A380530 Positions of records in A380528.
1, 4, 10, 42, 366, 3246, 37266, 631266, 11563926, 271591926
Offset: 1
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For n = 108 = 2^2 * 3^3, it is 2 that is the smallest prime factor p satisfying p^p | 108, thus a(108) = 2.
Array[If[IntegerQ@ #, #, 1] &@ First@ SelectFirst[FactorInteger[#], #1 <= #2 & @@ # &] &, 120] (* Michael De Vlieger, Oct 01 2019 *)
A129252(n) = { my(f = factor(n)); for(k=1, #f~, if(f[k, 2]>=f[k, 1], return(f[k, 1]))); (1); }; \\ Antti Karttunen, Oct 01 2019
A129252(n) = { my(pp); forprime(p=2, , pp = p^p; if(!(n%pp), return(p)); if(pp > n, return(1))); }; \\ Antti Karttunen, Feb 09 2025
From _Antti Karttunen_, May 09 2025: (Start) 10 = 2*5 is a term, as it is squarefree, and A380459(10) = 54 = 2 * 3^3, thus the prime factor 3 overflows, i.e., has an exponent at least as large as that prime. 5117046 = 2*3*11*31*41*61 is a term, as it is squarefree, and A380459(5117046) = 2 * 3^2 * 5^4 * 7^11 * 11^22 * 13^36 * 17^31 * 19^8, thus the least prime factor which overflows is 7 [= A380528(5117046)]. 31203546 = 2*3*11*31*101*151 is a term, as it is squarefree, and A380459(31203546) = 2 * 3^2 * 5^4 * 7^6 * 11^28 * 13^37 * 17^56 * 19^18 * 23^2, thus the least prime factor which overflows is 11 [= A380528(31203546)]. (End)
186 = 2*3*31 and A276086(186/2) = 2058 = 2 * 3 * 7^3, A276086(186/3) = 3 * 7^2, A276086(186/31) = 5, whose product = 2^1 * 3^2 * 5^1 * 7^5 = 1512630 = A380459(186), and as all the exponents are less than the corresponding primes, the product is in A048103, and because there are no any smaller number with three prime factors satisfying the same condition (of A380468), 186 is the term a(3) of this sequence. Note that A049345(A003415(186)) = 5121, where the digits are the exponents in the product read from the largest to the smallest prime factor. See also the example in A380476 about 4686.
4686 = 2*3*11*71 and taking subproducts of three primes at time, we obtain 2*3*11 = 66, 2*3*71 = 426, 2*11*71 = 1562, 3*11*71 = 2343. Then A380459(4686) = A276086(66) * A276086(426) * A276086(1562) * A276086(2343) = 1622849599205985150 = 2^1 * 3^2 * 5^2 * 7^6 * 11^9 * 13^1, and because all the exponents are less than the corresponding primes, the product is in A048103. Considering the primorial base expansions of the same summands (subproducts), we obtain 2100 = A049345(66) 20100 = A049345(426) 73010 = A049345(1562) 101011 = A049345(2343) ------ 196221 = A049345(A003415(4686)), with the summands adding together cleanly without any carries. Note how the primorial base digits at the bottom are the exponents in the product A380459(4686) given above, read from the largest to the smallest prime factor
for(n=1,oo,if(issquarefree(n),print1(A380528(n),", ")));
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