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A180346
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Primes that divide every circular permutation of the digits of at least one number of the form 123...(n-1)(n) (see A007908), where n is 3 digits long (that is, for some n in the range 99<n<1000).
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1
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3, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 53, 61, 67, 73, 83, 97, 101, 107, 127, 163, 211, 271, 277, 1009, 18973
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OFFSET
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1,1
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COMMENTS
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Every a(i) divides at least 192 permutations of the digits of an element belonging to [A007908]. Skipping the trivial case a(1)=3, the most recurring elements are a(2)=7 and a(10)=37. The occurrences in our 1386450 terms set are the following [A181373]:
a(2) | 7 ⇒ n=100+14*v (v=0,1,2,...,64)
a(3) | 11 ⇒ n=106+22*v (v=0,1,2,...,40)
a(4) | 13 ⇒ n=120+26*v (v=0,1,2,...,33)
a(5) | 17 ⇒ n=196+272*v (v=0,1,2)
a(6) | 19 ⇒ n=102+114*v (v=0,1,2,3,4,5,6,7)
a(7) | 23 ⇒ n=542
a(8) | 29 ⇒ n=400
a(9) | 31 ⇒ n=181+155*v (v=0,1,2,3,4,5)
a(10)| 37 ⇒ n=123+d(v),
(where d(v)=0,12,25,12,25,12,25... for v=0,1,2,3,...,47)
a(11) | 41 ⇒ n=216+205*v (v=0,1,2,3)
a(12) | 43 ⇒ n=372+301*v (v=0,1,2)
a(13) | 53 ⇒ n=127+689*v (v=0,1)
a(14) | 61 ⇒ n=616
a(15) | 67 ⇒ n=399
a(16) | 73 ⇒ n=196+584*v (v=0,1)
a(17) | 83 ⇒ n=118
a(18) | 97 ⇒ n=516
a(19) | 101 ⇒ n=416+404*v (v=0,1)
a(20) | 107 ⇒ n=884
a(21) | 127 ⇒ n=106
a(22) | 163 ⇒ n=576
a(23) | 211 ⇒ n=306
a(24) | 271 ⇒ n=936
a(25) | 277 ⇒ n=174
a(26) | 1009 ⇒ n=960
a(27) | 18973 ⇒ n=903
N.B.
Every coefficient of "v" is a multiple of i. This is a general property of [A007908], valid for an arbitrary fixed digits interval of the parameter "n" (10^k-1<n<10^k).
a(28) >= prime(10^6) if it exists. - Chai Wah Wu, Nov 12 2015
a(28) > prime(2.3316*10^9) if it exists. Conjecture: 18973 is the last term. - Chai Wah Wu, Oct 09 2023
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REFERENCES
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Vassilev-Missana and K. Atanassov, “Some Smarandache problems”, Hexis, 2004.
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LINKS
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FORMULA
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For n<10 the only a(i) is 3. If 9<n<100 an a(i) is 37 (the other one is a(1)=3), which divides all the circular permutations of Sm(21).
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PROG
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(PARI) isA180346(p)={ isprime(p) & p!=2 & p!=5 & for(n=100, 999, my(S=eval(concat(vector(n, i, Str(i)))), L=#Str(S)-1); S%p & next; for(k=1, L, (S=[1, 10^L]*divrem(S, 10))%p & next(2)); return(n)) } /* returns the least corresponding n or 0 if not in this sequence */ \\ M. F. Hasler, Jan 23 2011
(Python)
from itertools import islice
from sympy import nextprime
def A180346_gen(startvalue=1): # generator of terms >= startvalue
p = max(startvalue-1, 0)
while (p:=nextprime(p)):
c, q, a, b = 0, 1, 10, 10
for m in range(1, 1000):
if m >= b:
a = 10*a%p
b *= 10
c = (c*a + m) % p
q = q*a % p
if m>99 and not (c or (q-1)%p):
yield p
break
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CROSSREFS
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KEYWORD
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base,fini,hard,nonn
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AUTHOR
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STATUS
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approved
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