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%I #16 Sep 05 2023 12:21:39
%S 1,12,6266169,6931848
%N Numbers n such that n = sigma(sum of digits of n)*sigma(product of digits of n).
%C No further terms up to 10^26. - _Chai Wah Wu_, Nov 28 2015
%C No other terms below 10^100. The sequence is likely finite and complete. - _Max Alekseyev_, Sep 05 2023
%e 12 = 4*3 = sigma(1+2)*sigma(1*2).
%e 6931848 = 56*123783 = sigma(6+9+3+1+8+4+8)*sigma(6*9*3*1*8*4*8).
%t Do[h=IntegerDigits[n]; s=Apply[Plus,h];p=Apply[Times,h];If[n== DivisorSigma[1,s]*DivisorSigma[1,p],Print[n]],{n,2000000000}]
%Y Cf. A038369.
%K base,more,nonn
%O 1,2
%A _Farideh Firoozbakht_, Oct 26 2008
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