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A069514
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Numbers n such that sigma(reversal(n)) = reversal(sigma(n)). Ignore leading 0's.
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3
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1, 2, 3, 4, 5, 7, 14, 41, 124, 194, 333, 421, 491, 1324, 4231, 13324, 17054, 17571, 42331, 45071, 120530, 138465, 386650, 564831, 1130324, 1216360, 1333324, 1727571, 1757271, 1757571, 1787871, 2249422, 4230311, 4233331, 4369634
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OFFSET
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1,2
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COMMENTS
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For an arithmetical function f, call the arguments n such that f(reverse(n)) = reverse(f(n)) the "palinpoints" of f. This sequence is the sequence of palinpoints of f(n) = sigma(n).
If n is in the sequence and 10 doesn't divide n then the reversal of n is also in the sequence. - Farideh Firoozbakht, Aug 31 2004
Comments from Farideh Firoozbakht, Jan 16 2005. "The largest term that I found is M=(58*100^687 - 157)/33; the length of M is 1375. I proved the following facts about this sequence:
"I : If p=(58*100^n - 157)/99 is prime then 3*p is in the sequence, the sequence A102285 gives such n's.
"II : If p=(59*100^n - 257)/99 is prime then 3*p is in the sequence, I found only two primes of this form the first for n=3 and the second for n=27, next such n is greater than 3400.
"III : If both numbers p=10^n - 3 & q=5*10^n - 9 are primes then both numbers 2*p & q are in the sequence, q is reversal of 2*p. I found only two such n's, n=1 & 2.
"IV : If both numbers p=(10^n-7)/3 & q=(127*10^(n-1)-7)/3 are primes then both numbers 4*p & q are in the sequence, q is the reversal of 4*p, the sequence A102287 are these terms of A069514, I found only four such n's, n=2,3,4 & 6."
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LINKS
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EXAMPLE
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Let f(n) = sigma(n). Then f(194) = 294, f(491) = 492, so f(reverse(194)) = reverse(f(194)). Therefore 194 belongs to the sequence.
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MATHEMATICA
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rev[n_] := FromDigits[Reverse[IntegerDigits[n]]]; f[n_] := DivisorSigma[1, n]; Select[Range[10^6], f[rev[ # ]] == rev[f[ # ]] &]
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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