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A260705
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Least integer k such that the set of the divisors of k contains exactly n pairs of numbers having the following property: for each pair of two distinct divisors, the reversal of one is equal to the other.
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3
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84, 168, 336, 1008, 3024, 5544, 11088, 16632, 33264, 49896, 99792, 182952, 365904, 249480, 498960, 1097712, 2162160, 3359664, 1846152, 3027024, 5538456, 6054048, 9081072, 9230760, 14270256, 19891872, 20307672, 25197480, 33297264, 45405360, 55135080, 71351280
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OFFSET
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1,1
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COMMENTS
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It seems that a(n)==0 (mod 84).
Additional terms with n > 75: a(77) = 15455984544, a(80) = 27719972280, a(83) = 22439977560, a(84) = 18479981520, a(86) = 28559971440. - Lars Blomberg, Jan 04 2016
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LINKS
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EXAMPLE
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a(4)=1008 because the set of the divisors {1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 18, 21, 24, 28, 36, 42, 48, 56, 63, 72, 84, 112, 126, 144, 168, 252, 336, 504, 1008} contains 4 pairs (12, 21), (24, 42), (36, 63) and (48, 84) with the property 21 = reversal(12), 42 = reversal(24), 63 = reversal(36) and 84 = reversal(48).
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MAPLE
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with(numtheory):nn:=10^8:
for n from 1 to 16 do:
ii:=0:
for m from 1 to nn while(ii=0) do:
it:=0:d:=divisors(m):d0:=nops(d):
for i from 1 to d0 do:
dd:=d[i]:y:=convert(dd, base, 10):n1:=length(dd):
s:=sum('y[j]*10^(n1-j)', 'j'=1..n1):
for k from i+1 to d0 do:
if s=d[k]
then
it:=it+1:
else fi:
od:
od:
if it=n
then
ii:=1:printf("%d %d \n", n, m):
else fi:
od:
od:
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PROG
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(PARI) nbr(vd) = {nb = 0; for (j=1, #vd, da = vd[j]; rda = eval(concat(Vecrev(Str(da)))); rrda = eval(concat(Vecrev(Str(rda)))); if ((da != rda) && vecsearch(vd, rda) && (da == rrda), nb++); ); nb/2; }
a(n) = {k=1; while (nbrp(divisors(k)) != n, k++); k; } \\ Michel Marcus, Dec 27 2015
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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a(7), a(19), a(20) corrected and a(21)-a(32) added by Lars Blomberg, Jan 04 2016
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STATUS
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approved
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