A069514 Numbers n such that sigma(reversal(n)) = reversal(sigma(n)). Ignore leading 0's.

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

Offset: 1

Author

Joseph L. Pe, Apr 15 2002

Keywords

• base
• nonn

Comments

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."

Examples

			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.
		

Crossrefs

Cf. A102285, A102286, A102287.

Programs

Mathematica

rev[n_] := FromDigits[Reverse[IntegerDigits[n]]]; f[n_] := DivisorSigma[1, n]; Select[Range[10^6], f[rev[ # ]] == rev[f[ # ]] &]

Extensions

More terms from Farideh Firoozbakht, Aug 31 2004