841,961,1369,1681,1849,2209,2809}
> rank 3
> {5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101},
> rank 5
> {6,13,14,17,22,29,37,38,41,46,53,61,62,73,86,89,94,95,97,101,109,113,118},
> rank 7
> {9,13,19,31,37,43,61,67,73,77,79,97,103,109,127,139,151,157,163,181,193},
> rank 11{21,25,31,33,41,51,61,71,78,91,101,111,131,138,141,151,161,181,191,201,211}
> ,
> rank 13{14,18,37,38,45,55,61,62,73,86,97,109,115,134,153,156,157,158,181,187,193},
> rank 17
> {10,12,21,26,28,41,44,58,69,73,74,76,77,89,92,93,97,106,113,122,124,133},
> rank 19
> {27,37,63,73,109,117,127,163,181,199,220,266,271,279,307,379,387,397,433},
> rank 23
> {30,67,70,89,119,121,199,290,331,353,397,406,419,426,427,463,470,493,581},
> rank 29{86,98,113,142,197,245,254,281,337,421,422,449,478,617,637,673,701,757,758}
> ,
> rank 31{11,25,41,61,71,101,131,151,181,191,211,241,251,271,281,311,331,401,421,431
> },
> rank 37
> {30,38,54,73,109,117,126,135,138,181,246,254,326,354,397,398,433,459,495},
>  
> More if needed.
> 
> Best regards,
> Peter.
> Well, that went over like a lead balloon!  Here's what I guess to be the
> next three members of A000048:  954437120, 1857283155, 3616814565.
> 
> This gives  4096, 2, 4  for the next three members of the suggested
> sequence.  Conjecture: in this sequence, if  n  is 3-smooth. then
> a(n)  is a power of 2.  [probably easy to prove.]
> 
> A000048  is a sort of pseudo-divisibility sequence.  E.g.
> 
> 2 has ranks of apparition  4, 9 and 25 (and 49, ...???) in the
> sense that  A000048(n) is even just if  n  is a multiple of
> 4 or 9 or 25 (or 49 or ???)
> 
> 5 has ranks  6, 13, 14, 17, 22, ???
> 
> 7 has ranks  9, 13, 19, ??
> 
> 11 has ranks 21, 25, 33, ??
> 
> 13 has ranks 14, 18, 37, 38, ??
> 
> 17 has ranks 10, 12, 21, 26, ??
> 
> 31 has ranks 11, 25, ??
> 
> However,  3  does not divide  A000048(15), so maybe I'm barking
> up the wrong tree.  Check?  Mod 2^15 + 1, there is a 2-cycle
> {10923,-10923}, a 6-cycle {3641,7282,14564,-3641,-7282,-14564},
> three 10-cycles of which one is
> {2979,5958,11916,-8937,14895,-2979,-5958,...}, and 1091
> 30-cycles.    R.
> 
> On Thu, 19 Apr 2012, Richard Guy wrote:
> 
> > Would an editor more competent than I like to enter the following
> > sequence into OEIS, if it's not there already (I'm not a good looker) ?
> >
> > [check & extend.  These are only hand calculations.  A000048 could
> > also easily be extended]  For  n = (0) 1 2 3 ...
> >
> > (0),0,0,2,0,2,4,2,0,8,4,2,16,2,4,38,0,2,64,2,16,134,4,2,256,32,4,
> > 512,16,2,1084,2,0,2054,4,159,
> >
> > It's the total length of all cycles which are strictly less than
> > the full length of  2n.
> >
> >    2^n  -  2 * n * A000048(n)
> >
> > a(2^k) = 0,  a(prime) = 2,  a(2p) = 4.
> >
> > There's a simple formula using the Moebius function (v. A000048).
> >
> > Let me know if I've made errors.   Thanks!   R.
> >
> 
> 
>