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Minimum value of k such that sum_{i=k..k+n-1} i = sum_{i=k..k+n-1} i’, where i’ is the arithmetic derivative of i.
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%I #13 Mar 17 2018 04:06:20

%S 0,16,390,30,156,93052,1884,2910

%N Minimum value of k such that sum_{i=k..k+n-1} i = sum_{i=k..k+n-1} i’, where i’ is the arithmetic derivative of i.

%C a(9) > 10^10. - _Giovanni Resta_, Jul 10 2013

%e n = 1, k = k’. The minimum solution is k=0 because k’=0. Other non-minimal solutions are listed in A051674.

%e n = 2, k + (k+1) = k’ + (k+1)’. The minimum solution is 16 because 16 + 17 = 33 and 16’ + 17’ = 32 + 1 = 33.

%e Other non-minimal solutions are 108, 3294172, 7975979, etc.

%e n = 3, k + (k+1) + (k+2) = k’ + (k+1)’ + (k+2)’. The minimum solution is 390 because 390 + 391 + 392 = 1173 and 390’ + 391’ + 392’ = 433 + 40 + 700 = 1173.

%p with(numtheory); ListA227390:= proc(q) local a,b,k,i,n,p;

%p for n from 1 to q do for k from 0 to q do a:=n*(k+(n-1)/2); b:=0;

%p b:=add((k+i)*add(op(2,p)/op(1,p),p=ifactors(k+i)[2]),i=0..n-1);

%p if a=b then print(k); break; fi; od; od; end: ListA227390(10^12);

%Y Cf. A003415, A051674.

%K nonn,more

%O 1,2

%A _Paolo P. Lava_, Jul 10 2013