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a(n) = Sum_{i=0..m} d(i)*5^i, where Sum_{i=0..m} d(i)*6^i is the base-6 representation of n.
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%I #41 Jun 29 2019 01:39:09

%S 0,1,2,3,4,5,5,6,7,8,9,10,10,11,12,13,14,15,15,16,17,18,19,20,20,21,

%T 22,23,24,25,25,26,27,28,29,30,25,26,27,28,29,30,30,31,32,33,34,35,35,

%U 36,37,38,39,40,40,41,42,43,44,45,45,46,47,48,49,50,50,51,52,53,54,55

%N a(n) = Sum_{i=0..m} d(i)*5^i, where Sum_{i=0..m} d(i)*6^i is the base-6 representation of n.

%H Seiichi Manyama, <a href="/A303788/b303788.txt">Table of n, a(n) for n = 0..10000</a>

%e 16 = 24_6, so a(16) = 2*5 + 4 = 14.

%e 17 = 25_6, so a(17) = 2*5 + 5 = 15.

%e 18 = 30_6, so a(18) = 3*5 + 0 = 15.

%e 19 = 31_6, so a(19) = 3*5 + 1 = 16.

%o (Ruby)

%o def f(k, ary)

%o (0..ary.size - 1).inject(0){|s, i| s + ary[i] * k ** i}

%o end

%o def A(k, n)

%o (0..n).map{|i| f(k, i.to_s(k + 1).split('').map(&:to_i).reverse)}

%o end

%o p A(5, 100)

%o (PARI) a(n) = fromdigits(digits(n, 6), 5); \\ _Michel Marcus_, May 02 2018

%Y Sum_{i=0..m} d(i)*b^i, where Sum_{i=0..m} d(i)*(b+1)^i is the base (b+1) representation of n: A065361 (b=2), A215090 (b=3), A303787 (b=4), this sequence (b=5), A303789 (b=6).

%Y Cf. A037465.

%K nonn,base

%O 0,3

%A _Seiichi Manyama_, Apr 30 2018