A369953 - OEIS

%I #73 Feb 20 2024 10:40:10

%S 0,3,24,63,264,1374,3114,8937,60663,94863,545793,1989417,5477133,

%T 20736417,82395387,260191833,706399164,2428989417,9380293167,

%U 28105157886,99497231067,538479339417,1974763271886,4472135831667,14106593458167,62441868958167,244744764757083,836594274358167

%N a(n) is the least integer k such that the sum of the digits of k^2 is 9*n.

%C 3|a(n).

%H Zhining Yang, <a href="/A369953/b369953.txt">Table of n, a(n) for n = 0..40</a> (terms 19..40 from Zhao Hui Du)

%H Shouen Wang, <a href="http://bbs.mathchina.com/bbs/forum.php?mod=viewthread&amp;tid=2060076">Chinese BBS: How many of these A's are there?</a>

%F a(n) = A067179(4n).

%e a(3)=63 because k=63 is the least integer k such that the sum of the digits of k^2 = 3969 is 9*3 = 27 (3+9+6+9 = 27).

%t n=1;lst={};For[k=0,k<10^8,k+=3,If[Total[IntegerDigits[k^2]]==9*n,AppendTo[lst,k];n++]];lst

%o (Python)

%o n=1

%o lst=[]

%o for k in range(0,10**8,3):

%o     if sum(int(d) for d in str(k*k))==9*n:

%o         lst.append(k)

%o         n=n+1

%o print(lst)

%o (PARI) a(n) = my(k=0); while(sumdigits(k^2) != 9*n, k+=3); k; \\ _Michel Marcus_, Feb 17 2024

%Y Cf. A004159, A008591, A067179, A369955.

%K nonn,base

%O 0,2

%A _Zhining Yang_, Feb 06 2024

%E a(19)-a(27) from _Zhao Hui Du_, Feb 09 2024