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A293832 Smallest of a run of n consecutive integers in A061910, i.e., for which the sum of digits of their square is again a square.

%I A293832 #32 Apr 25 2025 06:28:57
%S A293832 1,1,1,9,9,9,9,46045846,302260461719025
%N A293832 Smallest of a run of n consecutive integers in A061910, i.e., for which the sum of digits of their square is again a square.
%C A293832 a(9) > 10^11. It can be expected that a(n) exists for all n, at an order of magnitude a(n) ~ 10^m with m/log_10(m) ~ n.
%C A293832 A061910 considers only positive integers, and here we also consider this case. If 0 were allowed, then the first 4 terms would be 0.
%H A293832 User 'mathlove', <a href="https://math.stackexchange.com/questions/1093266/">Numbers n such that the digit sum of n^2 is a square</a>, Mathematics StackExchange, Jan. 6, 2015.
%e A293832 For k = 1, 2 and 3, k^2 is a single-digit number and thus equal to its sum of digits, which therefore is a square. Therefore a(n) = 1 starts the first run of n consecutive integers with this property, for n = 1, 2 and 3.
%e A293832 However, the square of k = 4 has digit sum 7 which is not a square, and the same is the case for k = 5, 7 and 8. (Only k = 6 would have the required property.)
%e A293832 The consecutive integers { 9, 10, 11, 12, 13, 14, 15 } have squares 81, 100, 121, 169, 196, 225 which all have a digit sum (9, 1, 4, 16, 16 and 9) which is a square. Therefore a(n) = 9 starts the first run of n consecutive integers with this property, for n = 4 through 7.
%e A293832 (Actually, 10^(3m-2)^2-1 starts a run of 7 such numbers, for any m >= 1.)
%e A293832 The first run of 8 such numbers is (46045846, ..., 46045853), whence a(8) = 46045846.
%o A293832 (PARI) A293832(n,k=0)=until(,forstep(j=n,1,-1,issquare(sumdigits((k+j)^2))||next(2+!k+=j));return(k+1))
%Y A293832 Cf. A007953, A061910, A068833.
%K A293832 nonn,base,hard,more
%O A293832 1,4
%A A293832 _M. F. Hasler_, Oct 16 2017
%E A293832 a(9) from _Giovanni Resta_, Aug 27 2018