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A240580 Numbers k such that DigitSum(5^k) > DigitSum(5^(k+1)).

%I A240580 #18 Jun 19 2025 20:57:13
%S A240580 4,10,11,12,16,18,24,27,28,32,34,36,39,44,45,49,51,52,57,58,60,61,62,
%T A240580 63,64,69,75,76,77,78,80,83,84,87,88,90,91,94,96,97,100,103,106,107,
%U A240580 108,113,114,115,118,119,124,129,130,132,135,138,139,142,143,144,149
%N A240580 Numbers k such that DigitSum(5^k) > DigitSum(5^(k+1)).
%H A240580 K. D. Bajpai, <a href="/A240580/b240580.txt">Table of n, a(n) for n = 1..1000</a>
%e A240580 The digitsum(5^4) = 13 > 11 = digitsum(5^(4+1)). Hence, 4 appears in the sequence.
%e A240580 The digitsum(5^11) = 38 > 28 = digitsum(5^(11+1)). Hence, 11 appears in the sequence.
%p A240580 a := proc(n) local a,b,d,e; a:=5^n; b:=add( i,i = convert((a), base, 10))(a); d:=5^(n+1); e:=add( i,i = convert((d), base, 10))(d); if b > e then RETURN (n); fi; end: seq(a(n), n=1..300);
%t A240580 k=Table[Total[IntegerDigits[5^n,10]],{n,1,300}];Flatten[Position[Greater@@@Partition[k,2,1],True]]
%t A240580 Select[Range[150],Total[IntegerDigits[5^#]]>Total[IntegerDigits[5^(#+1)]]&] (* _Harvey P. Dale_, Jun 04 2025 *)
%Y A240580 Cf. A186775 (digitsum(2^k) > digitsum(2^(k+1))), A239935 (digitsum(3^k) > digitSum(3^(k+1))).
%K A240580 nonn,base
%O A240580 1,1
%A A240580 _K. D. Bajpai_, Apr 08 2014