

A230007


Numbers n such that sigma(n) = reversal(n+7).


0



62, 206, 6002, 6439562, 60000002, 6004356002, 29355232298, 60000000002, 292369967108, 600439956002, 643956439562, 6303723993362
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

If p=3*10^m+1 is prime then 2*p is in the sequence.
Also if p=322*10^m219 is prime then 2*p is in the sequence. 206, 6002 and 6439562 are of this form. It is interesting that 6002 = 2*(3*10^3+1) = 2*(322*10^1219) is of both two forms.
Let f(r,s,t) = 3*10^(t*(s+4)+2*r+1)+10^(r+1)*(22*(10^(s+2)1)*(10^(t*(s+4))1)/ (10^(s+4)1))+1, if p=f(r,s,t) is prime and (i) r, s and t are nonnegative integers or (ii) s=t=0 and 2r is a positive integer or (iii) r=t=0 and s=3, then 2*p is in the sequence. Note, very interestingly, this pattern is a generalization of the previous two forms mentioned above. In fact we have 3*10^(m+1)+1 = f(m/2,s,0) and 322*10^m219 = f(0,m3,1).  Farideh Firoozbakht, Dec 17 2013


LINKS



FORMULA

a(1)=2*f(0,s,0), a(2)=2*f(0,3,1), a(3)=2*f(1,s,0), a(4)=2*f(0,1,1), a(5)=2*f(3,s,0), a(6)=2*f(2,0,1), a(8)=2*f(9/2,s,0), a(10)=2*f(2,2,1), a(11)=2*f(0,1,2), where f(r,s,t) = 3*10^(t*(s+4)+2*r+1) +10^(r+1)*(22*(10^(s+2)1)*(10^(t*(s+4))1) / (10^(s+4)1))+1.


EXAMPLE

sigma(206) = 312 = reversal(213) = reversal(206+7).


MATHEMATICA

Do[If[c=FromDigits[Reverse[IntegerDigits[7+n]]]; c>n && DivisorSigma[1, n] == c, Print[n]], {n, 500000000}]


CROSSREFS



KEYWORD

nonn,base,more


AUTHOR



EXTENSIONS



STATUS

approved



