A105324
Numbers n such that 2*reversal(n)=sigma(n).
Original entry on oeis.org
6, 73, 483, 4074, 4473, 4623, 7993, 42813, 69855, 253782, 799993, 7999993, 46000023, 426000213, 4600000023, 6718967838, 42600000213, 46000000023, 79999999993, 426000000213
Offset: 1
253782 is in the sequence because reversal(253782)=287352; sigma(253782)=574704 & 2*287352=574704.
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reversal[n_]:= FromDigits[Reverse[IntegerDigits[n]]]; Do[If[2* reversal[n]== DivisorSigma[1, n], Print[n]], {n, 1000000000}]
Select[Range[8*10^6],2*IntegerReverse[#]==DivisorSigma[1,#]&] (* The program generates the first 12 terms of the sequence. *) (* Harvey P. Dale, Oct 29 2022 *)
A104907
Numbers n such that d(n)*reversal(n)=sigma(n), where d(n) is number of positive divisors of n.
Original entry on oeis.org
1, 73, 861, 7993, 8241, 799993, 7999993, 44908500, 82000041, 293884500, 6279090751, 8200000041, 62698513951, 79999999993, 82000000041, 374665576800, 597921764310, 7999999999993, 8200000000041
Offset: 1
Let p=8*10^n-7 be a prime so d(p)=2; reversal(p)=4*10^n-3 and sigma(p)
=8*10^n-6 hence d(p)*reversal(p)=sigma(p) and this shows that p
is in the sequence. 73,7993,799993 and 7999993 are such terms.
Also let q=(2*10^n+1)/3 be a prime, so 123*q=82*10^n+41; reversal
(123*q)=14*10^n+28; d(123*q)=8 and sigma(123*q)=168*q+168=112*10^n
+224 hence d(123*q)*reversal(123*q)=sigma(123*q) and this shows
that 123*q is in the sequence. 861,8241 and 82000041 are such terms.
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reversal[n_]:= FromDigits[Reverse[IntegerDigits[n]]]; Do[If[DivisorSigma[0, n]*reversal[n] == DivisorSigma[1, n], Print[n]], {n, 1125000000}]
Select[Range[8*10^6],DivisorSigma[0,#]IntegerReverse[#]==DivisorSigma[1,#]&] (* The program generates the first 7 terms of the sequence. *) (* Harvey P. Dale, Jan 31 2023 *)
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