hihi - i'll write a script for that, to show the largest a[n] and first missing acceptable number for values of n up to nmax for various nmax; it does look like the sequence eventually fills in all of them meanwhile, here are the first 100 values - again, not yet completely verified, but i think pretty close: it took me a while to find a good (easily computable) criterion for unacceptability, so here is what i used: for a prospective value b for a[n], if every prime dividing b also divides the previously computed a[n-1], then there would be no acceptable a[n+1], so b is unacceptable (otherwise b is acceptable, since the a[n+1] can use only the primes not also in a[n-1]) the program generates values up to n=10000 in just under a minute on a 5 year old commodity pc running freebsd, so if we get some conjectures, a lot of experimentation can be done f = gcd(a[n], a[n-1]) is prime a lot a[1] = 1 a[2] = 2 a[3] = 6 f = 2 a[4] = 15 f = 3 a[5] = 35 f = 5 a[6] = 14 f = 7 a[7] = 12 f = 2 a[8] = 33 f = 3 a[9] = 55 f = 11 a[10] = 10 f = 5 a[11] = 18 f = 2 a[12] = 21 f = 3 a[13] = 77 f = 7 a[14] = 22 f = 11 a[15] = 20 f = 2 a[16] = 45 f = 5 a[17] = 39 f = 3 a[18] = 26 f = 13 a[19] = 28 f = 2 a[20] = 63 f = 7 a[21] = 51 f = 3 a[22] = 34 f = 17 a[23] = 38 f = 2 a[24] = 57 f = 19 a[25] = 69 f = 3 a[26] = 46 f = 23 a[27] = 40 f = 2 a[28] = 65 f = 5 a[29] = 91 f = 13 a[30] = 42 f = 7b is unacceptable a[31] = 30 f = 6 NOT prime a[32] = 85 f = 5 a[33] = 119 f = 17 a[34] = 56 f = 7 a[35] = 24 f = 8 NOT prime a[36] = 75 f = 3 a[37] = 95 f = 5 a[38] = 76 f = 19 a[39] = 36 f = 4 NOT prime a[40] = 87 f = 3 a[41] = 145 f = 29 a[42] = 50 f = 5 a[43] = 44 f = 2 a[44] = 99 f = 11 a[45] = 93 f = 3 a[46] = 62 f = 31 a[47] = 52 f = 2 a[48] = 117 f = 13 a[49] = 105 f = 3 a[50] = 70 f = 35 NOT prime a[51] = 58 f = 2 a[52] = 261 f = 29 a[53] = 111 f = 3 a[54] = 74 f = 37 a[55] = 68 f = 2 a[56] = 153 f = 17 a[57] = 123 f = 3 a[58] = 82 f = 41 a[59] = 80 f = 2 a[60] = 115 f = 5 a[61] = 161 f = 23 a[62] = 84 f = 7 a[63] = 60 f = 12 NOT prime a[64] = 155 f = 5 a[65] = 217 f = 31 a[66] = 98 f = 7 a[67] = 48 f = 2 a[68] = 129 f = 3 a[69] = 215 f = 43 a[70] = 100 f = 5 a[71] = 54 f = 2 a[72] = 141 f = 3 a[73] = 235 f = 47 a[74] = 110 f = 5 a[75] = 66 f = 22 NOT prime a[76] = 147 f = 3 a[77] = 133 f = 7 a[78] = 152 f = 19 a[79] = 72 f = 8 NOT prime a[80] = 135 f = 9 NOT prime a[81] = 175 f = 5 a[82] = 112 f = 7 a[83] = 78 f = 2 a[84] = 143 f = 13 a[85] = 187 f = 11 a[86] = 102 f = 17 a[87] = 86 f = 2 a[88] = 301 f = 43 a[89] = 189 f = 7 a[90] = 90 f = 9 NOT prime a[91] = 88 f = 2 a[92] = 209 f = 11 a[93] = 171 f = 19 a[94] = 96 f = 3 a[95] = 92 f = 4 NOT prime a[96] = 253 f = 23 a[97] = 165 f = 11 a[98] = 108 f = 3 a[99] = 94 f = 2 a[100] = 329 f = 47 more later, chris On 2020-08-16 15:02, Allan Wechsler wrote:
Chris, what is the smallest eligible (non-prime-power) number your list leaves out?
On Sun, Aug 16, 2020 at 5:55 PM Christopher Landauer <topcycal@gmail.com> wrote:
hihi, all - i wrote a program to generate terms to n=1000 (i still need to verify it, but it matches what was in this e-mail)
it is a curious fact that the gcd of successive terms is a prime, up to a point, and mostly in any case
i'll have more in a few days
more later, chris
On Sun, Aug 16, 2020 at 6:04 AM Fred Lunnon <fred.lunnon@gmail.com> wrote:
Already posted on August 11th ?!
WFL
On 8/16/20, Neil Sloane <njasloane@gmail.com> wrote:
Obviously this is an inverted version of the Yellowstone sequence A098550 ! The name Enots Wolley is for personal use only, it must not be mentioned in the OEIS! We frown on such made-up names.
Definition: Lexicographically earliest sequence {a(n)} of distinct positive numbers such that, for n>2, a(n) has a common factor with a(n-1) but not with a(n-2). 1, 2, 6, 15, 35, 14, 12, 33, 55, 10, 18, 21, 77, 22, 20, 45, 39, 26, 28, 63, ... The original idea was due to Scott, with a different sequence, but this is my (canonical!) version.
Could someone please prove the conjecture that this is a permutation of the set {1, all numbers with at least two distinct prime factors} ?
I can't even prove that every number 2*p (p prime) appears, or that there are infinitely many even terms (although I've found a dozen false proofs). It's a slippery problem.
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-- dr. christopher landauer topcy house consulting thousand oaks, california _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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