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A373177
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Integers k such that 2k + 1 and 4k + 3 are anagrams of k.
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0
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15632, 126530, 130265, 150632, 152630, 156329, 162530, 163025, 1265030, 1265300, 1265309, 1300265, 1302650, 1302659, 1500632, 1502630, 1506329, 1526300, 1526309, 1563299, 1566332, 1625030, 1625300, 1625309, 1630025, 1630250, 1630259, 1656332, 12650030
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OFFSET
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1,1
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COMMENTS
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The terms of this sequence begin with decimal digits 1 or 2, otherwise 4*k + 3 has more digits than k and cannot be an anagram. The first term whose first digit is 2 is a(3931) = 2055114278.
This sequence has infinitely many terms, since 1500*10^m + 632 is a term for all positive integers m.
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LINKS
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EXAMPLE
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15632 is a term, since 2*15632 + 1 = 31265 and 4*15632 + 3 = 62531 are both permutations of the digits of 15632.
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MAPLE
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filter:= proc(n) local L;
L:= sort(convert(n, base, 10));
sort(convert(2*n+1, base, 10))=L
and sort(convert(4*n+3, base, 10))=L
end proc:
R:= NULL: count:= 0:
for d from 1 while count < 100 do
for x from 10^(d-1) + 7 by 9 to (10^d-3)/4 while count < 100 do
if filter(x) then R:= R, x; count:= count+1 fi
od od:
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MATHEMATICA
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sid[n_] := Sort[IntegerDigits[n]]; Select[Range[13000000], sid[#] == sid[2*# + 1] == sid[4*# + 3] &] (* Amiram Eldar, May 27 2024 *)
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PROG
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(Python)
from itertools import count, islice
def agen(): # generator of terms
for e in count(1):
for k in range(10**(e-1), 10**e//4):
if sorted(str(k)) == sorted(str(2*k+1)) == sorted(str(4*k+3)):
yield k
(PARI) isok(k) = my(d=vecsort(digits(k))); (d == vecsort(digits(2*k+1))) && (d == vecsort(digits(4*k+3))); \\ Michel Marcus, May 28 2024
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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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STATUS
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approved
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