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a(n) = sum of digits of k^4 as k runs from 1 to n.
0

%I #12 Jun 08 2021 17:08:40

%S 1,8,17,30,43,61,68,87,105,106,122,140,162,184,202,227,246,273,283,

%T 290,317,339,370,397,422,459,477,505,530,539,561,592,619,644,663,699,

%U 727,752,770,783,814,841,866,903,921,958,1001,1028,1059,1072,1099,1124,1161

%N a(n) = sum of digits of k^4 as k runs from 1 to n.

%C Partial sums of A055565.

%F a(n) = a(n-1) + sum of decimal digits of n^4.

%F a(n) = sum(k=1, n, sum(m=0, floor(log(k^4)), floor(10((k^4)/(10^(((floor(log(k^4))+1))-m)) - floor((k^4)/(10^(((floor(log(k^4))+1))-m))))))).

%F General formula: a(n)_p = sum(k=1, n, sum(m=0, floor(log(k^p)), floor(10((k^p)/(10^(((floor(log(k^p))+1))-m)) - floor ((k^p)/(10^(((floor(log(k^p))+1))-m))))))). Here a(n)_p is a sum of digits of k^p from k=1 to n.

%e a(3) = sum_digits(1^4) + sum_digits(2^4) + sum_digits(3^4) = 1 + 7 + 9 = 17.

%t f[n_] := Block[{s = 0, k = 1}, While[k <= n, s = s + Plus @@ IntegerDigits[k^4]; k++ ]; s]; Table[ f[n], {n, 50}] (* _Robert G. Wilson v_, Nov 18 2004 *)

%t Accumulate[Table[Total[IntegerDigits[n^4]],{n,60}]] (* _Harvey P. Dale_, Jun 08 2021 *)

%Y Cf. k^1 in A037123, k^2 in A071317 & k^3 in A071121.

%K nonn,easy,base

%O 1,2

%A _Yalcin Aktar_, Nov 16 2004

%E Edited and extended by _Robert G. Wilson v_, Nov 18 2004

%E Existing example replaced with a simpler one by _Jon E. Schoenfield_, Oct 20 2013