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A371885 a(n) is the least number k such that the sum of the final digits of the prime-power divisors of k (including 1) is n.

%I A371885 #12 Apr 16 2024 13:59:15
%S A371885 1,11,2,3,33,5,4,7,15,12,21,20,9,28,8,76,49,24,36,27,16,54,32,48,135,
%T A371885 80,64,112,240,192,336,320,144,216,128,1216,784,384,576,432,256,864,
%U A371885 512,768,2160,1280,1024,1792,3840,3072,5376,5120,2304,3456,2048,19456,12544,6144,9216,6912,4096,13824
%N A371885 a(n) is the least number k such that the sum of the final digits of the prime-power divisors of k (including 1) is n.
%C A371885 a(n) is the least k such that A367502(k) = n.
%H A371885 <a href="/index/Rec#order_20">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,16).
%F A371885 G.f.: (1 + 11*x + 2*x^2 + 3*x^3 + 33*x^4 + 5*x^5 + 4*x^6 + 7*x^7 + 15*x^8 + 12*x^9 + 21*x^10 + 20*x^11 + 9*x^12 + 28*x^13 + 8*x^14 + 76*x^15 + 49*x^16 + 24*x^17 + 36*x^18 + 27*x^19 - 122*x^21 - 393*x^24 - 232*x^33)/(1 - 16*x^20).
%F A371885 a(n + 20) = 16 * a(n) for n >= 15.
%e A371885 a(17) = 49 because the prime-power divisors of 49 are 1, 7 and 49, the sum of their final digits is 1 + 7 + 9 = 17, and 49 is the least number that works.
%p A371885 f:= proc(n) local F,i,j,t;
%p A371885   F:= ifactors(n)[2];
%p A371885   1 + add(add(F[i,1]^j mod 10, j = 1 .. F[i,2]),i=1..nops(F))
%p A371885 end proc:
%p A371885 V:= Vector(100): count:= 0:
%p A371885 for n from 1 do
%p A371885   v:= f(n);
%p A371885   if v <= 100 and V[v] = 0 then V[v]:= n; count:= count+1;
%p A371885       if count = 100 then break fi fi
%p A371885 od:
%p A371885 convert(V,list);
%Y A371885 Cf. A000961, A367502.
%K A371885 nonn,base,easy
%O A371885 1,2
%A A371885 _Robert Israel_, Apr 10 2024