This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A280798 #21 Jan 09 2017 04:20:03 %S A280798 1,2,13,16667 %N A280798 a(n) is the smallest integer m such that sumdigits(m^2) = 4^n. %C A280798 See the Mathematical Reflections link for a proof that a(n) exists for all n. %C A280798 a(4) > 300*10^6. %C A280798 From _Jon E. Schoenfield_, Jan 08 2017: (Start) %C A280798 As shown below, 264575131106460 <= a(4) <= 13663784168010583. %C A280798 The maximal sum of the last three digits of a square is 25, which occurs only when those digits are 889 (e.g., 83^2 = 6889), so since sumdigits(a(4)^2) = 256, the sum of the digits that precede the last three must be at least 256 - 25 = 231, thus a(4)^2 >= 69999999999999999999999999889 (call this number j), so a(4) >= ceiling(sqrt(j)) = 264575131106460. (This bound could easily be improved; e.g., since j is not a square, m^2 cannot be less than the smallest number greater than j whose digit sum is 256 and whose last three digits are 889, i.e., m^2 >= 78999999999999999999999999889, so m >= 281069386451104.) %C A280798 The last digit of a square m^2 is maximized (at 9) iff the last digit of m is 3 or 7. The sum of the last two digits of m^2 is maximized (at 17) iff the last two digits are 89, which occurs iff the last two digits of m are 33, 83, 17, or 67. For k >= 3, it appears that the sum of the last k digits of m^2 is maximized (at 9k-2) iff the last k digits are all 9s except for the two 8s immediately before the final 9, which occurs iff the k-digit suffix of m takes one of eight values, as shown in the table below; the line extending upward from each suffix of more than one digit connects it to the suffix from which it inherits all but its first digit. %C A280798 . %C A280798 k k-digit suffixes of m that maximize sumdigits(m^2 mod 10^k) %C A280798 = =============================================================== %C A280798 1 3 7 %C A280798 / \ / \ %C A280798 / \ / \ %C A280798 / \ / \ %C A280798 / \ / \ %C A280798 / \ / \ %C A280798 / \ / \ %C A280798 / \ / \ %C A280798 2 33 83 17 67 %C A280798 |\ |\ |\ |\ %C A280798 3 833 333 583 083 417 917 167 667 %C A280798 |\ |\ |\ |\ %C A280798 4 1833 6833 0583 5583 9417 4417 8167 3167 %C A280798 |\ |\ |\ |\ %C A280798 5 41833 91833 10583 60583 89417 39417 58167 08167 %C A280798 |\ |\ |\ |\ %C A280798 6 041833 541833 010583 510583 989417 489417 958167 458167 %C A280798 |\ |\ |\ |\ %C A280798 7 5041833 0041833 8010583 3010583 1989417 6989417 4958167 9958167 %C A280798 |\ |\ |\ |\ %C A280798 . ... ... ... ... ... ... ... ... %C A280798 . %C A280798 Since numbers m ending in these suffixes have squares m^2 such that sumdigits(m^2 mod 10^k) is maximized, their full digit sums sumdigits(m^2) tend to be larger than those of nearby numbers with other suffixes. A search over a range of prefixes using the 10-digit suffix 4168010583 found that m = 13663784168010583 has sumdigits(m^2) = 256, which yields an upper bound for a(4). (End) %H A280798 Mathematical Reflections, <a href="https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2014-05/mr_4_2014_solutions.pdf">Solution to Problem J307</a>, Issue 5, 2015, p. 1. %e A280798 a(1)=2 since 2^2=4 with sum of digits 4. %o A280798 (PARI) a(n) = my(k=1); while (sumdigits(k^2) != 4^n, k++); k; %Y A280798 Cf. A000302, A007953, A061912. %K A280798 nonn,base,more %O A280798 0,2 %A A280798 _Michel Marcus_, Jan 08 2017